#geometry-and-trigonometry | Mathematics | Page 292

upper karma Jun 20, 2020, 3:36 PM

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Otherwise someone like me wouldn't have understood

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Other lines seem okay! 👍

nova prawn Jun 20, 2020, 3:37 PM

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If you are avaliable

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can you chech #help-0

pine oar Jun 20, 2020, 8:08 PM

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https://i.imgur.com/T0ckEKu.png

Imgur

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What is the name of this?

upper karma Jun 20, 2020, 8:15 PM

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Uh hold on for the name

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Tho it is used to get the area of a triangle

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Wait a sec

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Welp i don't think there is an specific name

pine oar Jun 20, 2020, 8:18 PM

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dang

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yeah its used to find triangle area

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i wanted to see some application

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on a video

silent plank Jun 20, 2020, 8:20 PM

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trig formula for area of a triangle probs gets you what you need

pine oar Jun 20, 2020, 8:23 PM

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ok

upper karma Jun 21, 2020, 6:28 AM

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Yes

naive harbor Jun 21, 2020, 2:22 PM

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Hello. I have a couple of identically sized rectangles like such:

📎 unknown.png

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If i were to calculate the x distance between the centers of both rectangles would the following expression be correct?

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(2 x distance from rectangle center to edge) + x seperation.

upper karma Jun 21, 2020, 2:26 PM

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That's not a formula, that is an expression

naive harbor Jun 21, 2020, 2:27 PM

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Sorry expression, i mean.

upper karma Jun 21, 2020, 2:27 PM

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If you mean "x separation" as the white part between those 2 rectangles, i think you are right

naive harbor Jun 21, 2020, 2:30 PM

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Ah okay. so assuming that the distance between a rectangle's center and an edge is 0.5 units, and the y seperation is 0.8 units, to get an x separation that is 0.8 I should calculate it as (2 x 0.5) + 0.8?

upper karma Jun 21, 2020, 2:34 PM

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Iff the y separation and x separation are both 0.8, the general expression for the distance between 2 rectangles that form a horizontal or vertical is (2*0.5)+0.8

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But if you choose 2 rectangles, which are located like diagonally, the expression will obviously change

naive harbor Jun 21, 2020, 2:39 PM

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Ah okay, thank you .

upper karma Jun 21, 2020, 2:40 PM

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Np

drifting sorrel Jun 21, 2020, 8:52 PM

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I am doing a practice test (not graded) and I'm stuck on this problem. I mostly just guessed because I'm pretty lost on how they'd get any of these answers.

So far I've tried breaking out tanx+cotx to see if those simplify more and getting them under an LCD which gave me sin^2x + cos^2x. I tried breaking that up in a few different ways, but didn't get any of their options.

📎 unknown.png

silent plank Jun 21, 2020, 9:03 PM

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which gave me sin^2x + cos^2x
could you make that clearer? what's that supposed to be for?

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@drifting sorrel

drifting sorrel Jun 21, 2020, 9:11 PM

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Sorry, it's hard to write these kinds of problems in Discord. Let me try to take a picture of the paper.

upper karma Jun 21, 2020, 9:13 PM

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Ight

drifting sorrel Jun 21, 2020, 9:13 PM

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📎 IMG_20200621_171328.jpg

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Taking pictures is hard apparently lol

silent plank Jun 21, 2020, 9:14 PM

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, rotate

somber coyoteBOT Jun 21, 2020, 9:14 PM

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📎 220517274845577216.png

drifting sorrel Jun 21, 2020, 9:15 PM

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I got there and the rest of my scratch paper is just trying every combination of identities to try to get to one of their options

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Kind of a crap shoot

silent plank Jun 21, 2020, 9:15 PM

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you cant just multiply the numerators by something like that

drifting sorrel Jun 21, 2020, 9:16 PM

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Oh wait, they'd both be over sinxcosx

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Dur that was the point of multiplying them lol

silent plank Jun 21, 2020, 9:17 PM

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need to multiply to both the numerator AND denominator

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(otherwise you aren't multiplying by 1, and are changing the value of the expressions)

drifting sorrel Jun 21, 2020, 9:19 PM

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Right, I was trying to get the denominator to the LCD of (sinx)(cosx) and didn't write it out after multiplying the numerator. So I got the same thing, but just over (sinx)(cosx) now. Let me try playing with this a bit.

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Turns out I'm best guesser in all the west and the one I had selected is correct - haha! Thank you for your help, that was a dumb mistake not to put the denominator under it after. I got (cscx)(secx) in the end. :)

somber coyoteBOT Jun 21, 2020, 9:34 PM

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ramonov:

📎 220517274845577216.png

drifting sorrel Jun 21, 2020, 9:36 PM

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Yeah, that's what I got! Then 1 over sin x 1 over cos can be written as csc x sec via reciprocal identity 👌

rugged timber Jun 22, 2020, 1:57 AM

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Is anyone available for help? Specially need help for f=asin/cos(b(x-c)+d

lunar plover Jun 22, 2020, 2:57 AM

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Can help me solve something?

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📎 image0.jpg

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I need to find the area of this triangle

dark sparrow Jun 22, 2020, 2:59 AM

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what have you tried so far

novel flax Jun 22, 2020, 2:59 AM

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TW is an altitude to hypotenuse

lunar plover Jun 22, 2020, 3:00 AM

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Well, i dont even know where to start!

dark sparrow Jun 22, 2020, 3:00 AM

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if you knew TW, would you be able to do the problem?

lunar plover Jun 22, 2020, 3:00 AM

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Whats TW?

dark sparrow Jun 22, 2020, 3:00 AM

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....the length of that line segment

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look at the diagram

lunar plover Jun 22, 2020, 3:01 AM

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Oh, i dont

dark sparrow Jun 22, 2020, 3:01 AM

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so you could not do your problem even if you knew how long TW was?

lunar plover Jun 22, 2020, 3:01 AM

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Yes i could

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I know how to get the area of a triangle knowing the base lenght and its height

dark sparrow Jun 22, 2020, 3:02 AM

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yes ok that's what i wanted to hear

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idk why you said no earlier

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anyway ok

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so why not attempt to find TW then

lunar plover Jun 22, 2020, 3:02 AM

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Because i dont know where to start

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Hmmm

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Ok, imma call the TW x

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So i think i can do 9/x = x/16 right?

novel flax Jun 22, 2020, 3:06 AM

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yes @lunar plover

lunar plover Jun 22, 2020, 3:08 AM

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So the height would be 12?

novel flax Jun 22, 2020, 3:08 AM

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yes

lunar plover Jun 22, 2020, 3:08 AM

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And it would be 300 cm^2

dark sparrow Jun 22, 2020, 3:08 AM

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well there you have it don't you

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except... 1/2 (25 * 12) is not 300

lunar plover Jun 22, 2020, 3:09 AM

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Oh right

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I forgot about that xD

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Lets do another one

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📎 image0.jpg

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So i have this

dark sparrow Jun 22, 2020, 3:11 AM

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uh huh

lunar plover Jun 22, 2020, 3:11 AM

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So first off, i need to calculate the area

dark sparrow Jun 22, 2020, 3:12 AM

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do you?

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all you're asked for here is BC as it seems

lunar plover Jun 22, 2020, 3:12 AM

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OH

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So i first need to fond BD

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Then i use pythagore to find BC!

dark sparrow Jun 22, 2020, 3:13 AM

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you don't NEED to, but that's definitely a possible path

lunar plover Jun 22, 2020, 3:14 AM

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I am actually not as bad as i thought

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So the BC is 80 right?

novel flax Jun 22, 2020, 3:20 AM

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that's right

lunar plover Jun 22, 2020, 3:23 AM

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So i would basically do the same for this one right?

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📎 image0.jpg

dark sparrow Jun 22, 2020, 3:25 AM

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well it's exactly the same setup isn't it

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just with the numbers changed

lunar plover Jun 22, 2020, 3:25 AM

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Yeah

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2385 cm

weary heart Jun 22, 2020, 7:14 AM

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Given the ellipse:

x^2 + y^2 - 2axy = b^2 ,
Find the length of semi major and semi minor axes for the above ellipse. (a and b are constants)

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<@&286206848099549185>

upper karma Jun 22, 2020, 7:19 AM

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you should have waited to ping the helpers

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anyway what have you tried?

weary heart Jun 22, 2020, 7:28 AM

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@upper karma
Hey man, I was trying to do this since morning.
Only to realize just now that I was making a silly mistake.

upper karma Jun 22, 2020, 7:29 AM

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alright

idle bloom Jun 22, 2020, 7:32 AM

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Top two ways to realize your mistakes

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1: turn in the test and leave the room

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2: ping helpers

pine oar Jun 22, 2020, 7:33 AM

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https://i.imgur.com/HzGuMJz.png

Imgur

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hey guys

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for this question

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i know that you need the distance formula to get the sides

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does it matter what qualifies as x1 and x2 and y1 and y2 when you plug into distance formula?

dark sparrow Jun 22, 2020, 7:35 AM

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no, distance is symmetric

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d(A,B) = d(B,A)

raw pagoda Jun 22, 2020, 1:28 PM

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hi, how should i approach this problem

📎 howshouldi.png

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im not as fluent in geometry and i have no idea what i should do first, i tried trig and tried summing up the triangles and squares

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but i got stuck

wooden current Jun 22, 2020, 1:38 PM

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@raw pagoda Consider the smaller triangles formed by the squares and the sides of the main triangle and their angles

📎 unknown.png

raw pagoda Jun 22, 2020, 1:47 PM

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thx man

wooden current Jun 22, 2020, 1:51 PM

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GWjiangPepeThumb

dusky surge Jun 22, 2020, 2:47 PM

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I think yes.

upper karma Jun 22, 2020, 3:31 PM

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what two graphs?

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there's only one graph

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and a unit circle

tender kite Jun 22, 2020, 3:45 PM

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@pure pendant For the circle one, imagine drawing a radius and x is equal to the angle the radius makes with the horizontal axis

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in the regular sine graph x is equal to the x-axis value

blazing panther Jun 22, 2020, 4:40 PM

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a square floor is covered with congruent square tiles, the tiles on the diagonals are black, all the rest are white. If there are 101 black tiles, how many are there in total?

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How do i do this

dusky surge Jun 22, 2020, 4:53 PM

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Hello @blazing panther

blazing panther Jun 22, 2020, 4:53 PM

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Hello

dusky surge Jun 22, 2020, 4:54 PM

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So, diagonals are black tiles.

blazing panther Jun 22, 2020, 4:54 PM

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Yes

dusky surge Jun 22, 2020, 4:54 PM

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Lemme draw it out.

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Ouch, my drawing sucks

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📎 1592844921374.jpg

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So, I make the middle red

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That is where the 2 diagonals meet

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So, can you count how many black tiles are there in 1 diagonal?

blazing panther Jun 22, 2020, 4:56 PM

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Completely across or until the red dot

dusky surge Jun 22, 2020, 4:57 PM

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Completely across

blazing panther Jun 22, 2020, 4:57 PM

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51

dusky surge Jun 22, 2020, 4:57 PM

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Yep

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Since it's the diagonal, every tile move down and right from the top left.

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That is horizontal and vertical of the whole square would be 51 tiles wide and 51 tiles long

blazing panther Jun 22, 2020, 4:59 PM

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Oh but that wouldnt work in a regular square right

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Because these are tiles so you can easily see how the diagonal tile number equals that width

dusky surge Jun 22, 2020, 5:00 PM

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Well... If the diagonal becomes a line with no width, then you're right, it's hard to find it like that.

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But for Area type question, you can always count tiles :)

blazing panther Jun 22, 2020, 5:00 PM

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Thanks so its 51^2?

dusky surge Jun 22, 2020, 5:00 PM

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Yea

blazing panther Jun 22, 2020, 5:00 PM

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Ok i got confused because i tried to do it like a regular geometry problem

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So i said it was 51 and i used the pythagorran theorem instead of just matching up the tiles

dusky surge Jun 22, 2020, 5:01 PM

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More like a tricky question XD

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Aha

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Anyways there ya go

blazing panther Jun 22, 2020, 5:01 PM

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Thank you

dusky surge Jun 22, 2020, 5:02 PM

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You're welcome!

upper karma Jun 22, 2020, 5:21 PM

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📎 unknown.png

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how is the answer b and not c

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o

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i got it

minor field Jun 22, 2020, 5:25 PM

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if theta is between pi/2 and pi then sine is positive but tangent is negative

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(because cosine is negative)

upper karma Jun 22, 2020, 6:54 PM

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📎 image0.jpg

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I’m very lost here

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If someone could help

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doing what? lol

upper karma Jun 22, 2020, 7:36 PM

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👻

upper karma Jun 22, 2020, 8:32 PM

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@upper karma Apply the distance formula

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That’s to show that they are congruent

upper karma Jun 23, 2020, 2:33 AM

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@upper karma YOO I PLAY THE EUPHONIUM.

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LOW BRASS GANG.

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POG.

upper karma Jun 23, 2020, 4:14 AM

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Ayyy

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Low brass rules the world

wraith drum Jun 23, 2020, 6:42 AM

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Guys, how do i get the real and imaginary parts of sin^-1(a+bi)?

versed river Jun 23, 2020, 7:15 AM

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Have you seen cos(ix)=cosh(x) and sin(ix)=isinh(x) before?

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Although my simultaneous equations from my method seem pretty fuckin messy without wolfram alpha catThink

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I found this on the net http://mathonweb.com/help_ebook/html/complex_funcs.htm but it doesnt exactly reveal how those formulas came to be. Dont feel like try figure it out but it looks like it vaguely follows catshrug not gonna bother further with this one

nimble horizon Jun 23, 2020, 9:01 AM

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i had a thought

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i wanted to find out when cos(x) = tan(x)

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there are clearly solns to that but the only solns im getting are complex numbers

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my strategy was this

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$cos(x) = tan(x) = \frac{sin(x)}{cos(x)} => cos^2(x) = sin(x)$

somber coyoteBOT Jun 23, 2020, 9:02 AM

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funnynumberatyahoo.com:

📎 453650288851812354.png

nimble horizon Jun 23, 2020, 9:03 AM

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and then

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since $cos^2(x) = 1 - sin^2(x)$

somber coyoteBOT Jun 23, 2020, 9:03 AM

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funnynumberatyahoo.com:

📎 453650288851812354.png

nimble horizon Jun 23, 2020, 9:03 AM

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you get

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$1 - sin^2(x) = sin(x)$

somber coyoteBOT Jun 23, 2020, 9:03 AM

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funnynumberatyahoo.com:

📎 453650288851812354.png

nimble horizon Jun 23, 2020, 9:04 AM

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then that's a quadratic in sin(x), solving for sin(x) you get the golden ratio

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then i tried doing sin^-1(golden ratio) but i got a complex number

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and actually when i plot that function

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it coincides exactly with the values

📎 unknown.png

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but then when i evaluate sin^-1(golden ratio) i get completely incorrect results

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i guess it probably has something to do with complex values not being able to translate to the real plane, but id like a second opinion

open rune Jun 23, 2020, 9:15 AM

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How do I solve this?

📎 image0.jpg

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I know i would have to solve the triangle first

dark sparrow Jun 23, 2020, 9:17 AM

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uh

eternal crag Jun 23, 2020, 9:17 AM

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i'm pretty sure this has infinitely many solutions

dark sparrow Jun 23, 2020, 9:17 AM

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this looks like WAY too little info

eternal crag Jun 23, 2020, 9:17 AM

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i.e.^

open rune Jun 23, 2020, 9:17 AM

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Apparently that was all that was given

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I’m just as confused

dark sparrow Jun 23, 2020, 9:18 AM

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are you absolutely sure this is all you have

open rune Jun 23, 2020, 9:19 AM

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Yes

nimble horizon Jun 23, 2020, 9:19 AM

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wHO ping

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*pinged

dark sparrow Jun 23, 2020, 9:23 AM

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then it is impossible to determine the values of x and y.

supple onyx Jun 23, 2020, 9:26 AM

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i personally cant figure out any unique solutions yeah

wraith drum Jun 23, 2020, 9:27 AM

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@open rune Your teacher screwed up

open rune Jun 23, 2020, 9:28 AM

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it was on a test I took but idk how to solve it

supple onyx Jun 23, 2020, 9:28 AM

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i highy doubt it's possible to solve

wraith drum Jun 23, 2020, 9:29 AM

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Do you know if it involves trig?

upper karma Jun 23, 2020, 9:29 AM

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by "solve" do you mean get numerical values for x and y?

wraith drum Jun 23, 2020, 9:30 AM

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or is it "y" as a function of x?

open rune Jun 23, 2020, 9:30 AM

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it involves geometry and yes to solve for x and y

wraith drum Jun 23, 2020, 9:30 AM

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k

open rune Jun 23, 2020, 9:30 AM

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i'll figure something out

wraith drum Jun 23, 2020, 9:30 AM

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is x perpendicular to y

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like the lines

upper karma Jun 23, 2020, 9:31 AM

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it's impossible, you can "stretch" y however you want

open rune Jun 23, 2020, 9:31 AM

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im not sure..

supple onyx Jun 23, 2020, 9:32 AM

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@wraith drum i assumed they were perpendicular and i personally couldnt solve it

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as @upper karma said y can be stretched as far as you want

wraith drum Jun 23, 2020, 9:42 AM

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I've got the area as 1/2 x^2 + 3y - 3x + 9/2 , I might be wrong though...

versed river Jun 23, 2020, 10:05 AM

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@nimble horizon i apologise if im missing something here but your quadratic sin^2x+sinx-1=0 should have had 2 solutions, one of which was sinx=-goldenratio(x not real), and the other one was (sqrt(5)-1)/2, which if you take inverse sine of should give you your answer

nimble horizon Jun 23, 2020, 10:05 AM

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yeah ive done that too

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it doesn't match up

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i get -0.666

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no intersection at that point

versed river Jun 23, 2020, 10:06 AM

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Bruh you forgot to take arcsine of it

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Or sorry i mean

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You have the wrong sign

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It shoulld be positive

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📎 Screenshot_20200623-201034_Chrome.jpg

nimble horizon Jun 23, 2020, 10:09 AM

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(1-sqrt(5))/2 is the negative solution, no?

versed river Jun 23, 2020, 10:10 AM

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No the negative solution is sinx= (-sqrt(5)-1)/2

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You made a mistake solving the qquadratic somewhere theb

nimble horizon Jun 23, 2020, 10:11 AM

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i don't think so

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Let me check

versed river Jun 23, 2020, 10:11 AM

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📎 Screenshot_20200623-201303_Chrome.jpg

peak mesa Jun 23, 2020, 10:11 AM

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hey anyone knows how i can make a square with sin/cos? i cant find nothing in internet

nimble horizon Jun 23, 2020, 10:11 AM

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I swear i remember it as (1-sqrt(5))/2

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let me run it through mathematica

peak mesa Jun 23, 2020, 10:12 AM

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https://gifer.com/en/8ri9
https://gifer.com/en/CXSu
I found this in internet but i think is not a help

versed river Jun 23, 2020, 10:13 AM

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Ok, do you agree the quadratic is 1-x^2=x?

nimble horizon Jun 23, 2020, 10:13 AM

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*wolframalpha

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sorry internet is iffy

versed river Jun 23, 2020, 10:14 AM

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Do you agree thats the right quadratic

nimble horizon Jun 23, 2020, 10:14 AM

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yeah it is

versed river Jun 23, 2020, 10:14 AM

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Yeah no worries

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Ok then

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📎 Screenshot_20200623-201607_Chrome.jpg

nimble horizon Jun 23, 2020, 10:14 AM

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i see

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so i need to find my mistake then

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here's what i did

versed river Jun 23, 2020, 10:15 AM

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You mustve messed up a sign in the quadratic somewhere

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Lemme see

nimble horizon Jun 23, 2020, 10:15 AM

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$x^2 + x - 1 = 0$

somber coyoteBOT Jun 23, 2020, 10:15 AM

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funnynumberatyahoo.com:

📎 453650288851812354.png

nimble horizon Jun 23, 2020, 10:15 AM

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$x = \frac{1 \pm \sqrt{(1^2 - 4(1)(-1))}}{2}$

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hopefully this works

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lmAo oof

versed river Jun 23, 2020, 10:15 AM

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\sqrt

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Plaintext if youd like is perfectly fine though

nimble horizon Jun 23, 2020, 10:16 AM

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aaaaa

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you get the idea tho right

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OH yeah i see it wait

versed river Jun 23, 2020, 10:16 AM

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Braces aroubd whatev needs to be sqrted

nimble horizon Jun 23, 2020, 10:16 AM

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i see it i see it

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should be -1 in the beginning part

somber coyoteBOT Jun 23, 2020, 10:17 AM

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funnynumberatyahoo.com:

📎 453650288851812354.png

versed river Jun 23, 2020, 10:18 AM

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Should be negative 1 at the start

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Oh you figured it out

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Do you see how to get the general solution from there and everything?

nimble horizon Jun 23, 2020, 10:20 AM

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yeah i do

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the sign was messing me up ty

versed river Jun 23, 2020, 10:20 AM

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No worries mate. Have a good one

nimble horizon Jun 23, 2020, 10:20 AM

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mhm right back at u

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wait wait hold up a second

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now the other one is the wrong sign

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the (-1 - sqrt(5))/2 one

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and the (-1 + sqrt(5))/2 is slightly off

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tho it's closer now

versed river Jun 23, 2020, 10:24 AM

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I mean the (-1-sqrt5)/2 one is complex

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📎 Screenshot_20200623-202633_Chrome.jpg

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Its pretty exact for me

nimble horizon Jun 23, 2020, 10:25 AM

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this is what im getting

📎 unknown.png

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the black one is slightly off

📎 unknown.png

versed river Jun 23, 2020, 10:25 AM

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You are forgetting the arcsin

nimble horizon Jun 23, 2020, 10:25 AM

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OHH lmfaooo

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sorry

versed river Jun 23, 2020, 10:25 AM

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Nice meme

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Bahaha dont worry about it

nimble horizon Jun 23, 2020, 10:26 AM

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wait so it looks like this isn't the golden ratio

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it just looks very close to it

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hM

versed river Jun 23, 2020, 10:27 AM

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Sometimes (sqrt(5)-1)/2 is reffered to as the second golden ratio

nimble horizon Jun 23, 2020, 10:27 AM

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or its actually the negative of the golden ratio woah

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ohhh i see

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oh wait im reading on wikipedia

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apparently people call it the silver ratio

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woah cool

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ty

hushed spear Jun 23, 2020, 12:08 PM

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hello

dusky surge Jun 23, 2020, 12:08 PM

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hello

hushed spear Jun 23, 2020, 12:08 PM

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how are you

dusky surge Jun 23, 2020, 12:09 PM

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Question(s) on geometry-trigonometry?

hushed spear Jun 23, 2020, 12:09 PM

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yes

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thank you

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my friend needs help he just needs to be in the server for 10 minutes to talk

she*

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so i will speak for her

upper karma Jun 23, 2020, 12:10 PM

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JustAsk

hushed spear Jun 23, 2020, 12:11 PM

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okay so basically

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i walk forward on a bearing of x to point B

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then i turn and walk forward on another bearing to point C

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I start on point A

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how would i find the bearing of point C from A?

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thank you in advance

dusky surge Jun 23, 2020, 12:16 PM

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it really depends on the angle. I made an example.

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the red angles are the same, the green angles add up = 180 degrees.

📎 unknown.png

fast burrow Jun 23, 2020, 12:17 PM

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wow thank you (i'm the friend)

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🙂

silent plank Jun 23, 2020, 12:18 PM

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are you given the distances walked?

fast burrow Jun 23, 2020, 12:18 PM

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yes

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A --> 20km --> B on a bearing of 100 deg

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B ----> 15km -----> C on a bearing of 190 deg

silent plank Jun 23, 2020, 12:20 PM

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so yeh, construct a proper bearing diagram
and find the information needed by applying stuff like the sine and/or cosine laws

fast burrow Jun 23, 2020, 12:20 PM

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okay

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so we know that triangle is right angled, right? how do you prove that? do you use what i learnt two years ago with co-interior and different types of angles adding up to 180deg?

silent plank Jun 23, 2020, 12:22 PM

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which triange are you referring to?

fast burrow Jun 23, 2020, 12:22 PM

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the one made up by the three points

silent plank Jun 23, 2020, 12:23 PM

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1 sec

fast burrow Jun 23, 2020, 12:23 PM

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okay

silent plank Jun 23, 2020, 12:24 PM

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you can conclude it from the bearing diagram

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NS and WE axis are perpendicular

fast burrow Jun 23, 2020, 12:25 PM

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yes

silent plank Jun 23, 2020, 12:25 PM

#

NS axes are parallel to each other
similar with EW axes

fast burrow Jun 23, 2020, 12:25 PM

#

so you can use cointerior angles

silent plank Jun 23, 2020, 12:26 PM

#

yes, that could be applied here

fast burrow Jun 23, 2020, 12:27 PM

#

wow! thanks for your help

silent plank Jun 23, 2020, 12:28 PM

#

note that if the angles weren't as nice, you would probably need to apply those laws stated

fast burrow Jun 23, 2020, 12:28 PM

#

yes

#

like minutes and seconds

#

disgusting

silent plank Jun 23, 2020, 12:29 PM

#

don't even need to involve minutes and seconds

#

eg if the bearing from B was 180° instead of 190°.

#

you won't get a right angle at B, and would involve more calculations

fast burrow Jun 23, 2020, 12:31 PM

#

wow

#

thank you so much for your help!

#

appreciate it

unique junco Jun 23, 2020, 1:00 PM

#

https://i.imgur.com/BdpwT2P.png

Imgur

#

I do not know how radians work

#

could someone help?

dark sparrow Jun 23, 2020, 1:01 PM

#

radians are just a unit of angle

#

there are 2π radians in a full circle

#

really all you need here is a calculator with a radian mode and an understanding of basic trigonometry

unique junco Jun 23, 2020, 1:02 PM

#

well I would need to know what to press on that calculator

dark sparrow Jun 23, 2020, 1:02 PM

#

are you familiar with the basics of trigonometry? what one might call "SOHCAHTOA", after the mnemonic central to it?

unique junco Jun 23, 2020, 1:02 PM

#

Yes I am

dark sparrow Jun 23, 2020, 1:03 PM

#

so is it clear that $d = \frac{16}{\cos(0.19)}$?

somber coyoteBOT Jun 23, 2020, 1:03 PM

#

Ann:

📎 183668144404037632.png

unique junco Jun 23, 2020, 1:03 PM

#

Yes

dark sparrow Jun 23, 2020, 1:03 PM

#

and you don't know how to evaluate cos(0.19) using your calculator?

#

if so then that is an issue with your lack of knowledge about your calculator and not the problem itself

unique junco Jun 23, 2020, 1:04 PM

#

oh so ican uust plug in .19 radians into cos?

dark sparrow Jun 23, 2020, 1:04 PM

#

how else do you think you would evaluate cos(0.19) if not by entering cos(0.19) into your calculator?

unique junco Jun 23, 2020, 1:05 PM

#

I thought that I would have to do some more math to find what .19 was equal to

dark sparrow Jun 23, 2020, 1:05 PM

#

what do you mean "find what 0.19 was equal to"

unique junco Jun 23, 2020, 1:05 PM

#

in degress

dark sparrow Jun 23, 2020, 1:05 PM

#

your calculator almost certainly has a radian mode

unique junco Jun 23, 2020, 1:06 PM

#

it does, ive solved the problem now

dark sparrow Jun 23, 2020, 1:06 PM

#

BUT, if you wanted to convert 0.19 radians to degrees, that's a matter of multiplying by the number of degrees in a radian

#

i.e. 180/pi

unique junco Jun 23, 2020, 1:06 PM

#

0.19 times 180/pi?

dark sparrow Jun 23, 2020, 1:06 PM

#

converting between radians and degrees is no different than converting between two units of the same physical quantity like feet and inches

#

yes that's what i said

unique junco Jun 23, 2020, 1:07 PM

#

Cool, thanks

upper karma Jun 23, 2020, 1:09 PM

#

Hei can i agsk a qestion?

cosmic pebbleBOT Jun 23, 2020, 1:09 PM

#

From

#❓how-to-get-help

Rule 1

The help channels are solely for help with math, so feel free to post your question. Asking whether you can ask a question or if anyone knows about some specific topic is unnecessary, so please try to avoid questions of that nature.

upper karma Jun 23, 2020, 1:10 PM

#

Go with it

#

Ill take that as a yes?

#

Of course.

#

So i made up this wacky question, and im tryna find a and b

📎 image0.jpg

#

So the given information is two points, and the angle between them.

#

And I want to see if I can find out the point where the angle starts, the lengths of a, b, etc

#

Idk if its possible tho.

#

Maybe 2 points an angle is too little?

#

(So the actual points used to be (2.3, 0) and (-1.6, -1.4), but I added some numbers to make it harder since the original points give away length a)

#

So should be a = 2.3, and b = sqrt(4.52)

#

But yeah idk how i'd solve it just given information (5.8, 7.9) (4.2, 4.2) and 140 degrees

glad anvil Jun 23, 2020, 1:21 PM

#

There's too little information. Remember your circle geometry? If you mark two points on the circle, and then draw lines from both those points to a 3rd point on the circle, the angle is the same no matter where the 3rd point is

upper karma Jun 23, 2020, 1:22 PM

#

Yea, tho id see how that'd help?

glad anvil Jun 23, 2020, 1:25 PM

#

📎 unknown.png

#

Your intended solution is point C

#

But point D, or point E, or any other point on the circle also works

upper karma Jun 23, 2020, 1:26 PM

#

Ah right yea

#

But idk how this could be used to find C, or AC or BC

glad anvil Jun 23, 2020, 1:28 PM

#

True, it doesn't show a way to calculate the coordinates of the point

upper karma Jun 23, 2020, 1:28 PM

#

I think linear algebra can probably be used to fill in the missing point, and solve for it.

#

Using cosine rule for dot product or something

glad anvil Jun 23, 2020, 1:29 PM

#

Yea, I think that probably works

upper karma Jun 23, 2020, 1:30 PM

#

Which was how I constructed the problem in the first place, so it feels a bit like cheating

glad anvil Jun 23, 2020, 1:30 PM

#

xD

upper karma Jun 23, 2020, 1:30 PM

#

AB, can be found really easily, so I was hoping there was some clean nice geometric solution

wraith drum Jun 23, 2020, 1:34 PM

#

Hey peeps, I have a question, prove that sin(nx)/1+cos(nx)=tan(nx/2) for all natural values of "n"

#

i haven't tried complex analysis, but i'm scared it will get complicated

upper karma Jun 23, 2020, 1:36 PM

#

we already proved this few days ago

dark sparrow Jun 23, 2020, 1:36 PM

#

$\frac{\sin(nx)}{1} + \cos(nx) \neq \tan(nx/2)$

somber coyoteBOT Jun 23, 2020, 1:36 PM

#

Ann:

📎 183668144404037632.png

wraith drum Jun 23, 2020, 1:38 PM

#

no

#

like (1+cos(nx))

dark sparrow Jun 23, 2020, 1:38 PM

#

ah

#

is there a tax on parentheses where you're from?

wraith drum Jun 23, 2020, 1:38 PM

#

i thought it would be obvious

#

but nvm

dark sparrow Jun 23, 2020, 1:38 PM

#

it is not

wraith drum Jun 23, 2020, 1:38 PM

#

sry

#

is there a tax on parentheses where you're from?
@dark sparrow 🤣

dark sparrow Jun 23, 2020, 1:39 PM

#

well, i'd go with "are you allergic to parentheses" on any other day, but that would probably make you write sinnx / 1+cosnx = tannx/2 or something to that effect

wraith drum Jun 23, 2020, 1:40 PM

#

@dark sparrow that spacing is bigger than my future

dark sparrow Jun 23, 2020, 1:41 PM

#

it was exaggerated on purpose.

#

anyway, let θ = nx/2 and apply double angle identities on the left hand side

wraith drum Jun 23, 2020, 1:43 PM

#

oh got it, thanks

dusky surge Jun 23, 2020, 1:58 PM

#

@upper karma About your question. There is not enough information to find a and b. What I was taught in school is that you need at least 3 of the 6 information of a triangle, namely the length of the 3 sides, and the 3 angles, to find the remaining information.
I am not sure if this can help you.

upper karma Jun 23, 2020, 2:11 PM

#

Right, so that question only gives 1 side and 1 angle, so its not enough. That makes sense

fluid moat Jun 23, 2020, 2:12 PM

#

Hi

#

is ther e a place i can learn the trig functions in depth

#

i only really know 2 things

#

trig ratios

#

and this other thing on the calculator

#

im a nob

dark sparrow Jun 23, 2020, 2:13 PM

#

khanacademy

upper karma Jun 23, 2020, 2:13 PM

#

idk unit circle mebe?

fluid moat Jun 23, 2020, 2:14 PM

#

ok

#

the algrebra 2 book i got isnt really that helpful

#

it just has questions

#

lol

#

ill use khan academy

#

thanks

supple onyx Jun 23, 2020, 2:23 PM

#

I mean id probably serahc for a lot of practice questions

#

to really learn the identities

fluid moat Jun 23, 2020, 2:26 PM

#

okay, thanks

quaint quarry Jun 23, 2020, 9:14 PM

#

Find the midpoint of the segment that has endpoints at (a+b, a-b) and (a,b)

upper karma Jun 23, 2020, 9:15 PM

#

Do you know the midpoint formula? @quaint quarry

#

Sure

wooden current Jun 23, 2020, 9:26 PM

#

GWchadMEGATHINK

upper karma Jun 23, 2020, 9:33 PM

#

Great got ghosted

#

👻

#

Lmaooooooooo

graceful narwhal Jun 23, 2020, 9:35 PM

#

😦

quaint quarry Jun 23, 2020, 10:00 PM

#

Yeah

#

Sorry I was busy

#

X+X divided by 2 equals the x coordinate for midpoint

#

And just replace the x’s with y’s

upper karma Jun 23, 2020, 10:03 PM

#

What

plucky marlin Jun 23, 2020, 10:38 PM

#

GWseremePeepoThink

upper karma Jun 23, 2020, 10:46 PM

#

@quaint quarry i mean yeah but that's way too vague

#

Are you still here?

minor field Jun 24, 2020, 1:19 AM

#

Again

#

😔

brave drift Jun 24, 2020, 4:36 AM

#

I am trying to prove that if the an n-sided convex polygon has at least one side parallel to x and at least one side parallel to y. Then, n is the multiple of 4.

📎 unknown.png

dark sparrow Jun 24, 2020, 4:38 AM

#

if you have two sides that are perpendicular to one another then some multiple of the external angle will be 90°

brave drift Jun 24, 2020, 4:40 AM

#

if you have two sides that are perpendicular to one another then some multiple of the external angle will be 90°
@dark sparrow This is what I am unable to understand.

#

In order for one side to align to the vertical and another side to the horizontal, the sum of the exterior angles of some k consecutive sides must equal 90.

#

Why sum of the exterior angles of some k consecutive sides must be equal to 90?

dark sparrow Jun 24, 2020, 4:44 AM

#

i could give you either an informal explanation of this fact or an alternative proof that isn't handwavy

brave drift Jun 24, 2020, 4:45 AM

#

The one which will help me to solve some more similar questions would be the best.

dark sparrow Jun 24, 2020, 4:46 AM

#

hhh

#

that's so vague

#

anyway w/e

upper karma Jun 24, 2020, 4:47 AM

#

yus mission accomplished

I successfully proved law of cosines independently... I just had to look at a triangle for 10 minutes before being able to do anything. Aaaand my proof is more algebraic than geometric so yus

dark sparrow Jun 24, 2020, 4:47 AM

#

take your pair of perpendicular sides

#

connect their midpoints to the center of the polygon

#

the resulting angle at the center will be 90° by assumption

#

but on the other hand, if you considered the angle subtended by one side at the center, that'd be (360/n)° bc your polygon is regular

#

ok i think it's time i made a picture

#

or not bc i'm lazy

brave drift Jun 24, 2020, 5:00 AM

#

📎 unknown.png

#

I will check that if (sum of exterior angles + 90) = 180 in this case. If it is then, I am happy to move on.

#

Proofs should be easy because they are just logic. ;_;

#

3x+x+90 = 180
So, 4x = 90. That is the sum of the exterior angles should be 90 degrees.

📎 unknown.png

upper karma Jun 24, 2020, 5:35 AM

#

Next up on my list of things to prove:: i have no idea whatsoever but I think perhaps Euler’s identity that e^pii = -1.

#

wait either that or binomial theorem...

#

Either seem reasonable

#

prove/construct real numbers and its arithmetic operations

#

its seems pretty reasonable

#

;^)

dark sparrow Jun 24, 2020, 8:31 AM

#

@upper karma try e^ix = cos(x) + i sin(x)

unique junco Jun 24, 2020, 12:17 PM

#

https://i.imgur.com/XLOUO60.png

Imgur

wooden current Jun 24, 2020, 12:17 PM

#

What's the problem?

unique junco Jun 24, 2020, 12:17 PM

#

https://i.imgur.com/n1jmULh.png

Imgur

versed river Jun 24, 2020, 12:21 PM

#

I think tilde means "whats the problem you have" more specifically

unique junco Jun 24, 2020, 12:21 PM

#

Oh ok

#

I honestly dont know what the specific problem is, I have absolutely no clue how 1 is the answer

silent plank Jun 24, 2020, 12:24 PM

#

have you ever worked with radians before?

unique junco Jun 24, 2020, 12:24 PM

#

Yes

somber coyoteBOT Jun 24, 2020, 12:24 PM

#

HoboSas:

📎 310725130097786880.png

silent plank Jun 24, 2020, 12:25 PM

#

and do you know the ratios for trig functions of special angles?

unique junco Jun 24, 2020, 12:25 PM

#

I definitly learned them, probably just forgetting what they are

silent plank Jun 24, 2020, 12:25 PM

#

revise them

unique junco Jun 24, 2020, 12:25 PM

#

like 2pi = 360 and stuff liek that?

silent plank Jun 24, 2020, 12:25 PM

#

they're very common

unique junco Jun 24, 2020, 12:26 PM

#

r u talking about the angles and radians of the trig circle?

silent plank Jun 24, 2020, 12:26 PM

#

both

#

sin(0° ) = sin(0) = ?
sin(30°) = sin(pi/6) = ?
etc...

unique junco Jun 24, 2020, 12:27 PM

#

Yea

silent plank Jun 24, 2020, 12:27 PM

#

and the numerical value of something like
sin(pi/2) = sin(90°) is something you should definitely know

unique junco Jun 24, 2020, 12:27 PM

#

Oooh so in the problem I posted the pi/2 was radians

#

I was trying to plug it in for opposite/hypoeneus

silent plank Jun 24, 2020, 12:28 PM

#

plug what in for opp/hyp

unique junco Jun 24, 2020, 12:28 PM

#

sin(o/h) sin(pi/2)

silent plank Jun 24, 2020, 12:29 PM

#

oh no.

#

pi/2 is the angle

wooden current Jun 24, 2020, 12:29 PM

#

If it's easier for you to think in terms of degrees, you can convert radians to degrees by multiplying the radians by 180/pi.

silent plank Jun 24, 2020, 12:30 PM

#

and isn't the ratio of the opposite and the hypotenuse

arctic vortex Jun 24, 2020, 3:41 PM

#

i dont understand how to do 2) and 3)

📎 Annotation_2020-06-24_113930.png

upper karma Jun 24, 2020, 3:43 PM

#

same

arctic vortex Jun 24, 2020, 3:46 PM

#

oh

dusky surge Jun 24, 2020, 3:46 PM

#

Hello kurax

#

For number 2, can you find any real life situation that you'll have to use the scalene triangle?

#

Like can you find a real life object that is of triangular shape and it's useful for you?

#

@arctic vortex

arctic vortex Jun 24, 2020, 3:49 PM

#

the flatiron building?

#

hmm

#

idk

dusky surge Jun 24, 2020, 3:49 PM

#

that's a good example

#

Why does it need to be in triangular shape?

arctic vortex Jun 24, 2020, 3:50 PM

#

a hanger?

#

idk

dusky surge Jun 24, 2020, 3:50 PM

#

That's a good one too

arctic vortex Jun 24, 2020, 3:50 PM

#

but its not scalene

dusky surge Jun 24, 2020, 3:50 PM

#

Hmmm...

#

the flatiron building?
Why does it need to be in triangular shape?

arctic vortex Jun 24, 2020, 3:51 PM

#

it said to do it in a scalene triangle shape

dusky surge Jun 24, 2020, 3:51 PM

#

Okay, is the flatiron building a scalene triangle?

arctic vortex Jun 24, 2020, 3:51 PM

#

no

#

i cant think of any examples

#

can u help me

dusky surge Jun 24, 2020, 3:52 PM

#

Let's think together :)

#

Look around your room

#

See if there's anything triangular

arctic vortex Jun 24, 2020, 3:52 PM

#

ok

#

no nothing triangular :(

dusky surge Jun 24, 2020, 3:52 PM

#

Ouchy

#

How about a deformed hanger😂

upper karma Jun 24, 2020, 3:53 PM

#

real life triangles are too regular 😦

dusky surge Jun 24, 2020, 3:53 PM

#

Artificial stuff~

arctic vortex Jun 24, 2020, 3:53 PM

#

yes

#

i cant find a scalene triangle

#

am i allowed to say deformed in a school assignment

dusky surge Jun 24, 2020, 3:54 PM

#

If you can point out why a deformed one is useful, then yes

arctic vortex Jun 24, 2020, 3:54 PM

#

ok then ill do deformed hanger

#

hmm or maybe ill say distorted hanger

dusky surge Jun 24, 2020, 3:55 PM

#

Like if you have to hang different stuff, or need to use the hanger to do other things

silent plank Jun 24, 2020, 3:55 PM

#

doorstop?

dusky surge Jun 24, 2020, 3:55 PM

#

Oh, nice one

arctic vortex Jun 24, 2020, 3:55 PM

#

oh yeah

#

📎 Annotation_2020-06-24_115606.png

#

how do i explain mathematically

#

sorry english isnt my first language

dusky surge Jun 24, 2020, 3:56 PM

#

Do you recall the triangle inequality?

arctic vortex Jun 24, 2020, 3:57 PM

#

yes

dusky surge Jun 24, 2020, 3:57 PM

#

Use that

arctic vortex Jun 24, 2020, 3:57 PM

#

sum of 2 angles must be greater than the third side?

dusky surge Jun 24, 2020, 3:57 PM

#

Sum of 2 sides

arctic vortex Jun 24, 2020, 3:58 PM

#

yea

dusky surge Jun 24, 2020, 3:58 PM

#

Done?

arctic vortex Jun 24, 2020, 3:59 PM

#

i still dont understand it :(

dusky surge Jun 24, 2020, 3:59 PM

#

Ohhhh

#

Hint: use a ruler to measure all three sides

arctic vortex Jun 24, 2020, 3:59 PM

#

oh ok

dusky surge Jun 24, 2020, 4:00 PM

#

And then try to calculate all 3 situations of the triangle inequality

arctic vortex Jun 24, 2020, 4:01 PM

#

how do i use a ruler in microsoft word

dusky surge Jun 24, 2020, 4:02 PM

#

Hmm.....

#

Place the ruler on the screen, I do that when I am doing measuring myself before I print it out

arctic vortex Jun 24, 2020, 4:03 PM

#

i dont have a ruler in real life :)

#

:(*

dusky surge Jun 24, 2020, 4:03 PM

#

But make sure that the paper in the screen fits the paper in real life

#

Oh....

#

There's another method, use the insert line function, and draw a line directly on top of the given line of the triangle

#

Microsoft words would tell you the horizontal span and vertical span of the line. Use Pythagoras theorem to calculate the length of the line

arctic vortex Jun 24, 2020, 4:05 PM

#

📎 Annotation_2020-06-24_120517.png

#

oh ok

#

wait how do i make it tell me

dusky surge Jun 24, 2020, 4:07 PM

#

Double click your drawn line

arctic vortex Jun 24, 2020, 4:07 PM

#

ok

dusky surge Jun 24, 2020, 4:07 PM

#

It should show on the top right

arctic vortex Jun 24, 2020, 4:08 PM

#

oh i see it

#

height: 0.02
width: 3,31

dusky surge Jun 24, 2020, 4:08 PM

#

Yea, that's the dimension.

arctic vortex Jun 24, 2020, 4:08 PM

#

ok

dusky surge Jun 24, 2020, 4:09 PM

#

Now move it and reshape it till it fits your triangle

#

One side at a time.

arctic vortex Jun 24, 2020, 4:09 PM

#

oh ok

#

like this?\

📎 Annotation_2020-06-24_121024.png

dusky surge Jun 24, 2020, 4:10 PM

#

Hmmm....

#

Do you know how to reshape your line so that it is exactly the same with one side, let's say 'a'

arctic vortex Jun 24, 2020, 4:11 PM

#

yea

dusky surge Jun 24, 2020, 4:11 PM

#

That's good!

arctic vortex Jun 24, 2020, 4:12 PM

#

like this one

📎 Annotation_2020-06-24_121221.png

dusky surge Jun 24, 2020, 4:12 PM

#

Yea, seems like it

arctic vortex Jun 24, 2020, 4:13 PM

#

ok

dusky surge Jun 24, 2020, 4:13 PM

#

Remember to jot down your results!

arctic vortex Jun 24, 2020, 4:13 PM

#

\

📎 Annotation_2020-06-24_121335.png

dusky surge Jun 24, 2020, 4:13 PM

#

Nice!

arctic vortex Jun 24, 2020, 4:13 PM

#

ok so do i measure them?

dusky surge Jun 24, 2020, 4:14 PM

#

Do you know Pythagoras theorem?

arctic vortex Jun 24, 2020, 4:14 PM

#

pythagorean

#

?

dusky surge Jun 24, 2020, 4:14 PM

#

Yea

arctic vortex Jun 24, 2020, 4:15 PM

#

yeah i know that

dusky surge Jun 24, 2020, 4:15 PM

#

Since height and width is given, you can use Pythagorean to solve the length by

arctic vortex Jun 24, 2020, 4:15 PM

#

ok

dusky surge Jun 24, 2020, 4:15 PM

#

$\sqrt{Height^2+Width^2}$

somber coyoteBOT Jun 24, 2020, 4:15 PM

#

Biscuit:

📎 681190496273432663.png

arctic vortex Jun 24, 2020, 4:16 PM

#

ok

#

3.8825

#

is what i got for A

dusky surge Jun 24, 2020, 4:18 PM

#

Please continue for b and c and mark down in your file 😁

arctic vortex Jun 24, 2020, 4:18 PM

#

ok

#

📎 Annotation_2020-06-24_122031.png

#

THIS

#

this*

dusky surge Jun 24, 2020, 4:22 PM

#

Did you take the square root?

arctic vortex Jun 24, 2020, 4:22 PM

#

OH OOPS

#

📎 Annotation_2020-06-24_122310.png

dusky surge Jun 24, 2020, 4:25 PM

#

Good! Remember for sides of triangles, we keep our letters in lowercase

#

a=1.97
b=2.9
c=3.92

arctic vortex Jun 24, 2020, 4:25 PM

#

ok

#

what now?

dusky surge Jun 24, 2020, 4:26 PM

#

Now we have to compare sum of 2 sides is larger than the remaining side

#

We have 3 cases

arctic vortex Jun 24, 2020, 4:27 PM

#

thats true

#

everything is true

#

a + b > c

#

b + c > a

#

a + c > a

#

b*

dusky surge Jun 24, 2020, 4:27 PM

#

Yea, you just have to type that into your document like

#

a + b = 1.97+2.9 = 4.87 which is bigger than c=3.92

arctic vortex Jun 24, 2020, 4:28 PM

#

is this good

📎 Annotation_2020-06-24_122836.png

dusky surge Jun 24, 2020, 4:29 PM

#

That's pretty good, but we still got to add in the numbers to 'show' it is really 'greater'

arctic vortex Jun 24, 2020, 4:29 PM

#

o ok

#

i have to eat lunch now

dusky surge Jun 24, 2020, 4:30 PM

#

Sure

arctic vortex Jun 24, 2020, 4:30 PM

#

ill be back in like 10 minutes

#

tysm

dusky surge Jun 24, 2020, 4:30 PM

#

See ya later~

arctic vortex Jun 24, 2020, 4:50 PM

#

back @dusky surge

dusky surge Jun 24, 2020, 4:51 PM

#

You're fast

arctic vortex Jun 24, 2020, 4:51 PM

#

thanks

📎 Annotation_2020-06-24_125132.png

#

do i just say the formula

#

like how i used the formula

dusky surge Jun 24, 2020, 4:53 PM

#

I think you'll have to use the formula

#

For the explanation part

arctic vortex Jun 24, 2020, 4:55 PM

#

wait whats the difference betwen lines and angles?

dusky surge Jun 24, 2020, 4:56 PM

#

Huge difference.

#

Basically they are not the same.

#

Have you ever heard of Law of sines?

arctic vortex Jun 24, 2020, 4:58 PM

#

no

#

wait 1 second

#

is this good?

📎 Annotation_2020-06-24_125819.png

#

im going to submit it

dusky surge Jun 24, 2020, 5:00 PM

#

Let's see

#

It's not perfect I can say. There are some points that can be clarified.

arctic vortex Jun 24, 2020, 5:02 PM

#

oh

#

what can be clarified

dusky surge Jun 24, 2020, 5:02 PM

#

But it's good enough for an assessment I guess😁

arctic vortex Jun 24, 2020, 5:02 PM

#

oh

dusky surge Jun 24, 2020, 5:03 PM

#

Oh, you want me to point out or just submit?

arctic vortex Jun 24, 2020, 5:03 PM

#

can u point it out pls?

dusky surge Jun 24, 2020, 5:03 PM

#

Sure

#

For Q3, you might wanna do the addition for your teacher since this would truly show that you did the math.

arctic vortex Jun 24, 2020, 5:04 PM

#

oh ok

dusky surge Jun 24, 2020, 5:04 PM

#

the teacher wouldnt know you did the math if you don't write it down

arctic vortex Jun 24, 2020, 5:05 PM

#

ok

#

i will do that

#

thanks!

#

can u also help me review for a test

#

"parallel lines intersected by a transversal"

dusky surge Jun 24, 2020, 5:05 PM

#

Q4, the Pythagoras just help you find the length of the sides but but not the angles, so it's not right.

arctic vortex Jun 24, 2020, 5:06 PM

#

oh

dusky surge Jun 24, 2020, 5:06 PM

#

Angles should be marked with CAPITAL LETTERS like A,B,C. And sides should be in lowercase letters like a,b,c

#

Law of sines can be used to find the angles, but I am not sure if you have learnt that.

arctic vortex Jun 24, 2020, 5:07 PM

#

oh

#

wait wahts the difference between angles and lnes?

dusky surge Jun 24, 2020, 5:08 PM

#

can u also help me review for a test
And for this, I don't know if I can do it because I gonna sleep soon😂

#

Have you learnt 'sine'?

arctic vortex Jun 24, 2020, 5:08 PM

#

no :(

#

and thats ok

dusky surge Jun 24, 2020, 5:10 PM

#

You can use direct comparsion on the 3 angles to see which is the smallest

arctic vortex Jun 24, 2020, 5:10 PM

#

oh

dusky surge Jun 24, 2020, 5:10 PM

#

You can say, use fingers to 'measure' them as a tool of comparing

arctic vortex Jun 24, 2020, 5:11 PM

#

wait so

#

is that what i say for 3)

#

for 4)*

dusky surge Jun 24, 2020, 5:11 PM

#

Like making an angle with 2 fingers, and find that the smallest one the fingers will be closest together

arctic vortex Jun 24, 2020, 5:12 PM

#

oh

dusky surge Jun 24, 2020, 5:13 PM

#

That's one method of estimation for elemetary school

#

I am not sure what your teacher wants, but IMO it's something that you'll have to figure out.

#

Which you did use your understanding and concepts to tackle this problem, which is already better than most of the students

#

pandaHugg

arctic vortex Jun 24, 2020, 5:15 PM

#

rlly o wow

dusky surge Jun 24, 2020, 5:16 PM

#

Just wanna say you did great. By asking, and by trying it out yourself.

arctic vortex Jun 24, 2020, 5:16 PM

#

:D

#

ok i will submit it now

#

can i add u as a friend

dusky surge Jun 24, 2020, 5:16 PM

#

Sure

arctic vortex Jun 24, 2020, 5:16 PM

#

yay

upper karma Jun 24, 2020, 6:14 PM

#

Thank you @dark sparrow for that sliver of info.

#

pretty straightforward proof.

#

now to prove that e^ix = cos(x) + isin(x)

upper karma Jun 25, 2020, 2:06 AM

#

📎 20200623_225549.jpg

#

this is a normal equation, but I didn't understand what my teacher did

#

She subtracted 30 degrees from both sides.

#

$x+30-30=y-10-30 \ x = y-40$

somber coyoteBOT Jun 25, 2020, 2:07 AM

#

leviosa:

📎 604465962053533709.png

upper karma Jun 25, 2020, 2:08 AM

#

Why the y is after the equal sign?

#

?

#

What's the original problem?

#

I was taught that letters are before the equal sign

#

x is a letter?

#

When you're looking for the value of a "letter" or otherwise known as a variable.

#

You want to look for what that specific variable is equal to.

#

he isolated the variable x

#

So

#

Exactly. And that gave him the x value.

#

Alternatively, y=x+40.

#

Idk what the problem is asking for etc.

#

I didn't know that moving numbers between the first and second member counted how to isolate the variable, I thought that what counted how to isolate was to move another variable.

#

I understood it now

#

Sorry for my english, I'm using the translator for hard words XD

#

Nah it's fine.

#

How school level are you?

#

I'm 16 years old.

#

I'm 15

#

I turned 16 6 days ago KEKW

#

I am Nine year (1 year until high school

#

I did 15y yesterday lol

#

Ooo.

#

Nice.

#

Happy birthday lol.

#

I'm going to 11th grade in a few months.

#

After the summer is over.

#

Nice

#

u smart for a 16 yr old @upper karma

#

I dumb asf.

#

But pls clarify.

#

@upper karma

#

u be answering bare question with ease

#

and im pretty sure you answered some difficult questions before

#

xDoubt

#

But thanks lol.

#

yikes

#

i might be mixing u up with someone else tbh

#

LOL.

#

Thanks for the ego boost ~~and unboost sadotato ~~

#

im sorry fit the unboost 😦

#

nah but i have a feeling u be solving hard ass questions

#

yikes

upper karma Jun 25, 2020, 3:36 AM

#

Awesome

upper karma Jun 25, 2020, 4:13 AM

#

yo so is sin squred of lets say 5x plus cos squred of 5x equal to 1

versed river Jun 25, 2020, 4:14 AM

#

sin^2(5x)+cos^2(5x)=1 yes

dark sparrow Jun 25, 2020, 4:18 AM

#

"squred"

upper karma Jun 25, 2020, 4:19 AM

#

Ok

pearl lava Jun 25, 2020, 1:10 PM

#

https://gyazo.com/fb58cfbe15c190a58a230980e2b61887
how would you do this?

Gyazo

silent plank Jun 25, 2020, 1:23 PM

#

multiple applications of pythagoras and properties of addition

pearl lava Jun 25, 2020, 1:25 PM

#

I got 28.9, is that correct?

silent plank Jun 25, 2020, 1:27 PM

#

not what i'm getting. show work

paper vale Jun 25, 2020, 1:29 PM

#

u only need one application, from projection of C onto side AD

#

it is arctan(6/14)

pearl lava Jun 25, 2020, 1:30 PM

#

oh

#

i get it

#

thanks

silent plank Jun 25, 2020, 1:30 PM

#

wait. sry misread the question

#

thought it was asking for CD,

#

so just one application of pythag,
properties of addition
and inverse trig

#

but still doesn't get 28.9 (°) for me

#

no, not 6/14

pearl lava Jun 25, 2020, 1:33 PM

#

nvm I got 27.6

silent plank Jun 25, 2020, 1:34 PM

#

yeh. looks good

#

,w arctan(6/(23- 9 - sqrt(6.5^2-6^2))) * 180°/pi

somber coyoteBOT Jun 25, 2020, 1:36 PM

#

📎 wolf.png

Display results online and refine query

pearl lava Jun 25, 2020, 1:39 PM

#

why is it multiplied by 180 degrees over pi?

silent plank Jun 25, 2020, 1:39 PM

#

conversion to degrees
since the default output on wolfram is in radians

pearl lava Jun 25, 2020, 1:40 PM

#

ok

silent plank Jun 25, 2020, 1:40 PM

#

if your calc is already set in degrees, you don't need to

pearl lava Jun 25, 2020, 1:40 PM

#

kk

upper karma Jun 25, 2020, 6:37 PM

#

Hello, I was hoping someone could help explain how this is working?

I'm meant to find AC, but I have no idea why they're able to multiply both sides by sin(B)

📎 unknown.png

#

why not?

#

what's wrong with multiplying both sides by sin(B)?

#

I don't know how they managed to go from

📎 unknown.png

#

to

📎 unknown.png

#

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

somber coyoteBOT Jun 25, 2020, 6:42 PM

#

HoboSas:

📎 310725130097786880.png

upper karma Jun 25, 2020, 6:42 PM

#

But why? I don't get it

#

you multiply both sides by sin(B)

#

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

somber coyoteBOT Jun 25, 2020, 6:44 PM

#

HoboSas:
Compile Error! Click the errors reaction for details. (You may edit your message)

📎 310725130097786880.png

warm basin Jun 25, 2020, 6:44 PM

#

Sir zebra, do you know what the sign = exactly means?

#

or rather the word, Equality

upper karma Jun 25, 2020, 6:45 PM

#

ofc, I just don't see how they're equal
It just says to do something without explaining why it works

warm basin Jun 25, 2020, 6:45 PM

#

okay now listen

upper karma Jun 25, 2020, 6:45 PM

#

?????

warm basin Jun 25, 2020, 6:45 PM

#

let's say 5 = 5 ok?

#

if you multiply BOTH sides by 2

#

you would get 10 = 10

#

right?

#

So you see the equality still holds?

upper karma Jun 25, 2020, 6:45 PM

#

📎 Cattura.PNG

warm basin Jun 25, 2020, 6:45 PM

#

They're still equal to each other.

#

Both sides

#

This is the same thing that happens there.

#

if you multiply BOTH sides of the equality, the expression on the left will still be equal to the expression on the right

#

Do you get it now? or

upper karma Jun 25, 2020, 6:49 PM

#

That's not the issue, it's a Khan Academy trigonometry thing I'm following and up until this point they've covered the stuff that's come up in the questions beforehand explaining why everything works the way it does. I got to this question and was stuck, I can see that they're equal but I don't know why.

warm basin Jun 25, 2020, 6:49 PM

#

📎 unknown.png

#

oh so its more of a figure thing?

upper karma Jun 25, 2020, 6:49 PM

#

so you have never heard of law of sines?:

#

a/sin(A) = b/sin(B) = c/sin(C) I think
I don't see how

b = a(sin(B)/sin(A))

comes into it

#

you forgot the c/sin(C) part

#

and you multiply both sides by sin(B)

#

What? I just copied what was on the screenshot

#

I meant that you leave it in the bin

warm basin Jun 25, 2020, 6:53 PM

#

(we dont need it)

#

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

somber coyoteBOT Jun 25, 2020, 6:54 PM

#

trece:

📎 720843757989396500.png

warm basin Jun 25, 2020, 6:54 PM

#

if we find b from here

#

we can plug it back in the equation to see if the sides are equal

#

$b = \frac{5\cdot\sin(108)}{\sin(31)}=9.23$

somber coyoteBOT Jun 25, 2020, 6:55 PM

#

trece:

📎 720843757989396500.png

warm basin Jun 25, 2020, 6:55 PM

#

$\frac{b}{\sin(B)}=\frac{9.23}{\sin(108)} = 9.7\
\frac{a}{\sin(A)}=\frac{5}{\sin(31)}=9.7$

somber coyoteBOT Jun 25, 2020, 6:56 PM

#

trece:

📎 720843757989396500.png

warm basin Jun 25, 2020, 6:56 PM

#

as you can see both give 9.7

#

i mean thats really just what law of sines says

#

📎 unknown.png

#

the law of sines tells us that those will be equal to each other.

#

so you dont need to doubt it

upper karma Jun 25, 2020, 7:05 PM

#

Thanks I guess
I'll sleep on it because I'm lost here

#

we are not getting your question

#

lmao

#

whats the question?

#

scroll up

#

it was an interesting discussion

#

i just skimmed thru it and damn i'm lost for words

nocturne dove Jun 25, 2020, 7:46 PM

#

I dont know if this fits in this channel, but could anyone help me with this?
https://cdn.discordapp.com/attachments/701658660526620702/725791761049452634/unknown.png

upper karma Jun 25, 2020, 7:48 PM

#

It does fit here.

#

well if x + y + z = 180 then the angle BCD = angle x

#

and angle BAZ = angle y

#

@nocturne dove ^

#

^

#

You can develop loads of stuff from there

#

^

#

Lol

#

^

nocturne dove Jun 25, 2020, 7:56 PM

#

aite thanks guys

upper karma Jun 25, 2020, 7:56 PM

#

np

chrome lake Jun 26, 2020, 6:06 AM

#

Hmmm

upper karma Jun 26, 2020, 7:32 AM

#

Hmmm

sick rover Jun 26, 2020, 1:17 PM

#

Can someone please tell me what my mistake is? The result should be 5dm2 but I got 6,9dm2

📎 image0.png

silent plank Jun 26, 2020, 1:30 PM

#

triangle RPQ isn't a right triangle

#

and you can't use "pythagoras"

#

to get x

upper karma Jun 26, 2020, 1:31 PM

#

I'll use a meme

silent plank Jun 26, 2020, 1:31 PM

#

use the formula indicated above. (i.e. apply the cosine law)

upper karma Jun 26, 2020, 1:31 PM

#

Pythagoras. Doesn't. Work. On. A. Non-right. Triangle

#

@upper karma

silent plank Jun 26, 2020, 1:32 PM

#

also not a fan of using h_m and h_f to indicate the respective altitudes of your triangles.

#

wait 1 sec.

#

that formula looks questionable

#

either that and/or the diagram

sick rover Jun 26, 2020, 1:39 PM

#

I guess the diagram
But like, I‘m not sure

That‘s just what I got to work with

And thank you for helping me
My teacher didn’t answer my Mail with the question

silent plank Jun 26, 2020, 1:41 PM

#

for the formula to be correct, the epsilon should be marked at angle RPQ

#

which takes quite a bit of effort to determine

upper karma Jun 26, 2020, 1:45 PM

#

@upper karma ahahah

silent plank Jun 26, 2020, 1:45 PM

#

also "general square"...

upper karma Jun 26, 2020, 1:45 PM

#

📎 Cattura.PNG

sick rover Jun 26, 2020, 1:49 PM

#

Sorry I translated it from German

#

I had to use google translator, I‘m not familiar with English math vocabulary at all

#

So, I guess I tried what you said but either I made a mistake again or idk

📎 image0.png

silent plank Jun 26, 2020, 1:51 PM

#

the english term would be (convex) quadrilateral

sick rover Jun 26, 2020, 1:51 PM

#

Uhh
Okay

silent plank Jun 26, 2020, 1:52 PM

#

but as mentioned the formula/diagram arent consistent and/or wrong

sick rover Jun 26, 2020, 1:53 PM

#

So, I should just not try to solve it and tell my teacher that it‘s not solvable?

silent plank Jun 26, 2020, 1:54 PM

#

technically its solvable. is that the original diagram as given?

sick rover Jun 26, 2020, 1:54 PM

#

Yes

silent plank Jun 26, 2020, 1:55 PM

#

to be consistent with the formula

#

epsilon should be here:

#

📎 unknown.png

#

and you can determine possible values of that with the use of the sin law.

sick rover Jun 26, 2020, 1:56 PM

#

Yeah well the diagram is given but I made the labels that aren‘t black

silent plank Jun 26, 2020, 1:57 PM

#

...

#

which was why i asked...

#

anyway,

sick rover Jun 26, 2020, 1:57 PM

#

Sorry I thought that was clear because it stated that one should label the diagram

#

Okay so
When epsilon is there
What would I have to do then?

#

Because like
No matter where it is
It is given that it‘s 32 degrees

And I calculated the Formular with that

silent plank Jun 26, 2020, 2:01 PM

#

angle RQP is given as 32 but that isn't the angle you should be using to find x

sick rover Jun 26, 2020, 2:02 PM

#

Okay, so the Formular itself is wrong?
Which one should I be using?

silent plank Jun 26, 2020, 2:02 PM

#

the formula is fine with a correctly labelled diagram

#

you can first apply the sine law to determine possible values of angle QRP

sick rover Jun 26, 2020, 2:04 PM

#

How do I do that?

silent plank Jun 26, 2020, 2:04 PM

#

start with:

#

$\frac{\sin(\angle QRP)}{312mm} = \frac{\sin(32\deg)}{277mm}$

somber coyoteBOT Jun 26, 2020, 2:05 PM

#

ramonov:

📎 220517274845577216.png

sick rover Jun 26, 2020, 2:08 PM

#

Okay I‘ll try

#

Wait but wouldn’t according to that epsilon be at yet another place
Like
Where R is instead of P like you drew it

#

Or no
Sorry

#

I was getting sth wrong

#

Or am I
Now I‘m not sure anymore

#

Yeah
Like
When you take 312 which is f, the angle would have to be at R wouldn’t it?

silent plank Jun 26, 2020, 2:23 PM

#

that is a bit unclear,
but the ratio of the sin of that angle (unambiguously referred to as angleQRP) and the side(f) is represented on the left side of the equation above

wraith drum Jun 26, 2020, 2:26 PM

#

Guys, I'm having problems with a question. Let ABC be a triangle right-angled at B, If the opposite side to an angle "theta" has a length (x^2)-4 and the adjacent side to "theta" has a length 4x, then what will be the smallest value attainable by the ratio between the area of the triangle and sin(theta).

dark sparrow Jun 26, 2020, 2:28 PM

#

have you made a diagram yet

wraith drum Jun 26, 2020, 2:31 PM

#

tried it, but the ratio is a polynomial and the minimum value is negative infinity

dark sparrow Jun 26, 2020, 2:31 PM

#

can you show me the polynomial?

#

also keep in mind that 4x and x^2 - 4 both have to be positive

wraith drum Jun 26, 2020, 2:31 PM

#

oh damn

#

give me a min

#

the polynomial is 2x^3+8x

#

i got it...32 was the answer

finite vigil Jun 26, 2020, 4:36 PM

#

i need help

#

on geometry

cosmic pebbleBOT Jun 26, 2020, 4:38 PM

#

From

#❓how-to-get-help

Rule 1

The help channels are solely for help with math, so feel free to post your question. Asking whether you can ask a question or if anyone knows about some specific topic is unnecessary, so please try to avoid questions of that nature.

upper karma Jun 26, 2020, 4:52 PM

#

Ask :)

charred zephyr Jun 26, 2020, 9:00 PM

#

show that cos(sin(tan(1))) = 1

#

degrees

upper karma Jun 26, 2020, 9:01 PM

#

1 what

#

Ok

#

Use arcos on both sides

eternal crag Jun 26, 2020, 9:02 PM

#

that is simply not true

upper karma Jun 26, 2020, 9:03 PM

#

That's kinda spoiling

eternal crag Jun 26, 2020, 9:03 PM

#

no it's not...

weary drift Jun 26, 2020, 9:04 PM

#

,calc cos(sin(tan(1)))

somber coyoteBOT Jun 26, 2020, 9:04 PM

#

Result:

0.54037772137207

silent plank Jun 26, 2020, 9:04 PM

#

it being a show question implies that its true,
when its not one needs to point out there is likely and error with the problem

upper karma Jun 26, 2020, 9:05 PM

#

is that in degrees though? @weary drift

weary drift Jun 26, 2020, 9:05 PM

#

,calc cos(sin(tan(pi/180)))

somber coyoteBOT Jun 26, 2020, 9:05 PM

#

Result:

0.9998476796922

silent plank Jun 26, 2020, 9:05 PM

#

regardless of whether the argument of tan is in degrees or radians,
its still false

eternal crag Jun 26, 2020, 9:05 PM

#

~~it is exactly approximately true~~

upper karma Jun 26, 2020, 9:06 PM

#

That's unlucky

eternal crag Jun 26, 2020, 9:06 PM

#

that was a copswing bait rokabe

charred zephyr Jun 26, 2020, 9:06 PM

#

input it on your calculators with degree mode

eternal crag Jun 26, 2020, 9:07 PM

#

calculator is not precise

upper karma Jun 26, 2020, 9:07 PM

#

Neither is wolfram

eternal crag Jun 26, 2020, 9:07 PM

#

wolfram is precise enough

upper karma Jun 26, 2020, 9:08 PM

#

Mmmh

charred zephyr Jun 26, 2020, 9:08 PM

#

then what should i do with the question

silent plank Jun 26, 2020, 9:08 PM

#

can you show exactly how the question is being presented

eternal crag Jun 26, 2020, 9:09 PM

#

problem: show that 1 = 2

#

what would you do with this question?

#

it doesn't make sense, just as the last one, you're asked to prove a false identity

upper karma Jun 26, 2020, 9:09 PM

#

Apply (mod 0) both sides

charred zephyr Jun 26, 2020, 9:10 PM

#

can you show exactly how the question is being presented
@silent plank
show that cos(sin(tan(1°))) = 1

silent plank Jun 26, 2020, 9:10 PM

#

ok. that's definitely false

#

and whoever made that question should be reported

upper karma Jun 26, 2020, 9:11 PM

#

Are you using Taylor's approx?

silent plank Jun 26, 2020, 9:11 PM

#

if it were: $\cos(\sin(\tan({\color{red}{0}}))) = 1$ then it would be true. in which case you can simplify stuff from the inside.

somber coyoteBOT Jun 26, 2020, 9:11 PM

#

ramonov:

📎 220517274845577216.png

charred zephyr Jun 26, 2020, 9:12 PM

#

why should i report him

#

my teacher wrote this question

upper karma Jun 26, 2020, 9:12 PM

#

Report him

charred zephyr Jun 26, 2020, 9:12 PM

#

i mean it's just a math question

#

wtf what

upper karma Jun 26, 2020, 9:12 PM

#

just a math question

silent plank Jun 26, 2020, 9:13 PM

#

might be a bit of an exaggeration,
but if your teacher asked you to prove 1+1 = 3 (something trivially false)

#

would you be confident in their ability to teach math?

upper karma Jun 26, 2020, 9:13 PM

#

A small mistake can crash a rocket into a planet

charred zephyr Jun 26, 2020, 9:13 PM

#

but 1+1 is approximately equal to 3

upper karma Jun 26, 2020, 9:14 PM

#

a large 2 is basically 3

charred zephyr Jun 26, 2020, 9:14 PM

#

you can say 10 is approx equal to 20 just so you can know

eternal crag Jun 26, 2020, 9:14 PM

#

just so you can know, that is incredibly dumb to say

charred zephyr Jun 26, 2020, 9:14 PM

#

approximations are always true unless you put that weird sign

eternal crag Jun 26, 2020, 9:15 PM

#

are you high or trolling now

charred zephyr Jun 26, 2020, 9:15 PM

#

lol absolutely not

#

in the question the teacher gave cos(....) is approx equal to 1

#

so stop saying i should report him

eternal crag Jun 26, 2020, 9:16 PM

#

cos(....) is approx equal to 1
that was never mentioned

silent plank Jun 26, 2020, 9:16 PM

#

$=$ does NOT mean approximately equal to

somber coyoteBOT Jun 26, 2020, 9:16 PM

#

ramonov:

📎 220517274845577216.png

charred zephyr Jun 26, 2020, 9:16 PM

#

i know right

#

it's the ~ thing

silent plank Jun 26, 2020, 9:16 PM

#

and when one writes =, one would assume it means equal to

charred zephyr Jun 26, 2020, 9:16 PM

#

i know that

#

what's the relation

silent plank Jun 26, 2020, 9:17 PM

#

or if you cant find a symbol for it, explicitly state it in words

charred zephyr Jun 26, 2020, 9:17 PM

#

anybody help

#

can i use the tan = sin/cos

eternal crag Jun 26, 2020, 9:18 PM

#

for what?

charred zephyr Jun 26, 2020, 9:18 PM

#

for my question

#

substitute it

eternal crag Jun 26, 2020, 9:19 PM

#

jesus fucking christ it's NOT A TRUE IDENTITY

silent plank Jun 26, 2020, 9:19 PM

#

what class are you in and what are you allowed to use?

#

he backtracked and stated its meant to be $\approx$

somber coyoteBOT Jun 26, 2020, 9:19 PM

#

ramonov:

📎 220517274845577216.png