geometry-and-trigonometry

[acoustic pollen]

https://www.youtube.com/watch?v=Dsf6ADwJ66E

[spark vale]

tbh that line doesnt make sense

cuz when the angle is close to 90 degrees, the line goes to infinity

i am talking about tangent func

[acoustic pollen]

yes because tan90 is undefined

[spark vale]

lets think of tan89.999999999999...

it's not undefined

then think of the line drawing the tangent line

in my case, i cannot imagine it

[acoustic pollen]

actually if 89.999... goes on forever, then it is equal to 90

so it's still undefined

[spark vale]

that's fk weird imo

bizarre

uncomfortable

[acoustic pollen]

i mean you can prove the relations (like tan = sin/cos) with similarity and in a bunch of other ways

so that might help you

to understand it

but yeah the fact 0.999... = 1 is weird

[spark vale]

but this video is very helpful to see what is trigonometry functions!

also i like the music choice.

[spark vale]

yea lol

[acoustic pollen]

@spark vale right here the thing i said about proving the relations with similarity

[spark vale]

yea that one was cool.

impressive thing when learning math

Tan(135) = Tan(45)

[acoustic pollen]

tan 135 = -tan45

[spark vale]

because of that yellow triangle?

this makes more sense now

[acoustic pollen]

you know tan = sin/cos, but on the left side, cos is negative, so tan = sin/-cos, that's why it's negative

[acoustic pollen]

yeah basically

[kind quail]

Sorry I pinged the wrong person

[spark vale]

When the angle is 135 degrees, people often use a triangle with a 45-degree angle in the second quadrant. Is that because it is similar to the 45-degree triangle in that quadrant?

[kind quail]

Idk how that even happened

I'm sorry

My discord was acting up

[acoustic pollen]

yes

[spark vale]

now this makes sense!

I was stressed out about this.

but u resolved it! thanks

I am now understanding arctan

[acoustic pollen]

youre welcome

[spark vale]

which is the inverse function of tan

but this is the only one thing I know about arctan

[kind quail]

Yes I belive it is law of sine. I was just confused because I didn't know how to solve for the other two angles

[cunning lion]

arctan is the inverse function of tan with its domain restricted to (-pi/2, pi/2)

[acoustic pollen]

yeah im pretty sure you can use both of them and get the answer, but the answer doesn't seem cool at all 😔

[spark vale]

why is it restricted to that domain tho

[cunning lion]

it's not one-to-one otherwise

[spark vale]

sorry but not understanding it

[kind quail]

Wait what do you mean?

To get the angles

[acoustic pollen]

yes

[kind quail]

Ohhh

[acoustic pollen]

you know law of cosine?

[cunning lion]

tan(0) = tan(pi) therefore tan is not one-to-one, and we need a function to be one-to-one to have an inverse function

[acoustic pollen]

a^2 = b^2 + c^2 - 2bc(cos a)

[kind quail]

Yes I do! But for this assignment he wanted me to just use law of sines and I was confused

I was just confused because using law of sines for this seemed impossible

[spark vale]

this makes sense! u clearly understand the tan func

[acoustic pollen]

i dont think you can

you need at least one other angle

[kind quail]

Really?!?!?

Hm...

[acoustic pollen]

yeah you can't do anything with something like

[kind quail]

Ok I will use law of cosine then

[acoustic pollen]

4/sin a = 9/sin b

that won't get anywhere

[kind quail]

I swear my teacher wants me to fail

Ok thank you!

[spark vale]

🙂

[formal geyser]

Hello, how do i solve this problem

I'm totally clueless

[obsidian harness]

Let the base of the triangle be x, so x/6 = 6/h

Now what is xh/2?

[warm shuttle]

call something x and spam various identities and theorems

works for virtually any geometry problem

[formal geyser]

18

Thanks

[obsidian harness]

npnp!

yeah amazing how it works out

[frank ore]

can u all teach me trigonometry

im learning it soon

[silent plank]

Its not likely that people will be able to teach from scratch.
There are plenty of free online resources like khanacademy.
People can guide you on specific math questions you have.

[next gazelle]

my guess is this is a geometry q. any help on how to derive this?

[formal geyser]

Hello

[formal geyser]

How do i prove that all this angles are 45 degrees?

If i prove it, then i will be able to solve it

[twin surge]

Did you try using properties or a square? That the diagonal always splits the 90 degrees in half?

[formal geyser]

Yeah

But how does it help me here

[trail tendon]

are they 90 degrees?

[formal geyser]

I meant the red angles

[trail tendon]

I meant 45

Lol

they seem to be, the square seems to be symmetrical, but I don't think you can guarantee it

If you rotate I could see a way that they aren't 45 degrees

[formal geyser]

So can we prove that red angles are 45?

[twin surge]

Yeahh symmetry of squares can be useful in this

[trail tendon]

I don't see why we could. I'm not a professional but even tho they seem 45, I don't see why they have to be

[twin surge]

Try to take a look and identify the orientation of the square

[trail tendon]

But how do we know the square isn't more tilted to the side than it appears?

:l

unless this is drawn to scale (proportionally)

[twin surge]

Yeah it needs to be drawn to scale for that

[trail tendon]

I think the only way to solve this problem is too pretty much assume it's symmetrical and assume those angles are 45 😂

[twin surge]

Lmao Fr

Thankfully I don’t have to deal with these kinda dumbass proofs 💀🙏

[trail tendon]

💀

[twin surge]

Proof by induction and contradiction>>>>

[trail tendon]

I like proof by inspection 😎

[twin surge]

lmao you’re at university?

[trail tendon]

"seems right, les go"

[twin surge]

Ikr

[trail tendon]

eh

[trail tendon]

sure ☠️

[twin surge]

Sure is crazy 💀

[pliant kettle]

.💀

[twin surge]

Cuz your roles say undergraduate math

[trail tendon]

Well I'm doing undergrad math through a cc but technically still in hs

[twin surge]

interesting, which grade are you in?

[trail tendon]

I don't know whether that's a yes or no ☠️

[trail tendon]

11

[twin surge]

same here

[trail tendon]

Noice

[twin surge]

Prolly a no since that role is for people who are doing an undergraduate course 😭

[trail tendon]

well I am doing undergrad courses 💀

[twin surge]

yeahh where are you from

[trail tendon]

U.S. 😂

[twin surge]

Correction:- the role is for people who are doing an undergraduate course AT university

[twin surge]

Nice nice APs?

[trail tendon]

Eh not if u look at the rules

[trail tendon]

Nah like I go to a community college dual enrollment

[twin surge]

well, that changes things, but anyway, which courses are you taking rn

[twin surge]

woah that’s really cool

[trail tendon]

In math multi and diff eq

[pliant kettle]

A lot of highschoolers have the role here but like what does it matter

[twin surge]

Multivariable calc is so awesome

[trail tendon]

I assume u mean math cuz this math channel

[trail tendon]

Yeh XD

[twin surge]

Might as well change it

[trail tendon]

Yea

[trail tendon]

Fr

[twin surge]

Especially if you visualise it on a graph with parametric equations multivariable calc is genuinely one of the most elegant math topics

[pliant kettle]

Ye

[trail tendon]

Yes, desmos 3d and calc plot 3d go brr 😂

[twin surge]

Goes brr don’t forget python

[trail tendon]

I don't use 💀

[twin surge]

💀

You’re missing out Fr

[trail tendon]

💀

[pliant kettle]

💀

[twin surge]

💀

[pliant kettle]

I thought everyone just uses desmos

Or some other graphic calc

[twin surge]

Desmos /graphic calcs are the most common since it’s already there you just gotta plug equations/numbers in

But coding it out is genuinely satisfying

[pliant kettle]

Ook

[trail tendon]

Oh yeah I forgot to ask what u studyin 💀

[trail tendon]

Yea is easier 💀

[pliant kettle]

Coding it takes time

[twin surge]

Oh I’m doing IBDP lol but I’m self-studying many topics on YouTube for fun, like multivariable calc, differential equations in engineering, financial mathematics like stochastic calculus, Brownian motion and real analysis

[pliant kettle]

Cool

[trail tendon]

Yea, it's for cs geeks 😂

[pliant kettle]

Fr tho

[trail tendon]

yo noice, idk what IBDP is

[twin surge]

IB is a really popular and prestigious international curriculum, which is across two years (grade 11 and 12) and you take 6 subjects. The final IB score is out of 45.

[pliant kettle]

I read it as integration by differentiable parts 😂

[twin surge]

haha I guess I classify as a CS geek

[twin surge]

🤣🤣🤣

[pliant kettle]

Coz like IBP exists

[twin surge]

Integration by parts yessir

[pliant kettle]

Yes so what's it

The full form

[twin surge]

International Baccalaureate and DP is basically the first and second year of IB

So together it’s called IBDP

[pliant kettle]

Ook ic

[old elbow]

8th

[split venture]

IS THIS CORRECT? 😵😵

IS THIS CORRECT? 😵😵

[zenith glade]

only the square

[split venture]

I dont understand the little rectangle and triangle question

[zenith glade]

the image seems a bit confusing

mainly by this

not the number

[split venture]

Yeah it is but it’s due tomorrow and 😞 i have a test tomorrow

[zenith glade]

this intersection

[split venture]

yeah the clip and overlap

[livid root]

Reduce it to a printing error and carry on

Idk

[zenith glade]

yes, probably the most reasonable thing to do

[livid root]

Ye

[zenith glade]

would be too hard to calculate this small piece, and I think it lacks on information for so

[livid root]

Wait there are small cuts on the top left and right sides of the rectangle

[split venture]

Yeah

[livid root]

And they’re equal to x

[zenith glade]

they measure x i think

[livid root]

Yeah but

Then how

Is the bottom length

[zenith glade]

the question says

[livid root]

7x-9

It sort of looks like a 45-45-90 triangle

[zenith glade]

its his mistake

[livid root]

Oh

[zenith glade]

you have to do pythagoras

not just subtract 2x

@split venture

want me to do it or youre gonna try again?

[split venture]

I’m gonna try again

[zenith glade]

k

[livid root]

Venus do you know about 45-45-90 triangles

[split venture]

Never heard of it i dont think

[livid root]

Bruh

Well with these special triangles, Pythagorean theorem discovers a relationship between side lengths

[split venture]

Ohhhh

[livid root]

Its a side length ratio of 1: root2

You can work through the math behind it if you’d like

[zenith glade]

this is basically pythagoras, but saves a bit of time

[split venture]

I’m only in 8th grade I dont understand lol 😭

[livid root]

So you’re given two side lengths

They’re equivalent

[zenith glade]

@livid root it looks its gonna be a bit ugly the lower side

its gonna have sqrt(2)

since the cut is equals to x

[livid root]

What

[split venture]

Can you do the question and explain 😴

[zenith glade]

wait, ill grab my pen

[split venture]

i think im getting more confused

I’m also listening to a mongol empire documentary while doing this so 👨‍🦲

[zenith glade]

itll definetely make things harder

[livid root]

Okay

[zenith glade]

k

got my pen here

@split venture

[livid root]

I think it’s the idea of a 45-45-90 triangle that’s tripping you up

[split venture]

Yeah

ive never heard of it

[livid root]

I mean you know what degrees are right

[split venture]

Yeah

And I know Pythagorean Theorem

[livid root]

Triangles are made up of 180 degrees

I don’t know exactly why but I have a sort of like intuitive grasp on it

Idk

Later on you’ll learn more about these ratios through trigonometric functions

[split venture]

😵

[livid root]

But you should understand that

Like

For something to be a triangle, it must add up to 180 degrees

And

[split venture]

Yeah

[livid root]

I dont know it’s hard to explain

☠️

[split venture]

I’ts hard to understand 🫠

[zenith glade]

@split venture

[split venture]

[zenith glade]

I called alpha the lower side, which is where we having trouble at

[zenith glade]

btw, whats that 6 in the side of the triangle?

[livid root]

Why are you using alpha in geo☠️

[zenith glade]

?

just needed a variable name

[livid root]

Y

[split venture]

I’m not sure

[zenith glade]

idk, here in brazil we usually call angles theta, beta and alpha

[zenith glade]

k

[zenith glade]

so we need to calculate the lower side

and its not just sum 2x

[livid root]

Oh

[zenith glade]

cause these x are on diagonal

and we need the horizontal

@split venture

I called the lower side alpha

understood?

or do you need further explanations on the alpha? if dont ill proceed to show how to calculate y

[split venture]

Uh

Is your x the perimeter?

[zenith glade]

T_T

[split venture]

I’m so confused

😳

[zenith glade]

perimeter? x is the thing that is in the question

[split venture]

I have to find the perimeter

[zenith glade]

I know, youll have to sum all sides, but now we're discovering the lower side

[split venture]

Ohhh

[zenith glade]

@split venture

understood?

[split venture]

Yes

[zenith glade]

why alpha is 5x-9 + 2y

[split venture]

Yea

[zenith glade]

k

so lets findout y

@split venture

@split venture ?

[slim fulcrum]

hey all, i need help with some trigonometry for my lineman class

[faint pasture]

whats a lineman class

[slim fulcrum]

power lineman

i need a equation in trigonometry that shows how long and wide a crossarm for a power pole is

im not that good at math

[faint pasture]

what do we know

[slim fulcrum]

well hold up

i grabbed this offline but i need the equation and shape for it

[faint pasture]

i need to know something in order to manipulate that to get what you want

[slim fulcrum]

like the shape?

hold up

[faint pasture]

anything

like some lengths

angle

and something

[slim fulcrum]

lets do something lile that

[faint pasture]

ohkay

[slim fulcrum]

the lengths were in the other screenshots

[faint pasture]

do you know about something called tan()?

[slim fulcrum]

ehh not really ive heard a little

[faint pasture]

um without that its gonna be almost impossible to explain i guess you should try to learn from like khan academy or smthn

[slim fulcrum]

i can learn it and ill get back to you

ohh ik what it is 😭

its like the adjacent sides and everything

[faint pasture]

💀 wdym by everything

[sly fable]

I want to know....is there any book that have lots of difficult gemotry problem with some explanation?? Without any explanation is also fine

[zenith glade]

I only know books in portuguese

well I mean

the math is gonna be the same

but

yk

[sly fable]

Well....can I take a look at it??

Or any pdf version of it??

[turbid reef]

Is this wrong if so what am I doing wrong?

[zenith glade]

3,1 * 2,25

area

6,2 * 1,5

circumference

well

the question doesnt say the value of pi theyre using

but basically, using 3,14, you just forgot to round the answers

[turbid reef]

Ohh

[zenith glade]

round to 9,4 and 7,0

[turbid reef]

9,4 for circumference and 7,0 for area?

Tysm

[zenith glade]

yup

[cursive bone]

This is mathematics

Very mathematical

[stray marsh]

Is anyone can solve this?

[obsidian harness]

translation pls?

[stray marsh]

Its ask (AFC)

[obsidian harness]

I know but the top two lines?

[turbid reef]

Does anyone know what I do for these 2 questions?

[stray marsh]

Two triangles congruent and isosceles

[obsidian harness]

question 1: add all the distances for 'how far will Abby ride her bike altogether'

then it should be clear which sides are the legs and which side is the hypotenuse, so you can do A = 1/2 * b * h

[stray marsh]

E point b and c are linear

[obsidian harness]

question 2: use the formula V = base * height = pi r^2 h
and calculate the volume for both the drums

[turbid reef]

What answers should I get cuz I think I might forget

[obsidian harness]

1a) 2.3 + 3.4 + 4.1 = 9.8 mi
1b) 1/2 * 2.3 * 3.4 = 3.91 mi^2

2. first drum: pi * 2^2 * 3 = 12pi ft^3
   second drum: pi * 3^2 * 2 = 18pi ft^3

so second drum can hold more oil

[turbid reef]

Tysm

[obsidian harness]

npnp!

[stray marsh]

Can you solve it?

[obsidian harness]

ehhhhh I can't sorry

Turkish geometry questions are hell

[stray marsh]

Ah okay

[zenith glade]

@stray marsh

could you please translate?

[upper karma]

all its saying is given below the shape

asking for afc

[zenith glade]

k

[stuck sleet]

can someone prove this?

[obsidian harness]

try simplifying sqrt(1 + sin(2u)) by converting 1 + sin(2u) into R cos(u + alpha) form

I mean, prove $1 + \sin 2u = 2 \cos^2 (u - \pi/4)$

[somber coyote]

south

[faint pasture]

Oh

Or

1 + 2 sin(x/2)cos(x/2)

Write the 1 as

Sin^2(x/2) + cos^2(x/2)

And youre done

[obsidian harness]

oh smart

that's nice

[faint pasture]

More like ive done this exact problem earlier 💀

[jagged minnow]

To prove this equation, you need to square both sides to eliminate the square root, use the fundamental identity sin^2(1/2x) + cos^2(1/2x) = 1 to simplify terms, and then apply the double-angle formula 2sin(1/2x)cos(1/2x) = sin(x), thus arriving at the correct equality

[obsidian harness]

late

[jagged minnow]

No problem, good luck! 🙂

[storm zephyr]

saw this on wikipedia, shouldn't it be more like:

If and only if a + b = 180 degrees, then line 1 and line 2 are parallel?

[acoustic pollen]

no

it works both ways, if two lines are parallel then a+b = 180 and if a+b = 180 then the lines are parallel

there's no need to put an "if and only if"

[storm zephyr]

but doesn't it just mean the same thing?

if and only if p then q is the same as if p then q, and if q then p?

[acoustic pollen]

if you say "if and only if" that means two lines are parallel only if a+b = 180

but

what also happens is that if the lines are parallel then a+b = 180

those aren't the same

I mean technically they are but putting "if and only if" could cause the text to be more ambiguous and the reader could possibly think that just the lines being parallel does not imply on a + b = 180

[storm zephyr]

ok well I mean is this correct?

(a + b = 180 degrees) ↔ (line 1 and line 2 are parallel)

this goes both ways and is equivalent and bi-conditional right?

[acoustic pollen]

yes

but don't put the "if and only if" stament before the a+b

[storm zephyr]

ok, yeah i guess that makes it harder to read

but the point is that they're still equivalent, and bi-conditional right?

[acoustic pollen]

yeah

[storm zephyr]

ok got it, thanks

[wheat granite]

What reason could I give that ΔABC is an enlargement of ΔADE from point A? Im pretty sure it is

[jagged minnow]

When we say that triangle ABC is an enlargement of triangle ADE from point A, it means that triangle ABC is a scaled version of triangle ADE, with the scalling centered at point A. In other words, the two triangles are similar, and the sides of triangle ABC are proportional to the corresponding sides of triangle ADE, all emanating from point A. To justify this, you could check if corresponding angles in triangle ABC and triangle ADE are equal. You shoul also check if the sides of triangle ABC are proportional to the sides of triangle ADE, meaning: AB/AD=AC/AE=BC/DE. In these conditions hold, then triangle ABC is indeed an enlargement of triangle ADE from point A

[wheat granite]

so it isnt always true?

[jagged minnow]

Exactly, it is not always true that one triangle is an enlargement of another. For triangle ABC to be an enlargement of triangle ADE from point A, certain conditions must be met: 1. The triangles must be similar, meaning thei correspondin angles must be equal and 2. The sides must be proportional, meaning AB/AD=AC/AE=BC/DE. If these conditions aren't met, then the triangles are not an enlargement of each other.
So, it's only true under specific conditions, and not automatically for any pair of triangles.Example of when it's not true:
If the sides of
ΔABC don't maintain the same ratio as the sides of ΔADE, or if their corresponding angles aren't equal, then ΔABC is not an enlargement of
ΔADE.

[wheat granite]

so are they always similar?

I had assumed that since they seemed to be similar they were dilations

nvm forgot the definition of similar shapes

Alr I figured out why they have the same angles thanks

[jagged minnow]

Similar shapes are shapes that have the same form but may differ in size. The key characteristics of similar shapes are:
Equal corresponding angles: All angles in one shape are equal to the corresponding angles in the other shape.
Proportional sides: The lengths of corresponding sides are in the same ratio.
For example, two triangles are similar if their corresponding angles are equal and the lengths of their sides are proportional, even if one triangle is larger or smaller than the other.

[karmic relic]

looks the same

idkk

[jagged minnow]

Yes, all triangles with the same angles are similar. This is known as the Angle-Angle (AA) similarity criterion, which states that if two triangles have two angles that are equal, then the triangles are similar, meaning they have the same shape but may differ in size. In your case, if DE and BC are parallel, then the corresponding angles formed with a transversal will indeed be equal, establishing the similarity of the triangles.

Do you want to discuss a specific exercise?

[wheat granite]

No I understand the point, I just forgot the corresponding angles way of doing it, thank you so much

[jagged minnow]

Welcome, If you need any other answers, I am waiting for the questions.

[chilly reef]

A diagram of part of a broken stained glass window is shown below.
To prevent further cracking the school has requested the window be cut into four sections of more than 0.1m^2 with at least one section not being a triangle

Write a report explaining how the window can be divided.

Find length of the pane required to repair crack

Demonstrate it can be divided into four sections of more than 0.1m^2 and show one possible way of doing this.

the shape isn't parallel (it's supposed to be two triangles that are uneven i just suck at drawing)

triangle 1 (left)

0.51m - 0.84m

Triangle 2 (right)
0.56m - 0.78 m with an angle of 84.8 degrees shown in the middle corner

so far i've got this

c^2 = (0.56)^2 + (0.78)^2 - 2(0.56)(0.78) cos(84.8)

sqrt (0.56^2+0.78^2-2 (0.56) (0.78) cos (84.8pi/180) = 0.918

idk what to do from here

these are the formulas i have been given

[wheat granite]

uh this is probably a stupid idea but

why not just make 3 squares that are like .2 area

and then a weird funky shape

[chilly reef]

idk how to do that ☹️

[wheat granite]

uh

just make a bunch of squares on a side with length sqrt .1 on sides

?

[tropic reef]

what's the difference between degree and radian?
I missed the class so i don't know much about radian

not asking for homework help but just concepts

[stuck sleet]

what identity is that?

[stuck sleet]

but when proving the trigonometric identities, it is necessary to only manipulate only one side of the equation, right?

[stuck sleet]

is there a difference between a trigonometric ratio and a trigonometric function???

[warm shuttle]

2π radians = 360°

that's all

[elder tundra]

i need help with this

[hoary prism]

Ur already given BF cause it’s the height, now use AD and CD to calculate the length of BD, then use BF and BD to calculate the length of DF

[iron plaza]

I need help with trigonometric identities

[junior shore]

can someone explain how they got this awnser

[iron plaza]

Can you send me the link of this quiz please

[junior shore]

?! me

[obsidian harness]

both triangles are isosceles for question 1

so the base angle of the triangle with double marks is 180 - 2 * (180 - 109) = 38

and so x = 180 - 38 = 142

question 2 is super similar

just find the base angle of the triangle with double marks again

by looking at the first isosceles triangle

[lavish halo]

Help what does the ^2 thingie mean here

I missed 6 classes worth of beginning areas of trigonometry so woo

Full question qas what full number values can A be

[silent plank]

$\sin^2(x) = (\sin(x))^2$

[somber coyote]

ℝαμOmeganato5

[lavish halo]

Ic ty

[outer valve]

Do you need trig to learn calculus, specifically limits, tangent lines, chain rule, area under a curve, and reverse chain rule. None of the functions themselves involve trig.

[cunning lion]

you need to know some trigonometry to do calculus on trigonometric functuons

[outer valve]

None of it would involve trig funcs, so no? This is for a physics class I’m trying to take

[tall coral]

Just learn in regardless anyways

Worst case scenario, you know some trig

[outer valve]

K thx

[acoustic pollen]

another way to measure angles

[serene pond]

guys how would I find a midpoint between 3 or more points on a plane

so that when the point is placed, it is equidistant between every point

[warm shuttle]

what you are asking for is the circumcenter of the triangle formed by those 3 points

https://youtu.be/wVH4MS6v23U?t=102&feature=shared

(timestamped for circumcenter explanation and construction)

[warm dagger]

how do i get better with proofs and identities in trigonometry?

[acoustic pollen]

gotta know a couple of relations and practice in general

[civic ridge]

do algebraic trig problems

[wise zodiac]

How to prove [EFGH] > [ABCD] with trigonometry or calculus

This is the solution but i’m sure this can be done way better

[stuck sleet]

how do I bypass arcsin(2) ????

[unkempt gulch]

Help finding BC

[past thicket]

wouldn't it be 60? Correct me if I'm wrong .

I'm currently in geometry as well

[onyx prairie]

its 60

[past geyser]

consider the fact that ABCE is a square, and you can find EC using trig

does that help?

[obsidian harness]

ABCE is only a rectangle

[amber helm]

struggling with this problem from the last 2 hours

[obsidian harness]

use power of a point

$30 \cdot 50 = PT^2$

$PT^2 = (PA)(PA + 35)$

[somber coyote]

higher's secret brother

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[lavish halo]

(Cos^2(20))/(1-sin20) If sin20 is A what is this equal to in terms of A What do i do here

[amber helm]

im new here mb

[amber helm]

use trigo identity(cos^2 (x) = 1 - sin^2 (x)

[lavish halo]

(1-x^2)/(1-x) which is 1+x?

[amber helm]

yeah

[lavish halo]

Ty

[acoustic snow]

How to solve number 4 I got y=2x+5

Answer key says -1/2x+5

!ticket

[fervent oar]

Is this about right?

I’ll take this radial silence as a yes

And be on my merry way

[crystal tiger]

if one were to denote the radius of a cylinder with z and the height with a, then the formula for how much pizza you have becomes clear: pi(z*z)a

[terse sun]

nah, I don't eat the crust so it's less than that.

probably like π * (z-2) * (z-2) * a inches

[crystal tiger]

still tho

pizza

[trail tendon]

who said there was crust 👀

[acoustic snow]

Someone please answer bruh

[obsidian harness]

yes, it would just be y = 2x + 5 again

if you have any vector on that line and rotate it by 180 degrees, it will go in the opposite direction but it will still be on the line

[lavish halo]

(Cot335+1)/(cot25+tan155) if tan 25 is a what is this equal to in a

[terse sun]

science ;-;

[terse sun]

so is the equation:
0 = (cot(335) + 1)/(cot(25)+tan(155)) = 0?

[lavish halo]

Dont think its =0 the question was to convert it to A

(Cot335+tan75.cot75)/(Cot385+tan155)

Tan25=a

[celest pasture]

How do you solve for x
cos (2 tan (Arctan x))?

[unborn drum]

How does this find angle delta and not gamma?

Book:

Using law of cosines cos(gamma) = a^2 + b^2 - c^2 / 2ac

[obsidian harness]

take the negative of both sides of $\cos \gamma = \frac{\alpha_1^2 + \alpha_2^2 - x^2 - y^2}{2a_1 a_2}$ by the cosine rule

[somber coyote]

higher's secret brother

[obsidian harness]

$-\cos \gamma = \cos(\pi - \gamma)$

[somber coyote]

higher's secret brother

[obsidian harness]

and the right side matches with the book

[last ocean]

In abcd rectangle with 24,30cm sides, 30Cm side becomes radius and goes up and cuts other rectangle size, and removes that sector now there is small piece left in that side, and in that corner there is square, find area of the square

Need help

[unkempt gulch]

I need help understanding why OR is √3 - 1 and not 2. The only information given is that this is an Isosceles right triangle and it’s perimeter is √12 + √24

and that M is the midpoint of the hypotenuse

[obsidian harness]

what have you tried so far?

[obsidian harness]

oh wow

,w (2 sqrt3 + 2 sqrt 6)/(2 + sqrt2) simplify

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by sooouth.

[obsidian harness]

right so a mnemonic for a 30-60-90 right triangle is that the sides are in the ratio 1 : sqrt(3) : 2

so $OR = \sqrt{6} - \sqrt{6} \cdot \frac{1}{\sqrt 3} = \sqrt{6} - \sqrt{2}$

[somber coyote]

higher's secret brother

[obsidian harness]

oh wait I see what happened, so you thought that PRM's perimeter was √12 + √24

I also thought the same

the question assumes the big triangle, PRQ's perimeter is √12 + √24

so the perimeter of PRM is 1/sqrt(2) that of PRQ by similar triangles, PR/PQ = 1/sqrt(2) for example

[obsidian harness]

that means we divide this by sqrt(2) to get sqrt(3) - 1 as desired

[unkempt gulch]

So all we have to do is that?

[obsidian harness]

yes

[unkempt gulch]

But, what did I do wrong in my calculations exactly?

[obsidian harness]

$\frac{\sqrt3}{3} = \frac{MO}{PM}$ correct

[somber coyote]

higher's secret brother

[obsidian harness]

but you found PM = sqrt(6)

so you get $\frac{\sqrt{3} \sqrt{6}}{3} = MO$ or $MO = \frac{\sqrt{3} \sqrt{3} \sqrt{2}}{3} = \sqrt{2}

[somber coyote]

higher's secret brother
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[obsidian harness]

and then the 2nd mistake was thinking PRM's perimeter was √12 + √24

luckily there's a fix for this as I described above

also you seemingly forgot that PM = RM as PRM is also isosceles

[unkempt gulch]

Yeah you are right

Thanks

[obsidian harness]

no worries!

[acoustic snow]

Oh yeah

Thanks

[jade sierra]

what am i doing wrong

😭

i have a test tomororw

and im cooked

[dreamy vessel]

I think you're correct but check the x intercepts

[jade sierra]

wdym

[dreamy vessel]

You find the period by dividing two pi by the number in front of x. The interval between each x is one forth of the period, since the period is 2pi then you can find the distance between the x intercepts

But they gotta be in terms of pi because 2pi/4=pi/2

[jade sierra]

erm isent finding peroid 2pi/b

[dreamy vessel]

Ye

[jade sierra]

so do i just change my peroid to pi/2?

[dreamy vessel]

Nonon period is 2pi/2 because number in front of x is 2

Assuming that the x axis is in terms of pi it looks like you would be correct

[jade sierra]

well i tried submiting twice and its wrong

so il just try 2pi/2

well il be dammed

[dreamy vessel]

Scrumptious

Luscious

Delectable

Palatable

[jade sierra]

i hate deltam ath so much

i get it every night

[dreamy vessel]

😦

If you don't like math face my wrath

[jade sierra]

i like math

but not deltamath

its like ultra specific

[dreamy vessel]

Good

[jade sierra]

where if you write 6/2 instead of 3, youd get wrong

[dreamy vessel]

Is that deltamath

[modern root]

Can anyone help me with bearings?

[dreamy vessel]

Bear

[jade sierra]

yup

[dreamy vessel]

Yikes

[jade sierra]

😭

[dreamy vessel]

I have never heard of bearings in my life

[jade sierra]

idk this looks wrong

[modern root]

I know understand them at all

😭

[dreamy vessel]

Search up yt video

Or go to a help channel

[jade sierra]

my math teacher is always really quiet so i just accepted that ima be watching a youtube guide every night

[dreamy vessel]

Find period, find midline

[jade sierra]

midline easy

peroid not as much

cuz like what is 2pi times 5/6

😭

[dreamy vessel]

Liar liar Tunisia step father on fire

2pi/(5/6)

That

Which is equal to 2pi*(6/5)

[jade sierra]

god been so long since i did kcf

[dreamy vessel]

Yooo

[jade sierra]

never really liked fractions so i usally convert to decimals 😭

[dreamy vessel]

That's what my teacher called it

What a sigma

[final rose]

I cannot escape it, even here

[jade sierra]

😭

[dreamy vessel]

You can't escape the sigma for its due diligence and necessity to sigma you is imperative to its existence.

[final rose]

I only use sigma when reffering to sigma notation or standard deveation(i cant spell)

its everywhere, I cannot escape improper usage of the letter sigma

[jade sierra]

is this correct

[charred herald]

can anyone help with my 2nd grade math homework

please?

[final rose]

hi

[charred herald]

sorry grimlock i ddint want to bother you

thats why i didnt ask for mroe help

[jade sierra]

its fine

[dreamy vessel]

I don't see graph

But also I can't help you on everything because on the test you need to be CONFIDent

[jade sierra]

well hadnet made on yet cuz idk if my peroid is right 😭

just need to know if my peroid is right first than id graph

[dreamy vessel]

2pi/(1/4)=8pi

[jade sierra]

oh

😭

[final rose]

@charred herald skibidisigmalandon really?

[charred herald]

yes

[dreamy vessel]

My nad

Bad

[charred herald]

You seem to have a graph with a much smaller range and different amplitude than the one expected. Make sure your graph starts at
𝑦

2
y=2, goes down to
𝑦

−
2
y=−2, and completes one full cycle over the range
[
0
,
𝜋
]
[0,π].

You should adjust the x-axis to reflect a period of
𝜋
π and ensure the y-values range from 2 to -2. Try plotting again with these guidelines.

[jade sierra]

it was wrong but when i tired 2pi4 it was right

[charred herald]

I have to go to sleep it's 8 pm already

[dreamy vessel]

The the heck

[charred herald]

hope it helps a little i'm not that good at math

[dreamy vessel]

Is 2pi4=2*4pi?

[charred herald]

goodnight guys

[jade sierra]

errm probaly

ig

[dreamy vessel]

Why sleep when you can do matg

Get Landon back here

[charred herald]

my mom will hit me again and the purple spots hurt when i go to school

[dreamy vessel]

?

[charred herald]

so i have to go guys goodnitght

[dreamy vessel]

Well worth the math I tell you

[final rose]

Concerningjustalittlemabye?

[dreamy vessel]

That's what I am saying

Let the man do the math

[charred herald]

my mom fell asleep haha

I can stay awake longer yay

[dreamy vessel]

Math's night out

[charred herald]

?

[dreamy vessel]

Math time

[charred herald]

yeah

does anyone want to help me with my math homework?

[dreamy vessel]

I can but for 10

[charred herald]

what do you mean?

[dreamy vessel]

Bc after 10 min I have to do my math

[charred herald]

okay

go to messages?

[charred herald]

egg what useful and very nice

[tropic reef]

hi all
how to memorize those extra trigo ratios like sec and cot?

[trail tendon]

sec = 1/cos
cot = 1/tan

😁

[tropic reef]

there should be now 6 except sin cos tan

[trail tendon]

wdym?

[vagrant snow]

sec = 1/cos
csc = 1/sin
cot = 1/tan
?

[tropic reef]

total count

[trail tendon]

there are 6 basic trig functions which are sin, cos, tan, csc, sec, and cot

so you meant including?

[warm shuttle]

haversine be like

[trail tendon]

'basic'

._.

[tropic reef]

ya

but i didnt learn the remaining 3s when i was younger due to weird curriculum

[trail tendon]

😔

[warm shuttle]

tbh everything is just cos

sin(x) = cos(x - pi/2)
tan(x) = cos(x - pi/2)/cos(x)
sec(x) = 1/cos(x)
csc(x) = 1/cos(x - pi/2)
cot(x) = cos(x)/cos(x - pi/2)

You only need any one of them

oh and exp(x) is the true one function to rule them all

including the hyperbolic ones

[cunning lion]

sin(x) = (exp(ix) - exp(-ix))/(2i)
cos(x) = (exp(ix) + exp(-ix))/2)
tan(x) = (exp(ix) - exp(-ix))/(i(exp(ix) + exp(-ix)))
sec(x) = 2/(exp(ix) + exp(-ix))
csc(x) = 2i/(exp(ix) - exp(-ix))
cot(x) = i(exp(ix) + exp(-ix))/(exp(ix) - exp(-ix))

[tropic reef]

how to find a reference angle? since the textbook didnt mention how to find clearly

i forgot all of them after studying in secondary school

[vernal pilot]

It’s /2i not /2 for sin(x)

[flint marlin]

Hi , anyone knows how to find theta here?

[neat geyser]

if tan(x)=2/3 and x is in quadrant 3, find cosx, hi can somebody explain the answer?

[obsidian harness]

what did you get for the radius?

[flint marlin]

I didnt calculate

[obsidian harness]

then calculate R first

you should get that the central angle COA = 75 * 2 = 150

so gamma + theta = 30

ah so by power of a point, $12^2 = \left( \frac{12}{\sin 75} - R \right) \left(\frac{12}{\sin 75} + R \right)$

[somber coyote]

higher's secret brother

[obsidian harness]

by similarity, you can just assume that opp = 2 and adj = 3
so hyp = sqrt(2^2 + 3^2)

cos x will be negative in quadrant 3 so the answer is just -3/sqrt(13)

[obsidian harness]

@rotund stump https://www.bbc.co.uk/bitesize/guides/zx2sb82/revision/2
here is a nice summary of all the different geometrical constructions with a compass and ruler

for question 4, a and c are both angle bisectors

b is the perpendicular bisector

follows by RHS congruence

[rotund stump]

thank you so much!

[sleek stratus]

Anyone know as to why law of tangents isn’t taught alongside the law of cosine and sine? I know that it exists but why is it ignored and not taught alongside the other laws

[flint marlin]

Ok but how does it help us to get gamma- theta?

[trail tendon]

less useful? ig

[obsidian harness]

idk man

[flint marlin]

...?

[obsidian harness]

I seriously don't know the answer

[flint marlin]

Ok ill give it a try

[obsidian harness]

that's super interesting cause I've never heard of the law of tangents

note that the law of cosines corresponds to SAS and SSS congruency

and the law of sines corresponds to ASA and AAS

Wikipedia also says that the law of tangents is equivalent to the law of sines

[restive hornet]

anyone have any tips for how to do proofs better?

[sudden seal]

I want it too

[upper karma]

hey there, i have a fill in the blank and i wasnt at school this day so i dont know what to do with this. someone sent me to this server (may their pillow always be cold)

[winged yarrow]

good day everyone, was hoping to get some help on a physics related problem that applies some trig, on the help 26 channel

just hoping to understand how it works better

[frozen cobalt]

why is it when i put 2/cos(50) it gives me 2.07? when the right answer is 3.11

[sleek stratus]

Probably because you converted 50 to radians

[cyan steppe]

Yea I checked it too, it's coming about 3.125 when u take the angle in degrees

[flint marlin]

@obsidian harness How?

[upper karma]

wait

what got me here

sorry I thought this was discussy

[dim wren]

hii

[rich scroll]

Hello i need help with assignment cuz our teacher gave us one without any lecture . Numbers 1,4,6,8,14,18,19

[rotund wedge]

how to reflect

..?

[eternal marsh]

it depends what you’re reflecting over

[ionic coral]

can anyone teach

[obsidian harness]

not the right place brother

if you have questions just ask

otherwise look for a tutor in your area

or if you don't mind YouTube teaching you then org chem tutor and so on

[honest estuary]

Do we have a separate channel for 3D-Geometry?

[obsidian harness]

nope

if it's about 3D vectors then you can also ask in #multivariable-calculus though

[honest estuary]

It's about Line and Plane

[obsidian harness]

#multivariable-calculus

[honest estuary]

sure

[timid gale]

hey i need help with pythagorean theorean

[trail tendon]

what do you need help with, more specifically

[timid gale]

alr so i completely forgot what we was doin in class the other day

its literally the basics

find the hypotnuse

and theres a couple where they have the hypotnuse and 1 leg and i need to find the other leg

[trail tendon]

do you know how to find the hypotenuse given the two legs?

[timid gale]

do i square both the legs and then thats the hypotnuse?

[trail tendon]

you square them and add them and thats the square of the hypotenuse

if a and b are the legs, and c is the hypotenuse,
a^2 + b^2 = c^2

[timid gale]

alr

[trail tendon]

so c = sqrt(a^2 + b^2)

then if you just want to find one leg, say you want to find leg a

just get a alone, so a^2 + b^2 = c^2 becomes
a^2 = c^2 - b^2
-> a = sqrt(c^2 - b^2)

does that make sense?

[timid gale]

i think

so

to find the leg

i need to square the ones i have

minus the leg i have to the hypotnuse

and thats the missing leg?

i tried watching a video on youtube about it but my brain didnt comprehend it

[trail tendon]

hypotenuse^2 - leg^2 = (otherleg)^2

[timid gale]

that makes more sense

alr ima try a couple problems with this

and ill prob be back later for other things

I still feel i did somthing wrong

it said find x

so x should theoretically be 32 right?

[obsidian harness]

no

you can tell cause the leg can't be longer than the hypotenuse, 9

[wooden sandal]

nope

[obsidian harness]

7^2 + x^2 = 9^2 right

[timid gale]

wait

do i have to

square root it

[obsidian harness]

yep!

[wooden sandal]

yes secret

[timid gale]

alr so i t should beeeeeeeee

like 5.7 give or take

[wooden sandal]

yes if rounded up

√32 is what i would leave it in tho

[obsidian harness]

I mean preferably you could simplify that to be $4 \sqrt{2}$

[somber coyote]

higher's secret twin brother

[gritty panther]

@storm yarrow

[storm yarrow]

hmm

lemme try

doesn't look too bad

[gritty panther]

I keep hitting a brick wall at sin(u)+sin(u)cos(u)

[upper karma]

2csc(u)

[storm yarrow]

ok oogabooga

calm down buddy'

[upper karma]

is it right?

im not sure

[gritty panther]

it is right

[storm yarrow]

nah just surprised at your computational speed

💀

[upper karma]

nice

[storm yarrow]

did you do that in your head @upper karma

wtf

[upper karma]

no

i used paper

[storm yarrow]

oh ok

BRUH

[upper karma]

and pen

[gritty panther]

I found out the answer in desmos a while ago but I can't really figure my head around the theory

[storm yarrow]

IT STILL TOOK U LIKE

20 seconds

do u have trig identities meomrized

[upper karma]

i just have high fps and low ping

[gritty panther]

i crave to have your skills

[upper karma]

i know the pythagoras one

[storm yarrow]

lol

nice

[gritty panther]

this question in particular annoys me bc 1+cos(u) keeps firing the "1-cos^2(u)" receptor in my head

[gritty panther]

how did you get to this btw

if you dont mind

[upper karma]

step 1: rewrite each term to have a common denominator
step 2: combine the two terms
step 3: expand and simplify
step 4: apply Pythagorean identity sin^2(u) + cos^2(u) = 1
step 5: cancel out common factors

[gritty panther]

thank you

[upper karma]

hi can someone tell me if this is correct

[vagrant snow]

Bro did the speed math

[upper karma]

already know all the questions. i'm struggling w only q 4 now

because according to the diagram 360-320=40 but 90-40=50 so u use 50 degrees right cuz it's in the triangle?

[vernal pilot]

what tf

@trail tendon

[trail tendon]

😭

[trail tendon]

☠️

wait why we posting this in the geometry channel

💀

[vernal pilot]

oh damn, mb, i was in it for some reason

[trail tendon]

💀

imagine some poor guy sees this and thinks its geometry and gets traumatized

💀

[pseudo nebula]

you can integrate on non-continuous functions?

actually, dear god why is there a max in an integrand

[faint hollow]

$u = cos^x(x)$

[somber coyote]

Black_Gold

[faint hollow]

$du = cos^x(x)[ln(cos(x))-xtan(x)]dx$

hence answer would be $cos^x(x)$

[somber coyote]

Black_Gold

Black_Gold

[faint hollow]

$\int{}{}cos^x(x)[ln(cos(x))-xtan(x)]dx = \int{}{}du$

[somber coyote]

Black_Gold

[tall coral]

I might be slow, but something is going wrong. When I double check my answers with arctan(-4/3), I am getting the opposite sign of what I got for sin

[lavish fern]

Can anyone help with number 4

[obsidian harness]

classic trig problem

let the base of the triangle with hypotenuse a be x

in fact, x = 21/tan(48) from right angled trig

so you can set up a similar equation with the angle 31 degrees, the opposite side 21, and the base which is x + b = 21/tan(48) + b

can you continue?

[lavish fern]

Yes

Thanks

[obsidian harness]

no worries, also like you can find a at any time

it was b and c you were stuck on I was assuming

[obsidian harness]

also please don't ask in multiple places

[lavish fern]

My bad I won’t do it again

Also thanks for the help again

[pallid wagon]

i hate vectors

[mellow quest]

I hate vectors in planimetry

in stereometry I sometimes use

[jade sierra]

i dident take geomtry

😭

[restive hornet]

oh shit is that geometry

[trail tendon]

no 😭😭

[trail tendon]

"ACK" is so accurate

[jade sierra]

Yup

[trail tendon]

anyway iirc it should be 15

[jade sierra]

I finished the test a while ago

Got 283 I think

yup

[trail tendon]

😂

noice noice

[azure helm]

where can I do that test

[flint marlin]

I need some urgent help

Anyone available?

[lost cosmos]

nice! I finished mine today and got a new personal best of ||314||

[prime fable]

https://tenor.com/view/stone-gif-26569950

[prime fable]

it's called map testing and u cant do the real test u can do the guest version basically the same

[azure helm]

give me the link

[jade sierra]

its map growth test

issued by nyc schools

[prime fable]

also pak

[jade sierra]

what pak

pakistan?

[prime fable]

ye

[jade sierra]

how i do this 😭

i got D

[prime fable]

expain how u got D

[jade sierra]

if cosa = square root 5/3

[prime fable]

k

[jade sierra]

and sin a = o/h

[prime fable]

ye

[jade sierra]

the only answer there h on the bottem is d

[prime fable]

wait but

would this not be one of the values?

that would be divided with something else

because it says that a alone meaning one value is underoot5/3

[jade sierra]

well yuh

yea

[prime fable]

so whats the other value

[jade sierra]

idk but the one on the bottem should be the hyp

and no other has hyp

[prime fable]

mmm

try and ask one of teh mods

[runic pasture]

,w -sqrt(1-(sqrt(5)/3)^2)

[runic pasture]

you can use the identity [ \sin^2A + \cos^2A = 1 ]

[somber coyote]

bacc the sigma😔🤞

[runic pasture]

then solve for sine

you will get two solutions for sin(A) but you know which to pick because of tan(x) < 0

[upper karma]

do yall understand this

[ember inlet]

What do you not get?

[upper karma]

how to get distance traveld

and speed

[ember inlet]

It’s the area under the velocity curve no?

And speed is given by the graph

Speed at time t = value of graph at time t

[upper karma]

yeah but whats the equation to use in this scenario

[runic pasture]

use geometry

[upper karma]

cuz i swear it keeps changing

[runic pasture]

it's a triangle

[ember inlet]

What do u mean

[runic pasture]

or trapezium

[ember inlet]

What’s the equation

Yeah

[upper karma]

ok in a sense i guess what im asking is

[ember inlet]

You want to find the area under the graph

It is a trapezium

[upper karma]

how do i solve for the distance traveled and final speed

[ember inlet]

what’s the formula for

[upper karma]

whats the area

[ember inlet]

Area of trapezium?

U tell me…

[upper karma]

could you explain

[prime fable]

tri = 1/2 a+b sin a
trap = 1/2 (a+b) x h

i think

[upper karma]

im having diffuculty

[runic pasture]

you can divide it into a triangle and rectangle too

[upper karma]

understanding how to calculate the area under the graph

[ember inlet]

Yeah u can get the formula in lots of different ways

[upper karma]

ok wait this helps a lil

[ember inlet]

and the final speed?

[upper karma]

so i do base times height for distance traveled?

[runic pasture]

that's the area under the curve that you calculate

[ember inlet]

Yep

[upper karma]

so what

2 times 24

divided by 2

orrr

[runic pasture]

kash rules everything around me

[upper karma]

?

[jade sierra]

I swear my teacher never taught that 😭

[runic pasture]

probably because it's trivial

girmlock you need to lock in

you got this

[jade sierra]

Well if I never learn it how am I suppose to do it

.-.

[runic pasture]

oh uhm

do you know the unit circle

[upper karma]

how does the formula for trapezium rule actually work

[jade sierra]

probaly not

[wheat granite]

Does anyone know why whenever I try and look up a proof for the parallel transversal angle theorems theyre all self referential, use Eulers 5th or just make it a postulate? I could be wrong but cant you prove it another way

[maiden brook]

that’s why Euclid’s (not Euler) is an axiom

[maiden brook]

there’s no way of doing it without it being self-referential at some point, why is why Euclid has it there as an axiom

It can seem odd it’s an axiom, yes

[wheat granite]

Sorry I meant euclids

[wheat granite]

I can assume you can prove it if you somehow had the triangle sum theorem, but is it impossible to get that without those?

[maiden brook]

actually, it’s usually the other way around. That axiom can be used to quickly show the sum of the angles in a triangle is 180

[wheat granite]

I know about that, Im just asking if you could prove it with that theorem, and if so why cant the sum theorem be proved without the transversal angle theorems?

[maiden brook]

if you could prove what with that theorem? If you assume the sum of the angles in a triangle are 180 instead of Euclid’s postulate, it should be fine also, but things would be slightly different. I don’t get everything u r asking though

[wheat granite]

Okay so thanks for the info, but Im just asking why you cant prove the sum theorem if you didnt just make it a postulate or have euclids

[maiden brook]

I still find the way u phrased it unclear. It can be proved by Euclid’s 5th postulate, or Euclidean geometry should even work out nicely if u swapped the fifth one by saying that the sum of the angles are 180, but Euclid’s postulates are the convention

[wheat granite]

alr I should prob just show you what Im thinking of here
If F is the midpoint of the hypotenuse, then couldnt you say that ΔEFB and ΔAEF are congruent, as well as ΔCDF and ΔFDA are congruent. So wouldnt that say that angle C + angle B = 90 degrees?

[maiden brook]

well u would need to prove generally that the sum of the angles is 180

Not just in a right triangle

[wheat granite]

yeah but non right triangle triangles are made of right triangles so the angles add up to all the angles - 180 since the line that makes the right angle doesnt exist

and I dont see how thats needed to prove the transversal theorems

[hoary prism]

@wheat granite what’s like the precise statement you’re trying to prove