#geometry-and-trigonometry

1 messages · Page 328 of 1

brittle gyro
severe hinge
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Thx @brittle gyro

mighty zodiac
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Can someone help me with my trig homework?

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There’s a couple questions that I don’t know how to do

silent plank
humble pulsar
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Just post the question

tender prawn
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is the answer 2?

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theres like 2 values of nq where sin nq = c right

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and like

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n is already determined so there's 2 values for q?

shadow wadi
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Guys a question, anybody have any ideas how to get this input equation equal to the one on right (1 ) equation i have tried dividing everything by cos^2 x and transforming cos to sin and then dividing by cos but just cant get this result please help, thanks in advance

📎 20201125_185216.jpg
tender prawn
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um

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(sinx + cosx)/cos^2 x into cosx/cos^2 x + sinx / cos^2 x?

shadow wadi
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Sorry full pic didnt send

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Ill send again but be more specific

mighty zodiac
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Can someone help me with this

jaunty marsh
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$cos^2(2x)=2cos^2(x)-1$

somber coyoteBOT
tender prawn
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that means cos sqrt11 = A

shadow wadi
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Marked equation into (a-1)tan^2(x)...

mighty zodiac
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I don’t get it

jaunty marsh
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$a*sin^2(x)+b(1-2sin^2(x))=1$

somber coyoteBOT
shadow wadi
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No sorry

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I mixed it

mighty zodiac
jaunty marsh
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$cos(theta)/sin(theta)=2sin(theta)cos(theta)/(1-(1-sin^2(theta))$

somber coyoteBOT
jaunty marsh
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$cos(theta)/sin(theta)=2sin(theta)cos(theta)/(sin^2(theta))$

somber coyoteBOT
jaunty marsh
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Wait

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Bottom is supposed to be 2sin^2(theta)

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$cos(theta)/sin(theta)=2sin(theta)cos(theta)/(2sin^2(theta))$

somber coyoteBOT
jaunty marsh
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$cos(theta)/sin(theta)=cos(theta)/sin(theta)$

somber coyoteBOT
jaunty marsh
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And it’s proven

mighty zodiac
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I’m confused

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So it works?

jaunty marsh
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Yeah

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But you have to prove it

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So use double angle identities

subtle snow
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Can some help with some geometry

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???

silent plank
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where are you stuck?

subtle snow
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For the first one I’m not sure what number it is to figure m

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@silent plank

silent plank
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have you learned about properties of parallel lines?

subtle snow
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I don’t remember

silent plank
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are terms like alternate angles, corresponding, etc familair to uyou?

subtle snow
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I think so but I don’t remember

silent plank
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review those

subtle snow
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What video should I watch to help would be helpful?

silent plank
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plenty of vids around to help with math

subtle snow
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Ok for the second one

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I think R=18 q=90 and s=72

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Is right? @silent plank

silent plank
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lowercase r but yeh

subtle snow
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Ok what should I type to learn how do question number one?

silent plank
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properties of parallel lines

subtle snow
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Ok I’ll let you know what I get

raven ruin
subtle snow
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Oh what i think I know what properties of parallel lines is

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@silent plank

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But it’s still doesn’t make sense

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On how to figure out m

silent plank
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upon review are you able to identify stuff like alternate angles now?

subtle snow
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Yes

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I know how to find alternative angles

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@silent plank

silent plank
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can you see that the angles with measures:
68-m and 5m-10 are alternate angles?

subtle snow
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Yes

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Oh

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So n=5m-10?

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@silent plank

silent plank
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well that is true but not up to that part yet

subtle snow
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Ok what should I do then?

silent plank
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well they're alternate angles on parallel lines so their measures will be the same

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68-m = 5m-10
would be the equation to set up to determine m

subtle snow
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Oh ok let me do it

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So m=3

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??

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@silent plank

silent plank
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no

subtle snow
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What did do wrong?

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@silent plank

silent plank
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78/6 isn't 3

subtle snow
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Oh so 13?

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@silent plank

silent plank
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yes

subtle snow
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Ok I’ll see what I can fill in

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I’ll let you know what I get

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Ok so n=55 m=13 and p=62.5?

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@silent plank

silent plank
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what are you doing to get p?

subtle snow
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Since one side 55 I subtracted that out of 180 and then divided it by two to get the two sides which is 62.5

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But now I’m not sure what should I do?

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@silent plank

silent plank
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youre doing random things

subtle snow
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Ok what should I do?

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@silent plank

silent plank
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stopping pinging me every time ...
i'm here...

subtle snow
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Ok sorry

silent plank
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do you see those congruence markings at 5m-10 and the angle next to it?

subtle snow
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Yes but I forgot what it means

silent plank
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it means that they're congruent

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and will have equal measures

subtle snow
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So it’s also 55

silent plank
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yes

subtle snow
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So p = 70

silent plank
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yes

subtle snow
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Ok

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One last thing I don’t know if you would happen to know what to do here?

hollow raven
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Wonder why it says exam 3 at the top pandaThink

west basin
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deletes it right after

hollow raven
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What timing

west basin
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like a boss

subtle snow
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That was a difficult thing oops

west basin
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its ok i passed geometry like 5 years ago

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i can do it

subtle snow
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Can you help me?

west basin
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not on an exam

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which clearly it was even tho you deleted it

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so i dont think i can

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sorry

subtle snow
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Ok thanks anyways

west basin
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academic dishonesty can ruin your entire career

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even tho it seems like its no big deal

hollow raven
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Don't ask for help on exams in the future mate you're just playin urself

subtle snow
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Ok my bad sorry

mighty zodiac
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Can someone help me with this step by step I’m having trouble

humble pulsar
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$2\sin^2{x} + 9\sin{x} +4 = 0$ ?

somber coyoteBOT
humble pulsar
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@mighty zodiac

mighty zodiac
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Yes

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@humble pulsar

humble pulsar
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it's a quadratic

mighty zodiac
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Yes

humble pulsar
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do you know how to solve quadratics?

mighty zodiac
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Yeah but not in 0,2pie

humble pulsar
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that's only relevant for sinx

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solve 2u^2 + 9u + 4 =0

mighty zodiac
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Ok so I got 7pie/6 and 11pie/6

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Can I get help with this too?

round isle
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@mighty zodiac assume cos(3*theta) = x and replace it in the equation and solve

tame tartan
round isle
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was this question taken from an IMO in 1800s?

slate blade
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What's the stuff on the left and right of the diagram the 1/2 and 1?

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$$ \frac{1}{2}$$

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noice

upper karma
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It's the radius of the arc

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I'm assuming

hoary trellis
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what shape is on the bottom? I got the volume of the square. Need the other one thanks

📎 unknown.png
upper karma
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square pyramid

hoary trellis
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ty

coral dagger
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For the area of a triangle formula to work, does it have to be trig function of Sine that is used?

For this example 1/2•base•aSinC is used but what happened if C isn’t available (or for some reason I can’t use C)

📎 image0.jpg 📎 image1.jpg
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And for a question I have to use another angle (arrowed in photo, let’s call it F)

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so is 1/2 • base • aCosineF = 1/2 • base • aSineC?

upper karma
royal citrus
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hey could somone help me with this.

humble pulsar
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combine sec^2 and csc^2

royal citrus
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please hapen to ping me for an answer

humble pulsar
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@royal citrus

royal citrus
humble pulsar
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add the fractions together

royal citrus
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fractions???

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where did fractions come from

humble pulsar
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sec is 1/cos

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csc is 1/sin...

royal citrus
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oh those

west basin
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gotta know yo trig identities son

humble pulsar
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^

royal citrus
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,, \frac{1}{cosA.sinA}

somber coyoteBOT
royal citrus
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now what do I do?

round isle
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lol

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do you not see the solution?

west basin
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trig identities again dude

round isle
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I suggest you review your trig identities

west basin
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remember

royal citrus
humble pulsar
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try the fact sec=1/cos

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and csc=1/sin

compact saffron
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can someone help with a trig question?

deft blaze
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post the question

compact saffron
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It’s a practice test

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And I’m stumped :/

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<@&286206848099549185>

rotund bay
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any intuitive reason for why principal radii are on orthogonal planes?

dark sparrow
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what's a principal radius :?

upper karma
still laurel
swift summit
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@still laurel just use either side for the equation. cosine(34)=5 or 9/x

still laurel
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@swift summit right but what is the logic behind that, I get the calculation but how

swift summit
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because cosine=adjacent/hypotenuse. so by plugging the information in you end up with that

still laurel
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is there logic in this question behind how you pick which side is adjacent

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thats all

swift summit
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oh no either side works just as well as the other

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wait

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you do know what adjacent means, yes?

still laurel
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yes

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thanks

swift summit
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ok np

mystic dune
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I don't know how to do it but I got 13.5 m

spare plaza
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what am i doing wrong? idk how to do veritcal stretch

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isnt it 2(3)?

silent plank
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@Sarim1k#6558 the cosine law:
$$c^2 =a^2 + b^2 -2ab\cdot\cos(C)$$

somber coyoteBOT
deft blaze
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because cosine=adjacent/hypotenuse. so by plugging the information in you end up with that
@swift summit you only use this with right angles

somber coyoteBOT
earnest echo
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Since the question itself says all permissible values of x, I don't think you have to list out non-permissible values

somber coyoteBOT
quick timber
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so, I appearantly have all the info I need to say that E,F,A and B are on the same circle... tho I'm failing to see how they are... all I know is that AFB and AEB are the same size, basically.

upper karma
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can you send the original problem?

quick timber
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if you can read dutch 😅

rotund bay
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You mean like that they're on the same plane?

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Is it a 3D question?

quick timber
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2D

upper karma
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what is the goal?

rotund bay
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What does them being on the same circle mean?

quick timber
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proof that ABF and E are on the same circle

rotund bay
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And define "being on the same circle"

upper karma
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so ABFE is cyclic

quick timber
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it meanst that EAB + AFB =180

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idk english terms 😅

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lemme, translate

upper karma
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quadrilateral ABFE is cyclic

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we got it

quick timber
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oh alr

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google translate calls it "cord quadrilateral"

upper karma
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I think you already got the solution

quick timber
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thats what the answers say, but im not seeing it XD

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thats my issue rn

upper karma
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ok I see

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|| so angle ADE = EAD = y because triangle ADE is isosceles, same thing for triangle FCB (angle FCB = CBF = x). For the exterior angle theorem we have AEB = 2y and AFB = 2x, but notice x = y (angle ADB = ACB) because they are on the same segment AB, so we conclude AEB = AFB, which makes the quadrilateral cyclic. ||

quick timber
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id see it if EAB and FBA were the same size, but I havent proven that.. right?

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Im so confused, I got all that information but im just not seeing how u can say its cyclic rn

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😅

upper karma
quick timber
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I had all that, but I just dont understand the last bit

rotund bay
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AEB and AFB are both right angles right?

upper karma
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no

upper karma
rotund bay
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Doesn't AED have to be a right angle though

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2 same sides

upper karma
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you could think of it as angles AEB and AFB being on the same segment AB

quick timber
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but in class they said, AEB + ABE = 180 makes it cyclic

upper karma
quick timber
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OH

quick timber
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XD

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seeing it now LMAO

upper karma
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that's another condition

quick timber
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yeahh, makes sense now I think about it

upper karma
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there are several ways to prove that a quadrilateral is cyclic

rotund bay
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Triangle ADE is a right triangle, AED and AEB are both right angles

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Same on other side

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No?

upper karma
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???????????

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they are isosceles

rotund bay
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Side AE and DE are equal length

autumn vortex
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just isosceles

upper karma
quick timber
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atleast I got everything but the conclusion myself... LMAO

rotund bay
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NEVERMIND

sly elm
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do both sin(90-a) and sin(90+a) = cosa?

midnight ice
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Hi everyone

round isle
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hi

dusky surge
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Hi

deft blaze
obtuse tapir
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some good questions pls

obtuse tapir
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im stumped

mystic dune
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16

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I just put it in desmos and found which value it equals 64, I don't know how to do it algrebraically

obtuse tapir
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I tried it in wolfram alpha

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Doesnt work

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Maybe i used the wrong notation

mystic dune
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I tried doing it but I gave up then I used desmos

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to see if x is equaled to a good number

obtuse tapir
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Still stumped

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Show solution

silver spruce
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can someone help me by walking me through how to evaluate cos(-225) with a answer in fraction form.

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I think it has something to do with the unit circle and the fundamental trigonometric identity

umbral snow
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Can you pinpoint, at least approximately, where -225° lands on the unit circle?

silver spruce
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oh yea

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from (0,1)?

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does that awnser ur question?

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dunno

umbral snow
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Is it at 9 o'clock?

silver spruce
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3

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lol im dumb

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(1,0)

umbral snow
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No you're not dumb haha, but you'll want to know what the unit circle is and how to find where an angle lands on it

silver spruce
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i mean like it starts at (1,0)

umbral snow
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Basically, you always start at 3:00 or (1,0) yeah

silver spruce
#

and like the angle goes from there

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yea

umbral snow
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The negative means "go clockwise"

silver spruce
#

yea

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but like the cos in trhowing me off

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throw

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ing

umbral snow
#

180 would be half way around, 270 would be 3/4 way around

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But you have 225, so it lands half way between those two, somewhere around 10 o'clock

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That part make sense?

silver spruce
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225/360 =5/8

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yea

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so like 5/8 the way

umbral snow
#

Yeah haha, if you want to say it that way, that is good too

silver spruce
#

is that then (7pi)/6 whatever thing?

umbral snow
#

Big thing to realize, it's 45° away from the y-axis!

silver spruce
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like distance

umbral snow
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So you'll want to relate it to something to do with cos(45)

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Which is - you'll want to relate it to √2 / 2

silver spruce
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so how do we figure this out?

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is it just that 225/45 =5

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so that the 45 degree ange on the unit circle like the √2 / 2

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or like

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dunno

umbral snow
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What's the sign of cos(-225)? That is, what's the sign of the x-value when the terminal arm is near 10 o'clock?

silver spruce
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its negative

umbral snow
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You got it

silver spruce
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so like because 45 is a factor of 225

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and like 45 degrees is the √2 / 2 thing on the unit circle

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and that 225 is on the negative side of x

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its just -√2 / 2

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?

umbral snow
#

Well, -225 is

silver spruce
#

yea

umbral snow
#

But yes exactly

silver spruce
#

so the awnser would be -√2 / 2?

umbral snow
#

You got it

silver spruce
#

but how would we apply this to a diffrent problem?

umbral snow
#

Sure, have another example?

silver spruce
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because liek some problems arent a factor of 45

hollow raven
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@upper karma omg

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thats a really good question

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and there is a genius answer

silver spruce
#

uh

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cos(-150)

umbral snow
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Same idea first - get a sense of where it lands approximately

silver spruce
#

the x is still negative

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wait

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yea

umbral snow
#

It's almost -180, but 30° short

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So somewhere around 8o'clock

silver spruce
#

so √3 / 2

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wait no

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-√3 / 2

umbral snow
#

You got it

silver spruce
#

so its more logic than like math?

umbral snow
silver spruce
#

liek you jsut need to figure out where it is then its easu

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*easy

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*liek

umbral snow
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Then two blend together a lot haha

silver spruce
#

****like

hollow raven
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Kaynex, you are certainly cool

silver spruce
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yea

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lol thx

umbral snow
#

It's not all symbol pushing

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That's the "hip" term

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@hollow raven
And thank you but as my name implies, I am not-cool

hollow raven
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thonkeyes I see otherwise

silver spruce
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wait

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what happens when its sin instead of cos

umbral snow
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@silver spruce
Then you want to focus on the y-value, not the x-value

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Otherwise the question is the same

silver spruce
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thx

royal citrus
#

Hey how do u do this?

somber coyoteBOT
silent plank
#

apply a^4 = a^2*a^2 and pytahgorean trig identities

somber coyoteBOT
silent plank
#

parentheses, also assuming that you actually had those,
you aren't distributing properly

silk patio
round isle
#

triangle OPQ is isosceles

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so you can find the angle POQ

royal citrus
round isle
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thus finding the length of arc PQ

silk patio
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How does that help find the angle

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No trig allowed too

round isle
#

huh?

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what do you mean no trig allowed?

silk patio
#

Purely algebraic solution

round isle
#

trignometric equations aren't algebra?

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are you absolutely sure about trig not being allowed?

silk patio
#

I don’t like transcendental functions in my basic geometry problem

round isle
#

oh I see

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so you don't want to use trig?

silk patio
#

It can be done without reference to angles

round isle
#

I don't think you can but if you do then I guess good for you

silk patio
#

I created it

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It can

round isle
#

ok god

unkempt wasp
#

hello. I have an exercise with imaginary numbers and I don't know how to do it. The +- is the difficult part for me. Why is it here? What am I supposed to do with it? Would be really glad if someone could explain.

📎 unknown.png
round isle
#

for example

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a +- b

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means you have two possible answers

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a + b
a - b

unkempt wasp
#

but there's only one answer in the answers sheet.
Although if this is all it means then thank you.

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I'll try to solve it with both

unkempt wasp
#

but I keep getting 2

silent plank
#

how are you getting 2

unkempt wasp
#

sqrt((1/2)^2+(sqrt(3)/2)^2)

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but that's the length. the total answer I get is 8

silent plank
#

doesnt tell me how you're getting 2 because that's 1

unkempt wasp
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oh

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omg

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it is. sorry

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I somehow managed to get 2 in this

earnest sphinx
#

"If the tangent vector of a parametrized curve is constant, the image of the curve is part of a straight line."

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what does that mean? xd

round isle
#

huh?

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if the tangent is the same at all points

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wouldn't that mean the curve is just a line?

earnest sphinx
#

if we are in R^3 a line is a plane, or a line is a line?

round isle
#

it would be a plane in R3

earnest sphinx
#

okk ty

weary drift
#

no it’s a line in any R^n

earnest sphinx
#

ahh thanks

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a vector being constant means all its components are constants right?

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what's the tangent vector of a plane?

weary drift
#

a vector being constant means all its components are constants right?
yes
what's the tangent vector of a plane?
doesn’t make sense to speak of one. rather we speak of a plane characterized by a normal vector

upper karma
#

help

upper karma
#

do you know synthetic division?

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if not, i'd look up a video on how to do it

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it's basically long division, but with polynomials

ashen brook
#

how do i find the angle for the vector W= -2i-2sqrt3j

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could anyone help me with complex numbers

slate blade
#

sure wassup with ur complex numbers

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@ashen brook For your first question of vectors

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I'm assuming you're asking between vector and x axis?

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$$ W = -2i -2\sqrt{3}j $$

somber coyoteBOT
slate blade
#

Thus, you know that your angle between the x axis and the singular 2D vector is equal to this

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$$ let \theta = angle
\therefore \theta = \arctan{(\frac{-2\sqrt{3}}{-2})} $$

somber coyoteBOT
versed river
#

text goes outside the dollar signs if you want spaces and dont want it to be italics

slate blade
#

ah my bad sorry still getting used to the bot lol

versed river
#

$\frac{\pi}{3}$

somber coyoteBOT
slate blade
#

yeah that too

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wait what's the difference between single and double $$

versed river
#

display math mode and normal math mode, you can google it

slate blade
#

ok ty

obtuse tapir
#

OA=AB=1
Find PQ
@silk patio thats ez

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two equations can be formed

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x+y=2, x^2+pq^2/4=1 y^2+pq^2/4=2

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PQ=root(7)/2

brisk latch
silk patio
#

@obtuse tapir not right, close tho

brisk latch
#

kinda need an answer asap

upper karma
#

Why

brisk latch
#

its for my hm

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hw

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and it needs to be in in 20m

upper karma
#

We don't give away answers here

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Can you post your work so that we can try to spot your mistake

brisk latch
#

i dont have anywork rn

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idk how to work out the diameter/base

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BD

upper karma
#

Consider drawing triangle OBC

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Can you think of a way of getting angle BOC

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Using a specific circle theorem

brisk latch
#

no?

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let me go check

upper karma
#

note that angles BOC and BAC correspond to the same arc

brisk latch
#

idk srry

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first time attempting these questions

upper karma
#

Look at the first one

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Are you able to get angle y considering the theorem i just sent

brisk latch
#

um,

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y angle is 90?

upper karma
#

Why 90

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Did you used the theorem i sent above

brisk latch
#

ye

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OH shit

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its the A to 0 right

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or is it B to 0

upper karma
#

What?

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Where do you see a 0

brisk latch
#

O

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the middle

upper karma
#

What about it

brisk latch
#

the y angle

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is 5

#

56

upper karma
#

Yes, 56°

brisk latch
#

now wut.

upper karma
#

If we split the triangle like this

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Let me get a better pic

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Look at the green triangle

#

Remember we splitted the triangle in 2, so the angle BOP will be half of 56°

#

Now look at the 5th theorem here, according to this, on the green triangle we get a right triangle with 2 known angles and one known side

#

@brisk latch understood everything i've just said so far?

brisk latch
#

hm

upper karma
#

Please don't say "yes" if you haven't understood it

#

We are here to learn

brisk latch
#

sorry im learning from 5 different peoople at onc

upper karma
#

Huh

deft blaze
upper karma
#

It's better to learn from 1 person decently than from 5 people unefficiently

#

Focus on one

brisk latch
#

its 4 different people in 1 server and u

upper karma
#

Come back whenever you are able to follow me efficiently

brisk latch
#

got the answer

#

its DB = 10

#

well DB = 10.01

upper karma
#

What do you expect me to say

obtuse tapir
silk patio
obtuse tapir
#

whats the solution

dark sparrow
#

<@&268886789983436800> advertiser spam

#

@upper karma no advertising. and definitely no spam.

upper karma
#

this is not the place to do so

dark sparrow
#

and not the manner either

obtuse tapir
upper karma
#

,rccw

somber coyoteBOT
obtuse tapir
#

Thanks

#

@upper karma Perhaps you could tell me my error

dusky surge
#

What are you calculating? @obtuse tapir
Is PB=1?

scenic goblet
#

I need my program to make assumptions based on the contours of a surface for the path of a rope moving across it. Any chance I can find that on this server?

round isle
#

please post your question

scenic goblet
#

I’m able to ascertain the 3D coordinates of points along an uneven surface, but I don’t have any kind of math background, so I can’t use the data. My main goal is to determine location of the peaks and troughs and the distance between them, in the context of the programming environment I’m using.

#

Essentially, I’m looking for a mathematician to join my development team (me), to develop a rope physics simulation.

round isle
#

I see

#

so you want to know where the local maxima and minima are

#

based on data points

#

so your surface is just a collection of 3D points

#

there are many ways to do that if you don't know the actual function that parametrizes your surface

#

one thing you could do is

#

define a neighborhood of size 10 units (for example)

#

look for the point that has the highest/ lowest height in this neighborhood

#

in computer vision we use a technique

#

where we convolve a surface with the Sobel filter

#

to detect all the local maxima

#

you could probably try something like that too

scenic goblet
#

That sounds very likely to be true, but I don’t know how to do much of that, simply because I am using brute force to program, up to this point. I’m very versed with the tools, but lack all but the most basic math ability.

round isle
#

use this video as guidance

#

replace the filter in this with the Sobel filter

scenic goblet
#

That’s interesting.

#

So, it averages the image and reduces it to a lower resolution?

round isle
#

?

#

the Sobel filter?

scenic goblet
#

What’s the use case for the method on the video?

#

What’s it for?

round isle
#

this video is to help you understand how 2D convolution works

#

so you could do it for your program

#

replace the filter in this video by the sobel filter

#

which will look for local maxima on your surface

#

since your surface is represented as a grid with value in each grid representing the height of the surface at that location

scenic goblet
#

What’s a sobel filter?

round isle
#

it's a filter that can find local maxima

#

we use it in image processing to do edge detection

scenic goblet
#

Here’s an example surface. I want my rope to understand where the peaks and troughs are so it can lay across it naturally. How would I apply a sobel filter to this, would you say?

📎 image0.jpg
round isle
#

you should convert this to a 2D grid where (x, y) is the location of a point on the surface and the value at a grid square at some (x, y) location represents the height of the point on the surface

#

then convolve that grid with Sobel operator

scenic goblet
#

What is the Sobel operator? How would I apply it here?

round isle
#

?

#

have you not been reading anything I just said?

#

scroll over and read what I just wrote

obtuse tapir
scenic goblet
#

Sorry, I don’t know how to replace the filter in the video with a Sobel filter.

obtuse tapir
#

@dusky surge

scenic goblet
#

@round isle Ok, I googled it and now I see that these filters are in a common format. So, I can apply this within my program, no problem, but what will the resulting values represent?

#

It will be reduced to a smaller grid, for one thing.

obtuse tapir
round isle
#

to prevent your grid from getting reduced, you can pad it with a border. The new grid, after convolution with sobel operator, will be the same size. You'll get two grids: one after convolving with the horizontal Sobel filter and the other due to the vertical Sobel operator. For each corresponding location, you square and add the Sobel values and take square root. The new grid will now contain a lot of large values (representing minima and maxima). Now, you can define another window where the highest value will be highlighted only and you can use that to check the y value in the original grid (before applying the Sobel operator)

#

you can also try the Canny Edge detector

#

you better do some reading on these algos on your own time

scenic goblet
round isle
#

why are you not padding the sides

#

this will let you get a grid of the same size

#

after applying sobel operator

scenic goblet
#

Because I have no idea what I’m doing. I just needed to finish this in a basic form, now I can expand it.

round isle
#

ok

#

add padding to get a resulting grid of same size

scenic goblet
#

K

#

That’s pretty much already there, because I didn’t limit the movement of the filter

#

And I drop the value cubes with a button press

#

So, at each location on the grid I square the horizontal and vertical and then add them and then “take the square root”?

signal wolf
#

I have a rather hard time figuring this problem out, experimentation with GeoGebra shows me nothing particularly useful

📎 unknown.png
scenic goblet
#

@round isle In the video he places the sum of the kernel in the top left. Where does the sum of each kernel position go in the Sobel kernel?

round isle
#

the middle of the kernel

scenic goblet
#

What's the most straightforward way to find the square root if you only have basic calculator functions? (I told you, I'm terrible at math.)

round isle
#

?

#

what?

#

I have no idea what you mean

scenic goblet
#

How do I get the square root? Your instructions have me doing that.

round isle
#

?

#

what?

#

just take the square root...

#

?

scenic goblet
#

I have zero math ability. Basically handicapped.

round isle
#

what do you mean you have zero math capability

scenic goblet
#

I don't understand it

round isle
#

how old are you?

scenic goblet
#

Very old

round isle
#

huh?

scenic goblet
#

40

round isle
#

you didn't go to school?

scenic goblet
#

I didn't finish middle school

round isle
#

blimey

#

that's not good

#

you are in what country?

scenic goblet
#

US

round isle
#

???

#

wtf

#

I thought every kid is supposed to complete high school in US

#

legally required to

scenic goblet
#

Not at all

glacial dawn
#

in some states I think compulsory education is up to middle school

round isle
#

wait a second

#

you don't know basic math

#

but you know how to program?

#

how is that even possible

scenic goblet
#

I'm using brute force methods and a very friendly programming environment

round isle
#

what language are you programming in?

scenic goblet
#

And I've put a lot of time in

#

It's not a language.

round isle
#

?

scenic goblet
#

It's a platform, I guess. It's called Dreams.

round isle
#

oh wait

#

that PlayStation game

scenic goblet
#

Yes

round isle
#

bruh, is it really possible to make stuff in it without coding?

scenic goblet
#

That's no coding, per say. It's all visual nodes that you wire together.

round isle
#

so, you don't know what a square root is?

scenic goblet
#

No

round isle
#

well

#

I could tell you what it is

#

but before that

#

please do something about your education

#

you won't get far with the kind of stuff you're doing

#

without math

#

so, now

#

square root:

scenic goblet
#

That's why I'm here

round isle
#

2 x 2 = 4

#

when we take square root of 4

#

we get 2

#

because when we multiply 2 with itself

#

we get 4

#

so, taking square root is breaking a number down into its multiplicative components

scenic goblet
#

Ok, that makes sense

#

And how would I get the square root with basic calculator functions?

round isle
#

well you don't

#

calculators have square root button

#

let me guess

#

Dreams doesn't have square root?

scenic goblet
#

No it doesn't

round isle
#

ok

#

so, square root is not necessary

#

you can just square the x and y results and add them

#

that alone would be fine

scenic goblet
#

I don't need it?

round isle
#

no you don't

oak citrus
#

^(1/2) works right

round isle
#

I don't think that's available in Dreams

#

which he is using

oak citrus
#

oof

scenic goblet
#

Dreams has "power of". Is that what that is?

simple drum
#

Yes

glacial dawn
#

If u take ur number to the power 0.5 u will get the square root

upper karma
#

Use a calculator bro lmao

scenic goblet
#

So the square root of 150 is 12.247? That's what I'm getting.

simple drum
#

That sounds accurate

round isle
#

ha

#

didn't know Dreams has power of

#

what exactly are you trying to build in Dreams anyway?

scenic goblet
#

I'm trying to build a system that gives the program some general awareness of the shape of the terrain.

Lots of uses for that, but currently I hope to make a rope react naturally as it goes across terrain and over ledges.

#

I want high points to make that portion of the rope resist being dragged, and that kind of thing.

scenic goblet
neon hamlet
#

How would I solve this? Or just set this up

round isle
#

what have you tried so far

neon hamlet
#

Not much, still confused on how I would start off.

humble pulsar
#

What's the relationship between <FAD and <CAB

neon hamlet
#

Vertical Angles.

#

So congruent

humble pulsar
#

wait

#

nvm i was thinking the wrong thing lol

round isle
#

well you have two important straight lines here

#

CD and BF

#

sum of all angles on a straight line is 180

#

so use that here

neon hamlet
#

Well since im setting up a system of equations,
9x + y = 180
and I need something else.

hollow raven
#

Good thing to note here

#

90-4x+36 = 5x, because 4x + <FAE = 90, so <FAE = 90 - 4x, and 5x is an opposite angle of 36 + <FAE, so 5x = 36 + <FAE

#

Since both angle summations are equivalent, 90-4x+36 = 5x

neon hamlet
#

Gotcha, thanks for the help. Solved it now.

dusky surge
neon jolt
#

Solving for theta Cos(Theta/2) Does theta = 2ArcCos(Hyp/Adj)?

#

can you actually pull the one half out or is this purely a half angle

neon jolt
#

I assume that this would be some kind of Cos^-1 of a Half angle?

round isle
#

I can't understand your question

#

cos(theta/2) = ?

neon jolt
#

Correct

#

cos(theta/2)

round isle
#

use half angle formula

neon jolt
#

you mean set it equal to half angle and solve for theta?

round isle
#

???

#

what is your question exactly

#

cos(theta/2) = some number?

neon jolt
#

Well I am looking at this quaterion math for complex number and I am trying understand how the half angle become 2ArcCos

#

I'll link it

round isle
#

and?

neon jolt
#

so

#

q = cos(angle/2) + i ( x * sin(angle/2)) + j (y * sin(angle/2)) + k ( z * sin(angle/2))

#

at Theta = 0 degrees

#

imaginary part becomes 0

round isle
#

imaginary part?

neon jolt
#

because sine theta/2 = 0

round isle
#

what imaginary part?

neon jolt
#

i ( x * sin(angle/2)) + j (y * sin(angle/2)) + k ( z * sin(angle/2))

#

^

round isle
#

lol

#

have you studied vectors?

neon jolt
#

yes

round isle
#

good

#

do you know unit vectors along x, y, and z axis

neon jolt
#

yes

round isle
#

good

#

there you go

neon jolt
#

so check this out

round isle
#

check out what?

neon jolt
#

when the angle is at 0

#

q = 1 + i 0 + j 0 + k 0

round isle
#

that is not the imaginary part first of all

#

the i, j and k are unit vectors along x, y and z axis

neon jolt
#

umm

#

I am aware

#

but a quaterion has a real part and imaginary part

round isle
#

your high school is teaching you quaternions?

neon jolt
#

the imaginary part is the unit vectors

round isle
#

crap

#

I am out of my depth here then

#

I have not studied quaternions

#

best you wait for someone who has

neon jolt
#

Ignore it for a moment

#

I am just rusty

#

at the trig

#

all you need to know is as theta -> 0 you get Q = 1 + 0

#

W = real part

#

W = cos(theta/2)

#

so

#

q = 1 + i 0 + j 0 + k 0
gives us this angle = 2 * acos(qw) = 0

#

so working back from above equation qw = 1

round isle
#

wait what?

#

2 * acos(qw) = 0

#

you are solving this?

neon jolt
#

I am trying to see how they get 2*acos(qw) = 0

#

yes

round isle
#

angle = 2 * acos(qw)

#

they are saying angle is 0

#

so just put that in there and solve for qw

neon jolt
#

ha

#

hmm

#

QW = 1 at 0

#

but how did they solve for theta

#

thats the fundamental thing i am forgetting

#

I assume convert it to half angle

#

and then solve for theta?

#

then take the Arc Cos

#

I see it now

#

thanks

#

QW = 1 Because the one side is 1

#

it has to be

outer trench
#

im confused on this

#

y it upside down

#

if you know that two sides of a triangle have lengths 4 cm and 6 cm, what specifically do you know about the length of the third side if the triangle is a right triangle?

silent plank
#

consider theorems related to right triangles

outer trench
#

i mean it 90 degrees

silent plank
#

well one of the angles will be 90°

#

ever heard of pythagoras?

#

if the triangle is right, the sides will satisfy pythagoras' theorem

dire scaffold
#

Formula pls

#

thx

upper karma
#

If theta is on radians

#

Very important to note

#

290° isn't on radians

#

@dire scaffold do you know how to manage from degrees to radians?

dire scaffold
#

no

upper karma
#

Ok then that's not the formula you're looking for

dire scaffold
#

o

west basin
#

sorry about that kinda just glanced quick

upper karma
#

The formula you are looking for is $A_s=\frac{\theta}{360}πr²$ where $\theta$ is in degrees

somber coyoteBOT
dire scaffold
#

alright

#

thx

blazing pond
#

yo

#

can someone help me?

charred osprey
#

2cos(75-β)-sinβ=0
How could I solve this?

crisp copper
#

Wow I just discovered math discord

#

😭😭😭

#

This is awesome

umbral snow
#

Hey Samwise, you were my favorite star wars villian.

crisp copper
#

Wow dude

crisp copper
#

After all, I killed Yoda

upper karma
# charred osprey 2cos(75-β)-sinβ=0 How could I solve this?

First, I'd get the 75 degrees into radians. Following up, split up cos(5π/12 - β) using subtraction identity for cosine.

After you applied the identity, manipulate the formula in terms of acos(x) + bsin(x). If you are having trouble with getting it in said form, then I'd recommend you to completely expand it and factor out the cos(x) and sin(x).

Using another important identity, acos(x) + bsin(x) = Rcos(x - c), where R is the sqrt(a² + b²) and c = arctan(b/a).

Setting Rcos(x - c) = 0, we have x = nπ + π/2 + arctan(b/a), where n is the set of integers. Don't forget to state this if your teacher wants it to be solve generally.

charred osprey
#

Thanks!

plucky pebble
#

Hey guys, How do i solve this problem. Am I allowed to draw halves?

📎 Capture.PNG
round isle
#

you have to draw the 3D structure

#

try to visualize

plucky pebble
#

Am I allowed to draw between two points

#

I tried a ton of times

#

My teacher couldn't solve it

ashen brook
#

what would be the equation for finding the six sixth roots of z=-1

#

"Find the six sixth roots for z=-1. Leave your answers in trigonometric form."

weary drift
#

de moivre

ashen brook
#

i think its like = (cos (180) + i sin (180)) but im not for sure what to put infront

#

nvm

#

i dont think its that

limpid gust
#

can someone explain what I need to do to find the missing angles

📎 unknown.png
stuck dragon
#

#8: what is the relationship between <1 and 105 degrees?
#9: what is the relationship between <2 and 105 degrees?
#10: what is the relationship between <3 and <7?
#11: what is the relationship between <4 and 80 degrees?
#12: what is the relationship between <6 and <3 or <7?
#13: what is the relationship between <7 and 105 degrees?
#14: think about the sum of the angles of a triangle and the vertical angles around the triangle.
15#: what is the relationship between <9, <8, and 80 degrees?

#

you might want to rearrange the order you go for these problems

upper karma
upper karma
#

is this correct?

#

Just wanna make sure

olive plaza
#

Can anyone help with trig?

humble pulsar
#

@olive plaza yes, i can help with that question you totally posted

olive plaza
#

I can't figure out how to post it

#

anyways

#

the question is estimate a value for cot 140

humble pulsar
#

just post the picture... or type it out

#

If it's just estimate i'd probably average the values of cot(120) and cot(150)

limpid gust
umbral snow
#

@ashen brook
Still looking for it?

ashen brook
umbral snow
#

@ashen brook
Note that you care about when
z⁶ = -1 = cos(180 + 360k) + isin(180 + 360k)

#

Where you can put any k in, and that gives a coterminal angle. That bit make sense?

ashen brook
#

yeah i understood that part im just not sure what to put in cis form for example x

#

x cis 90

#

ah i think i got it

vast python
ashen brook
#

@umbral snow thankyou, i was able to complete the problem!

vast python
#

I did the first one but I don’t understand it I get that sin is y and cos is x but how do I use that for this problem

slate blade
#

for 58?

#

@vast python basically you know that within your cartesian plane, each basic trig operation (sin, cos, tan) will be positive in 2 areas

#

Sorry for the bad drawing btw

#

Now for 58, you know that Sin is positive, and Cos is negative. There is only one quadrant where this is possible and it can be solved through a process of elimination

#

Only quadrant 2 overlaps these two properties thus the answer is II

#

hope this helps

somber coyoteBOT
junior light
#

Please ping me when you answer.

slate blade
#

I think it's a really badly worded "arrowhead rule"

#

you know this one

#

@junior light

junior light
#

Aah okay, that makes sense.

slate blade
#

angle at the centre = 2*angle at the circumfrence

junior light
#

The wording is kinda absurd indeed. Thank you!

slate blade
#

but yeah that wording tho

#

np happy to help

vast python
#

Sorry I’m late at answering but this helped me a lot @slate blade TYSM

outer trench
#

if you know that two sides of a triangle have lengths 4 cm and 6 cm, what specifically do you know about the length of the third side if the triangle is a right triangle?

#

i dont understand how to do this

#

ik u use pythag theorem

#

but like how

umbral snow
#

Do you know that the two given sides are not the hypotenuse?

outer trench
#

no

#

do i plug those two for all 3

#

then say x, y, or z?

umbral snow
#

I'll call that unknown side length x. It's either that x is the hypotenuse, or 6 is

#

So there's only two possible triangles here

outer trench
#

ok

umbral snow
#

It can't be the case that 4 is the hypotenuse, as 6 is a longer side

#

And the hypotenuse is always the longest of course

outer trench
#

tru

summer linden
#

where is the basic algebra channel?

umbral snow
outer trench
#

wait so i did 6² + 4² = c²

#

is that one of them

#

simplify to 2root13?

umbral snow
#

Yep! That's if the unknown side is the hypotenuse

outer trench
#

ok then theres one more right?

umbral snow
#

Yeah, the case where 4² + c² = 6²
is possible

outer trench
#

ohh

#

ok

#

then i'd say (that answer) or 2root13?

#

for my answer

umbral snow
#

Yeah. There's no other triangles you can make

#

So you know the other side has to be one of those two

outer trench
#

third side = 2root5 or 2root13?

umbral snow
#

Ye

outer trench
#

ok thx for help

umbral snow
#

Np. Feel free to ask if you have anything else!

orchid perch
#

what does m∠C mean? triangle c or angle c?

glacial dawn
#

measure of angle C

orchid perch
#

ty

cold siren
scenic bolt
long falcon
#

having some trouble

round isle
#

what have you tried so far

gritty sail
#

chain rule probably

kindred rapids
#

I got through...

long falcon
#

there’s not much of it I know how to do

inland canyon
#

is the arc 2x the central angle?

#

cuz i feel like i remember seeing that

#

and if not

#

how does one

#

get said arc

upper karma
#

@inland canyon $s = r\theta$, where s is arc length bounded by the central angle, r is the radius, and $\theta$ is the central angle in radians

somber coyoteBOT
inland canyon
#

@upper karma

#

my nigga i have 0 idea what the fuck u just said bruh

upper karma
#

yea what's up?

inland canyon
#

that made no sense

#

at all

#

the fucking a thta

#

theta

#

and the fuck is a money sign = 4

#

r*

upper karma
#

it's for the math

#

look at what the @somber coyote said

inland canyon
#

well ik that

upper karma
#

ah ok

#

so

#

s is arc length

#

and it is equal to r (the radius)

#

and the central angle theta

#

theta is a variable

inland canyon
#

wait

upper karma
#

which typically represents some sort of angle

inland canyon
#

so the arc is = to the radius

upper karma
#

arc is = to the radius times the central angle in radians

inland canyon
#

dawg imma be real i dont even know what a radian is

upper karma
#

lol i appreciate the honesty

inland canyon
#

is that like

upper karma
#

do you have this question for homework or an assignment?

#

a radian is a unit of angle measure

inland canyon
#

its just a warm up im doing

upper karma
#

ah ok

inland canyon
#

because like i was behind on work

#

so im catching up

upper karma
#

all good 👍

#

props

#

so

inland canyon
#

ok so like

upper karma
#

you know degrees right

inland canyon
#

hold on

upper karma
#

that cool?

#

oh

#

ok

inland canyon
#

ye ik how degrees works

#

like angles

upper karma
#

360˚ in a circle?

inland canyon
#

yes

upper karma
#

ok great

#

now

#

by definition

#

360˚ is equal to 2π radians

#

so

#

π radians = 180˚

#

it's weird

inland canyon
#

dawg this is still making no fucking sense

#

hold up

#

lemme like

#

try to make sense of this

#

so

upper karma
#

😅

#

go ahead 👍

inland canyon
#

say i have point a and b