#geometry-and-trigonometry | Mathematics | Page 171

steady sky Apr 1, 2018, 6:01 PM

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right?

keen pilot Apr 1, 2018, 6:06 PM

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2*(r^2)*pi is wrong

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You need to rotate it by r^2*pi

robust rampart Apr 2, 2018, 3:29 PM

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can anyone help me with geometry?

waxen gorge Apr 2, 2018, 3:36 PM

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HALLO

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OKAI

robust rampart Apr 2, 2018, 3:36 PM

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hi

waxen gorge Apr 2, 2018, 3:36 PM

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WATCHU WANT

robust rampart Apr 2, 2018, 3:36 PM

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so here

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If triangle MAP is congruent to triangle FUN and the measure of angle M is 90 degrees and the measure of angle P is 40 degrees what is the measure of angle U?

waxen gorge Apr 2, 2018, 3:37 PM

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Ok

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one is 90

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One is 40

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What's 90+40

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130

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180-130 = 50°

robust rampart Apr 2, 2018, 3:37 PM

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a. 50 b. 90 c. 40 d. 140 e. cant tell

waxen gorge Apr 2, 2018, 3:38 PM

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U know two angle measurement s

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So u do 180-them added to get hte third

robust rampart Apr 2, 2018, 3:38 PM

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so A right?

waxen gorge Apr 2, 2018, 3:38 PM

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Yes

robust rampart Apr 2, 2018, 3:38 PM

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next one

waxen gorge Apr 2, 2018, 3:39 PM

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Ask someone else brb

robust rampart Apr 2, 2018, 3:39 PM

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ok

wide path Apr 3, 2018, 4:47 AM

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Hello...

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Anybody

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Hello

chrome fiber Apr 3, 2018, 5:09 AM

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What is it?

analog dust Apr 3, 2018, 10:34 AM

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This is #germoertyr

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Anyone has geomerty problems for me?

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4 mee

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Pleaz

mint sandal Apr 3, 2018, 10:41 AM

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@analog dust problem [045]

analog dust Apr 3, 2018, 10:42 AM

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Whats the deifference between R and R^2...

mint sandal Apr 3, 2018, 10:43 AM

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R is the real number line

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R^2 is the euclidean plane

analog dust Apr 3, 2018, 10:43 AM

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Oh

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Why not just say in a plane

mint sandal Apr 3, 2018, 10:45 AM

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same thing

wide path Apr 3, 2018, 1:09 PM

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This is sad

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Rip geometry

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I swear how is it that bad its the easiest one!

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The formulas are easy

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It may look hard with all the "PIr2+(b1,+b2)

wide path Apr 3, 2018, 2:01 PM

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Ps

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They need a trigonometry section

knotty shell Apr 3, 2018, 6:08 PM

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📎 Naamloos221312.png

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Can someone explain to me why the distance from P to -q is r+d and the distance to +q = r-d

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It seems so simple but can't figure it out.

shadow anvil Apr 3, 2018, 6:26 PM

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@knotty shell Still stuck?

knotty shell Apr 3, 2018, 6:26 PM

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Yeah kinda

shadow anvil Apr 3, 2018, 6:26 PM

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kinda

knotty shell Apr 3, 2018, 6:27 PM

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Can you help me out? thonker

shadow anvil Apr 3, 2018, 6:27 PM

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What does kinda mean

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Does that mean you've part way got it

knotty shell Apr 3, 2018, 6:27 PM

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Not really

shadow anvil Apr 3, 2018, 6:27 PM

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ok

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Well

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I'm not sure if there's anymore complications to it

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But it looks like r is being treated like a vector

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So if you choose your positive direction

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which is left

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You've got r

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Which is all positive in left direction

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and then you're going d in the left direction

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So there's no negatives

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but the other direction

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you're going forwards by negative d (backwards)

knotty shell Apr 3, 2018, 6:28 PM

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Yeah

shadow anvil Apr 3, 2018, 6:29 PM

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That's why it's negative

knotty shell Apr 3, 2018, 6:29 PM

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Yeah but I don't get why it's r+d and r-d

shadow anvil Apr 3, 2018, 6:29 PM

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because one is moving in positive d

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the other is going backwards by d

knotty shell Apr 3, 2018, 6:29 PM

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I get the minus signs

shadow anvil Apr 3, 2018, 6:29 PM

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Oh

knotty shell Apr 3, 2018, 6:30 PM

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But why is the length -q -> P for example r+ d

shadow anvil Apr 3, 2018, 6:30 PM

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I assumed that -q and q were charges and to do with the other topics here which I'm not really familiar with

knotty shell Apr 3, 2018, 6:30 PM

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I don't see the actual geometry

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Yeah that doesn't matter rn

shadow anvil Apr 3, 2018, 6:31 PM

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From p to -q

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That should be -r I think

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Because you want to go in the opposite direction

knotty shell Apr 3, 2018, 6:31 PM

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Wait what

shadow anvil Apr 3, 2018, 6:31 PM

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r

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is labelled

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to go the opposite direction

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so from -q to p

knotty shell Apr 3, 2018, 6:31 PM

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But you have -q is further away from P than +q right?

shadow anvil Apr 3, 2018, 6:31 PM

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what

knotty shell Apr 3, 2018, 6:32 PM

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thonker

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Isn't that right?

shadow anvil Apr 3, 2018, 6:32 PM

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Wait

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I'm not sure what you're stating

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-q looks closer to q than p

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but idk if this is to scale

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Like if it's not it depends

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It can be, or it might not be

knotty shell Apr 3, 2018, 6:34 PM

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Yeah but t's about -q -> p and q -> p

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not q-> -q

shadow anvil Apr 3, 2018, 6:36 PM

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Ok?

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You're asking

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=tex P\to -q

charred spearBOT Apr 3, 2018, 6:36 PM

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📎 latex.png

knotty shell Apr 3, 2018, 6:36 PM

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Yeah

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I think it also assumes d<<r

shadow anvil Apr 3, 2018, 6:37 PM

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So from P to the point between q and -q

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is -r

knotty shell Apr 3, 2018, 6:37 PM

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Yeah

shadow anvil Apr 3, 2018, 6:38 PM

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then +d

knotty shell Apr 3, 2018, 6:39 PM

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You can't just do that right?

shadow anvil Apr 3, 2018, 6:39 PM

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Why not

knotty shell Apr 3, 2018, 6:40 PM

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That's a vector that groes from P to the middle of -q and +q and then from there to -q

shadow anvil Apr 3, 2018, 6:40 PM

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yes

knotty shell Apr 3, 2018, 6:40 PM

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Which would be longer than the actual vector

shadow anvil Apr 3, 2018, 6:40 PM

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No

knotty shell Apr 3, 2018, 6:40 PM

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I want the distance between P and -q

shadow anvil Apr 3, 2018, 6:40 PM

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Then you find the magnitude of the resulting vector

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I've no idea what these vectors actually are

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But being vectors

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You can use vector addition like that

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Because they're not lengths

knotty shell Apr 3, 2018, 6:41 PM

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Wait really?

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But you have to get d as a vector too right?

shadow anvil Apr 3, 2018, 6:41 PM

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Yes

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Is d not a vector?

knotty shell Apr 3, 2018, 6:42 PM

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It is I guess

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Okay this kinda makes sense now

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Thanks 😃

strange crow Apr 3, 2018, 6:54 PM

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um

knotty shell Apr 3, 2018, 7:04 PM

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@shadow anvil nvm I don't get it. Even if these 2 had a 90 degree angle the distance is always gonna scale with a square root since when you take the magnitude it gives you something like $$\sqrt{r^2 + d^2}$$

charred spearBOT Apr 3, 2018, 7:04 PM

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📎 latex.png

knotty shell Apr 3, 2018, 7:05 PM

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Which wouldn't be r+ d right?

shadow anvil Apr 3, 2018, 7:07 PM

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Well

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It's not a distance

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It's a vector

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If they're distances

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Then yes

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You need to use some trig

knotty shell Apr 3, 2018, 7:08 PM

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Oh god

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I see it

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My bad haha

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it says $$\theta =0 $$ on top

charred spearBOT Apr 3, 2018, 7:09 PM

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📎 latex.png

shadow anvil Apr 3, 2018, 7:09 PM

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Lmao

knotty shell Apr 3, 2018, 7:09 PM

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lol thanks anyway

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for your time

shadow anvil Apr 3, 2018, 7:09 PM

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But it might be worthwhile trying to imagine it with another value for theta

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Just to understand it

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I just want to make a distinction too between vectors and vector quantities @knotty shell as I'm now realising that might have been what you meant

knotty shell Apr 3, 2018, 7:11 PM

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is a vector quantity just the magnitude?

shadow anvil Apr 3, 2018, 7:11 PM

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No

knotty shell Apr 3, 2018, 7:11 PM

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of a vector

shadow anvil Apr 3, 2018, 7:11 PM

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Newtons is a vector

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but it's a vector quantity

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like 5N

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Isn't a true vector

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It's linear a vector

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but a true vector

knotty shell Apr 3, 2018, 7:12 PM

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I don't really kow what linear in the sense of vector means

shadow anvil Apr 3, 2018, 7:12 PM

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=tex \begin{pmatrix}a\cr b\cr c\end{pmatrix}

charred spearBOT Apr 3, 2018, 7:12 PM

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📎 latex.png

knotty shell Apr 3, 2018, 7:12 PM

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same for tru

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true*

shadow anvil Apr 3, 2018, 7:12 PM

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Well

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It's only working in one plane

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Like 4n

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can only go two directions

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positive

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and negative

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there's no upwards

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downwards

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back or forwards

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or anything like that

knotty shell Apr 3, 2018, 7:14 PM

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But when is something a true vector then?

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When it's defined in multiple dimensions?

shadow anvil Apr 3, 2018, 7:14 PM

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Yes

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Those vectors we need to split up into components

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So

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=tex \overrightarrow{Pq}=\sqrt{\left(d+r\cos\left(\theta\right)\right)^2+\left(r\sin\left(\theta\right)\right)^2}

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fuck

knotty shell Apr 3, 2018, 7:17 PM

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This doesn't feel right

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where's the r

shadow anvil Apr 3, 2018, 7:18 PM

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forgot it

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whoops

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should be one infront of sin too

charred spearBOT Apr 3, 2018, 7:18 PM

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📎 latex.png

shadow anvil Apr 3, 2018, 7:18 PM

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and the first pq should be |pq|

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but w/e

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that's the length

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now theta equals 0

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which works in this case

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All I've done

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Should be -d too

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fml

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=tex |\overrightarrow{Pq}|=\sqrt{\left(r\cos\left(\theta\right)-d\right)^2+\left(r\sin\left(\theta\right)\right)^2}

charred spearBOT Apr 3, 2018, 7:19 PM

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📎 latex.png

shadow anvil Apr 3, 2018, 7:19 PM

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Ok there we go

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So all I've done

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Is taken the sum of horizontal components

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for the first part

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the horizontal component of r is r*cos(theta)

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Because of trig

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soh-cah-toa

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Then r is only existant in the horizontal plane

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so there's no trig just minus r

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So that's our horizontal movement from p to q

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now the veritcal is the same principle

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just the vertical component of r

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using trig

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now we just use pythagoras

knotty shell Apr 3, 2018, 7:22 PM

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And this is the distnace to +q?

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right?

shadow anvil Apr 3, 2018, 7:24 PM

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Yes

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-q is the same idea

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just you will get +d instead of -d

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because you're not going back on yourself

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and then that will work with any value of theta

knotty shell Apr 3, 2018, 7:26 PM

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Okay thanks

upper karma Apr 4, 2018, 12:01 AM

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https://cdn.discordapp.com/attachments/328766453027307522/430879380978401280/image.jpg

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have fun with these guys

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last two ones at leasat

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I ahd fun doing them

chrome fiber Apr 4, 2018, 12:29 AM

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So easy though

upper karma Apr 4, 2018, 12:30 AM

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yeah

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i was doing it really early

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and got dumb

nocturne cave Apr 4, 2018, 11:17 AM

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hello, I have a question about finding the value of a trig function (sin, cos, etc.) for a given angle. I understand that a triangle is made with the given angle, but I'm not sure how to find the two other acute angles of the triangle in order to find side lengths: take the example of 330 degrees, how would I find the two other angles of the triangle from that?

mint sandal Apr 4, 2018, 11:17 AM

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look up unit circle

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no triangle has an angle of 330 degrees

nocturne cave Apr 4, 2018, 11:18 AM

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I know that, 330 degrees is not an angle of a triangle

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I have to find the two other acute angles of the right triangle formed from 330 degrees

mint sandal Apr 4, 2018, 11:19 AM

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how do you form a triangle from 330 degrees?

nocturne cave Apr 4, 2018, 11:19 AM

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http://www.dummies.com/education/math/calculus/how-to-calculate-values-for-the-six-trigonometric-functions/

dummies

How to Calculate Values for the Six Trigonometric Functions - dummies

In pre-calculus, you need to evaluate the six trig functions — sine, cosine, tangent, cosecant, secant, and cotangent — for a single angle on the unit circle. For each angle on the unit circle, three other angles have similar trig function va...

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like this

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but this page doesn't explain it clearly

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which is why I'm asking here

mint sandal Apr 4, 2018, 11:20 AM

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well the axes are at angles 0, 90, 180 and 270

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330 passes the bottom axis, so one of the angles is 330-270 = 60

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another angle is 90-60 = 30

nocturne cave Apr 4, 2018, 11:22 AM

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ahh ok so the axis in counter clock wise is what's referenced

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thank you

fair spade Apr 4, 2018, 8:12 PM

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I got a terrifying problem. Any triangle ABC. BD bisects AC, CE bisects AB, AP is perp to BD, AQ is perp to CE. Prove PQ || BC

white harness Apr 4, 2018, 8:18 PM

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Image ?

fair spade Apr 4, 2018, 8:29 PM

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I don't have it digital

analog dust Apr 4, 2018, 8:30 PM

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Such a beautiful drawibg

📎 JPEG_20180404_222923.jpg

fair spade Apr 4, 2018, 8:30 PM

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Yep, pretty much

analog dust Apr 4, 2018, 8:31 PM

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We didnt kbow where P and Q was..

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Were

fair spade Apr 4, 2018, 8:32 PM

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not fully accurate as AQ and AP are segments

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but close enough

analog dust Apr 4, 2018, 8:33 PM

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?

fair spade Apr 4, 2018, 8:33 PM

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AQ/AP don't intersect BC

analog dust Apr 4, 2018, 8:33 PM

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Yea its just libes

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Lines

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That i drew

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Cuz why not

fair spade Apr 4, 2018, 8:34 PM

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There is also a point F where BD crosses CE

analog dust Apr 4, 2018, 8:34 PM

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Ok

fair spade Apr 4, 2018, 8:39 PM

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The only things I can easily prove is the opposing angles around F are equal and that in APFQ, angles Q and P are both 90. One needs to somehow prove angle BCQ = CQP

wide path Apr 4, 2018, 8:52 PM

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Beatifully Barely Readable

sterile stag Apr 5, 2018, 12:44 AM

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could someone pls help me w a trig problem

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?

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why is no one helping me? 😢

hazy delta Apr 5, 2018, 1:18 AM

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@sterile stag where is the problem?

sterile stag Apr 5, 2018, 1:19 AM

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📎 Screen_Shot_2018-04-04_at_8.16.02_PM.png

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@hazy delta

hazy delta Apr 5, 2018, 1:20 AM

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it will help to draw a picture

sterile stag Apr 5, 2018, 1:20 AM

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It's really unclear

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idek how to draw this

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@hazy delta do u think u can solve this ?

hazy delta Apr 5, 2018, 1:24 AM

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i am not sure how to help, sorry

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you must be able to draw a picture

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if you cannot do this, it suggests a severe lack of understanding

sterile stag Apr 5, 2018, 1:25 AM

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ok thanks anyways 🙏

keen aspen Apr 5, 2018, 1:36 AM

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Okay ill try

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anyways you have a trian

📎 Screenshot_10.png

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The triangle will look like that I believe

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So I would find the law of sines to find the distance from the observer to the peg

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and then make a triangle with that to find the horizontal distance

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@sterile stag

sterile stag Apr 5, 2018, 1:43 AM

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Ok I'll try that

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This is the answer I got but it's not the answer in the book

📎 JPEG_20180404_215129.jpg

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@keen aspen

keen aspen Apr 5, 2018, 1:54 AM

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No that is wrong

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You did the second part right

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but you didn't use the law of sines correct

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$$\frac{\sin(51)}{20}=\frac{\sin(57)}{x}$$

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?

charred spearBOT Apr 5, 2018, 1:55 AM

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📎 latex.png

sterile stag Apr 5, 2018, 1:55 AM

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where did u get 51 and 57 from

keen aspen Apr 5, 2018, 1:56 AM

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add 41+10

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to get that whole angle

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and then 180-51-72

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to get the 3rd angle

sterile stag Apr 5, 2018, 1:56 AM

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ohhhh i see

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I GOT IT

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THANK YOU SO MUCHHHH @keen aspen

keen aspen Apr 5, 2018, 1:58 AM

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No problem

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I am doing the same stuff in my trig class :v

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Applications using oblique triangles

past mantle Apr 5, 2018, 4:09 PM

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I'm about to get into probability

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In geometry

keen aspen Apr 5, 2018, 8:05 PM

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probability in geometry thonker

upper karma Apr 5, 2018, 8:43 PM

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^

upper karma Apr 7, 2018, 3:17 AM

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could I theoretically learn geometry using Euclid's Elements

upper karma Apr 7, 2018, 8:29 AM

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That would be pointlessly difficult

upper karma Apr 7, 2018, 2:54 PM

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@upper karma pointlessly difficult is my middle name

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Its not productive either

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Its a slow way of learning

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yeah I have all summer

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You wont get a better understanding if geometry by reading frok euclid

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Dont get that impression

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Its way faster and easier to learn from other sources

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And

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Other texts will give a perspective euclid didnt have

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@upper karma I also am gonna get another book just a bit after

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Fine

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but Euclids elements used to be the number 2 most printed book ever next to the bible

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like

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it was what people learned from for over two thousand years

dreamy meteor Apr 8, 2018, 6:27 AM

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people two thousand years ago bled themselves to death to cure disease

bronze harbor Apr 8, 2018, 2:19 PM

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help

📎 Screen_Shot_2018-04-08_at_8.46.28_PM.png

shadow anvil Apr 8, 2018, 6:03 PM

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@bronze harbor still stuck?

bronze harbor Apr 8, 2018, 11:35 PM

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nah i got it lol

waxen gorge Apr 9, 2018, 1:20 AM

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I mean u can obviously make a bunch of triangles and stuff but I need a fast way :(

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Jk that's the wrong question

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Here

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📎 JPEG_20180408_212134.jpg

blissful temple Apr 9, 2018, 1:49 AM

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Okay I’m gonna try this at home after dinner XD

sturdy bolt Apr 9, 2018, 3:21 AM

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@waxen gorge I'm not really familiar with these types of problems, so maybe there is another way, but I drew these lines:

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📎 2018-04-08_20_20_49-Geometry_-_GeoGebra.png

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The first step is to calculate that $$CF = \sqrt{r^2-1} + \sqrt{3}$$

charred spearBOT Apr 9, 2018, 3:22 AM

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📎 latex.png

sturdy bolt Apr 9, 2018, 3:24 AM

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Then, using the fact that $$GE = DE/2$$, calculate that $$CF = DE \frac{\sqrt{3}}{2} + \sqrt{r^2 - DE^2/4}$$

charred spearBOT Apr 9, 2018, 3:24 AM

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📎 latex.png

sturdy bolt Apr 9, 2018, 3:24 AM

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Then set those two expressions equal to each other and solve for DE

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This gives a formula that matches one of the multiple choice options you showed earlier

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Sorry my labels don't match yours

sturdy bolt Apr 9, 2018, 7:12 AM

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@waxen gorge AoPS did have a solution hidden away: https://artofproblemsolving.com/community/c5h595722p3534408

cyan saffron Apr 9, 2018, 12:38 PM

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Out of curiosity, when is a class of this type of geometry taught? Highschool? I'm doing calc 3 and I still haven't seen geometry of this complexity.

white swift Apr 9, 2018, 12:47 PM

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depends on the classes you take

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you can have geometry anywhere from grade 5 to 12 to uni etc

sturdy bolt Apr 9, 2018, 7:13 PM

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None of my gradeschool geometry or trigonometry classes had problems as complex as the one above. I didn't learn that kind of geometry in a class, really. Usually people learn it separately, by studying the problems from mathematics contests and their solutions. @cyan saffron

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Some universities have problem solving clubs which do exactly that.

analog dust Apr 9, 2018, 7:23 PM

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I think there must be an elegant solution to the problem..

sturdy bolt Apr 9, 2018, 7:51 PM

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The one on AoPS is pretty elegant @analog dust

analog dust Apr 9, 2018, 7:51 PM

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I wanna solve it myself so plz no spoil

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Ill check it out after

sturdy bolt Apr 9, 2018, 7:52 PM

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cool

waxen gorge Apr 9, 2018, 8:34 PM

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@analog dust check it out already >_>

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I tried the formula for the solution and plugged it in with the given values in the question and it said the answer was wrong

sturdy bolt Apr 9, 2018, 8:36 PM

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iirc the answer I got was sqrt(7)/2, is that what you got?

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@waxen gorge

waxen gorge Apr 9, 2018, 8:40 PM

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Yes

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Sometimes he messes up the answers

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He'll probably notice and fix it during the class

sturdy bolt Apr 10, 2018, 1:05 AM

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Where are you stuck and what have you tried so far?

upper karma Apr 10, 2018, 6:30 PM

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could someone please help in #help-3

waxen gorge Apr 10, 2018, 8:35 PM

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@upper karma did u get it already

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I'm assuming it was already due uwu

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BTW U SHOWED UR TEACHER AGAIN UGH HOW MANY TIMES DO I HAVE TO TELL U

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@ebon vapor can u delete the plebs image

upper karma Apr 11, 2018, 2:21 AM

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yo

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i took the test

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i think i did well

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so im g

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thanks tho

waxen gorge Apr 11, 2018, 2:22 AM

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Lol

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Ur pfp is just slightly making me cringe

upper karma Apr 11, 2018, 5:08 AM

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Heyyo.

upper sedge Apr 11, 2018, 5:11 AM

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Question?

upper karma Apr 11, 2018, 5:13 AM

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Not really.

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I just have a geometry class in high school.

brittle kraken Apr 11, 2018, 3:00 PM

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📎 IMG_20180412_005839.jpg

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Can someone help me with #89?

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B is the correct answer, but I need to learn how to solve it because my teacher expects me to show her solutions. I got that question wrong and I had to re-do it.

lavish haven Apr 11, 2018, 3:27 PM

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My answer I got 95 hope I helped

📎 image.jpg

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@brittle kraken

prisma gorge Apr 11, 2018, 4:01 PM

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Another useful identity is the law of sines: for any triangle with angle measures a, b, and c, with sides of length A, B, and C, where side A is opposite vertex a, side B is opposite vertex B, and side C is opposite vertex c, it can be said that:

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$$\frac{sin(a)}{A} = \frac{sin(b)}{B} = \frac{sin(c)}{C}$$

charred spearBOT Apr 11, 2018, 4:03 PM

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📎 latex.png

prisma gorge Apr 11, 2018, 4:05 PM

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So $$\frac{sin(52)}{(the length not in the picture that i'm not gonna figure out)} = \frac{sin(x)}{6.6cm}$$

charred spearBOT Apr 11, 2018, 4:05 PM

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📎 latex.png

lavish haven Apr 11, 2018, 4:22 PM

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Is that some rule

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Sine*

placid sandal Apr 11, 2018, 4:22 PM

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Do you mean the law of sines?

prisma gorge Apr 11, 2018, 4:29 PM

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Yuh it holds for all triangles in euclidean space

storm compass Apr 11, 2018, 7:25 PM

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Yes just beware. Of the AMBIGUOUS CASE!!!! Ambiguity may be among the midst of the illuminati.

dusk granite Apr 12, 2018, 4:29 AM

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yo

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does anyone know how todo this

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im stumped

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http://prntscr.com/j4465y

Lightshot

Screenshot

Captured with Lightshot

thorn talon Apr 12, 2018, 4:33 AM

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sine rule maybe?

zenith ember Apr 12, 2018, 4:41 AM

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Law of Cosines

brittle kraken Apr 12, 2018, 4:41 AM

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5.2 was the missing side measurement of the opposite side. My bad, @lavish haven

keen aspen Apr 12, 2018, 4:44 AM

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Yep Law of Cosines

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SAS case

brittle kraken Apr 12, 2018, 4:51 AM

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Was the method used by sweetcheaks on my question above the Law of Cosines method? I wasn't sure because idk the formula

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Okay so I made the triangle like what sweetcheaks did. But I don't know what to do next using the law of sines. Ik that I have to input, but I'm confused because idk the other two angles for either A, B, or C.

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📎 IMG_20180412_145708.jpg

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I know that I need to solve to find the other angle, but idk how to do that.

keen aspen Apr 12, 2018, 5:01 AM

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No for that you would use law of sines

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Which you have indicated on your paper

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When you have an angle with a corresponding side and another piece of given information you use law of sines

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If not, then you would most likely resort to the law of cosines

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There are multiple cases when using the law of cosines tho

vast pasture Apr 12, 2018, 9:48 PM

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uh

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📎 unknown.png

zenith ember Apr 12, 2018, 9:53 PM

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The support post is 12 ft tall. The length of the house is 25 ft. The diagram is a right triangle, with another right triangle dropped from the right angle to the hypoteneuse

keen aspen Apr 12, 2018, 9:53 PM

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^

zenith ember Apr 12, 2018, 9:54 PM

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If you consider the angle at the left side of that triangle...

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Then you can use trig identities, and the fact that the two triangles are similar to figure out all the distances.

vast pasture Apr 12, 2018, 9:55 PM

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but i have no idea where to start

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all i have is

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x * 12 = 12 * 25

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i dont think that's correct ;/

zenith ember Apr 12, 2018, 9:55 PM

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You are right. That is not correct.

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You know the trig identities, right? SOHCAHTOA?

vast pasture Apr 12, 2018, 9:56 PM

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no...

zenith ember Apr 12, 2018, 9:56 PM

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Hm.

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Ok.

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I think we're going to need labels to talk about this meaningfully.

vast pasture Apr 12, 2018, 9:57 PM

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it sayts

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geometry

zenith ember Apr 12, 2018, 10:00 PM

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📎 Right_Triangles.bmp

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Ok, so you know AB = 25, and CD = 12.

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You are looking for AD.

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Following me?

waxen gorge Apr 13, 2018, 6:43 PM

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I'm not really

storm compass Apr 14, 2018, 7:31 AM

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No

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Who stole my eyes 😵

urban dawn Apr 14, 2018, 1:50 PM

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question, suppose we have a quadrilateral ABCD, DAB & CBA are complementary angles, AD = BC, AB = 50 CM, CD = 20CM, how can I get the area of ABCD

upper karma Apr 14, 2018, 3:45 PM

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Any diagram...

keen aspen Apr 14, 2018, 4:30 PM

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It'll look like a trapezoid I believe

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Yeah, a regular trapezoid with bases 50 and 20

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And the bottom two angles are 45°

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Just find the vertical height of the trapezoid to find the area

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Just do the tan of it or use your 45 45 90 triangle knowledge to find the height

robust rampart Apr 15, 2018, 1:52 AM

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Bob is standing under the edge of a circular dome. He walks 10 feet from the edge towards the center, looks up and sees the bottom of the dome 50 feet above him. Looking from the edge of the dome he’d been standing under to the opposite end of the dome, his gaze sweeps out a 90∘ angle. How far across is the dome?

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: a. 260 b. 100 c. 60 d. 250 (all in feet)

keen aspen Apr 15, 2018, 4:32 AM

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Not sure, I'm having troubles understanding the word problem

zenith ember Apr 15, 2018, 4:45 AM

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I think it describes this, but I haven't tried to work through the problem.

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📎 Temp.bmp

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Close.

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But that doesn't give an answer that is one of the options.

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If you assume the vertical only goes to the diameter, and is 50, you -do- get one of the options for an answer.

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So, that's probably the right way to look at it. Poorly worded, though.

robust rampart Apr 15, 2018, 1:43 PM

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so what would i end up with?

dusky yarrow Apr 15, 2018, 3:48 PM

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Find the measure of the angle between the two straight lines whose equations are: 2x=6-3y and x-5=0

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<@&286206848099549185>

dusky yarrow Apr 15, 2018, 5:22 PM

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well im gonna post it in #prealg-and-algebra then

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lets see what'll happen

upper karma Apr 15, 2018, 5:51 PM

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Hey guys can someone help me?

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📎 30652781_729342464119926_5416003676365914112_n.png

copper valve Apr 15, 2018, 5:53 PM

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that looks spectacularly spooky

upper karma Apr 15, 2018, 5:53 PM

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It's simpler once I explain it

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so we have this cyclic quadrilateral, and the centers of the inner circles of the diagonals of the quadrilateral form a shape

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I've already realized that it is a rectangle

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but I can't prove it

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can someone give me any tips?

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nd the centers of the inner circles of the diagonals of the quadrilateral form a shape - what does this mean?

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if you connect the dots

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you get a shape

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Ahhh

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Ooh yeah - that's spooky

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yeah I gotta do it tho

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I thought I was close to the answer

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so I begged my math teacher to give me 3 extra days

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and now I'll bring shame to my family if I fail

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I mean all I can think of is to abstract the centers of those circles to some trig relationship and the showing that they must lie on parallel chords.

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Bring shame to your family?

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Doing hard drugs brings shame to your family, not not being able to do tough geometry questions. That's how you give yourself a complex.

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I said it jokingly but I have a feeling my math teacher thinks I'm not smart at all I only work hard for the gains

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so I want to prove it with doing the super hard exercises he gives us for A-s that nobody can do

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I don't care about A-s I care about my imag

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e

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Mmm, try to see this is an emotionally exhausting approach. But it's a good problem

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guess I'll just stare at it

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my theory involves manipulating the central and perimeteral(is that a word?) angles to and see where that takes me

wintry hatch Apr 15, 2018, 6:05 PM

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Owo

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That problem though Owo

upper karma Apr 15, 2018, 6:05 PM

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I get too many lines tho

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You need to derive the coordinates of those centers or atleast the chords that they lie on.

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After that, it's like a couple steps to showing they form a rectangle.

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we haven't learned coordinatal geomtry yet

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Wat

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they teach that in 11th grade here

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So you're doing that all with trig?

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yes

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trying to*

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Dayum.

wintry hatch Apr 15, 2018, 6:07 PM

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thonker

upper karma Apr 15, 2018, 6:08 PM

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well if anyone has an idea don't be shy 😄

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wait

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if I have two lines that cross each other, the opposite angles they form are going to be equal right?

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Mhmm

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I'm doing sth wrong

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lemme check in geogebra

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it's such a hassle to draw shapes in that thing

wintry hatch Apr 15, 2018, 6:11 PM

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Try to get some ratios with thales and thz tangents maybe?

upper karma Apr 15, 2018, 6:12 PM

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we haven't learned tangents yet so I'm not allowed to use that 😦

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we've only learned that for physics competitions and that's unofficial

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So what all can you actually use?

wintry hatch Apr 15, 2018, 6:13 PM

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I mean, the tangents to the circle

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Which are the diagonals

upper karma Apr 15, 2018, 6:14 PM

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Simson lines

wintry hatch Apr 15, 2018, 6:14 PM

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(well the diagonals and the sides too)

upper karma Apr 15, 2018, 6:14 PM

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angel of view

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angle*

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umm central and the other angles

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you know where AOB <) = 1/2 APB <)

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if that makes sense

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idk the English word for it unfortunately

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cyclic quadrilaterals

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mostly that

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but those exercises are usually logic based

wintry hatch Apr 15, 2018, 6:16 PM

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You know much more geometry than I do 👀

upper karma Apr 15, 2018, 6:17 PM

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oh I highly doubt that

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I'm just making it look much

wintry hatch Apr 15, 2018, 6:18 PM

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I just know thales and Pythagore 👀 👀

upper karma Apr 15, 2018, 6:19 PM

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those are very useful tools

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I have a theory now

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but it doesn't feel right

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Run us through it

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let me draw it in paint because my English mathematical dictionary is too lacking to describe it

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so I can prove that

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📎 30652781_729342464119926_5416003676365914112_n.png

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the angles described by the same color are equal

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if two angles of a rectangle are equal that means they are similar

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and that proves something because the center of the inner circle of a triangle always stays in the same place no matter how you change it's sides as long as the two triangles are similar

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it is as if I mirrored the similar triangles and then changed their sides

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aand

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that proves nothing

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damn

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It's a start.

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I have an idea

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turns out

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all four triangles are similar

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and from here if I use the green and the purple angles

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that could lead to somewhere

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I can appreciate that the opposite triangles are similar.

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it's as if I rotated the central points by the green and the purple angles

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But I am skeptical that all four triangles are similar.

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they always have a red-black angle and a red and a black angle

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if I'm correct

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Yeah

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I was going to point out one diagrammatic inconsistency.

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But you were getting somewhere with opposite triangles being similr.

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could you explain what diagrammatic inconsistency is? 😄

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please

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Sure, the two red angles inside one of the triangles are not equal unless the chord in the triangle is parallel to the bisection of the green angle.

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Or if the where the two diagonal lines happen to cross is the origin of the larger inscribing circle.

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they should be tho should they not

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since green=2black

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I mean

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purple=2black

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green=2red

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if we only look at two angles

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and from there you can go around with that and name every angle red or black or green or purple

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can you not?

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I'm just intuiting from the diagram.

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You have two diagonals that meet right? That gives you a green angle.

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Now only if you draw a line that intersects the two diagonals at points the same distance from where they diagonals meet, will you get two equal angles.

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This is not necessarily the case with these triangles.

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oh of course I'm so stupid

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The opposites triangles are similar but the two angles aren't.

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indeed

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And of that, I am also dubious.

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the black and red angles are not at all equal

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like nowhere

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I messed it up

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okay I have no idea 😄

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It's a good place to start.

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I'd start by proving that the opposite triangles are similar though.

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I can't prove it because I made a wrong assumption in the beginning so it's uncertain the two opposite triangles are similar

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tho

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I could use the fact that the shape is a cyclic quadrilateral

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unfortunately I have no idea how to take that notion further

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hmmm unless

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nope

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I mean this is one hell of a problem to tackle with the cyclical geometry.

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with only cyclical geometry.

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I can use angles of view

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and simson lines

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but that's pretty much it

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we've only started learning about angles and stuff

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hmm

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I'll think about it until tomorrow, I usually manage to figure out though problems right before I fall asleep

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thanks for the help though @upper karma ^^

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I still adore your name

wintry hatch Apr 15, 2018, 7:08 PM

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You should really look up the tangents to the circles

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I've tried drawing your figure with Geogebra and it's very hard to keep up with the tangents things, so it means there are some precise conditions to them

dusk quartz Apr 17, 2018, 5:32 AM

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yo, is there a way to calculate the area of a shape that has 4 sides, but neither one is the same?

patent lion Apr 17, 2018, 5:33 AM

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What measurements about the shape are you given?

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Like side lengths and angle measures I mean

dusk quartz Apr 17, 2018, 5:34 AM

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well, the angle measures i don't think so, but i have all of the side lenghts, the left side, the right side, the top, and the bottom

patent lion Apr 17, 2018, 5:34 AM

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I think you need at least one angle measure

dusk quartz Apr 17, 2018, 5:35 AM

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actually, i think that all of the angle measures are 90 degrees

patent lion Apr 17, 2018, 5:36 AM

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Well, if they are all 90 then the shape is a rectangle

dusk quartz Apr 17, 2018, 5:36 AM

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well, maybe, but the shape's height isn't the same, for example, on the left side, it is 5, but on the right, its 7

patent lion Apr 17, 2018, 5:36 AM

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Do you have a picture?

dusk quartz Apr 17, 2018, 5:38 AM

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alright, just a sec, here i will make one. Also thanks for taking your time to help me :D

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here, something like this

📎 somethinglikethistoo.png

patent lion Apr 17, 2018, 5:41 AM

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Oh

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It's a trapezoid

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So

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The area

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Where the side on the right is b_1

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The side on the left is b_2

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And the one on the bottom (the same as the distance between b_1 and b_2) is h

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The area can be written as

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$$A = \frac{b_1 + b_2}{2} \cdot h$$

charred spearBOT Apr 17, 2018, 5:43 AM

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📎 latex.png

dusk quartz Apr 17, 2018, 5:43 AM

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thanks so much :D

patent lion Apr 17, 2018, 5:43 AM

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This is only the case when b_1 and b_2 are parallel

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And if the bottom side is slanted, then you need the distance between b_1 and b_2 for h

dusk quartz Apr 17, 2018, 5:44 AM

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ah, i understand, thanks so much again ^^

dusk quartz Apr 17, 2018, 6:15 AM

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sorry to plop in so quickly again, but how do i calculate a triangle's height given that i know it's 3 sides?

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sorry this is probably a noob question :P

patent lion Apr 17, 2018, 6:17 AM

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That's actually something I don't know offhand

dusk quartz Apr 17, 2018, 6:38 AM

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:(

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well, if anyone knows, please tell me

dusk quartz Apr 17, 2018, 7:19 AM

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alright, after a little bit of digging, i found this : https://www.mathopenref.com/heronsformula.html
this tells the area of a triangle without the height, and this is the solution i was looking for

dusk quartz Apr 17, 2018, 10:41 AM

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alright, sorry to blop in again, but how would i calculate the distance between b_1 and b_2?

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this has really been bothering me

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i can't figure it out at all :(

charred spearBOT Apr 17, 2018, 10:46 AM

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📎 latex.png

upper urchin Apr 17, 2018, 10:46 AM

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For any two points fo the form (x1,x2,x3,...,xn)

brittle kraken Apr 17, 2018, 11:38 AM

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📎 IMG_20180417_213757.jpg

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📎 Screenshot_20180417-140328.png

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Every time I keep trying to find sinB, I always get an error

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Anyone know what I'm doing wrong?

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The answer is supposed to be 95%

covert orchid Apr 17, 2018, 11:50 AM

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@brittle kraken i dont know a lot of english but the number that is inside the sen-1 must be in a range between -1 and 1

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If you try to calc the sen-1 of 1.00000001 you're gonna get an error

brittle kraken Apr 17, 2018, 11:51 AM

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Well damn

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I'm trying to get an angle that is 95 degrees, but I don't know what to do

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I've used the law of sines

covert orchid Apr 17, 2018, 11:53 AM

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Let me check...

thorn talon Apr 17, 2018, 11:54 AM

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is this the start of the question?

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or follow on from previously found values?

brittle kraken Apr 17, 2018, 12:23 PM

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Follow on, hold on, I'll take a picture of the question itself

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I drew that so I could see what law would work

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📎 1523967921427.jpg

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It is supposed to be 95 degrees, but I'm trying to show work

full mortar Apr 17, 2018, 12:26 PM

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can anyone help with #6... i never learned how to do this

📎 image.jpg

brittle kraken Apr 17, 2018, 12:26 PM

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I tried using the law of sines but idk if I did it correctly, since what I've done cannot be converted into an angle

shadow anvil Apr 17, 2018, 12:36 PM

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=tex \frac{\sin\left(52\right)}{5.2}=\frac{\sin\left(\theta\right)}{6.6}\\sin\left(\theta\right)=\frac{6.6\sin\left(52\right)}{5.2}\\theta=\sin^{-1}\left(\frac{6.6\sin\left(52\right)}{5.2}\right)

charred spearBOT Apr 17, 2018, 12:36 PM

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📎 latex.png

shadow anvil Apr 17, 2018, 12:37 PM

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=pup calc \frac{6.6\sin\left(\frac{\pi}{180}*52\right)}{5.2}

charred spearBOT Apr 17, 2018, 12:37 PM

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📎 result.png

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Query made by @shadow anvil
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=calc+\frac{6.6\sin\left(\frac{\pi}{180}*52\right)}{5.2}
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

shadow anvil Apr 17, 2018, 12:38 PM

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oh

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Ok

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That's interesting

brittle kraken Apr 17, 2018, 1:34 PM

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Yeah.... I need to get 95 degrees as my angle, but I can't seem to since my calculator always says I got an error.

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Can someone help me with this question please?

📎 1523967921427.jpg

brittle kraken Apr 17, 2018, 1:54 PM

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I did.

📎 IMG_20180417_213757.jpg

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Got up to that but I end up getting the same exact thing, not getting the answer that I'm supposed to have

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📎 Screenshot_20180417-140328.png

covert orchid Apr 17, 2018, 1:55 PM

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Wait

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@brittle kraken i did this

📎 IMG_20180417_100007.jpg

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The result says 83.16 degrees

brittle kraken Apr 17, 2018, 2:03 PM

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Damn... Sorry to say, but the answer HAS to be 95° as it was the correct answer when I scanned it.

covert orchid Apr 17, 2018, 2:06 PM

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I mean i think that you cannot do it with las of sines because i'm pretty sure that the sen-1 will never be more than 90 degrees

brittle kraken Apr 17, 2018, 2:06 PM

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I see

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Does law of cosines work?

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I tried it earlier but I get different results.

covert orchid Apr 17, 2018, 2:07 PM

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Let me see what you did

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What

brittle kraken Apr 17, 2018, 2:08 PM

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Dw I threw my work away using law of cosines as a form of frustration lol

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<@&286206848099549185> care to help with that please? I'm definitely confused.

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@celest obsidian it has to be possible because the answer was generated to be 95°>

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I have

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No

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📎 1523967921427.jpg 📎 IMG_20180417_213757.jpg

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What I've done

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It does?

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I could but my teacher wants specific details in order for myself to move on

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:/

charred spearBOT Apr 17, 2018, 2:27 PM

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📎 latex.png

brittle kraken Apr 17, 2018, 2:27 PM

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Right, thank you

covert orchid Apr 17, 2018, 2:27 PM

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Yay

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At least i did know what to do jajaja

upper karma Apr 17, 2018, 7:56 PM

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good for u

brittle kraken Apr 17, 2018, 10:27 PM

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Can it be CosB btw?

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Considering the angle for A is given already

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Unless I'm doing this incorrectly

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I'm getting .99994 and .620

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📎 Screenshot_20180418-083652.png 📎 Screenshot_20180418-083657.png

covert orchid Apr 17, 2018, 10:42 PM

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wat

brittle kraken Apr 17, 2018, 10:49 PM

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I'm trying to use the law of cosines, all I'm getting now is weird answers

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They don't show errors, but I always get .99999 or .6203

covert orchid Apr 17, 2018, 10:49 PM

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let me 5 min to recall the lay of cosines i'll help

brittle kraken Apr 17, 2018, 10:49 PM

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Unless I'm doing this cosine thing wrong

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Alright

covert orchid Apr 17, 2018, 10:52 PM

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okay

sturdy bolt Apr 17, 2018, 10:57 PM

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is your calculator set to degrees or radians @brittle kraken

brittle kraken Apr 17, 2018, 10:57 PM

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Degrees

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Is it supposed to be radians?

covert orchid Apr 17, 2018, 10:59 PM

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@brittle kraken

📎 IMG_20180417_185916.jpg

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It's okay in degrees

brittle kraken Apr 17, 2018, 11:00 PM

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Is that 0109 or 0.09?

covert orchid Apr 17, 2018, 11:00 PM

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0.09

brittle kraken Apr 17, 2018, 11:00 PM

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Oh okay thank you

covert orchid Apr 17, 2018, 11:01 PM

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Hahahah you're welcome, i'm sorry in latin America we use the comma instead of dot

brittle kraken Apr 17, 2018, 11:01 PM

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It's fine

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Also why does the negative symbol move up (it was at the denominator but now it's at the numerator)?

covert orchid Apr 17, 2018, 11:03 PM

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You can move up the negative if it's in the denominator

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If you want you can leave the negative symbol in the denominator

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You're gonna get the same result

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Try using the negative symbol in the denominator

upper karma Apr 18, 2018, 12:23 AM

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📎 unknown.png

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help

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needed

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asap

upper karma Apr 18, 2018, 1:10 AM

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google cosine rule and apply it

full mortar Apr 18, 2018, 2:51 AM

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What does this mean

📎 image.jpg

dusk quartz Apr 18, 2018, 5:32 AM

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hello everybody, i need some help, i have no idea how to calculate the distance between the blue line, and the red line. Please help! [also, the lines are not parrellel]

📎 somethinglikethistee.png

dusk quartz Apr 18, 2018, 6:00 AM

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anybody? please?

copper valve Apr 18, 2018, 6:04 AM

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the blue and red are not parallel?

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what information are you given

opal blaze Apr 18, 2018, 6:11 AM

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yes becase they are 90 degrees that means part of the circle the angles are spanning over is equal to 180 degrees or pi. Therefore it is the diameter of the circle. @full mortar

dusk quartz Apr 18, 2018, 6:15 AM

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@copper valve well im given the size of all of the lines, is that enough?

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@opal blaze what do you mean?

dusk quartz Apr 18, 2018, 6:35 AM

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?

opal blaze Apr 18, 2018, 6:46 AM

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Sorry wrong person @dusk quartz

dusk quartz Apr 18, 2018, 6:46 AM

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ah, okay

opal blaze Apr 18, 2018, 6:49 AM

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Well if the black lines are paralell and of the same length. That means the blue line and the red line have the same length.

dusk quartz Apr 18, 2018, 6:49 AM

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well, maybe, but how would i calculate the distance between them?

opal blaze Apr 18, 2018, 6:51 AM

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I am unsure if that is possible only given the length of 2 parlell lines.

dusk quartz Apr 18, 2018, 6:51 AM

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:(

opal blaze Apr 18, 2018, 6:52 AM

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Are there more information and does it have an awnser?

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Brb

dusk quartz Apr 18, 2018, 6:53 AM

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well, it sadly doesn't have a answer. Since this isn't a quiz, im using this for programming purposes. Also, it depends, since it's for programming purposes, maybe i can try and get the angels of the corners. But would that help?

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alright, tell me when you're back

dusk quartz Apr 18, 2018, 7:20 AM

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hello?

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well, i guess you're not coming back :(

opal blaze Apr 18, 2018, 7:34 AM

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I am in school atm. But if you think of this as vectors you need 2 to creat a new one. @dusk quartz

dusk quartz Apr 18, 2018, 8:33 AM

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?

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im not really sure what you mean?

dusk quartz Apr 18, 2018, 9:45 AM

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alrgiht, ping me when u get back

opal blaze Apr 18, 2018, 10:05 AM

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@dusk quartz you need a relation between the two lines.

dusk quartz Apr 18, 2018, 10:24 AM

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sorry for being a noob, but can you explain a bit further? sorry again. EDIT: Actually, i got another idea that i can try to solve my problem, thanks larzanda

wheat bramble Apr 18, 2018, 10:29 AM

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does the alternate segment theoreum work on all triangles inscribed in a circle?

waxen gorge Apr 18, 2018, 12:45 PM

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@wheat bramble ya

wheat bramble Apr 18, 2018, 12:57 PM

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how can I proof that two different angles in a circle subtended from the same arc has the same value?

raven robin Apr 18, 2018, 2:57 PM

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Could I have a few example problems on the formula rhombus, trapezoid, and triangle?

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I'm reviewing for a quiz

upper karma Apr 18, 2018, 4:59 PM

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Hi, do somebody know how to prove this property:

Let be 4 points A, B, C, D, if and only if [AC] and [BD] intersect in their middle then the quadrilateral ABCD is a parallelogram.

upper sedge Apr 19, 2018, 2:16 AM

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Since the question concerns how to prove it, I'll take that to mean "What would the structure of such a proof look like?" as I proceed.

It's an if and only if statement, so first break it into two parts.

a) if [AC] and [BD] intersect in the middle, then the quadrilateral ABCD is a parallelogram.

b) if the quadrilateral ABCD is a parallelogram, then [AC] and [BD] must intersect in the middle.

Very loosely: (Read: There is more rigor to apply here that I'm glossing over.)

a) Suppose I have lines [AC] and [BD], and suppose that they intersect at the midpoint, m, between all points. Now, take the line from m to the line [AB] such that it intersects [AB] at a right angle. If we continue the line in the other direction, note that [CD] is also intersected at a right angle. This shows that [AB] and [CD] are parallel. Do the same for [AD] and [BC] to complete this sub-proof.

b) This is essentially a) in reverse. Start by noting that the parallel lines in the parallelogram are parallel, and find the midpoints for the lines that form the edges of the parallelogram, and then show from those the midpoint of the parallelogram. Finally, show that it's the same as the point where the diagonals intersect.

From here, since will have shown both the if case and the only if case, the theorem will have been proven.

I'll reiterate, this is very loose, and I'm skipping over the algebraic part of it because the question seems directed at the structure of the proof.

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Does this help @upper karma ?

zenith ember Apr 19, 2018, 2:22 AM

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I did it by alternate interior angles and similar triangles. 😛

upper sedge Apr 19, 2018, 2:28 AM

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That works too!

upper karma Apr 19, 2018, 5:06 AM

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📎 image.jpg

upper karma Apr 19, 2018, 7:19 AM

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Which question?

remote kiln Apr 20, 2018, 12:07 AM

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hey all - i have a set of four points (some sort of quadrilateral) on a 3D plane...is there a way to find a clockwise ordering of these points, about the plane's normal?

zenith ember Apr 20, 2018, 4:44 AM

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I can think of a way to do it, but it's pretty tedious.

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Basically pick any three of the four points, and use them to generate an orthogonal basis for the plane. Then transform ABCD into that basis and convert to polar.

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Thinking a little more about it, if you pick one point, and take the three difference vectors to the remaining three points...

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The magnitude of the cross product divided by the magnitude of the vectors gives you the sin of the angle between them.

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I think you could use that to generate an ordering.

supple lotus Apr 20, 2018, 7:55 AM

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Hello, I have following function for shader to draw a cone:
p is Vector3 of xyz of starting point, not sure what c stands for exactly.
Could someone enlighten me of how to properly edit c?
If I change c[0] it changes the narrowness of cone, but c[1] affect it also

📎 unknown.png

zenith ember Apr 20, 2018, 2:47 PM

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I don't have enough information to guess what this is doing. Sorry

upper karma Apr 20, 2018, 3:57 PM

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help

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📎 IMG_20180420_115823.jpg

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nvm I got it

remote kiln Apr 20, 2018, 4:18 PM

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@zenith ember yep, the signed angle method is what i ended up doing..a bit tedious, as you said, but works!

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thank you!

zenith ember Apr 20, 2018, 4:25 PM

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yw

upper karma Apr 21, 2018, 6:55 AM

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Hey I’m not in geometry is the answer 54?

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I think it is because 26+82=108

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and 180-108 = 172

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I mean 72*

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So angle b is 72

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And then

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180-72=108

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108/2=54

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=pup solve 54*2

charred spearBOT Apr 21, 2018, 6:56 AM

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📎 result.png

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Query made by @upper karma
Data sourced from Wolfram|Alpha: http://www.wolframalpha.com/input/?i=solve+54*2
Do more with Wolfram|Alpha Pro: http://www.wolframalpha.com/pro/

upper karma Apr 21, 2018, 6:56 AM

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Ok I think I got it

charred spearBOT Apr 21, 2018, 8:42 AM

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Rendering failed. Check your code. You can edit your existing message if needed.

brittle kraken Apr 21, 2018, 12:09 PM

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http://prntscr.com/j87g44

Lightshot

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What do I have to do to solve this?

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Do I need to follow a theorem or something?

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Is the answer C?

brittle kraken Apr 21, 2018, 1:11 PM

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Yeah I doubt it has a theorem, thanks btw.

brittle kraken Apr 21, 2018, 1:59 PM

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http://prntscr.com/j88fri

Lightshot

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📎 IMG_20180421_235820.jpg

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I keep getting -22.3 as my x but it ends up being wrong.

upper urchin Apr 21, 2018, 2:00 PM

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what equation are you starting with?

brittle kraken Apr 21, 2018, 2:01 PM

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I started with 84.5 - 5x = 642 + 20x, then turned it into 84.5 - 25x = 642

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I tried going the other side to make it 84.5 = 642 + 25x, but I get the same result

upper urchin Apr 21, 2018, 2:04 PM

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are you multiplying the sides on the same triangle?

brittle kraken Apr 21, 2018, 2:06 PM

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No

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Am I supposed to?

upper urchin Apr 21, 2018, 2:06 PM

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no

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I'm just trying to work out how you got that equation

brittle kraken Apr 21, 2018, 2:06 PM

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Oh

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I followed the example that was given to me online
http://prntscr.com/j88iq3
http://prntscr.com/j88imj

Lightshot

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Lightshot

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upper urchin Apr 21, 2018, 2:07 PM

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oh I see

brittle kraken Apr 21, 2018, 2:07 PM

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But to no avail with a valid answer lol

upper urchin Apr 21, 2018, 2:08 PM

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you're not combining the fraction correctly

brittle kraken Apr 21, 2018, 2:08 PM

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Ah

charred spearBOT Apr 21, 2018, 2:08 PM

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📎 latex.png

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📎 latex.png

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📎 latex.png

brittle kraken Apr 21, 2018, 2:09 PM

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Ye I cross multiplied

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Then I got that equation

upper urchin Apr 21, 2018, 2:10 PM

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you cross multiplied wrong

brittle kraken Apr 21, 2018, 2:10 PM

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Oh

upper urchin Apr 21, 2018, 2:10 PM

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if what I'm saying

brittle kraken Apr 21, 2018, 2:10 PM

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Fuck

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YEAH

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I see it now lmfao

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My bad, thank you very much

upper urchin Apr 21, 2018, 2:10 PM

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np

brittle kraken Apr 21, 2018, 2:14 PM

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Okay now I don't know how to do this one

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http://prntscr.com/j88l5r

Lightshot

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Captured with Lightshot

upper urchin Apr 21, 2018, 2:17 PM

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well what do you need to be able to uniquely define a triangle?

brittle kraken Apr 21, 2018, 2:18 PM

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Theorems/postulates?

zenith ember Apr 21, 2018, 5:03 PM

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Think SSS

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Or SAS

stuck jewel Apr 22, 2018, 2:08 AM

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📎 unknown.png

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what do i do next?

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i cant isolate "a" with the inverse cosh

zenith ember Apr 22, 2018, 4:38 AM

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I would sub in the 1/2(e^x + e^-x) formula for cosh

stuck jewel Apr 22, 2018, 6:04 AM

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how would i log a sum of (e^x + e^-x)?

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actually, how do you log something simpler like this?

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=tex 2^3 + 2^4

charred spearBOT Apr 22, 2018, 6:06 AM

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📎 latex.png

zenith ember Apr 22, 2018, 4:28 PM

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You don't.

stoic adder Apr 22, 2018, 5:03 PM

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📎 math_circle1.PNG

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can someone prove to me why this is true?

zenith ember Apr 22, 2018, 5:39 PM

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@stoic adder I assume the inscribed triangle is equilateral?

stoic adder Apr 22, 2018, 5:39 PM

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yes

zenith ember Apr 22, 2018, 5:41 PM

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Ok. Lemme work on it a bit.

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📎 geogebra-export.png

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@stoic adder So, fundamentally, this problem comes down to one of similar triangles.

stoic adder Apr 22, 2018, 5:47 PM

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ok

zenith ember Apr 22, 2018, 5:48 PM

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What you're trying to show is that the radius of the larger circle is twice the radius of the smaller circle.

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The radius of the larger circle is DC

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And the radius of the smaller circle is DA_1

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Following me so far?

stoic adder Apr 22, 2018, 5:49 PM

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yeah

zenith ember Apr 22, 2018, 6:00 PM

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Ok.

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So the most interesting triangle is DCA_1

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And you can show, pretty easily, that DCA_1 is a 30-60-90 right triangle.

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Which means that DC is twice DA_1.

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And you're done. 😃

mortal briar Apr 22, 2018, 8:52 PM

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This is my problem

upper karma Apr 22, 2018, 8:55 PM

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?

mortal briar Apr 22, 2018, 8:56 PM

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I have a fan that is 4.7 x 4.7. Height is no issue. These fans need to be equally spaced across a block of aluminum with the remader of free space in-between each fan. Meaning the 2 most outer fans would be parallel vertically with the ends of the block.

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The block is 24 inches long

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does that make since?

upper karma Apr 22, 2018, 9:15 PM

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@rugged moat yay can help (let's bring question here as the other channel is busy)

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Mad scientist is here to help 😃

mortal briar Apr 22, 2018, 9:15 PM

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ok

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I was confused because the rooms just disappeared

upper karma Apr 22, 2018, 9:15 PM

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naag sry @mortal briar that's another question

rugged moat Apr 22, 2018, 9:16 PM

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I am here

upper karma Apr 22, 2018, 9:16 PM

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but you can ping Helpers so that helpers will notice your question @mortal briar

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well $$\dfrac{|b|}{|c|}$$ is what you want right?

charred spearBOT Apr 22, 2018, 9:16 PM

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📎 latex.png

upper karma Apr 22, 2018, 9:16 PM

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but what is $$|c|$$

charred spearBOT Apr 22, 2018, 9:16 PM

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📎 latex.png

rugged moat Apr 22, 2018, 9:16 PM

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📎 Screenshot_20180423-023625.jpg

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Yup

upper karma Apr 22, 2018, 9:16 PM

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or actually $$|c|^2$$

charred spearBOT Apr 22, 2018, 9:16 PM

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📎 latex.png

upper karma Apr 22, 2018, 9:17 PM

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it's $$c\cdot c$$

charred spearBOT Apr 22, 2018, 9:17 PM

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📎 latex.png

upper karma Apr 22, 2018, 9:17 PM

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but $$c=a+b$$

charred spearBOT Apr 22, 2018, 9:17 PM

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📎 latex.png

upper karma Apr 22, 2018, 9:17 PM

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so this becomes $$(a+b)\cdot(a+b)$$

charred spearBOT Apr 22, 2018, 9:17 PM

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📎 latex.png

upper karma Apr 22, 2018, 9:17 PM

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expand & Eurêka

mortal briar Apr 22, 2018, 9:18 PM

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Ummm what just happend?

rugged moat Apr 22, 2018, 9:19 PM

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What do you mean by expand?

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@upper karma

mortal briar Apr 22, 2018, 9:21 PM

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<@&286206848099549185>

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I have a fan that is 4.7 x 4.7. Height is no issue. These fans need to be equally spaced across a block of aluminum with the remader of free space in-between each fan. Meaning the 2 most outer fans would be parallel vertically with the ends of the block.
The block is 24 inches long

upper karma Apr 22, 2018, 9:23 PM

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@rugged moat things like $$(x+y)z=xz+yz$$

charred spearBOT Apr 22, 2018, 9:24 PM

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📎 latex.png

rugged moat Apr 22, 2018, 9:24 PM

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a2 +b2+2ab?

upper karma Apr 22, 2018, 9:27 PM

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yep

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but

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$$a\cdot a=|a|^2$$

charred spearBOT Apr 22, 2018, 9:27 PM

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📎 latex.png

upper karma Apr 22, 2018, 9:27 PM

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same for b

wary steeple Apr 23, 2018, 12:00 AM

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how do you calculate the 'minimum' tangent for a given point?

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from a line that extends from the middle of a circle

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📎 angle_of_tangent.png

upper karma Apr 23, 2018, 12:55 AM

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What is that 40 degrees measuring?

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Like what angle?

wary steeple Apr 23, 2018, 12:55 AM

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the point on the circlke

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if 0 is top

upper karma Apr 23, 2018, 12:56 AM

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Confused

wary steeple Apr 23, 2018, 12:56 AM

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how high does the line that goes straight up need to be to have diect line of sight with a point at 40 degrees

upper karma Apr 23, 2018, 12:57 AM

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What is your answer in terms of? Radians?

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Doesn't look like you were given any units

wary steeple Apr 23, 2018, 1:07 AM

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the radius is 2000

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40 / 90 is .44....

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so i took

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44 percent of 2000

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is 880

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so the slope is the ratio of 880 over 2000

sudden stream Apr 23, 2018, 1:38 AM

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📎 image.png

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I need help with both cause I don’t understand either

keen aspen Apr 23, 2018, 1:56 AM

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Do you know the relations of side lengths with secant lines?

sudden stream Apr 23, 2018, 2:43 AM

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@keen aspen Nope.

keen aspen Apr 23, 2018, 2:43 AM

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(A+B)*B=(C+D)*D

sudden stream Apr 23, 2018, 2:44 AM

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Cool. Now how is that applyable to Question 5?

keen aspen Apr 23, 2018, 2:44 AM

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(5+4)*4=(ZC+2)*2

sudden stream Apr 23, 2018, 2:45 AM

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I see.

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Could I have help with

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6?

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@keen aspen

keen aspen Apr 23, 2018, 2:47 AM

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Angle BMN=1/2(AP-BN)

sudden stream Apr 23, 2018, 2:47 AM

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We have to find AP

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We have BMN

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it's 12(degrees)

keen aspen Apr 23, 2018, 2:48 AM

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12=1/2(AP-33)?

sudden stream Apr 23, 2018, 2:48 AM

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yes

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Ohh

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I see.

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Thanks!

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@keen aspen so the final answer would be 57

keen aspen Apr 23, 2018, 2:50 AM

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yes

sudden stream Apr 23, 2018, 2:51 AM

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cool

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thanks.

upper karma Apr 23, 2018, 6:07 AM

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I miss geometry lol

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When math was so simple

upper karma Apr 23, 2018, 9:34 AM

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@ocean quartz Back to your problem

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so I edited the question where I explained furthur things I understood

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https://math.stackexchange.com/questions/2748606/determining-the-radius-of-a-circle-knowing-an-arc-length-and-a-distance?noredirect=1#comment5672674_2748606

Determining the radius of a circle knowing an arc length and a dis...

Here's the problem:
($2,5$ is $2.5$)

To determine $r$, I used Pythagoras and trigonometry to find that:
$\angle{BOC}=\dfrac{\beta}{r}$
$\tan{\dfrac{\beta}{r}}=-\dfrac{\sqrt{(\alpha-r)^2-r^2}}{r}...

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no one answered me, but someone commented but couldnt' find his ideas helpful

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never mind

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so at the end

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for your algorithm

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given any $$\alpha$$ and $$\beta$$

charred spearBOT Apr 23, 2018, 9:35 AM

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📎 latex.png

upper karma Apr 23, 2018, 9:35 AM

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you can determine $$r$$ as follows

charred spearBOT Apr 23, 2018, 9:35 AM

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📎 latex.png

upper karma Apr 23, 2018, 9:35 AM

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firstly

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check that $$\beta\in\left(0,\dfrac{\alpha\pi}{2}\right)$$

charred spearBOT Apr 23, 2018, 9:35 AM

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📎 latex.png

upper karma Apr 23, 2018, 9:36 AM

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because with this construction

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$$\beta$$ can't exceed $$\dfrac{\alpha\pi}{2}$$

charred spearBOT Apr 23, 2018, 9:36 AM

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📎 latex.png

upper karma Apr 23, 2018, 9:36 AM

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secondly

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once this condition is satified

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you consider the function

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$$f(x)=\beta-x\pi+x\arctan\left(\dfrac{\sqrt{(\alpha-x)^2-x^2}}{x}\right)$$

charred spearBOT Apr 23, 2018, 9:37 AM

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📎 latex.png

upper karma Apr 23, 2018, 9:37 AM

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this function will have exactly one solution

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that you can approximate using numerical methods

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the easisest method is this:

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https://en.wikipedia.org/wiki/Bisection_method

Bisection method

The bisection method in mathematics is a root-finding method that repeatedly bisects an interval and then selects a subinterval in which a root must lie for further processing. It is a very simple and robust method, but it is also relatively slow...

ocean quartz Apr 23, 2018, 9:54 AM

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@upper karma thank you very much! I actually just finished. I'll make a demo in a moment.

upper karma Apr 23, 2018, 10:02 AM

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You're welcome 😃

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that's cool!

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Would like to see the results!

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😉

ocean quartz Apr 23, 2018, 10:35 AM

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Probably will look broken in discord. Should work normally in browser.

📎 circle_param0.gif

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📎 circle_param1.gif

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@upper karma @upper sedge here is the loop I used, as promised: https://pastebin.com/ha8pTXWB
It's a shame I wasn't able to google proper implementation (looked for "binary search" instead of "bisection method"), so I stopped once it started to output proper result.

Pastebin

[C] int i = 0; while (f > 1e-5 && i < 64) { rad += step; ...

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@upper karma also I used yesterday's formula. Maybe will redo with later one.

brittle kraken Apr 23, 2018, 12:07 PM

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http://prntscr.com/j8yqos
For that kind of question, do I have to find a triangle that looks identical to the one above?

Lightshot

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Captured with Lightshot

eager pendant Apr 23, 2018, 12:09 PM

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do you know what similar means?

brittle kraken Apr 23, 2018, 12:09 PM

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Not in mathematical terms lol

fallow quail Apr 23, 2018, 12:09 PM

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For similar triangles ABC~DEF, with sides a,b,c and d,e,f respectively, they usually use stuff like:
a/d=b/e comparing the ratio of 2-2 sides of the different triangles, but can you use a/b=d/e for e.g? Checking the ratio of 2 sides in each triangle?

eager pendant Apr 23, 2018, 12:09 PM

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@brittle kraken look up what a similar triangle is in geomtry

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and make sure you can make the distinction

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between a similar triangles and congruent triangles

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@fallow quail yes. in fact, given you know you can use a/d=b/e, you can get to ae=bd, which means a/b=d/e

fallow quail Apr 23, 2018, 12:12 PM

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Ah, right, thank you. Just the fact that my math book always used only one of those, kinda confused me

brittle kraken Apr 23, 2018, 12:16 PM

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Jazza - Today at 10:09 PM
@Kyle look up what a similar triangle is in geomtry(edited)
and make sure you can make the distinction
between a similar triangles and congruent triangles(edited)```

Okay, so in using the information I got from similar triangles here: 
https://www.varsitytutors.com/hotmath/hotmath_help/topics/similar-triangles
My answer is A, right? since the shape is scaled down from its actual size

Similar Triangles

Two triangles are said to be similar if their corresponding angles are congruent and the corresponding sides are in proportion. In other words, similar triangles are the same shape, but not necessarily the same size.

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Or is it B?

fallow quail Apr 23, 2018, 12:19 PM

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You need to either have all 3 sides of the triangle be proportional to each other, or 2 sides and an angle between them being the same

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Neither A or B has that

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27/8 does not equal 45/19 and 15/6 does not equal 30/19

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@brittle kraken

brittle kraken Apr 23, 2018, 12:20 PM

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I see

fallow quail Apr 23, 2018, 12:21 PM

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But if you check C, 24/36 will be equal to 30/45

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So you can say C is similar

brittle kraken Apr 23, 2018, 12:39 PM

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Alright. And this should be A right? since the scale factor is 5/4.
http://prntscr.com/j8z7po

Lightshot

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Captured with Lightshot

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45/36 = 50/40 = 40/32 all scaled up to 5/4

upper sedge Apr 23, 2018, 1:39 PM

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Yep

upper karma Apr 23, 2018, 1:44 PM

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@ocean quartz Oh nice animations!

upper sedge Apr 23, 2018, 1:44 PM

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Animations?

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Did I miss something interesting?

upper karma Apr 23, 2018, 1:45 PM

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Shenzao RIP you were pinged xD

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look ahead

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oh notice that when alpha groes, sure r groes as well

upper sedge Apr 23, 2018, 1:45 PM

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Ah, I see the tag.

upper karma Apr 23, 2018, 1:45 PM

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but seems that r doesn't grow too much!

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in fact from the formula

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$$\beta=r\pi-r\arctan\left(\dfrac{\sqrt{(\alpha-r)^2-r^2}}{r}\right).$$