geometry-and-trigonometry

[somber coyote]

AMD:

[cold oriole]

I have soluti9on

Its like goniometric relations in german

like sin(2x)=2*sin(x)*cos(x)

[upper karma]

Does anybody has a pic of ALL the trigonometric identities, like that one

It'd help a lot to a lot of people

[sly grove]

i dont have a pick but i have a link to a website with them

https://www.purplemath.com/modules/idents.htm

[acoustic jungle]

https://www.mathsisfun.com/algebra/trigonometric-identities.html

[upper karma]

Thank you two c:

[bitter jetty]

how do i do his, the corners are not on exact points

[idle bloom]

those look to be half-way between points @bitter jetty

oh wait no

those are on points

look at the x and y axis scaling

[bitter jetty]

how do determine their exact coordinates

g and f are good but e and d arent on exact points

so how do i determine them

[devout harbor]

@bitter jetty kinda looks lke E's y coordinate is 17 and D's is -7

[daring solstice]

I have a quick question, in a 45-45-90 triangle is one of the congruent sides the heightv

[bitter jetty]

@devout harbor but if i dont use the exact numbers the website that the teeacher uses to give us assignments marks it off completely

so is there any way to figure it out

?

[devout harbor]

i dont think so @bitter jetty

[bitter jetty]

ok

thx

[lofty dust]

how would i do this?

[chrome quarry]

anyone know what this solid is called?

[dark sparrow]

looks like some sort of... irregular square antiprism i guess?

[acoustic jungle]

@lofty dust there is a lot of unecessary information in there I think

I found a with pythagorean in 2 steps

[loud fulcrum]

ik this is probably really simple for some of u guys but how would i solve this

[silent plank]

consider finding BD first

[spice flax]

rectangle $12 * 20$ mm

[somber coyote]

Da Storyteller:

[silent plank]

@loud fulcrum don't dm me like that

find BD using trig

[loud fulcrum]

i just learned trig like yesterday

wait i use the cos or sin right

[silent plank]

sure. use the appropriate trig function

[loud fulcrum]

ok thanks alot @silent plank \

[digital gulch]

If u were to make this net of an octahedron into an octahedron, which faces would be next to face number 3?

I can’t imagine this in my head

[paper mauve]

Well pick a face that you want to be on the bottom

Personally I’d pick 5

[digital gulch]

Ok

[paper mauve]

Imagine the 7 and everything else above that 5 folding upwards

[digital gulch]

Ok

[paper mauve]

Then you’re off. Just start slowly folding things piece by piece in your head

Orr you can construct this yourself with paper and scissors and pencil

[digital gulch]

Yeah constructing sounds like a good idea I’ll do that

[lofty dust]

how would i do this?

https://cdn.discordapp.com/attachments/326138757474680852/694949331974881330/preview.png

[upper karma]

look at the radii

what is the radius of this circle?

in fact im not sure this is even a possible diagram

[lofty dust]

i dont see a radius

[lofty dust]

ohhh

[silent plank]

you could apply chord bisection theorems directly

[lofty dust]

so would you do Pythagorean theorem with 5^2+3^2= x^2?

[upper karma]

there seems to be a issue with this problem

because if you take the legs, 5 and 4, you get r = sqrt(41)

but if you take the legs 3 and 7, you get r = sqrt(58)

[silent plank]

lol rip

[lofty dust]

yeah i got the same thing ig ill just ask my teacher about it

[pliant matrix]

help

[upper karma]

i believe its the base angles theorem

[pliant matrix]

why

[upper karma]

base angles theorem says that if you have an isosceles trapezoid the opposite angles are equal

[pliant matrix]

ahh ok

can you assist with this one too?

i just need help with this one

[unborn jacinth]

alright so I had question where it said z1 = -6+i and z2 = 6+i and I had to find what z1/z2 is in standard form

so I converted z1 to trig form and z2 to trig form

and used a theorem to get cis161.08

but if its no calc, I can't rlly convert that back to standard form

so how would I do it

[chrome quarry]

@dark sparrow thanks dude!

[acoustic jungle]

it is not a dude

I've been told today

[steep lion]

what are the practical aplications of radians

[rapid pollen]

I have a question, am I doing this right?

Or does this not even involve the law of sines? I think it’s a simpler approach to the problem lol I’m probably totally wrong.

[ancient jacinth]

you should need law of sines for this

[rapid pollen]

OHHHHH I JUST REaD THE directions

[ancient jacinth]

its just using the trig

oh

lFMAO

[rapid pollen]

LOL

XD

[ancient jacinth]

tis ok happens to me a lot

[rapid pollen]

Good to know I’m not the only one

[ancient jacinth]

ahah it happens to everyone

[upper karma]

24th

So I have solved it

But what's the absolute best method here

[high zephyr]

ok wait a sec

[upper karma]

Sure

[high zephyr]

i dont think i can help u on this one soz

i got stuck :(

[upper karma]

No worries,I appreciate you trying

[umbral snow]

Ooh these are a mess

[high zephyr]

:(

[rich wolf]

this is why wolfram alpha was invented

[vale nimbus]

wtf those are painful

[upper karma]

Lol this problem United people

I know but I have to solve it

And I feel like elegance is right around the corner

<@&286206848099549185>

[high zephyr]

question?

[visual mist]

for part b, can i say that because DC=CB=AB=DA and also <AEB=<DEA=<DEC=<CEB because they are corresponding angles in congruent triangles?

[acoustic jungle]

yes

[vivid parrot]

could someone explain to me how to do cos(sinx)-1=0

[high zephyr]

this channel is taken @vivid parrot

[vivid parrot]

wat do you mean

[high zephyr]

as in, somebody already asked for help first and they're still waiting for a solution

oh no wait

edited just then

ok yea

[visual mist]

@acoustic jungle thanks

[fresh pawn]

Anybody can help with circle theorem?

[high zephyr]

what are u trying to find?

angle a or x?

[fresh pawn]

if youre finding a then x is the same

so both i guess

any one first is going to get the same result

[high zephyr]

oh soz forgot to come back

by intuition, find the middle angle at O

[odd shuttle]

If someone could help me on #1 that would be great I’ve been stuck for around 10 min

[high zephyr]

what's the angle of straight line minus the 108 degree angle

@odd shuttle channel's taken

[odd shuttle]

Ok

[high zephyr]

@fresh pawn ?

https://cdn.discordapp.com/attachments/326138757474680852/695118527056707605/unknown.png

[fresh pawn]

180-108?

72

and i think if 72/2 = 36 apparently that's the answer?

[high zephyr]

do u know why tho?

thats important

[fresh pawn]

idk that's what i was missing out on

[high zephyr]

ok

180-108=72 degrees yep

and that's the middle angle at point O

you see the angle labelled x?

[fresh pawn]

yes

[high zephyr]

well that angle, you can stretch it so that it goes on top of O

so just imagine that the angle x is moved to top of point O along the circumference

[fresh pawn]

huh?

the angle overlaps O/

?

[high zephyr]

so just imagine that the angle x is moved to top of point O along the circumference in an anticlockwise fashion

ok only focus on angle x

for now

edited

[fresh pawn]

so x always is touching the circumference?

the point i mean

[high zephyr]

yep you got it

move it anticlockwise to the very top

so that it's above point O

[fresh pawn]

ok

[high zephyr]

do u mind drawing it?

the new shifted angle on a new diagram

the intersection of the two legs of the angle x and the circumference stays as it is

so it kinda becomes a pivot point when shifting the angle x along the circumference

😅 this is prob the best way i can word it...

[dark sparrow]

@acoustic jungle don't fucking call me "it"

[high zephyr]

@fresh pawn ?

[dark sparrow]

azrol left the server

[high zephyr]

:|

[onyx basin]

@high zephyr https://cdn.discordapp.com/attachments/326138757474680852/695118527056707605/unknown.png

in reference to azrol's question

if we're finding angle a, and O is a given center

[high zephyr]

azrol said it doesnt matter

but ye continue

[onyx basin]

isn't the triangle with angle a an isoceles

so we can just find angle a that way

[high zephyr]

i wouldnt assume that

[onyx basin]

since the triangle is formed at a center point and two radii

[high zephyr]

oh you meant the small one mb

[onyx basin]

ah so we're not assuming O is center point

wait what

[high zephyr]

azrol said it doesnt matter which angle so i just lead azrol into finding x

wifi lag

[onyx basin]

ah ok

[high zephyr]

so you're good?

[onyx basin]

yea I thought you were solving for a my bad

I was wondering why you wanted him to draw out stuff haha

nvm then

[high zephyr]

ok

[rich wolf]

Ok boomer

[lethal tangle]

Could anyone help me out with some year 12 trig questions

[rich wolf]

K

@lethal tangle

[dark sparrow]

until you post them, nobody can

when you do, there may be someone who can and will help

[dark sparrow]

alright that sure is a lot of questions

which one do you wanna start with

[lethal tangle]

Question 1

[dark sparrow]

ok great

question 1

what's giving you trouble here

@lethal tangle

[cold oriole]

you can help me? Its ALPHA 2.3° and BETA 65°. I Have calculated h1 but i also need h2 and b and idk

[lethal tangle]

I just not sure how to start off I'm confused

I think it's the way the teacher has shown us

How would you approach the question?

[mystic dock]

I dont want to spoil the answer to question 1 but notice that the angle is 45 degrees, thats fishy right? its another 45 to make 90.

[dark sparrow]

nah there's nothing special about the 45° angle here

@lethal tangle okay so there are many, many formulas for the area of a triangle

maybe one of them could apply here directly

another one a bit less so

[lethal tangle]

Wait would I be using area=1/2 bc sin A

Or finding the opposite, adjacent and hypotenuse

Then using cos, sin or tan

@dark sparrow

[dark sparrow]

well you want the area don't you

[lethal tangle]

True

How would I know which is A,B and C

[dark sparrow]

you... don't?

there's no names here

names don't matter anyway, what matters is that the angle should be between the two sides

which in your case, it is

[upper karma]

@cold oriole do you still want help?

[cold oriole]

yeah in y

channel

[steep temple]

how many sides does a circle have

[dark sparrow]

a circle does not contain any straight line segments which one could call sides

a circle is not a polygon

[languid swan]

Can anyone help me with the questions I circled?

[supple wedge]

yes

wat hv u tried? @languid swan

[languid swan]

I mean i tried figuring out all of them but I got no clue

[supple wedge]

ok q5

what have u tried

@languid swan

notice that it is a right angles isosceles triangle

so both sides r 17

km

then use pythagoras

[upper karma]

What is a sidewalk

A diagonal?

[languid swan]

You know my boi

I think if you gave the answers I’d appreciate it because if you did then I could probably figure out what I did wrong and what I was supposed to do

But yes thank you

What about the bottom?

[upper karma]

If sidewalk is the square diagonal just use Pythagore

[supple wedge]

we're not supposed to give answers

only help u get to them

[mild barn]

@supple wedge my teacher was very vague with my instructions can you help me with this problem

[supple wedge]

for teardrop

do

$\frac{3}{4}\pi r²+r²$

[somber coyote]

Hmm:

[mild barn]

Do you think I’m supposed to assign my own values?

Also I picked 2 and 3

[supple wedge]

i actually dont know lol

but i can give u the formula for area

[mild barn]

That’s how I’ve been doing the rest of the packet so lol I guess

[supple wedge]

id use 1 and 3

bc idk 2

LOL

[mild barn]

So for choice 2 how many numbers would I need to assign

Norm for #1 then

Norm for #1 then

[supple wedge]

scroll

[mild barn]

I don’t follow

Did you use 3 and 4 as your numbers

[bitter jetty]

can someone help me with this

[supple wedge]

@mild barn its 3/4 area since its 3/4 circle and a square

[bitter jetty]

o wait do i use law of cosines

?

[supple wedge]

yes

1 angle and 2 sides

[bitter jetty]

kk thx

[versed river]

cosine law works

[mild barn]

Wait so I’m confused 3/4 was your answer then???

[bitter jetty]

what do they mean second set of answers

[acoustic jungle]

@dark sparrow what??

[dark sparrow]

what

[acoustic jungle]

what was the message

"don't fucking call me it"??

@bitter jetty there are two possible values when you are given that information

[acoustic jungle]

ok.

[bitter jetty]

@acoustic jungle how do u find the second set of values

[acoustic jungle]

Here

[bitter jetty]

tf

ive never heard of that

can u explain how i wud do it

[acoustic jungle]

The angle B2 is basically H here

notice what the two angles add up to

the two possible angles

And it's not because the sin law is wrong, it's right but sin (theta)=sin(180-theta) however your calculator doesn't show the other result so you'll have to do that manually

Cold male?

[upper karma]

someone help please

[acoustic jungle]

The center of the circle is the midpoint of the diameter

The radius is half the diameter

axe

@upper karma

[upper karma]

so its the radius?

[acoustic jungle]

The endpoints of the circle are those points. 1. The center of the circle is the midpoint of those points, 2. and the radius is exactly half the diameter

1. tells us the position of the center (For LHS equation) and 2. tells us the distance (RHS)

[bitter jetty]

@acoustic jungle sry i had to help my brother

ya it kinda makes sense now

so i have to do

sin(180-38)

ok

[narrow plume]

https://i.imgur.com/Gx0EAsG.png

I tried solving this, got 4...

[acoustic jungle]

@bitter jetty no that is incorrect

angle "I" can't change because it's given in the question: https://www.youtube.com/watch?v=S1oDtGHC2iA

[dark sparrow]

is that meant to be $(f^{-1} \circ g \circ f \circ g^{-1})(0)$ @narrow plume

[somber coyote]

Ann:

[dark sparrow]

if so, your answer is correct

[narrow plume]

Yes, and thank you

https://i.imgur.com/35lWtV8.png Can anyone help me with this word problem?

budget = 9,000

cost of construction = $100/ft. for the bare walls + $100/ft. for bookcase.

Bookcase = half length of its wall

Not sure what I have to do with these values...

[ancient jacinth]

could i get some help here?

[quiet mason]

which angle is 60,70 and the y

[narrow plume]

https://i.imgur.com/qtwvzl5.png

Any1 know how to do this? I assume i plug in the value to x somehow

and find what is exlcluded

[dark sparrow]

can you talk to your teacher and tell them this document is corrupted

i have no idea if that G with an accent is a plus or a minus or something else entirely

[narrow plume]

Me? Yeah the T and G is super weird

T is the =

G is +

LOL

[dark sparrow]

okay so it's a plus

$f(x) = \frac{2x+5}{x+1}$

[narrow plume]

mhm

[somber coyote]

Ann:

[dark sparrow]

the domain of f consists of all real numbers except -1

the domain of (f o f) will consist of all real numbers except for -1 and all the solutions of the equation f(x)=-1

[upper karma]

x+1 =/= 0
And f(x) + 1 =/= 0

[narrow plume]

Ok, so basically if I graph this function (?) I will see that it includes all real numbers except -1?

[dark sparrow]

you're overthinking it

[narrow plume]

😆

[dark sparrow]

if a number x is such that (2x+5)/(x+1) = -1, you can't plug it into f o f

[narrow plume]

Ah kk

Thank you!

[ancient jacinth]

lol

[narrow plume]

Lmao all g

[haughty summit]

wat is dis

I know that it's geometry lol im just saying that question is so badly formed haha

one sec

what polygon am I looking at

I just see a circle

which is not a polygon

lmfao

My best guess? Add your sides up and you would get 45

round your answer

[quiet mason]

i guess you have to find the perimeter of the circumscribing polygon

no they havent said that 8.8

is the length of the entire side

[haughty summit]

ah I see

[quiet mason]

you gotta use tangent properties

but im not sure as its poorly stated

[haughty summit]

maybe it's propotional?

Bonchu is right about not stating the entire sides.

figure out the other half of the two black lines

oh

11.5x2+15.7x2=23+31.4=54.4

hmmm

yeah just take that and round it

thats my guess

[upper karma]

How do i go about this: (cos(70°)/cos(80°)) - cos(10°). By finding a common denominator?

Nvm, i figured it out

[acoustic jungle]

@upper karma use the property that when 2 tangent lines intersect their distance to their tangent point is the same

and yes it's 54.4

None of the answers are right

[summer spire]

because it's less arbitrary than degrees or gradians?

it's a simple enough conversion

just depends on convention

[loud pike]

"An isosceles trapezoid with bases 12 and 16 is inscribed in a circle of radius 10. The center of the circle lies in the interior of the trapezoid. Find the area of the trapezoid"

<@&286206848099549185>

I've been thinking maybe the radius has something to do with the height, but that doesn't seem right

[modest vine]

i have been trying to do this for a while but i am stuck

[iron rain]

What are you supposed to find?

[modest vine]

Surface area

[upper karma]

Yall I need help with 10 questions and I am so lost ne

Rn

[modest vine]

need to find the surface area have been trying for a while but i cant do it

[upper karma]

A lamp must be hung on an inclined wall. It is therefore suspended in two strings, as shown in Figure 7
below. The angle between the two inclined cords is denoted 𝑣 = 53 °. The angle between the vertical and one of the inclined cords is denoted 𝑤 = 142 °. The inclined wall has the angle 𝑢 = 45 ° to the horizontal.
When the lamp hangs still, the sum of all forces must be zero. The vertical vector is gravity having the length Ft = 100 𝑁, forces in the cords are designated F1 and F2, see Figure 8

a) Determine the magnitude of the forces F1 and F2

[upper karma]

im stuck

[modest vine]

im stuck need to find surface area

[void moat]

look at ur surface area equation

what is it for tianglar pyramid

Base area + 1/2(perimeter x slant height)

so ur slant height is 13.8

u need to find the perimter of ur base

and also the area

@modest vine

this is an ez problem

u just need to surface area formula

[steel thorn]

ik it’s an easy question but I’m stuck lol

[acoustic jungle]

use tan fomula

tan(theta+alpha)=(tan theta + tan alpha)/(1- tanthetatanalpha)

[glad comet]

<@&286206848099549185>

[upper karma]

is the arc circular

because if it is then Q would be the center of that circular arc

which means point A is on circle P and circle Q

now consider the radius of circle P and circle Q

what can I say about the two in relation

[glad comet]

im assuming so

that the arc is circular

[upper karma]

ok then what is the radius of circle Q and radius of circle P

what line segment specifically

[glad comet]

PQ

[upper karma]

correct

this would imply that the circles have the same length radii

now consider triangle PAQ

what are the angles of this triangle

[glad comet]

<PAQ & <PBQ

[upper karma]

sorry, what are the measures of this triangle

[glad comet]

45-45-90

[upper karma]

how do you know that

[glad comet]

because both circles are the same and share the same tangent making those arcs equal which makes the lengths equal

the radius is its own thing which could be the base

[upper karma]

but they aren't tangent

there are no tangents here

at least i dont see any

[glad comet]

im just making them tangent considering they are exactly 1 point

[upper karma]

Need help

<@&286206848099549185>

[chrome fiber]

dont ping if 15 mins havent passed

read #❓how-to-get-help

[vale nimbus]

@upper karma sin²(A) + cos²(A) = 1

[high zephyr]

conversion of degrees to radians

[vale nimbus]

yup

[high zephyr]

simply times degree by pi/180

for radians to degrees, times by 180/pi

[upper karma]

Can u do that for me I lost my calculator

Thank you

Nvm

Got it

[high zephyr]

what trig ratios would u use here

[upper karma]

Tan?

[high zephyr]

yeah correct

[upper karma]

Oh so tan(31)=h/25

[cunning night]

Is the tree 11 feet tall??

[upper karma]

Wrong

Wtf

@high zephyr 11 should be correct right

[high zephyr]

no...

tan(31)=h/25

what do u need to do to get the h?

[upper karma]

Ohhhhhhh

Thank you

[high zephyr]

np

[glad comet]

i still do not understand this

[high zephyr]

is the question asking for the cut-off length of arc?

[glad comet]

yeah

[high zephyr]

there are no numbers...im assuming they would be the same length

[glad comet]

yeah

they are both the same circles

[visual mist]

any help? im stuck

[high zephyr]

you can divide the rhombus into four triangles

[visual mist]

yea i did that

[high zephyr]

since the diagonals are bisectors of each other, AM=MC and DM=BM where M is pt where diagonals intersect

since it's a rhombus, all sides are equal (given)

therefore triangles are congruent (SSS)

[acoustic jungle]

@glad comet find the angle of APB then go from there. Try drawing the rest of the circle Q for it to make more sense

[high zephyr]

@visual mist

what do u notice when the triangles are congruent?

[visual mist]

is M where they intersect?

[high zephyr]

i said it

[visual mist]

0h oops didnt see

[high zephyr]

what do u notice when the triangles are congruent?

[visual mist]

idk

[high zephyr]

when they're congruent, their corresponding angles at the centre are equal

[visual mist]

yep

and it adds up to 360 therefore each is 90

[high zephyr]

angles of 4 triangles at centre are equal and adds up to 360 degrees

yep

[vale nimbus]

@high zephyr cant u use the law of cosines for this too?

[high zephyr]

whut

why would u when you can use simple trig ratio

[vale nimbus]

idk

we never used trig ratios to solve triangles

[high zephyr]

???

[vale nimbus]

[high zephyr]

trig ratios like sohcahtoa

i meant those

[vale nimbus]

yeah we didnt

where is my education

[high zephyr]

nani? it saves soo much time

[vale nimbus]

law of sin/cos

or pythagoras

[high zephyr]

ye u can but y

so much longer

[vale nimbus]

wish i knew

[high zephyr]

try googling trig ratios for sin cos and tan

[acoustic jungle]

You can figure it out by finding the height of the tree with tangent (yes) , finding the hypotenuous with pythagorean, then finding the height of the tree with cosine law

[vale nimbus]

i know what sin =

like the ratio

but we just never used it

we just learned it and then moved on

[acoustic jungle]

google

[proper zinc]

I know the radius is 3.7, but how would you find that out? I can't see any indication of radius/diameter here?

[high zephyr]

it's a hemisphere

the 3.7cm is the radius

the centre of a sphere to the bottom of it is the radius

and in this case it's half a sphere

[proper zinc]

ohhhhh

I swear I didn't see that because its 2am, I'm not stupid I swear xD

[high zephyr]

haha so you're good now?

[proper zinc]

Yes, thanks :)

[high zephyr]

np

[digital gulch]

How would you find the radius of the smaller circle?

[acoustic jungle]

GodGamer, what is the answer

Just trying to varify

@digital gulch

@digital gulch https://www.youtube.com/watch?v=f-ETwjwwOAM

[high zephyr]

what a legend

[acoustic jungle]

I just searched up the question lol

[digital gulch]

U did!

Woah

Lmao

[loud pike]

How do I find the area?

[acoustic jungle]

is this a trapzoid

[loud pike]

yep

i sketched it in paint

[acoustic jungle]

use trig, find the top value, use trig, find the height, then use the trapzoid formula

[loud pike]

yeah no i know to find the top value and height

but how

what do you mean "use trig"

[acoustic jungle]

draw a line down from the top to the bottom

[loud pike]

specifically

[acoustic jungle]

brb i'll draw this out

@loud pike

[plain pollen]

Heyo, I asked this question a while back but due to miscommunication with the kind person trying to help me I didn't get a chance to go through it. Would anyone be willing to help?

part b in particular

[devout harbor]

For the x coordinate, we have x = 1+t and it needs to be 0.

What's the value of t that makes this happen?

[plain pollen]

-1

[devout harbor]

Yep.

You can plug t=-1 in the equation of the line and find your point.

[plain pollen]

omg

i am extremely slow sometimes

thank you

[devout harbor]

np

[viscid ginkgo]

for y = 3 + 2tan(x/2 + pi/8) how can I find the period? using 2pi/B formula gets me 4pi / x

[devout harbor]

You should know that if y=Atan(Bx+C)+D, then the period of the function is pi(period of tanx) / b

so it's just pi / (1/2) or 2pi.

@viscid ginkgo

[viscid ginkgo]

someone else helped me, but thank you.

[devout harbor]

All g

[bitter jetty]

can someone help me with this

[silent plank]

construct line DP

[bitter jetty]

how?

[silent plank]

by drawing a line connecting the points D and P...

[bitter jetty]

ok

and then what

[silent plank]

what does the given information tell you about (m)angle DPA

[bitter jetty]

ceentral angle

[silent plank]

with the size of:

[bitter jetty]

its 40?

[silent plank]

degrees yes

[bitter jetty]

so how does that help me find x and y

[silent plank]

were there any angles you were able to find yourself?

[bitter jetty]

arc sc is 110

[silent plank]

eg. (m)angle APC?

[bitter jetty]

bc*

ya its 70

[silent plank]

now consider the properties of triangle DPC

[bitter jetty]

o crap

thx

i can do it

i got x = 40 and y =110

is that right

[silent plank]

show your calculations for x

[bitter jetty]

i got y

then i did 180-y

added the 2 angles and subtracted that form 180 to get x

[silent plank]

how did you get y?

[bitter jetty]

crap o did smth wrong

[silent plank]

you should be finding x first

[bitter jetty]

?

[silent plank]

what's the ? for?

[bitter jetty]

how do i find x first

[silent plank]

now consider the properties of triangle DPC

[bitter jetty]

p is 70

[silent plank]

what type of triangle is it?

[bitter jetty]

isosceles

so the 180-110/2

[silent plank]

parentheses

(180-110)/2

[bitter jetty]

ye

which is 35

[silent plank]

yes

[bitter jetty]

ok

thx

how do i split it

[acoustic jungle]

@bitter jetty it's a straight line

[bitter jetty]

im sry im the biggest ding dong ever

[languid pagoda]

yoooo

[flint heath]

Hello

I have a question gamers

[flint heath]

no

and i need help too

whatcha need

okay

yeah same lol

whats your review sheet look like brother

[flint heath]

can u do question one

thats just looking on your unit circle bruv

BOI

GOOGLE
\

okay google is kinda bad

so there are 4 quadrents in the unit circle

[upper karma]

mhm

[flint heath]

you go from the right side and rotate counter clockwise for a positive angle

[upper karma]

so 635 is in the 4th quadrant

wait

its in the 1st i think

[flint heath]

BIG DICK now you thinkin boi

if you take 635 and divide it by 90

[upper karma]

No

635 is in the fourth quadrant.

no

[flint heath]

im just dumb im sorry

[upper karma]

wait

if u go around the circle tho it adds up to 630

635 minus 360 is?

so the extra 5 would be in the 1st quadrant

275 is in the fourth quadrant.

[flint heath]

no he is rigght mayoo

[upper karma]

k

[flint heath]

if u take 635/90 you get 7.05 repeating

[upper karma]

[flint heath]

7 quadrents plus .05 repeating of the 8th

so quad 4

but thats how i do it

ree

[upper karma]

what about number 2

[flint heath]

so coterminal angles share the same end line

does that make sense

[upper karma]

ok

[flint heath]

so like a 90 degree angle looks the same as a 450 degree angle

[vale nimbus]

360 fits once into 630 then ur left with 270°

[flint heath]

which makes them coterminal

[upper karma]

so

i need to find 2 angles that are the same as 150

[vale nimbus]

you might aswell substract 2.360° from 630 then ur left with an angle of -90°

wait whats the question now

lagging behind i feel

[flint heath]

yup @upper karma

[upper karma]

coterminal angle for 150

2*

[vale nimbus]

yup

[upper karma]

how do i find the angles

[flint heath]

so if you add 360 degrees to 150 thats one coterminal angle

[vale nimbus]

ah cmon man dont just tell him

explain to him why

[flint heath]

thats the first

find the second

[vale nimbus]

[upper karma]

• 360

;o

[vale nimbus]

@americans ^^

[upper karma]

so do i -360

[vale nimbus]

yeah

you can either add 360° or substract 360°

because thats like adding or substracting a whole circle

which puts you back into the same place

if that makes sense

[upper karma]

so 510 and 210 are coterminals

-210

[vale nimbus]

yes

[flint heath]

151,350 degrees is a conterminal too

[upper karma]

why

[flint heath]

360*420 = 151,200

151,200 + 150

[upper karma]

whered u get 420

[flint heath]

Out of my ass

[vale nimbus]

funny weed number probably

[flint heath]

ding ding ding

[upper karma]

so 360 X anynumber + the value of the angle equals another coterminal

[vale nimbus]

yup

[upper karma]

bettttttttt

[flint heath]

okay now mayoo, find a negative coterminal angle

[vale nimbus]

put into symbols thats
α+k360° (k belongs to the whole numbers)

[upper karma]

360x 100

3600 - 150

[flint heath]

no

awit

[upper karma]

do i divide

[flint heath]

wait

[upper karma]

;0

[vale nimbus]

charley u shoulda put it in the right order

you have to do 150-3600

[upper karma]

o

[flint heath]

because you start at 150 degrees

[upper karma]

-3450

[flint heath]

my bad

[upper karma]

so -3450 would b a coterminal

[vale nimbus]

yup

[upper karma]

k

how do i express a radian measure

[flint heath]

https://www.desmos.com/calculator/qmzx2skkzy

desmos is also really nice

[upper karma]

do i need to use that

to do radian measure

[flint heath]

no

that was for the last one

[upper karma]

k

[flint heath]

so im pretty sure what that is about is converting your 300 degrees into radians

[vale nimbus]

radians = $frac{π}{180°}.degrees$

[somber coyote]

piece:

radians = $frac\{π}{180°}.degrees$
```Compile error! Output:

! Extra }, or forgotten $.
l.54 radians = $frac{π}
{180°}.degrees$
I've deleted a group-closing symbol because it seems to be
spurious, as in $x}$'. But perhaps the } is legitimate and you forgot something else, as in \hbox{$x}'. In such cases
the way to recover is to insert both the forgotten and the
deleted material, e.g., by typing `I$}'.

(/usr/local/texlive/2018/texmf-dist/tex/latex/ucs/data/uni-0.def
File: uni-0.def 2013/05/13 UCS: Unicode data U+0000..U+00FF

[vale nimbus]

why do you hate me bot

[upper karma]

yeah how do i convert 300 to a radian measure

[vale nimbus]

$frac{π}{180°}.degrees$

[somber coyote]

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

[flint heath]

is it

[vale nimbus]

work

[flint heath]

pi divided by 180 degrees

times your degree value

[vale nimbus]

yes

it is

[upper karma]

for radian ?

[vale nimbus]

if only i knew how the bot works

[upper karma]

can u just do 3.14 divided by 180

[flint heath]

are you on your phone?

google has pi on their calc

and the desmos app is a calculator as well if you only have a smartphone

[upper karma]

i found it

0.017

0.017453

[flint heath]

okay take that value and multiply it by the angle that you have

and that is a 300 degree angle in radians

[upper karma]

5.23598

[flint heath]

degrees and radians are like inches and centimeters

both measure the same thing but have a different value attached to them

[upper karma]

alr

[flint heath]

try figuring it out, look at your notes/book i'm working on a question right now myself

[vale nimbus]

@upper karma if an angle is 1 radian, the arc its faced at is equally as long as the radian of the circle

that should be enough of a hint to solve 4

[dull remnant]

am i supposed to put this in one of the questions channels?

[dusky surge]

https://images.app.goo.gl/xH1W9sWV3LAVP57D7

Does this help?

Usually there are 2 answer for the arc sin cos tan question within the range of 0≤θ<360°

[dull remnant]

yea ik the alt students take calculus thing

but

how would that help?

i dont know which method is correct

[dusky surge]

The table is correct. But since there are 2 answers (usually) you might have to use the one in suitable range or both

Maybe a real question or two will clarify your uncertainty

[dull remnant]

wdym by 2 answers

for the answer key there is 1 answer...

[dusky surge]

Hm..... The range of values of arc sin cos tan has specific range. It really depends on which answer you need. Like which range of angle you need.

For example. What is arcsin(1/2)

[dull remnant]

ive nvr heard of arc

is arcsin the same as inverse sin

[upper karma]

[dusky surge]

Yea

[dull remnant]

ok

arcsin(1/2) is

30 degrees

because the table says 1/2 ratio when entered in sin is 30 degrees

its postiive so it has to be in quad 1

so 30 works?

[dusky surge]

But remember that picture I send it out here? For quad 2, sin can also be positive

[dull remnant]

but it cant be there

it can only be in quad 1 and 4 right?

[dusky surge]

So 150° can also be a correct answer

Oh

You mean you have fixed it should be in quad 1 and 4

But then cos will be weird

[dull remnant]

well i have a picture

it says inv sine has to be in 1 and 4 same with tan but cos has to be in 1 and 2

[dusky surge]

Right

If it's a function, yes is it

[dull remnant]

so when it is a function the rules of 14 and 12 are still there, but when its not

it can be anywhere

[dusky surge]

That's kinda of principal ones

Right!

It can be 30, 150, 390, 510,...

[dull remnant]

ok i get that. but back to the question of how do i find the degree from the ratio of inverse sin, tan or cos,

wait why 30 to 150 to 390

[dusky surge]

Lemme find the table

[dull remnant]

do u want me to send it i have it

[dusky surge]

https://www.google.com/imgres?imgurl=https%3A%2F%2Fwww.siyavula.com%2Fread%2Fmaths%2Fgrade-12%2Ftrigonometry%2Fimages%2F2b56c3f89f685c430c6cc22358093016.png&imgrefurl=https%3A%2F%2Fwww.siyavula.com%2Fread%2Fmaths%2Fgrade-12%2Ftrigonometry%2F04-trigonometry-01&tbnid=PKrBu3yc1d1VUM&vet=1&docid=pSeGl_g1APDadM&w=666&h=666&q=sin cos change 180-x&source=sh%2Fx%2Fim

https://images.app.goo.gl/3cWP9inKAF88n19i7

[dull remnant]

that is not the table i have rip

[dusky surge]

Take a look at the second one, it's more clear

[dull remnant]

but i understand that concept

[dusky surge]

Cool

[dull remnant]

how did u get 30, 150, 390, 510

is it just multiples of 30

[dusky surge]

30,(180-30),(360+30),(360+180-30)

Since sinθ=sin(180°-θ)

[dull remnant]

ok i dont understand the second image now

what does x stand for

[dusky surge]

The angle

First one is graphically clear, 2nd one is mathematically clear

Just pick one that suits you

[dull remnant]

@dusky surge can we go from the basics

im really lost

[dusky surge]

Oh... Okay...

Like we have a basic table for sin cos and tan

0, 30, 45, 60 ,90 degrees

But after that we have many more

Like 120,135,150,180... And so on

[dull remnant]

ok

the teacher only gave us the 30 45 60 90 degree on

one*

[dusky surge]

So it's stupid to memorie them all because there are so rules

It's hard to explain why it works in a short period of time.

But you can try to search the relationship between a unit circle and trigonometric functions

[dull remnant]

i think my main problem is i dont know how to do questions like arcsin (1) or arctan(-1)

i dont know the method

of solving the problems

[dusky surge]

If your teacher says the range is within something, then just stick to the range and answer the questions.
E.g.
Arcsin , arctan will be Q1,4
Arccos will be Q1,2

[dull remnant]

ok so step one is to find what quadrant it is in

[dusky surge]

Yea, go on

[dull remnant]

step 2 is to

determine the quadrant

based on whether the ratio, (which is the thing in the parentheses?) is negative or positive

[dusky surge]

Yep

[dull remnant]

if the ratio is positive, than for sin and tan its in q 1 and not q4 and if its in cos its in q 1 and not q 2

so after that

step 3 is to just determine the ratio form the table?

im stuck there

or is it to find the reference angle

[dusky surge]

You're right

But the point you can't understand is why it doesn't work on cosine

Well, if you just want the answer, memorize cos(180-θ)=-cosθ

[dull remnant]

so u can do the table on tan and sin

but for cosine

u need to get the angle form the table and then subtract it from 180

[dusky surge]

Yes.

[dull remnant]

omg ok i understand now

ty

[dusky surge]

And the angle must be acute.

Which....

[dull remnant]

wait

what if the angle is 90

[dusky surge]

Will be the positive one

Lemme explain with a question

Cos60°=1/2

[dull remnant]

yes

[dusky surge]

arccos(1/2)=60°

[dull remnant]

yea

[dusky surge]

But what about arccos(-1/2)

We first determine arccos(1/2) and find the angle

[dull remnant]

-135?

[dusky surge]

Then use cos(180°-θ)=-cosθ

[dull remnant]

im p sure thats against the rules @ornate locust

[stuck torrent]

<@&268886789983436800>

[dusky surge]

For this example we first find 60°

[ornate locust]

SNITCHES

UR A SNITCH

[stuck torrent]

[ornate locust]

WOW

nobody likes u

[copper valve]

lol

[ornate locust]

deadass

[stuck torrent]

Ok

[ornate locust]

owner wait

[stuck torrent]

But you said whoever

[ornate locust]

dont ban me im sorry

i didnt know rules

[copper valve]

Ok

read em

#❓how-to-get-help

[dull remnant]

@dusky surge ok it is 60 degrees

[ornate locust]

ok thank you

[dull remnant]

so arccos(-1/2) is 120?

[stuck torrent]

4π/6

Or 2π/3

[dusky surge]

Yea

Let's do another question to make sure you understand

arccos(-1/√2)

[dull remnant]

ok

the ratio is actually

-sqrt2/2

which means it is in quad 2

also means its 45 degrees

so u want to fidn the ref angle

which means the answer is 45 degrees

[dusky surge]

The ref angle is 45degrees.

But the answer is 180-45

[dull remnant]

whattt

ok do u subtract the ref angle

from 180

or the original angle

[dusky surge]

Hmmm...

The terms we share may not be the same

How about just....

The angle that when the ratio is postive

And then 180 minus that angle for Q2

[dull remnant]

what is q2

[dusky surge]

Quad2

[dull remnant]

ok

[dusky surge]

Is it better?

[dull remnant]

can u give me another question please

[dusky surge]

Sure

arccos(-√3/2)

[dull remnant]

so its in quad 2

[dusky surge]

Yes

[dull remnant]

according to table it is 30 degrees so answer is 150?

[dusky surge]

Yep

[dull remnant]

ok

and this 180-angle

only works for cosine

[dusky surge]

Yes

[dull remnant]

can u give me a problem that is tangent or sine

[dusky surge]

Sure

arctan(-1/√3)

[dull remnant]

so the ratio is -sqrt 3/3

which means it is 30 degrees

its in quad 4

answer is 30?

-30*

[dusky surge]

Yep

You got it man!

[dull remnant]

ok i see, it seems pretty easy

its just 180-angle when it is cosine

[dusky surge]

For now yes

[dull remnant]

and tan and sine its just table

o no...

do u have time to answer 1 more question, its about the same concept

[dusky surge]

when the range expand to 0 to 360 degrees, it get messy

Sure

[dull remnant]

oo.

ok

so

2 contradicting methods again

for question 21 it is sin theta=-0.45; the angle has to be between 180 and 270

for question 24 it is tan theta=-5.3 angle must be from 90 to 180

i got 100.7 degrees for #24 by doing arctan(-5.3) and then doing 180+ the answer to that

[dusky surge]

When doing such questions, always first find the reference (acute) angle when the ratio is positive.

Then, find the Quad of the angle.

[dull remnant]

but it already gives u the quad

for #21 the method changes and now i added the answer of arcsin(-.45) to 180 instead of the bigger number

but #24 was adding the answer to the bigger number why is that?

[dusky surge]

Wait, I'm finding the picutr

The first picture I gave you

https://images.app.goo.gl/T5ibqs3NvcQgFEKdA

Then, find the Quad of the angle.
@dusky surge
After this, do what list on this table

θ will be the angle when the ratio is positive

[dull remnant]

i cant use the table though

because the angles are wild

[dusky surge]

Example time.

sin theta = 1/2
Theta between 90 and 180 degrees

What will theta be?

[dull remnant]

ok

um

150

180-30

[dusky surge]

Yep

[dull remnant]

yea

so u subtract form the top

largest number*

[dusky surge]

Now arctan(-1)
Theta between 90 and 180 degrees

[dull remnant]

oof

135

degrees

[dusky surge]

Yea

Let's try some Q3,4

[dull remnant]

its positive too rigth?

but