geometry-and-trigonometry

[drowsy elm]

should be-1

2nd q so - and pi/4 is 1

1/1 is stillv1/1 so cot is -1

[patent quarry]

lol

because we would help you in a time consuming problem after saying that

[trail magnet]

https://youtu.be/7NrR_F_ot5E

[royal citrus]

Here, AOB and CBO are alternate angles right?

[deft veldt]

yh they are equal

[placid parrot]

How do I solve?

[thorny current]

is theta and x interchangeable

[grave pond]

All variable letters are interchangeable.

[gleaming nova]

if only one appears in your question yes

they’re all basically just random letters

actually i guess even if they both appear they’re interchangeable

[grave pond]

That might be too uncompromising a statement, though. Some variables letters have conventional uses that will often be assumed without stating them explicitly. For example, you often won't find "I'm using x to denote the coordinate that grows towards the right, and y to denote the coordinate that grows upwards" stated in so many words, but if you're using them reversed (which is allowed, for sure), or if you're using p and q instead, you had better state that explicitly or risk being misunderstood.

[thorny current]

think i got the gist of it thanks

[near sand]

Okay, I'm not a fan of geometry but I was to show that x=5 and y=6.7. How can I write this conclusion in a proof format? I know I will use ASA for x, and SAS or ASA for y

[pseudo ore]

hey guys I wanted to ask how we can identify by an equation the type of ellipse it is whether it lies on x axis or y

[inner spoke]

Does anyone have some hard problems on trigonometry? Like they must be super hard

[pseudo ore]

sin x = ?

[signal swallow]

what do you mean by this

[inner spoke]

Idk lol

[shrewd osprey]

I’m so confused if anyone’s knows lmk

[inner spoke]

Please give me hard trigonometry questions

[pseudo ore]

see here

this is a table of two different types of equations as we know

but How can we identify what equation

this is from the table

a>b?

[signal swallow]

ah the major axis

yeah

36 > 16 so the x axis is gonna be the major axis

[pseudo ore]

Oo thank you!

[near sand]

Can someone help me?

[chrome sable]

Did I solve the first photo correctly? It’s a shadow problem and how do I solve the second problem

This is geometry H chapter 8

[chrome sable]

@violet nest

[forest drift]

any trig function except sine and cosecant

the person who calls others useless is useless

cuz nobody receives hate from people who have done more than them

there are good geometry books out there

[upper karma]

Wrong channel sorry

[sharp flame]

how to answer this? im confuse

"Area and Sector of a Circle"

A. Direction: Find the area of the following circle

a. A= _____________ if r = 10m

[upper karma]

Remember the formula for the Area of a circle?

[sharp flame]

area of sector = θ/360 x πr^2

[silent plank]

0?

if you want a lazy substitute for theta, just use t

[sharp flame]

θ

[silent plank]

you're also overcomplicating this

[sharp flame]

what formula should i use?
area of circle = πr^2?

[silent plank]

yes

you're being asked for the area of a circle so it would make sense to use the formula for the area of a circle

[sharp flame]

so it will gonna be Area of circle = π10m^2

[silent plank]

no

[sharp flame]

xd what im super confuse to this lesson

[upper karma]

It should be, without typing the "m"

[silent plank]

pi10m^2 doesn't correctly represent the area

[sharp flame]

ahh

[silent plank]

as 10m^2 isn't the same as (10m)^2

[upper karma]

the ^2 is only on the radius yes

[silent plank]

the radius is 10m,
and that should be squared properly

[upper karma]

πr² = π(10)² = 100π

[silent plank]

units are still important

units are given for the radius and the relevant unit should be in the final answer

[upper karma]

≈ 314 meters

[sharp flame]

a. A= _____________ if r = 10m

soo my a is 314m?

[silent plank]

no

Area of circle = π10m^2
you can't just blindly replace r with 10m like that
you need to add parentheses to maintain the value of your expression

i.e.
$$A = \pi \cdot \red{(}10m\red{)}^2$$
and simplify that

[somber coyote]

ℝamonov

[lapis adder]

what's cot (-π)

[dark sparrow]

@lapis adder in case you couldn't use a calculator to answer this question yourself,

it's undefined.

[lapis adder]

yeah

[upper karma]

Ye its undefined

Because cot=cos/sin

sin of a full 180 degree rotation becomes 0

Wether if π is negative

we really don't care about cos anymore

As if denominator is 0

Whatever the numerator is

The result is always undefined

[upper karma]

Ok then this one is something hard I guess if I take a cube and note I am saying in 3 dimensions and imagine rotating that cube on the z axis, fine! Makes sense! Now imagine a light veritcally over the cube projecting the cube on a 2D plane if I tell you angle theta of that cube what will be the area of its projection

[upper merlin]

assuming that it is being rotated about the centre this should be useful

[upper karma]

Thanks

And also

The length

Of the cube

Is 2

Ig I finally got it

In that sense

One side is equal to sqrt(1^2-sin2(theta))

The other one is 2 ofc

So it's sqrt(1-sin2(theta)) * 2

[upper karma]

That's nice

[thorn valve]

guys is there a trig identity for cos^2-sin^2-tan^2

[gleaming nova]

that’s not really an identity

[late lion]

does the order matter in naming the polygon

for example, does ABCD imply it's like this

rather than this

[wide pumice]

Go clockwise

I think

That is the way to order it

[late lion]

well it's the same thing either way

[misty knoll]

Nop, it's the first not the second

[wide pumice]

Its neither...

I don't think

You usually order clockwise

[misty knoll]

But people usually plot the point from A to D in the clock way

[wide pumice]

B should be top right then?

[misty knoll]

Yes

[wide pumice]

Then it is neither of those options

[misty knoll]

Yes but the first is technically not wrong, it only not the right direction

The second would be ACDB

[wide pumice]

ADCB

[misty knoll]

The second

[wide pumice]

Bruh

The second is completely wrong

[misty knoll]

Yes

[late lion]

no i mean counter clockwise and clockwise are the same

no?

for all intents and purposes

[misty knoll]

That is not the case

They are not the same

Well you can try to do a demo but good luck

[gleaming nova]

how are they the same

[misty knoll]

Since we are talking about normalization of mathematics and rule that have been put to make mathematics an universal language

Understandable in only one way to avoid confusion and put precision as the most important thing

[late lion]

icic

[low fable]

What do I put in for the equation

[upper karma]

I suppose those 3 are the angles

[low fable]

Yeah x is 5

I don’t know what I’m suppose to say in the equation

[upper karma]

I guess

It's simple

(19x-18) + (7x+1) + (10x-9) = 180

I mean all angles of a triangle sums up to 180 right @low fable

[low fable]

Yeah

Thanks for helping it’s so annoying when teachers don’t explain the instructions

[upper karma]

U can remove the brackets though

[upper karma]

Lmao 👍

[silent plank]

(19x-18) + (7x+1) + (10x-9) = 180
is not correct

[upper karma]

U need to simplify it though

[silent plank]

based on what i see in the diagram, 19x-18 isn't an angle in the triangle

[upper karma]

...

It has

The degree sign

I was confused

Too

[low fable]

19 times 5 -18 is 77

[upper karma]

If it's 180-(19x-18)

Like

If it's the opposite

[low fable]

I’m just confused what you put in for the equation

[upper karma]

Is (19x-18) a angle inside the traingle?

[silent plank]

no it is not

[upper karma]

Then ofc it's angle inside the traingle?

[silent plank]

what?

[upper karma]

Umm

I guess you should see the img

[silent plank]

i see the image, and 19x-18 is not the angle inside the triangle but the exterior angle

if you know the exterior angle theorem you can apply that directly
or go the long way and consider the angle sum on a line and angle sum of a triangle

[low fable]

You do 19x-18 equals 7x plus 1 plus 10x-9

[silent plank]

yes

[upper karma]

Ye

[silent plank]

19x - 18 = 7x + 1 + 10x - 9

and then solve for x

[upper karma]

Is 19x-18 inside or outside the traingle?

[silent plank]

outside...

[low fable]

Outside

I have x

[upper karma]

Then it's right

[silent plank]

as i've said and implied multiple times

[low fable]

I just need the equation

[silent plank]

and what's implied in the image

[upper karma]

To find the answer we need to understand the question though

[upper karma]

And he also is saying the same

[low fable]

We have x?

[silent plank]

you have your equation

19x - 18 = 7x + 1 + 10x - 9
and then solve for x

[low fable]

Yeah

[upper karma]

Ye

[low fable]

Ok thanks m8’s

[upper karma]

That's the equation

[red tapir]

can someone help solve these?

[low fable]

How to solve?

[grave pond]

The triangle is isosceles, which gives you a relation between two of the sides.

It's really just a camouflaged algebra problem.

[ebon thistle]

i need help

[dark oasis]

Use sin(30°)

[upper karma]

Use cosin(60°)

[upper karma]

cotan(30°), I think that will work

[upper karma]

tan(45°)

Then sqrt(7 * 7 + 7 * 7)

which is equal to: 9,899494936611665

[vernal grotto]

bro how am I supposed to find the angle of a sinusoidal function when all I'm given is the value of y, am I suppose to use a calculator? Because that would be easy but it doesn't say that I can use a calculator, and the number isn't on the unit circle at all

[jade kite]

So, my algebra teachers gives me a geometry challenge problem that idk how to do. This may seem easy to you, but plz help me

[finite gull]

What are the correct answer for this

??

[wide pumice]

For which? 4.?

[finite gull]

Both

If you can

Please

[wide pumice]

So the first one uses rigid transformations, meaning those 2 figures are equal. They are congruent figures

You have to use definition of a rhombus to solve this. Opposite angles in a rhombus are congruent. And same side ones add to 180

Or you could notice that only one of the options is obtuse LMAO

[finite gull]

so is it 110

??

[wide pumice]

@finite gull The second one is much easier, your answer was wrong because the lines are clearly not perpendicular, they would have to be marked with a right angle for that to be true. The correct answer is corresponding angles. So option D was correct.

[finite gull]

ahhhhh

[wide pumice]

I sent you a helpful video

[finite gull]

okay i ma watch it

but for number 4 is it 110

[wide pumice]

Yes

[finite gull]

kk thank you

https://tenor.com/view/smiling-dog-i-love-you-i-love-you-i-love-you-i-love-you-dog-gif-22568852

[wide pumice]

https://tenor.com/view/butta-dawg-butter-dog-butter-dog-gif-20600440

[silk charm]

is this a timed assignment/quiz?

[cerulean orbit]

no

[upper karma]

Is there an image that goes with the question?

[royal citrus]

RPQ = POS [alternate angles]
is that right?

[solemn hinge]

Can anyone explain me multiple and sub multiple angles ?

If u can pls do

U can just tag me and write about multiple and sub multiple angles

Or u can recommend me a vid in yt

[mild birch]

sin2(x) + cos2(x) =1
cotx = cosx/sinx

[gleaming nova]

people just don’t know how to use exponents do they

[mild birch]

i cba at times

[gleaming nova]

why yes, sin(2x)+cos(2x) does equal 1

[mild birch]

but you my friend, you’re being pedantic

[upper merlin]

yes

[mild birch]

yeah. and ROP is a straight line so 180. and ROS is an isosceles triangle because OR= OS because they’re radii. base angles in isosceles triangles are equal.

[royal citrus]

ye thanks

[upper karma]

https://media.discordapp.net/attachments/785856938536599552/940434403107143690/20220208_081457.jpg yo can anyone help me with qno 4

<@&286206848099549185>

Help

Qno 4

[unique oyster]

H

Help,

[brave narwhal]

hopefully I am not too late

11pi/8 is what theta is

you can achieve this by taking arccos of both sides

now 11pi/8 is the same angle as adding or subtracting 2pi from it

11pi/8 radians = 27pi/8

[unique oyster]

No, i am listening but i will review it tomorrow thanks for the help. I will take a picture of your explanation. Thanks

日本人ですか?

[brave narwhal]

分かりますけど、日本語は良くないです

ww

でもアメリカ人です

お前？

[unique oyster]

私も日本人じゃありませ。私はベネズエラ人です。少し日本語も分かります。

[brave narwhal]

is there another answer choice?

[unique oyster]

No, i think there is only four answer

[brave narwhal]

hmm

I am getting that none of them are right

[unique oyster]

Answer is C, but i just need simple explanation

[brave narwhal]

well what you would normally do is set all the answer choices equal to the original function

and solve for k

so

-75pi/8 +2pik = 11pi/8

-75pi + 16pik = 11pi

16pik = 86pi

k = 86/16

but that isn't an integer

so idk how that can be the right answer

[unique oyster]

Oh, okay no problem. Helped me a lot. Thank u tho lol

[prisma quartz]

sqrt30=sqrt(3^2+HF^2)

[deft veldt]

Are you familiar with $\tan(x) = \tan(x-180^{\cdot})$

[somber coyote]

azeem321

[solemn hinge]

There might be a chance of getting the answer , if u convert 330° to rad form

330°×π/180

As per my opinion

[near sand]

Is anyone familiar with non-euclidean geometry?

[unique oyster]

Okay, got it. 16pi/8 is the same as 2pi of unit circle. So, 16pi/8 x 4=64pi/8. Then, put it on C. Since, we know the answer of Cos(theta)=-cos(theta). -75pi/8+64pi/8=-11pi/8. So, it is the same as cos(theta)=cos(11pi/8)

[floral coyote]

help please with the rhombus question

[solemn hinge]

Convert it to radian

[solemn hinge]

Like if u mark a point "O" at the centre and apply Pythagoras theorem , you'll get the answer

You'll get the value of BO and multiply it with 2

That's it 👍

[unique oyster]

What radian? Isn't the question in radian form?

[unique oyster]

Maybe this, give it a try

[fallen sleet]

does there exist a 2D shape for which no circle exists such that: shares at least three points with the edge of the shape AND the area of the shape is a subset of the area of the circle

[grave pond]

A straight line segment?

[devout mango]

guys pls help me

u can use translator

nvm

[solemn hinge]

I misunderstood

@unique oyster sry

I was talking about @cerulean orbit 's ques

Sry my bad

[solemn hinge]

Is it cosec^2theta + cosec^4theta = cottheta - cot4theta ???

[loud fulcrum]

how do i graph one period of f(x)=5sin(2x+2)-3

[iron spoke]

okay well

you know how the sine function looks like and how to transform functions, so get the amplitude (which is 5), how much it is displaced vertically (-3 times), how much it is displaced horizontally (-2 times), and get the period, which is 2pi/2 = pi

now graph it with this information

[loud fulcrum]

yeah i figured it out i’m stupid

[loud fulcrum]

thanks tho

[unique oyster]

No, problem my brother. You got my approve.

[upper karma]

who can help me

[solemn hinge]

Lol

[upper karma]

What do you need help with

plsssss

[upper karma]

They are asking if line segment AC bisects the angle BCD bi means two so does line segment AC give two parts of angle BCD

[lone sandal]

Help

[snow kite]

Hey everyone, I'm not sure why this formula works to find the area of a triangle?

I understand the ones where you do 0.5ba*sin(phi)

Because the asin(phi) is supposed to represent the h. Which brings you back to the original formula of 0.5b*h

[grave pond]

Isn't your formula the same, just using different names for the sides?

Two sides of the triangle, times the sine of the angle between those two sides, divided by two.

[deft veldt]

The text states that [BGD]=[CGD] because the triangles share an altitute from G. But this line is not an altitute, it's a median right? So how do we know that it's necessarily perpendicular to BC?

[dark sparrow]

they're not talking about the line GD

they are talking about a perpendicular dropped from G onto the line that contains BD or CD (it is the same line for them both, hence the same perpendicular)

[deft veldt]

ohhh my bad. thanks

[royal citrus]

Here, can we say that $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ and $ \overrightarrow{BD} = \frac{1}{2} (\overrightarrow{AB} + \overrightarrow{BC})$ ?????

[somber coyote]

Inheritanc-e ♦

[whole bone]

From a vector perspective ab + bc = ac because its tail to head + tail to head if that helps

[loud fulcrum]

how do i find the period of 7tan(5x+10)

[fervent ember]

I made a guide covering perpendicular bisectors, medians, altitudes, and angle bisectors. Can anyone tell me if I am missing anything?

PERPENDICULAR BISECTOR
Perpendicular to the side
Does not touch opposing vertex
At the midpoint of the side
All perpendicular bisectors meet at the circumcenter

MEDIAN
Not perpendicular to the side
Touches opposing vertex
At the midpoint of the side
All medians meet at the centroid
The centroid is 2/3 of the distance from each vertex to the midpoint of the opposite side. Another way of stating this is that the distance from the opposing vertex to the centroid is twice the distance from the side to the centroid.

ALTITUDE
Perpendicular to the side
Touches opposing vertex
Not at the midpoint of the side
All altitudes meet at the orthocenter

ANGLE BISECTORS
Divides an angle in half
Perpendicular to all sides in equilateral triangles
Does not divide opposing side in half (except if perpendicular, that is in equilateral triangles and for the unequal angle in isosceles triangles)
Touches vertex
Lengths along the angle bisectors are equal
All angle bisectors meet at the incenter

[strange reef]

I constructed a triangle with side lengths 3, 3, and 5.9. I want to find the angle between the two sides of equal lengths. To do this, I used the special case of $a^2=b^2+c^2-2ab\textrm{cos}\theta$ where $a=b$ to get $-(\frac{l^2}{2s^2}+1)=\textrm{cos}\theta$, where $l$ and $s$ are the long and short sides. However, plugging in the side lengths, this got me $-2.93...=\textrm{cos}\theta$, which shouldn't be possible. What's wrong with my thinking?

[somber coyote]

person2709505

[strange reef]

Nevermind, solved for $\cos\theta$ wrong

[somber coyote]

person2709505

[near sand]

Can someone explain which do I use to solve this problem? Scalene inequality or triangle inequality?

[gleaming nova]

,

[near sand]

Oops

,rotate

[near sand]

Thank you

I hardly use this bot

We haven't proved that all angles equal 180, but we have proved that it cannot exceed 180

So the third angle can be 5 or less

[gleaming nova]

ok then i can’t help lol sorry

[gleaming nova]

i wasn’t aware that this could be a condition

[near sand]

I'm just confused about the inequality theorems

[gleaming nova]

oh wait

i can help

so it’s less than or equal to 5

[near sand]

Yeah it's called Saccheri-Legendre Theorem

[gleaming nova]

are you allowed to say that, for example, the side opposite to the second largest angle is the second longest side?

[near sand]

Which side?

[gleaming nova]

side RS is opposite to the second largest angle

[near sand]

That would prove it?

Cuz opposite of the largest angle is ST

Then that leaves RT being the smallest

[gleaming nova]

yeah

[near sand]

Ohh

[gleaming nova]

and you already know that the last angle is the smallest

no matter what

since that it must be less than or equal to 5 degrees

and the other two angles are greater than 5 degrees

[near sand]

Ohh okay thank you

[fallen sleet]

how to prove "is parallel to" is an equivalence relation?

[dark sparrow]

use the definition

[grave pond]

What is your definition of "is parallel to", though?

[fallen sleet]

Does this work: A line is parallel to another line if and only if: they are in the same plane and share a point if they share all points.

Transitivity seems the hardest to prove here because you need to break it down into 5 cases.

[lusty abyss]

Help - no 12

[dark sparrow]

you have a system of three equations in three variables: $$\begin{cases}x_A + x_B = 2 \ x_A + x_C = -8 \ x_B + x_C = 4 \end{cases}$$

[somber coyote]

Ann

[dark sparrow]

(you also have a similar system for the y coordinates but we can come back to those later)

this is a system of linear equations.

do you know how to solve systems of linear equations?

@lusty abyss

[grave pond]

Sounds good. That makes reflexivity and symmetry trivial. The claim that this relation is transitive fails in the hyperbolic plane, so it is actually equivalent to the parallel postulate (and is sometime stated as an alternative formulation of it).

[fallen sleet]

very interesting information! thank you!!

[lusty abyss]

We do simultaneous equations?

Confused

[untold forge]

Yeah, simultaneous equations are also called systems of linear equations

[dark sparrow]

i don't know what you mean by "do simultaneous equations"

[lusty abyss]

But here are 3 unknowns, how does simultaneous work here?

[untold forge]

When I first learnt systems of linear equations with just 2 equations they were taught as simultaneous equations, back in middle school

[lusty abyss]

What will be the next steps?

[dark sparrow]

well, one possible next step would be to add together all three equations that i mentioned

you will get $2(x_A+x_B+x_C) = -2$

[somber coyote]

Ann

[dark sparrow]

though of course before you do that you will have to understand that "simultaneous equations" does not refer only to systems of 2 equations in 2 unknowns.

[lusty abyss]

Is the x-coordinate of B correct?

[dark sparrow]

,rccw

[dark sparrow]

your handwriting sure does leave a lot to be desired

your x's barely look like x's

they look more like fucked up n or u

[lusty abyss]

ok

[dark sparrow]

and disregarding that, this line: $$2 - x_B + x_C = -8 - 2 + x_B = 6 + x_B$$ is somewhat unclear

[somber coyote]

Ann

[dark sparrow]

and you have a couple of arithmetic mistakes further down...

[median berry]

I need some help with vectors

[lusty abyss]

I think I’m failing. Fuck

[dark sparrow]

you don't have to isolate one of the variables in one of the equations, by the way.

i can tell you tried to do that.

[lusty abyss]

Is the x-coordinate of B correct now?

[dark sparrow]

let me take a look

[lusty abyss]

My x is still fucked up n, please stay patient

Thanks

[gleaming nova]

you can just write slowly if that helps

[dark sparrow]

ok yes this seems ok now

let me doublecheck

ok yeah checks out

now you can find the x coords of the other two points and do the same thing for the y coords

[lusty abyss]

Gotcha!

Tysm

[fallen sleet]

also just realized its supposed to be "share all points if they share a point"

[grave pond]

Yeah -- I think I instinctively read it as "iff".

For synthetic reasoning in plane geometry I think you could get through without too much pain if you start by proving that lines that have a common perpendicular are parallel, and that two parallel lines share all their perpendiculars.

[ancient elm]

hello can i have a quick question about euclid third postulate?

[grave pond]

https://dontasktoask.com/

[ancient elm]

what is the different of using any center only from the with any centre and any radius ? in the 3rd postulate. im in confuse with the term using and from with any centre

[tender shell]

Hey guys I keep getting confused with sine and cosine (of degrees). Is there anything I can really do to try to remember or should I just keep practicing?

[grave pond]

Are you saying you forget which of them is which?

[tender shell]

yeah i forget which one is opp. over hyp. and adj. over hyp.

[grave pond]

The silly mnemonic train-of-thought I've always used (which may or may not work for you, of course) is something like

when we say cosine we start earlier in the word than when we say just sine, therefore the cosine should come first in the pair (cos a, sin a), so therefore the cosine is the x-coordinate and the sine is the y-coordinate of the point on the unit circle that represents the angle.
You'd need to remember separately that the angle is measured counter-counterclockwise from "due right", though.

Even better, make up your own mnemonic device. The sillier the better, really, as long as it's memorable.

[tender shell]

oh okay thanks man

[silent plank]

i suppose you could start with the mnemonic that most people use
soh cah toa

[tender shell]

ohhhh yeah I've heard that one before

thx

[grave pond]

That one is a mystery to me, on the other hand. Why would it be easier to remember that particular nonsense word than e.g soh-coa-toe?

The only way I would be able to remember that would be by actually knowing how the trig functions work, then reconstruct it letter for letter by imagining little right triangles in my unit circle.

[silent plank]

cah toa not coa toe

its sufficient for basic right triangle trig

ofc later on you should learn/know the unit circle definitions

[grave pond]

My question was, how is that nonsense word supposed to be more memorable than the actual underlying math?

It's just a completely random-looking string of letters. That's the opposite of memorable.

[silent plank]

its the mnemonic describing the right triangle definitions of the basic trig functions

[grave pond]

I know what it describes.

I'm asking why is it a "mnemonic"?

[silent plank]

a whole lotta stuff reduced to 9 letters

[grave pond]

The point of a mnemomic is to help remembering something, and if the only way to remember the mnemonic itself is to already know the thing it was supposed to remind of, then it's a really shitty mnemonic.

[tender shell]

well i dont really know about the unit circle definitions yet since im in 8th grade so just a phonetic string of letters to remember which function does what. obviously I'll learn the underlaying math, but right now its a good mnemonic to know what function does what.

[grave pond]

But those nine letters would be exactly the same when given in a different order that doesn't describe the right thing.

[snow mesa]

i am also in 8th grade 🕶

[grave pond]

There's nothing in that particular random order that makes it easier to remember than a different random order that results in the wrong conclusion.

[silent plank]

evidently its common enough that a vast majority people use it and people recognise it at a shear mention of it

[grave pond]

It is not a functioning help to memory at all.

[silent plank]

i mean you can make up your own way to memorise if you want

[grave pond]

I do, as noted above.

[silent plank]

whatever floats your boat

[grave pond]

What I'm saying is that we shouldn't ask other people to use a shitty nonsense word that will not actually help memorizing anyting and tell them to spend effort on memorizing the random nonsense when that is not actually any easier than remembering something meaningful instead.

[silent plank]

whether the mnemonic is useful to them depends on them

i mean i could type out a whole paragraph stating what sin,cos,tan do in a right triangle

but how would you condense that info

and would they need to recall that paragraph verbatim every time they want to do a right triangle trig problem

[snow mesa]

guys i think soa cah toa is pretty useful personally. I’m in 8th grade and not there yet but i understand it to some level due to that

[tender shell]

yeah thats why i think I'll use soh cah toa for the time being

[grave pond]

"Condense" seems to me to be the opposite of what you'd want for making something memorable.

You'll want something that paints a vivid picture in your mind, not random syllables that could just as well have been other random syllables.

[tender shell]

I see what you're saying, but it is simpler to understand. I will definitely learn the "uncondensed version" but for right now soh cah toa makes more sense to me

[snow mesa]

precisely 🕶

[grave pond]

I completely fail to see how a nonsense string of letters is "easier to understand" that anything. There's nothing even TO understand there. There's no meaning at all, except the meaning you supposedly don't have present in your mind at the time you'd need a mnemonic trick for it.

[snow mesa]

soh (sin=opp./hyp) cah (cosin=adj./hyp) toa (tan=opp/adj)

[grave pond]

If you know that you don't need a mnemonic in the first place!

[snow mesa]

i know it because of the little trick

so i do need it

obviously

[tender shell]

That's what I was struggling on

[grave pond]

The premise of needing a mnemonic is that you don't remember the thing it is supposed to help you remember.

[tender shell]

I kept getting sin and cos confused

[silent plank]

well the knowledge should be there, mnemonics help you remember

[snow mesa]

it helped me remember now that i know becaus it helped me doesn’t revert it back to not a mnemonic

[silent plank]

and if soh-cah-toa works for whoever wants to use it, there's no reason to complain

[tender shell]

well yeah it does help me remember the difference now

[snow mesa]

same

[grave pond]

If the way you say you remember what the right nonsense word is, is "because that is the particular sequence of nine letters that gives the right result", then it hasn't actually helped you remember anything. You need to know the answer before you can reconstruct the word the will tell you what the answer is.

[snow mesa]

we know the construction because it may have been taught to us. By no means are we saying we created it

[grave pond]

Mnemonics only help remembering anything if the mnemonic is actually easier to remember than the thing it's supposed to help with.

[tender shell]

ohhhhh yeah I see what you're saying. But the thing is my school hasn't taught me how to construct and reconstruct stuff to get these functions besides showing us triangles

So I dont know how to break it down to these functions. As you said the information is just not in my brain

[grave pond]

And then how can you ever hope to remember whether it is soh or coh, either?

[silent plank]

the sounds when pronouncing them are also quite distinct

[tender shell]

I was just confused on which one was adjacent over hypotenuse and what was opposite over hypotenuse

[grave pond]

Exactly. And no matter which one it is, there will be a nonsense word describing it. So remembering "there is some nonsense word made up of these letters that gives the right correspondence" is actually useless as a way to remember what the correspondence was.

[tender shell]

So I wanted a way to remember it for the time being until I learn the right correspondence without needing these, "nonsense words"

[grave pond]

I mean, it's your funeral.

[tender shell]

And I'll be playing the coffin dance music

[silent plank]

i mean doing rotations and reflections and dilations for a basic right triangle problem and strictly applying unit circle definitions would be complete overkill

[grave pond]

Look, if you want a mnemonic device that's not based on the unit circle, you have my full blessing for that. Just come up with one that is ACTUALLY MEMORABLE instead of being based on a nonsense word that could just as well have been the opposite nonsense word, and therefore doesn't help with remembering anything, in any capacity, ever.

[snow mesa]

ok buddy

[grave pond]

"Actually memorable" means that if you remember it wrong you would automatically think "wait a minute, the story makes no sense this way around", no matter whether the story itself has nothing to do with the mathematics.

[silent plank]

i mean you could also use a phrase if you wanted

[tender shell]

Yeah that makes more sense than some letters. Like it makes more sense to logic your way into the answer instead of remembering something. That being said Soh Cah Toa is pretty easy to remember.

[silent plank]

there also may be a little bias if you've done trig enough that you don't require the use of that mnemonic

but yeh, use whatever you're most comfortable with and find most helpful

[tender shell]

okay thanks guys

[upper blaze]

Guys i have a fast inversio question thats is bugging me

If i have triangle and one of its vertacies is center of inversion will it map onto itself and not move at all or it maps into point at infinity even thpugh that concept isnt supported in euclidian geometry?

By definiion it should stay there but by intuition and knowing other geomezrys it makes sense to go to infinity

[dark sparrow]

what definition are you looking at that is making you conclude the center of inversion stays where it is?

@upper blaze

[upper blaze]

Hmm actualy I am tutoring one student and since I didnt have inversion on that subject few years ago I used definition they are provided with. And it stats "Given point O in plane A and real number k inversion is a map I: A/{O} -> A ..." and later it explains that when point M gets closet to center its image should go to infinity but it states also as I said that we dont do that infinity becouse its Euclidian geometry
@dark sparrow

[dark sparrow]

did you mean \ and not /?

if so, the definition explicitly says that O isn't in the domain of the inversion at all

inversion acts on all points EXCEPT the center

its action on the center is left undefined

[upper blaze]

Set difference xD And yes I thought the same but then comes the problem of Finding image of triangle that has one point being center of inversion. And I cant decide do I just leave it there or what? xD

Its image should not contain that point or it should thats my problem how to precisely define its image

in any neighbourhood of that point

[dark sparrow]

your messages are hard to read

[upper blaze]

😦 Okay 😦

[dark sparrow]

...

[upper blaze]

I will try to make it short and clear

[dark sparrow]

perhaps you could just ignore the one vertex that happens to be the center.

and invert everything else

[upper blaze]

Well I did that

[dark sparrow]

you should get an unbounded shape

enclosed by... i think it would be two rays and a circular arc

with the third vertex 'at infinity' except we don't actually plot it

[upper blaze]

Yes exactly. Thank you 😄

[rich parcel]

can I write cos(X-Y)= cos(X)-cos(Y) or can u not do that <@&286206848099549185>

[broken dock]

no

[serene vale]

no

[upper karma]

But is there a way you can do it?

[rich parcel]

idk

[serene vale]

$\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$

[somber coyote]

winston gergill

[upper karma]

Huh intresting

[rich parcel]

what about cos^2(x-y)

[upper karma]

Well.cant you just use the equilence aboce again

Sorry that was accidnetal

[serene vale]

$\cos^2(a-b) = (\cos(a)\cos(b) + \sin(a)\sin(b))^2$

[somber coyote]

winston gergill

[serene vale]

i dont have the energy to expand that/abuse mathematica lo siento

[rich parcel]

so it is not (cos a - cos b) (cos a - cos b)

[upper karma]

Juat assume it equals z and continue with your problem.

[serene vale]

yea $(\cos(a) - \cos(b)) (\cos(a) - \cos(b)) = \cos^2(a) - 2\cos(a)\cos(b) + \cos^2(b)$

[somber coyote]

winston gergill

[serene vale]

big headache

[young zephyr]

Hey

Anyone on

[wet violet]

I was solving some questions for fun
I need to get the yellow angle
Got all possible angles normally
Its not a cyclic quad
What is steps to get yellow angle
After getting all normal angles and that green is 150

[silent plank]

are you told whether that line is bisecting the green angle?

[wet violet]

Nope
Would be ended by now

[silent plank]

then unless you make that assumption,
the size of the yellow angle cant be determiend

[wet violet]

I constructed alot of things
But at a point
It seemed useless
Like my proof went that a clearly acute angle is 155

[wet violet]

So there is no other way?

[silent plank]

nope

without additional info/restrictions the top right point isn't fixed

[wet violet]

Ok ty
Gonna assume the problem is in given info
And that the line bisects green angle

[silk patio]

It’s not one green angle it’s two green angles

[silent plank]

they really need to do a better job at labeling

[sharp flame]

what is the formula for finding 3 geometric means of the sequence?

[stable forge]

F = kx

[upper karma]

what

[upper karma]

nth root when you multiply n numbers

[near sand]

Could someone give me an example of an isosceles triangle in taxicab geometry?

Just an example

[vagrant ore]

How do you solve this?

[signal spade]

@vagrant ore the answer is a=24.05cm you can tip me for more solution

[vagrant ore]

Oh ok thanks

[signal spade]

@near sand an isosceles triangle is a triangle with all the three sides and angles not equal

[near sand]

I know that lol I just needed an example in taxicab geometry but I got it after all

Thank you tho

[signal spade]

Ooh okay 👍

[vagrant ore]

In the law of sines do all angles in a triangle still add up to 180°?

[dark sparrow]

why wouldn't they

it's not like you suddenly stopped working in Euclidean geometry

[vagrant ore]

idk I'm running on 3 hours of sleep

Just asking so I can try and figure out the law of sines better

[dark sparrow]

the law of sines does not cancel whatever theorems came before it

[little terrace]

does anyone know how to do this¿

[upper karma]

hello can i get help, i've been stuck at this for 30 mins tt

[upper blaze]

Hmmm if I am correct and i am sure I am there is a mistake there. From $x^2+x=5x$ you get $x^2-4x=0$ so $x=0 or x=4$. Becouse US is lenght and must be positive you have $x=4 $and $US=40$. Also you get $HO=OU=14$ so$ HU=28$. So we have here P=14*40 which is 560. And you are correct. Anyone see something that I miss?

[somber coyote]

MotionMath

[glossy atlas]

No that's not true

[grave pond]

The sine function is not a logarithm.

[sly venture]

please

[grave pond]

Use similar triangles to find KM, then Pythagoras.

(All the three triangles in this diagram are similar).

[late hemlock]

the angle of elevation when you look at the top of the tower, which is 67° and the angle of depression when you look at the bottom of the tower, which is 45°. Using a tape measure, you went out and determined that the distance between your school and the tower is 80 feet. How tall is the tower?

Im confused as to why Elevation and depression are both different angles, I thought they were equal?

What should I do if the given are both given?

[grave pond]

Apparent you're watching from a point somewhat higher than the base of the tower.

So the angle of elevation is how much above horizontal you see the top, and the angle of depression is how much below horizontal you see the bottom.

[late hemlock]

Then what ways can I solve it? if it's ok can I get the formula?

[crystal quail]

So I'm working on this problem, full thing posted in #multivariable-calculus. I've gotten to a part that might just be geometry but I'm failing to see how it all comes together. In the picture posted above I have some vectors, r, r', and r-r'. And in my full problem, I have to find some way to evaluate sin(|r-r'|). If it's possible through geometric methods that's best, and I have the answer I'm expecting to probably see, but I don't know how to get there.

is this possible to show with vector geometry or should I keep searching for another way to solve my dilemma?

[grave pond]

Have you drawn a diagram? If not, that's always the first step.

[late hemlock]

I got it now thanks ty for the help

[late lion]

idk if this goes in geometry but

how come the volume of a parallelepiped determined by vectors a, b, and c is

$\Vec{a} \cross \Vec{b} \cdot \Vec{c}$

[somber coyote]

Leon

[late lion]

like, if it's base times height wouldnt it be

$(\Vec{a} \cross \Vec{b})|\Vec{c}|$

[somber coyote]

Leon

[crystal quail]

it the height was 90 degrees to the surface of axb then yes

but sometimes it's not

[late lion]

oop

u right

[grave pond]

That would give you a vector.

[late lion]

my bad I forgot to put || on a x b

okay but I don't see why $\Vec{a} \cross \Vec{b} \cdot \Vec{c}$ can give you the volume

[somber coyote]

Leon

[crystal quail]

my informal, non-rigorous answer is that axb gives you the area formed by a and b, and the "dot c" part tells you how much of that area can be projected (hence the dot product) into that 3rd dimension, and is entirely dependent on the angle. If the angle between axb and c was 90 degrees, then you're looking at a flat surface (no volume). If axb is parallel to c then the whole area of axb can get projected into the 3rd dimension and you just get the magnitude of what you wrote earlier, the axb*|c|

for a more formal answer you'd have to wait for one of the math guys to get to you

[late lion]

oh wait nevermind

the height is not |c|

its |c|cos(theta)

[crystal quail]

yep

[late lion]

which is why u use dot product

icic

[wide dragon]

I dont know what my next step should be. Absolutely confused cuz the most ive learnt was Tan (90 - Theta) = Cot Theta

[wintry tundra]

You can transform the tan function into cot

By subtracting the whole inside from 90°

[wide dragon]

So turn it into cot (3x + 30)’ = Cot (90 - 4)?

[serene widget]

yo just starting trigonomatery what do the symbols mean such as tan, cos, etc.?

[wintry tundra]

90-(x-4)

[wintry tundra]

Those are periodic functions

[serene widget]

?

[wintry tundra]

They have to do with side ratios of a triangle as you increase the reference anglr

Which will make more sense as u study trigonometry

[serene widget]

ok

math is in minecraft

[wide dragon]

Why does that work? How come I can just get the reciprocal of tan just cuz I did -90

[wintry tundra]

Bc of the periodicity of the functions

And their symmetry

[wide dragon]

Idk what that means 😩

[wintry tundra]

Trig functions are periodic

[wide dragon]

Periodic means repeats itself

[wintry tundra]

Yeah

Which all the trig functions do

[late hemlock]

wait what "(80) tan 45" is equal to 80ft?

I thought it it was 129 smth

im so confused

[grave pond]

tan(45°)=1

[sand verge]

Hi, could someone explain to me why the answer to question 3. shall be (d)

[upper karma]

Im stuck on a question, it is : A Electric engine that uses 750W active effect (P), has the power factor cosφ = 0,89, the voltage is 230V

What they want to find out with this information is: How large is the apparent effect?
I need to find apparent effect which is S but I dont know what formula to use to find it.

[near sand]

Does anyone know how to construct a midpoint on a segment? Like proving it?

[grave pond]

Generally, draw circles around each endpoint with the line segment as the radius; connect the two points where the circles intersect each other.

[near sand]

Not in that way, sorry I wasn't too specific

I meant where we are given a segment and we use the ruler postulate

[grave pond]

What is "the ruler postulate"?

[near sand]

Lemme send a picture

[grave pond]

Ah, then you're just after $\frac{a+b}{2}$?

[somber coyote]

Troposphere

[near sand]

I'll show what I got

So far

I have to prove that using this point that has coordinate q/2 is the midpoint and then prove that PM=MQ

[grave pond]

Isn't that as simple as noting that |q/2-0|=|q/2-q|, as by part 4 of your postulate?

[near sand]

Ah true

I guess I was overthinking this part

[quasi galleon]

can someone help me with this? ive been stuck at it for like 10 mins by now

[craggy river]

Start with the coordinates of A and B

Since it's linear, you can do the ratio bit componentwise, so just the x then just the y then combine them

Find the difference in the x positions, split that into (7+2) sections, then find the x value with 2 sections on the left and 7 on the right (which will be the same x value as one of the splits you just made)

Do the same for y

Then it's finished

[quasi galleon]

thanks!

[craggy river]

I got ||(-32/9,8/3)||

[vagrant ore]

Does anyone know the cos(30)?

[craggy river]

Degrees?

sqrt(3)/2

If that's 30 radians then I have no idea

[vagrant ore]

Ya 30°

I need it in decimal form

[quasi galleon]

do u have a calculator to find the cos?

[craggy river]

There u go

[vagrant ore]

Oh thanks

[sly drift]

hi guys

how do i get no.1 here?

[vale valley]

can someone help me w this? Im not sure how to solve it at all

[upper karma]

rip imma have to do this wierd ass shit next year

9th grade finna suck

[vale valley]

yeahh

it does

[upper karma]

parents gonna be on my ass and shit

"get good grades the colleges will see this"

[vale valley]

yup

[upper karma]

they be mad at me for getting c's

[vale valley]

dont expect highschool to be like the movies

it kind of sucks

[upper karma]

lol

i mean

it seems easy

as

u dont act stupid

and do ur hw

[vale valley]

now im on prozac

[vale valley]

theres so much of it

[upper karma]

what do u take in senior year

[vale valley]

whar

what

[upper karma]

yea im getting hw every night in 8th grade already

[upper karma]

what math u take in senior year

[vale valley]

dunno

calc

[upper karma]

yea same

i just cant wait to do that!

[vale valley]

i did algebra in 8th grade

imight do alg 2 as a summer course though

then id be another year ahead

i just suck at geometry

[vale valley]

same!

[upper karma]

tf is this shit

[pure void]

hi

[vale valley]

hi

[pure void]

can someone help me

pls

[upper karma]

i gotta take that

[vale valley]

uh

[pure void]

im in geometry

[vale valley]

w what

oh god

[upper karma]

sure if i can

[vale valley]

same

im bad

at it

[vale valley]

biology is nice ngl

[upper karma]

ok i cant

im dumb

[vale valley]

uh no clue im sorry

[upper karma]

and im in 8th grade

[pure void]

oh

im in 9th

[upper karma]

lmao

[vale valley]

same

[upper karma]

imma have to do that next year

[vale valley]

I am not passing geometry

so

[vale valley]

wait can u tell me if u know hiw to do smth in geometr

do you know how to do this

[pure void]

uh hm

[vale valley]

literally nobody has helped me in the help channels and its been hours

so

[upper karma]

;-;

rip

just ping helpers ig

[vale valley]

i did

[upper karma]

until one of them respons

[vale valley]

three times

[upper karma]

oh

;-;

[vale valley]

sobbing crying throwing up

[upper karma]

xd

i gotta take one of these each year

starting 9th grade

prob worse than math

[vale valley]

chem?

[upper karma]

yep

[vale valley]

chem isnt for freshman

unless u did bio this year

[upper karma]

yea

nah

i think

[vale valley]

physics is also a more advanced course i think

[upper karma]

we start out with advanced science

then bio

then chem

[vale valley]

ah

[upper karma]

then physics

[vale valley]

i did advanced science in 8th

so im in bio this year

[upper karma]

i did life science last year and earth science this year

[vale valley]

next year is space and or chem

[vale valley]

oh i did those last year

[upper karma]

this is for my 9th grade next year

[vale valley]

BRO

ONLY 30 HOURS??

I HAVE TO HAVE

150

[upper karma]

;-;

[vale valley]

BY SENIOR YEAR

[upper karma]

bruh

[upper karma]

idk there was a kid with like 1000 hours

once

at my school

gtg

gb

gn

[vale valley]

gbye gn

[deft veldt]

Just learning about prisms

From the information given, are we able to deduce that the other faces will be rectangles?

[buoyant palm]

4 rectangles and 2 squares, so you follow the general rules of each shape when answering those questions

[deft veldt]

all faces are rectangles

[buoyant palm]

That doesn't look like it from the shape on the right

Weird

[deft veldt]

ik, that's kind of my question. They state that all the faces will be rectangles, but why? A prism can have any kind of parallelogram for sides.

Since it's a rectangalur prism the bases

will definitely be rectangles

but the sides don't have to be

[buoyant palm]

That's a really odd diagram if they want to argue that the faces are all rectangles

Because the way I saw it, you'd effectively have the length of those square faces 1 to 1 and then the height for each of the rectangles

And there you go

You have all dimensions

[prisma glade]

what is the midline in a right trapezoid if we know that the inscribed circle divides the larger leg by m and n

answer is (sqrt(m) + sqrt(n))^2 idk how to prove it

oooh

i solved it

it was so easy bruh

[upper karma]

I have question about Jordan curves.

If I choose a point on a Jordan curve, can I always obtain an isosceles triangle with the vertex where the two equal sides are being the point, with any angle theta , 0<=theta<=180* ?

My intuition is that one can take a circle originating from the given point.
Start with r=0. This will give the 180* case as a Jordan curve can be approximated as a line at sufficiently small scales.

As the radius increases and we reach the radius where any increase will result in no intersection with the curve, we could generally have a few cases. In the simplest we get one point of intersection. In that case the triangles go from theta=180* to theta = 0* smoothly as the intersection of the circle and the curve move from the given point to the furthest point.

(Thus giving the desired result
aside from a few exotic curves I can come up with, but I am not sure they are Jordan curves, so I am satisfied with this reasoning)

We can get more than 2 intersection points at some point. I resolve this by requiring that we keep track of the original 2 intersection points that we had at r=0. In this case we'd also have to move the radius up and down as we decrease the theta

[dark sparrow]

you want your triangle to have all three vertices on your jordan curve, right

[upper karma]

yes

I also would like to model this problem. Unfortunately though geogebra does not support intersections with parametric curves 😦

[grave pond]

a Jordan curve can be approximated as a line at sufficiently small scale
Nobody guarantees this -- Jordan curves can be wild and fractal everywhere. (Just wrap a Weierstrass function around a circle, for example).

[upper karma]

I thought so

I am sorry, I just did not know exactly the definition of a Jordan curve

I was inspired to ask this question by the inscribed square conjecture

I thought it was about Jordan curves, but if all these exotic curves can be considered Jordan curves then I don't understand how the proof was at all successful for the inscribed rectangle theorem

[grave pond]

Usually a Jordan curve is a function from [0,1] into the plane which is continuous, and injective except that f(0)=f(1).

[upper karma]

Okay, I looked closer. The inscribed rect theorem concerns itself with simple closed curves

[grave pond]

"Simple closed curve" is usually the same as "Jordan curve".

[upper karma]

Thank you

But in any case, do you know a good software/site where I can model this?

with "well behaved" parametric curves?

[grave pond]

Sorry, no.

[dry hemlock]

hello

can i get some help?

[gleaming nova]

only if you ask your actual question

[mild birch]

sure. just use one of the free help channels

[dry hemlock]

could i message you privately rylo?

[gleaming nova]

there’s no need to

just ask your question

you’re wasting time already

[hybrid yoke]

Who has junior high school geometry notes?

Were just starting at geometry

[grave pond]

Don't you get a textbook?

[hybrid yoke]

Nop

[weary drift]

@edgy silo dont shitpost here

[edgy silo]

sad

[upper karma]

Just a quick one, can someone jog my memory: is the height of a right-angled isoceles = to the hypotenuse/2?

[edgy silo]

that makes sense ye

[humble pulsar]

If the leg is L, then the hypotenuse is sqrt(2)L

[upper karma]

Can you tell me where I'm going wrong in this calculation? I feel like it's where I square the hypotenuse.

[humble pulsar]

what is that suppose to be doing...?

but you have 0=0, so nothing's wrong

[upper karma]

IPythagorean theorem on top.

X would be a leg of the isoceles triangle.

sqrt(2)x is the hypotenuse.

[humble pulsar]

yeah...

you've written pointless math though, like it doesn't do anything useful

[upper karma]

I guess I'm trying to solve for x.

[grave pond]

The triangle can have any size.

So the equation you know doesn't tell you anything about what x can be

[upper karma]

right

[humble pulsar]

any right isoceles will have that equation hold

provided the length of the leg is x

[maiden totem]

uhh do I just like send da thing here '='

[upper karma]

I have no real reason to say anything but I know how to find x and the other side of the line if you'd like me to show you ( although I won't be much help)

[upper karma]

Help

[little sedge]

how can i get the 2 and 11? how do you get the measure of those lines?

[upper karma]

if you can help me please dm me

[next stag]

Can someone help me with these

[merry onyx]

At the second problem you know the area and the base of the triangle. You can use the formula:BC(the base of the triangle)*height/2=area

I am sorry,but I can't help you with the other problems,I haven't worked with these for years

[craggy river]

For problem 1, there's a square, and there's a circle

Finding the area of each individually shouldn't be that hard, as long as you're careful to get the radius of the circle right

Since the shaded region is everything in the square and not the circle (and technically because of the way they overlap), you can just subtract the circle from the square

Kori got problem 2 already

For problem 3, you'll use that same formula, but instead of finding the height, you'll use it to find the area. Do that for the two triangles facing you, then double both of those to account for the triangles on the back, because they're the same anyways

And the base is a rectangle with those given side lengths

[hearty cape]

I'm not actually sure if this should be in this channel, not sure what channel it should be in but

Is there any easy way to find lots of exponential equations between two points?

I'm looking to make a scaling cost for something and would like to be able to adjust exactly how exponential it is, while still falling between the two points

if that makes sense

[fallen sleet]

how in the world do you calculate segment GB without the use of trigonometric functions???

[grave pond]

Is AGD, CGF, etc? supposed to be collinear?

[upper karma]

We just learned about dilation vectors in class today, are these the type of vectors I’m thinking of?

Like there are so many things with the name of vectors is so confusing

Vector-spaces, vector dilations and vectors with magnitude and direction.

It’s confusing, can I get some light shed on my questions and confusion?

[cerulean socket]

that is the orthocenter i beileve

[cerulean socket]

it has similar triangles in it

[nocturne remnant]

That does not really look like orthocenter

BGE looks bent

[cerulean socket]

oh yeah

i thought they were straight

my bad

i dont think so

[nocturne remnant]

Yeh mb

[near sand]

Can someone help me finish this problem?

I'm trying to use the tangent theorem to solve it's corollary

Trying to prove PA congruent to PB and PO bisects angle APB

[nocturne remnant]

Try to prove that PAO and PBO are congruent

[near sand]

The triangles?

[nocturne remnant]

Ye

[near sand]

Well we know OA is congruent to OB

And PO is congruent to PO by reflexive

[nocturne remnant]

Yep!

[near sand]

Not sure after that

[nocturne remnant]

Check the angles

[near sand]

Can angle APO be congruent to angle BPO?

[nocturne remnant]

That’s what we’re trying to find so

Not that useful

[near sand]

Okay

I'm not sure

[nocturne remnant]

What is special about the tangent line of a circle

[near sand]

A tangent line to a circle can be perpendicular to a line from the center?

[nocturne remnant]

Ye

It’s always perpendicular to the radius which passes through the point of tangency

[near sand]

Okay

So because it's perpendicular

Is angle PAO and PBO 90?

[nocturne remnant]

Yes, exactly

[near sand]

So they're congruent

So by SsA, those two triangles are congruent

Thus angle APO is congruent to BPO

[nocturne remnant]

Ye
But normally we can use SSA only if the A is also 90 degrees

Which it is in this case

[near sand]

Ah

By angle bisector, PO bisects angle APB

Ooo

[near sand]

I think I'm going in the right direction but I'm not so sure. I stated KM is congruent to KN because they're equidistant. Same with MT and NT. Then I said KT is congruent to KT by reflexive. By SSS, Triangle MKT and triangle NKT are congruent

,rotate

[near sand]

My hint was to use the perpendicular bisector theorem

[upper karma]

195134404081940721644646182469

2 to the power of what

[grave pond]

Roughly 97.3?

[upper karma]

ok

[tall fractal]

i dont get this

i have to find the value of x and y

T-T

[wintry tundra]

i mean

y=x

[cerulean socket]

you need help?

M is

38

[tall fractal]

yes

[cerulean socket]

360-2(68)

divide 2

wait

let me check

[wintry tundra]

oh yeah bc AM is parallel to GE bc its a rhombus

[tall fractal]

how can you answer so fast i dont eve know what to do :')

[cerulean socket]

its just 224 right?

is it x plus y?

[wintry tundra]

well a rhombus is a parallelogram

so all the opposite sides are parallel to each other

[cerulean socket]

opposite angles are the same

and the diagonals split the angles

so x and y should be the same

[tall fractal]

so

x

and y

is equals to

um

56?

[cerulean socket]

no

[wintry tundra]

well because of side angle side both of the triangles are equal

[cerulean socket]

oops

[cerulean socket]

yes it is

i saw it wrong

sorry

[tall fractal]

ohh it's okay

THANKS FOR THE HELP

[cerulean socket]

nice job

[tall fractal]

ILL TRY AND ANSWER STUFF I DONT REALLY GET THIS LESSON

I DONT EVEN KNOW WHAT SHAPE 💀

[cerulean socket]

try learning about shapes and its properties

like a rhombus all sides are equal

[tall fractal]

ohh yeah that was adviced to us by our prof

i shouldve listened 😦

[cerulean socket]

keep trying and youll get it

[tall fractal]

i

what the