geometry-and-trigonometry

[dark sparrow]

wait so have you heard of inverse trig functions before

because without those this problem's impossible

[gritty spindle]

breifly

[dark sparrow]

you can use desmos if you dont have a physical calculator

(just make sure to set it to degree mode)

as an example $A = \cos^{-1}(0.31)$

[somber coyote]

Ann

[dark sparrow]

these angles are all given as acute so there's no edge case bullshit to worry about

[gritty spindle]

ok...?

[dark sparrow]

literally just evaluating the inverse trig functions at the relevant points

[gritty spindle]

Idk what that is

I legit just got the the quesiton as a warmup

[dark sparrow]

you told me you had heard of inverse trig functions

[gritty spindle]

once or twice

[dark sparrow]

maybe you saw arccos instead of cos^-1

[gritty spindle]

idk it

[dark sparrow]

bruh

[gritty spindle]

Bruh

I came here cuz im dumb

[dark sparrow]

i mean ok like

loosely speaking, arcsin(x) is the angle whose sine is equal to x.

ditto for arccos and arctan.

[gritty spindle]

wht the fuck

[dark sparrow]

?

[gritty spindle]

Bro

Im lost

[dark sparrow]

don't call me bro, please.

[gritty spindle]

ok

got any other prefrences

[dark sparrow]

my name is fine

[gritty spindle]

Ann?

[dark sparrow]

yes

anyway

okay youre having trouble grasping the concept so lemme try to explain it to you differently

you've heard of and are familiar with square roots, right?

(not related to this problem im just trying to springboard off this)

[gritty spindle]

yeah

ik square roots

[dark sparrow]

yeah ok well

arcsin is to sine what square roots are to squaring

[gritty spindle]

so arcsin is just sin but opposite?\

[dark sparrow]

in a sense.

it's called "inverse sine" sometimes

[gritty spindle]

oh

so i just take arcsine 0.31

[dark sparrow]

no, you're told cos(A) = 0.31, not sin(A) = 0.31

[gritty spindle]

oh

yeah

[dark sparrow]

you'll use arcsin for B

[gritty spindle]

is ther arccos?

[dark sparrow]

yes

and arctan too

same principle

[gritty spindle]

lol my next question

so i just do that and i get the answer?

[dark sparrow]

yes.

[gritty spindle]

wow

ur better than my teacher

Thanks

[mellow vector]

AG = d, <HED = beta, <FAH = 2alpha, ABCD is a deltoid, CO = m, and <C = 90.
I need to find the area of triangle AFH in terms of alpha, beta and d.
(That's only part 1 of the question, and in part 2 I need to use m aswell so I think m isn't related to part 1)

Can someone help me with this?

[upper karma]

"<A" is vague, can you specify whether it's FAH or FAG or GAH?

[mellow vector]

The whole angle

FAH

I got to go eat, if someone finds the way to solve this question please tell me and I will take a look when I come back.

[upper karma]

i have a solution and will help you in 1 or 2h, bc i have to do smth.

will tag you when so if you still haven't got helped.

[flint wolf]

Can someone help me with a couple circle problems

[mellow vector]

alright thank you very very very much!!!

[upper karma]

okay so

i'd have loved to make a tikz diagram but i'm not available of that free time rn.

[mellow vector]

WAIT UR SAYING IT HERE

ok

[upper karma]

yeah

[mellow vector]

I can find those angles

what next

wait

use sine law 2 times to find AH and AE...?

and then use them for the area..?

omg...

[upper karma]

hint: ||after finding all the "?" angles, our desired area will be bh/2, and we don't know what h or b are, and those are the ones we want to express in terms of alpha, beta and d, with those angles and some trigonometry you can archieve it. give it some attempts||

[mellow vector]

we can use a * b * sin(C)/2

we dont need h but I see u got a different method

[upper karma]

yeah whatever works for you it's fine

[mellow vector]

I can see how to find the h

from A to EH

but not EH

[upper karma]

you can use x_1 and x_2 added up to find EH.

[mellow vector]

I did it, ty!

[upper karma]

anyone have the time to confirm if i did this right

[upper karma]

nvm ive got it

[obtuse thicket]

can someone help me with this?

[wintry tundra]

"Find the area of the area of the triangle" what does that even mean

Also I'm not even sure we have enough info to solve it

[trim breach]

I think we are to assume those are regular shapes. You can then use a 30-60-90 triangle in the regular triangle by drawing an apothem. You can find the length the base by doubling the length of the leg adjacent to the hexagon. Then, split the equilateral triangle in half again and solve it for the height.

I cannot see another way to solve that problem with the assumption that they are regular polygons.

[trim breach]

New and improved image.

[gritty spindle]

@foggy pine lol u actually joined

[round flicker]

Hey, I was wondering if anyone has any similar questions to this that they could send me? Thanks

[zealous forge]

@solid needle

[solid needle]

ask away

[zealous forge]

i need confirmation on my answers i finished everything but i just need to know if they are right

i got
1=a
1=a

3=A
4=D
5=C

for this i got
6=a
7=b
8=idk

for this i got 9=a
10=c

[solid needle]

1 is wrong

2 is wrong

3 is good

4 is wrong

5 is wrong

ok

im gonna stop here

because i found a core gap in your understanding

sec, sketching something

[zealous forge]

ok

[solid needle]

what is the scale factor of these two triangles?

assuming they are similar

[wintry tundra]

Well u can see which sides correlate with each other

Make a ratio

And when u simplify the ratio u will get the scale factor

[solid needle]

@zealous forge

[wintry tundra]

Oh

Lol

I thought u were asking for help

[zealous forge]

lol um i was asking him

[upper karma]

anyone open to help me on a problem

[worn anvil]

Hey there, does anyone know how to get the sides of congruent triangles? I'm not begging for answers, I just need a few instructions since I got confused with my Teachers'

[upper karma]

send the problem

[upper karma]

what are you confused about

it's just oriented differently on the right

so like ca is 77

[worn anvil]

About how to find the missing sides

O h

Wait-

[upper karma]

look at the bottom triangle for example

FHG

triangles add up to 180

so u add 78 and 62

180 - 140

missing angle is 40

[worn anvil]

Oh my god I I've never felt so dumb before

Thanks man

[upper karma]

no problem

[jade badger]

how do i even start

nvm

i got it

nvm

[plush basin]

hi can someone help? <@&286206848099549185>

[icy dome]

how to reduce the power of sin(x) having non-interger values as power

[upper karma]

you mean something like (sin(x))^(1/2)?

it should be possible for rationals

[icy dome]

yes

is there no way to do it

?

[upper karma]

so if we want to express (sin(x))^(1/2) with an integer power we can substitute sin(u)=(sin(x))^(1/2)

[icy dome]

thanks for the heads uo

[zenith tangle]

I'm asking for help,
I have been trying to figure it out for hours and zero progress the information of the question is written on the photo.
Thanks for helping.

[upper karma]

ABCD is a square??

so i assume the problem means to say that figure ABCD is a quadrilateral

the first thing that i notice is that AGB and CGD are similar triangles

wait no we cant prove that yet

is this asking you to prove that line BD is a bisector of angle ABC?

[dapper estuary]

yes

[upper karma]

Hello! I need help understanding this material. I tried doing it but I didn’t understand it very well. It might seem easy but I don’t have the same level of understanding as my other peers.

[valid drift]

can someone help

[eager hatch]

Because all the lines are parallel to the opposite, you can use QR as a transversal to get that Q and R are supplementary, and use PQ to get that P and Q are supplementary, then because P and R are both supplementary to Q, they are congruent

[valid drift]

thank you

[eager hatch]

No problem

[dapper bronze]

Help

Please

[pale latch]

how would you answer this

[dapper bronze]

CAN SOMEONE HELP ME

[signal otter]

Can anyone explain in greater detail how we went from the second equation to the third one? I'm not clear on how that process went down.

[oblique hatch]

So, if you look at the right it says "multiply both sides by cos"

Therefore all terms on the LHS were multiplied by cos x - the way sec x seems to magically disappear is because recall: sec x = 1 / cos x

So, if you multiply cos x by 1 / cos x it reduces to 1

@signal otter

[trim breach]

The diagonals bisect each other and there is no right angle. That should be enough to define the quadrilateral.

[signal otter]

Okay, so they just didn't write out the step where they converted sec into 1/cos?

That makes much more sense

[oblique hatch]

@signal otter Exactly - but if you look at the LHS with that knowledge, you can see how every term has been multiplied by cos x

[signal otter]

If I may ask a follow up - I can see logically WHY we would multiply by cos, because we want to get into quadratic form, but could one theoretically choose any value to multiply by? Is there a rule regarding this?

[oblique hatch]

@signal otter Theoretically...but as you mentioned this way you get a nice quadratic to work with and that's an easy thing to handle...so, I suppose the answer would be you could...and suffer the consequences lol

[signal otter]

Sure, I understand why we wouldn't want to. Just struck me as slightly odd that I'm allowed to multiply by any value - I would have expected that there be a 'rule' about what you could multiply an equation of this form by

Thank you for clarifying that hidden step for me

[oblique hatch]

@signal otter Well, they've set this problem up to be academic - it's specifically designed to be solved in this way. As you progress, that may not be the best solution...This is just an academic exercise to show the learner that sec x can become 1 when multiplied by cos x because of the reciprocal identity. TL DR; it's just an academic exercise lol

[signal otter]

Makes sense. I've found myself struggling to memorize the fundamental identities, and having to constantly refer to my notes from previous sections any time I encounter a scenario where one might be used. Any advice or useful mnemonic devices you know of for getting those down?

[oblique hatch]

@signal otter The simple trig identities? Like, sec x = 1/cos x? Those weren't too bad for me because I just remembered that co-secant seemed like it should go with co-sine....but it doesn't! As for all the trigonometric identities like sin²x + cos²x = 1, sec² = tan²x + 1, etc.....muscle memory. Sorry that's useless!

[signal otter]

Yeah, all of these.

Its just been a weird semester. Thanks for your continued help.

[oblique hatch]

@signal otter Sorry, I was hunting something down I remembered - https://youtu.be/T7D1W1oD8wo

I remember using that video a couple of times until I got it down pat, then the rest I just did a lot of problems to drill it into my brain.

[signal otter]

Wow this is really helpful

[surreal jetty]

<@&681260373478735874>

hi i know how to do all the other steps just need help with 1

[stuck briar]

Did you read #❓how-to-get-help

[surreal jetty]

how do i know if its Sin Cos or Tan

no

[stuck briar]

You should read it

[surreal jetty]

o my bad

[nova rose]

wait so what's the point of this chat

[inland salmon]

someone help please]

[upper karma]

anyone know how to solve this?

[wintry tundra]

I can help

So c is the center of the circle

And the arc made by an angle with the middle point is $r\theta$ where $\theta$ is the angle

Oops

Sorry

Wrong rule

[upper karma]

i just need to find the measure of angle BDA lol.

[wintry tundra]

If you have an angle where the middle is in the center and there's another angle that shares the same ends but the middle is on the edge of a circle then that angle is half the first one

Like this

[upper karma]

thanks xoxo

[wintry tundra]

Np

[trim breach]

AAA only proves similarity. Without a length, you cannot prove congruency.

[hardy ibex]

hiii who knows about bearings??

i have a couple questions i need help with

word problems btw

heres an example of what they look like

[hardy ibex]

this is the question i need help with

2b) only

[rugged sage]

help

how do i "Use the unit circle to justify the fact that for all θ: (cosθ)^2+(sinθ)^2=1

[teal kayak]

Can someone help me with my hw?

I just need examples

1 & 8

[kind patio]

hi i really need help with my homework

[clear juniper]

can someone help

[dark sparrow]

okay, so i need a sanity check

is it true that a triangle is acute iff its circumcenter lies inside it?

[upper karma]

yeah that's true

[dark sparrow]

great ty

[jagged ore]

So I missed a few classes from moving issues and I need help learning how to do something like this

[wintry tundra]

Two lines who meet at a point and are tangent to a circle are equal

[jagged ore]

Yes ik that

They are both 17

But how do I get to the center point :P

[trim breach]

Unless I am missing something, you don’t have enough information to find the radius. If I am wrong, someone please correct me.

No other given angles for trig, and no other sides to complete the triangles.

[muted viper]

I'm a bit confused by notation. Why is sin^-1(x) considered the inverse of x? Instead of 1/sin(x). Why is there a special name for the reciprocal of sin called cosecant? This seems like an unremarkable function.
Is this some unfortunate accident of history. Or am I missing some connection here?
In programming languages we tend to have functions with names like acos and asin for the inverse of cosine and sin. And the languages often don't bother to have a function for csc sec which makes more sense to me.

[trim breach]

x^(-1) = 1/x

And yes, you could notate the trig functions in terms of sine and cosine, but having names makes things convenient.

[muted viper]

But isn't sin^-1(x) != 1/sin(x)?
Even though sin^2(x) = (sin(x))^2

sin^-1(x) = asin(x) right?

[silent plank]

stems from history

[muted viper]

Thanks. I wish sin^2(x) actually mean sin(sin(x)). Usually when I don't understand why something is the way it is, it's because I missed some key insight about math or about convenience.

I had a really hard time googling an answer to this question.

Finally convinced it to give me relevant search results with the query ambiguous notation inverse trig

[upper karma]

Well sin^2(x) is sin(x)sin(x) which makes completely sense

Just like x^2 = x*x

[muted viper]

But on the other hand f^2(x) usually means f(f(x))

[upper karma]

Uh, no

If you had f(x) = 3x

And wanna have f(x)^2

U get 9x^2

f(x)^2 = f(x)*f(x)

[silent plank]

f^2(x) is interpreted differently than f(x)^2

[upper karma]

f^2(x) doesn’t make any sense unless you choose f to be a variable like x, instead of a function of x

f(x)=y=3x | ^2
f(x)^2 = y^2 = 9x^2

Using y instead should make it clear

[silent plank]

But on the other hand f^2(x) usually means f(f(x))

[upper karma]

Then just write f^2(x)= ffx instead

[muted viper]

I can't change the notation everyone already uses. 🙂

[silent plank]

the notation is a bit ambiguous, its just a thing that exists

[upper karma]

Not f(f(x)) which looks like f of f of x

[upper karma]

This is not used

[silent plank]

i've seen it being used like that multiple times here

[upper karma]

You guys are thinking about composition of Functions which is a whole different thing

https://en.m.wikipedia.org/wiki/Function_composition

[silent plank]

wasn't that the whole point of this discussion?

directly from that page

[upper karma]

„Sometimes“

4 years of engineering at university and never seen this mess of a notation

But guess you were right then xD

[silent plank]

depends on location

as well

[upper karma]

Yeah probably

[silent plank]

think its more common in europe

[upper karma]

Hmm im from Europe 😄

[muted viper]

I'm reading lots of different discussions on stack exchange where people lean in all kinds of different directions on function composition notation. Some people say if you use f^n notation for function composition you should clarify that's what you're doing. Other people seem to treat it as the default. 😐

I'm going back to the comforting world of programming now where there is little to no ambiguity in my trig and non-trig functions.

[upper karma]

I support the clarification part lol

[rugged sage]

quick question

gg

[dark sparrow]

ok so

another sanity check is in order for me

ABC is an acute triangle => angle D is obtuse

this can be proved by extending BD and CD past D and using that one theorem about angles between chords, right

[wintry tundra]

mayyyybe

[dark sparrow]

what

look, im doing a problem that involves delaunay triangulations
all i want is a sanity check for like

basic geometry shit

[wintry tundra]

I'm going to be honest, I have no idea if this is true. The triangle thing, I mean.

[dark sparrow]

angle D should be equal to the average of arcs BAC and EF

[wintry tundra]

Is BC a diameter?

[dark sparrow]

no

ABC is an acute triangle

[tiny snow]

Intuitively, as long as ABC has the circle center in its inside ABC will be acute. The edge case being when one of its sides pass through the center. This is also the edge case for BDC, for which <BDC can reach its smallest value, a right angle, if D is placed on the circumference. I am assuming that D must remain in the circle and outside ABC.

(edited, wrong labeling fixed)

@dark sparrow ^

https://www.geogebra.org/geometry/ut82dvfz

Do you need a formal proof?

[dark sparrow]

nah not rly

i can and probably will get away with saying "this is highschool geometry and i hope its obvious to everyone"

[tiny snow]

Proof that "angle D should be equal to the average of arcs BAC and EF"

May very well be some much more straightforward way to see this, but...

@dark sparrow ^

[silent plank]

should be able to just quote inside angles theorem or whatever its called

[tiny snow]

Intersecting Chord Theorem, and here's a much simpler proof: https://www.jamieyorkpress.com/wp-content/uploads/2019/08/Circle-Geometry-Proofs.pdf

Ah well 🙂

[gray minnow]

pls simplify

[past geyser]

@gray minnow is that sin(2x)/(1+cos(x))

yeah it is

do you know the identity of sin(2x)

[storm portal]

$\frac{\sin(2x)}{1+\cos(x)}$

[somber coyote]

Lighthearted Sand Crab

[past geyser]

yeah i know

[storm portal]

📈

[past geyser]

📈

[storm portal]

📈

Even if you split it up into the most common identity for sin(2x), wouldn't you still need to multiply by the conjugate of the denominator?

oh yeah, then split up, and cancel what you can

[gray minnow]

ok

thxx

[boreal narwhal]

does anyone know how to solve this

[shrewd delta]

Try drawing a diagram, you should get a right angled triangle where you know one more angle and one of the short sides

[boreal narwhal]

like this?

[shrewd delta]

Which side is the redwood tree?

[boreal narwhal]

how would you do the diagram?

[shrewd delta]

So the tree is one side, the line along the ground from the base of the tree to the point 115 feet away is another side, and the hypotenuse would be the straight line from the point to the top of the tree

Assuming the tree is perfectly vertical, you'll see that one angle is 90°

[boreal narwhal]

like this?

[shrewd delta]

Yep

The bottom side is given to be 115

[shrewd delta]

Yes

[steep elk]

hey yoo

I only need help for number 4 and 6

[trim breach]

If two angles add up to 90 degrees, they are called complementary angles.

And because you are given BOD is a right angle, angle 1 and COD add to 90 degrees.

[boreal narwhal]

does anyone know how to solve this?

[storm portal]

Use Pythagorean Theorem for AC

And then you can use inverse trig functions for A, B, and C

@boreal narwhal

[boreal narwhal]

so 9^2+13^2

so would it be 15.8

[storm portal]

Careful of where the right angle is

@boreal narwhal

[boreal narwhal]

9^2 + b^2 = 13^2?

@storm portal

[storm portal]

👍

[boreal narwhal]

ok thanks

what about this one?

is there a formula?

[storm portal]

the arc is a PART of the full circumference

this may help - given that you know that the length of the arc is LESS than the circumference (less than 2πr)

[boreal narwhal]

so what would i do to solve

[storm portal]

can you figure out what portion of the circumference the arc length is equal to?

[boreal narwhal]

would the answer be 12pi

@storm portal

[upper karma]

yes.

@boreal narwhal

[outer mirage]

someone help me where to start?

[humble pulsar]

pythagorean theorem

[outer mirage]

thats it i thought i was supposed to use SOHCAHTOA @humble pulsar

[humble pulsar]

Nope, just pythagorean

[wintry tundra]

If u were to use trig you would need angles

Which are not given

[blissful willow]

anyone want to give me a hand with this one

I got 89 but idk if its right

[narrow seal]

can i get help on a pls

[trim breach]

For triangles where you don’t have all three side lengths, use Pythagorean Theorem to solve for the last side. To get the length ratios, you just need to identify which sides are which.

This might be useful if you are stuck naming the sides.

[livid moss]

right angle triangles

[sand zodiac]

Hey guys I need help with this question.
"A 40 foot ladder is set against the side of a house so that it reaches up 24 feet. If Jordan grabs the ladder at its base and pulls it 3 feet farther from the house, how far up the side of the house will the ladder reach now? (The answer is not 21 ft.) Round to the nearest tenth of a foot."

[humble pulsar]

draw a diagram

[sand zodiac]

Ok, thank you.

[humble pulsar]

then it's just use pythagorean twice

[sand zodiac]

Why would I use it twice?

@humble pulsar

[humble pulsar]

cause you need to find the starting horizontal distance

then the final vertical distance

[sand zodiac]

How do I do that?

[humble pulsar]

draw a diagram of the initial state of the ladder and the final, and solve for the unknown sides in each

[marsh quartz]

https://i.gyazo.com/thumb/1200/a6ae1c2c5eb77c2ffb733b364016b224-png.jpg

https://i.gyazo.com/thumb/1200/ecc5d034ef2d7399174f7769cea17a24-png.jpg

https://i.gyazo.com/thumb/1200/5dc1f5a8c99ccd716f0878c452d725bf-png.jpg

help please i forget my trig

[quartz torrent]

Solving for the x in the middle

[wintry tundra]

you can either use trig to find angles and eventually get to x with plug and chug or u can notice something about the angles

[upper karma]

calculate the area of the triangle in 2 different ways

[distant mortar]

finding surface area, please help

[lofty oxide]

just find a bunch of 2-d areas on a 3d shape

find the triangles

and then the rectangles

add em all up

and that is surface area

[distant mortar]

Would I use SA=B+1/2PL

For the triangle

I tried doing that and then the formula for a prism and adding them

But idk if that’s right

bc u cant rlly tell what the base is for the triangle “prism”

[lofty oxide]

is it me or do those traingles have right angles

[distant mortar]

Yes they do

[lofty oxide]

and i never seen that formula before so ill just leave this to someone else lol

[distant mortar]

I found the base of the triangle on the top using Pythagorean therom and got 9

oh ok nws

[lofty oxide]

so (9x12)/2?

i dunno

imma dip lol gl with the question

[distant mortar]

I have no clue LOL

thanksss

[lofty oxide]

ok im pretty dang sure the base is 9 and the height is 12

so (9x12) for both triangles on the top and bottom and add up the rectangles

cause if u think about it

u could have a square that is (9x12)

and then to find the triangle with a base 9 and height 12, you would divide by 2

and since we have two traingles

we can just say the area of both is 9x12

[distant mortar]

wait wouldn’t the square be 15 times 14

[lofty oxide]

yes for the one facing us

and then 12x14

and 9x14

[distant mortar]

9(12) + 15(14) + 12(14) + 9(14) = surface area

Right?

[lofty oxide]

yes i believe so

[distant mortar]

oh okay, thank u sm for the help

[lofty oxide]

ye

[lament kraken]

God help me

I've been banging my head against this for a solid couple hours at least

[jaunty rover]

<@&286206848099549185>

[pure bronze]

You still need help with that?

[lament kraken]

yes

[pure bronze]

I’ll try my best, although I’ve never encountered such an elaborate question.

I have no clue whatsoever on what the question desires, could you perhaps tell me what I am supposed to do?

[lament kraken]

prove QY and QX are the roots of the quadratic

[pure bronze]

Hmmm

Gimme a minute

I have no clue 😦

Sorry @lament kraken

[lament kraken]

welp

[digital rose]

does anyone know how to do this?

[pure bronze]

Pythagorean theorem.

Or trig

[humble pebble]

you could get the sides of the triangle using this formula

and then just do (base x height) / 2

[digital rose]

is the area 32

[pure bronze]

Yep

[inland mountain]

Would the distance be 25?

[pure bronze]

Not quite

It’d be 5

(You forgot to take the square root)

[inland mountain]

Ohhh okay thank you

In order from top to bottom qould it be true-false-true-true?

[pure bronze]

Well said bud

[inland mountain]

(5,-4)?

[upper karma]

correct

[inland mountain]

So I didnt understand this question on my homework

Like how do I tell the difference between Them

[pure bronze]

Linears that are parallel have the same x coefficient

Linears that are perpendicular, will always have their multiplication of coefficients equal to -1

[inland mountain]

Okay thanks

[pure bronze]

Np

[inland mountain]

Would this be 160 units?

[upper karma]

this open?

[narrow seal]

can i get help on number 1 pls

[wicked hawk]

Can someone please walk me through this

Sorry for the crappy handwriting

[trim breach]

You can use the Law of Cosines to get the last side.

Then, there are several ways to find the area from there, such as finishing the triangle to draw an altitude or using Herron’s formula.

[trim breach]

Your equation is right.

Solve for b.

[wicked hawk]

I get a negative though

[trim breach]

What are you doing to isolate b?

[wicked hawk]

multiply both sides by 25

[trim breach]

Yes. And the cosine of 36 degrees is positive.

So I am not sure where the negative is coming from.

Is your calculator in radian mode?

[wicked hawk]

I have no idea, how can i check?

Nvm it was lol

[trim breach]

Cool. If you do it now, it should be positive.

[wicked hawk]

20.225?

Sounds about right, thnxs man!

[trim breach]

Glad to help!

[narrow seal]

is the answer 130.03?

[trim breach]

I didn’t actually work it out myself yet. Give me a sec.

[vestal pendant]

i am super confused on how to begin this, can anyone help me please?

[trim breach]

What did you get for the last side?

My total area is a bit off of yours…I am not sure if it is because of rounding or one of us did something wrong.

[narrow seal]

i got 24.84 for the last side

someone told me i should do 1/2 (16) (19) sin (67)

i did that and that got me 130.03

do i have to add all the sides together and do

a2+b2=c2??

[trim breach]

It isn’t a right triangle, so you can’t use Pythagorean Theorem.

The Law of Cosines is what Pythagorean Theorem is based on though.

So I got 19.48 for the last side.

[narrow seal]

how did u get that

[trim breach]

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

[somber coyote]

Lidoh

[trim breach]

A is 67 degrees.

b and c are the sides you are given.

[narrow seal]

so it’s

67^2 = 16^2= 19^2

[trim breach]

a is the unknown side.

A is the known angle.

You know the value of uppercase A.

You are finding lowercase a.

,,a^2=16^2+19^2-2(16)(19)*cos(67)

[somber coyote]

Lidoh

[trim breach]

This is what it looks like when you plug the number in.

[narrow seal]

ohh ok i see

is the * multiplication

[trim breach]

Yes.

[narrow seal]

i think i did it wrong

but i got

310.14

i tried to solve it

[vestal pendant]

<@&286206848099549185>

Not a fan of pinging @ helpers, however i have a test on this and i dont rly understand it

[trim breach]

Did you understand the way I calculated the last side, a?

If you understood that step, I would like to know which way you calculated area.

[narrow seal]

no not really

i tried to solve it but i didn’t know how exactly

[trim breach]

I’ll see if I can work out a more visual way to convey this.

[narrow seal]

ok

[trim breach]

So the top part is very general. It is how you set up the Law of Cosines for any triangle.

[wintry tundra]

Sides are interchangeable too

[trim breach]

What I did in the bottom half is take that triangle and apply the numbers you were given in the problem.

A becomes 67 degrees

b becomes 16

c becomes 19

[narrow seal]

i get that part

idk how to work thru that problem tho

[trim breach]

Okay, so now we everything we need to find a.

We use the Law of Cosines:

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

[somber coyote]

Lidoh

[trim breach]

This is the general formula I wrote on the top half.

On the bottom half, I have plugged the numbers in that the problem gave me.

[narrow seal]

yes

[trim breach]

So now I have the value of a^2.

To get a, I take the square root of the other side.

In this case, I got a^2 = 379.43.

[vestal pendant]

channel still being used?

[trim breach]

So a = 19.48.

[narrow seal]

i don’t get the taking out the square root part

[trim breach]

We defined the missing side of the triangle as a. The Law of Cosine formula gives us a^2.

Algebraically, the way to isolate a from a^2 is to take the square root of both sides.

,,\sqrt{a^2} = a

[somber coyote]

Lidoh

[narrow seal]

i see

[vestal pendant]

Channel still in use?

[trim breach]

You can DM me if you have any more questions. I want to keep the channel less cluttered so tul can get some help.

[narrow seal]

ok thank u!

[upper karma]

be aware of the possible missinterpretation that may come by saying this. saying sqrt(x^2)=x is not true at all, sqrt(x^2)=|x|. but as side lengths can't be negative, you will end up with the right answer, but it doesn't justify that you can use what Lidoh said always, as x might or might not be a side length or whatever. be careful with going around saying sqrt(x^2)=x for each and every situation you encounter.

@narrow seal

[inland mountain]

Is this open?

[inland mountain]

Would this be 160?

[celest oar]

do you know the distance formula ?

@inland mountain

run this between adjacent points and add them to get perimeter

[outer mirage]

Dont know where to start with this

[wintry tundra]

Plus or minus

[trim breach]

Use the general equation of a circle. You will need to find its radius first and the center’s position away from the origin.

[inland mountain]

Yea I know distance formula thats what I used to get 160

[celest oar]

then ur good

[inland mountain]

What are the chords of the other endpoint?

So I got 11,38 is that even close?

[inland mountain]

<@&286206848099549185>

[silent plank]

coords not chords
also how are you getting (11,38)? also you should include those parentheses.

i would be inclined to say that you were close (but it depends a lot on what you did)

@inland mountain

[inland mountain]

Sorry fell asleep so I used distance formula

[silent plank]

ic. that seems to be the source of your error.

you can apply the midpoint formula directly

applying something like the distance formula here leads to some issues with signs

@inland mountain

[nimble kayak]

See this GeoGebra document😀

This is about angle subtended by an arc.

I have made this myself😀

I was unable to sign in GeoGebra, therefore, I have goven this file😀

[inland mountain]

Can anyone help explain how to do this to me

[trim breach]

I think the easiest way is use midpoint formula multiple times. Find the midpoint, take the midpoint; that becomes the new endpoint for the midpoint formula.

[inland mountain]

Im sorry im a lil confused on this

[trim breach]

The midpoint is (1/2) the distance between two points. Using midpoint formula gives you the coordinates of the point that is (1/2) the distance.

Actually, I think a picture would be helpful here.

Give me a sec.

[inland mountain]

Thanks for helping me out im behind in math rn and this is due soon

[trim breach]

So using midpoint formula once gives us point 1 on the line segment.

This point is (1/2) the distance between Q and T.

If we use midpoint formula, we know the coordinates of point 1.

Point 2 is (1/2) the distance between Q and 1 but it is also (1/4) the distance between Q and T.

[inland mountain]

So how do we find the coords of r from this

[trim breach]

Well, point Q has coordinates (30, -50), and T (4, 26).

Try finding the first midpoint, point 1.

[inland mountain]

So id use midpoint formula x1+x2/2 and y1+y2/2?

[trim breach]

Yes.

[inland mountain]

So 30+-50=-20/2=-10 and 4+26=30/2=15

So -10,15

[trim breach]

Not quite.

What you did was this:

(x1 + y1)/2 , (x2 + y2)/2

Remember that coordinates are written as (x, y).

[inland mountain]

Ohhh so 17,-12?

[trim breach]

Yes!

[inland mountain]

So thats the coords of 1?

[trim breach]

Yep. So the coordinates of Q are (30, -50), and the coordinates of point 1 are (17, -12).

That is the first midpoint.

Point 2 is the second midpoint.

Point 2 is (1/4) the distance between Q and T.

So, if we find the midpoint between Q and Point 1, we have point 2.

[inland mountain]

So 23.5,-36?

[trim breach]

I got (23.5, -31).

[inland mountain]

Omg im dumb i added 22 to 50 instead of 12

So 23.5,31

[trim breach]

Yes.

So we have the coordinates of point 2.

Point 3, which is R, is (1/8) the distance between Q and T.

We can find R by finding the midpoint between Q and Point 2.

[inland mountain]

So 26.75,-9.5?

[trim breach]

Yes.

[inland mountain]

So thats point r?

[trim breach]

Yes.

[inland mountain]

Ayy thanks alot lidoh thanks for taking the time to help me

[trim breach]

No problem.

Considering the question…that point might not be correct, but the general method is.

I am not a fan of the wording…but I see how the author of the question might have interpreted it.

If it is wrong and you need help again, just ping me.

[inland mountain]

Okay thanks my teacher is a very confusing one at that

[wintry tundra]

I think the most confusing part is it doesn't specify if the segment is QR or TR

[trim breach]

Yeah.

That’s what I just realized.

[wintry tundra]

Mhm

[trim breach]

There is no reference point.

[wintry tundra]

Poorly constructed question

[inland mountain]

This is the only other question I'm having trouble with

Its the last one too

[wintry tundra]

What

[trim breach]

I’m confused as to what that even means.

[wintry tundra]

What does it what you to find

Length?

This teacher is bad at making problems

[inland mountain]

Yea I know its kinda why I'm confused

I dont even know where to start

[wintry tundra]

I cant help you bc its an uninterpretable question

[inland mountain]

Okay thanks for trying im prolly just take a wild guess and hope my grade doesn't drop horrifically

[rotund sonnet]

can anyone help me with this

its not a test its practice for extra credit

[eager star]

i completely forgot how to do this

[silent plank]

convert 232° to radians then apply formula for arclength

[steady shoal]

Draw the figure on a sheet of paper and find the length of the diagonals using distance formula

[grim panther]

hi

does anyone know how to do this problem "the parent function 𝑓(𝑥)=2𝑥is reflected across the x-axis,then shifted 2 units left and 6 units down. Write an equation to represent the new function."

this for a practice because i am struggling a little

its for exponential and logarithmic functions

[wintry tundra]

Do you mean $f(x) = 2^x$?

[somber coyote]

visual of Petter's ascendance

[wintry tundra]

Also the thing u asked isn't geometry or trigonometry

[shy root]

can someone guide me through this?

[green forum]

Hi guys, I really need help with this question :/
(my sketch:https://www.geogebra.org/classic/hgyxdyxu )

Let ABC be an acute, non-isosceles triangle with D is any point on segment BC.

Take E on the side AB and take F on the side AC such that ∠DEB = ∠DFC.

The lines DF, DE cut AB, AC at M, N, respectively.

Denote (I1), (I2) as the circumcircle of DEM, DFN.

Let (J1) be the circle that internal tangent to (I1) at D and also tangent to AB at K,

let (J2) be the circle that internal tangent to (I2) at D and also tangent to AC at H.

Denote P as the intersection of (I1) and (I2) that differs from D and also denote Q as the intersection of (J1) and (J2) that differs from D.

(a) Prove that these points D, P, Q are collinear.

(b) The circumcircle of triangle AEF cuts the circumcircle of triangle AHK and

cuts the line AQ at G and L (G, L differ from A).

Prove that the tangent line at D of the circumcircle of triangle DQG cuts the

line EF at some point that lies on the circumcircle of triangle DLG.

[rough roost]

i don't understand how the answer for 4 a) is 9.17

[silent plank]

its not

who's saying it is

[upper karma]

should be ~14.7

s = $r \times \theta$

[somber coyote]

xi64

[plucky sedge]

Ive looked at this for a while now. I don't see how these two triangles are similar. I know that that they share C but I dont see how those 4 numbers are related

[silent plank]

determine whether certain sides are in the same ratio

[plucky sedge]

Tried that I dont see how they are a ratio

Guess im just not seeing it

[silent plank]

consider the ratios of the given sides in each triangle

simplify 21/27

simplify 14/18

[plucky sedge]

Ah

Alr

[brave zephyr]

Could someone help me out?

[peak flower]

The first one seems right to me.

[brave zephyr]

Could you explain why? I wanna understand this problem.

[peak flower]

https://en.wikipedia.org/wiki/Law_of_sines

[upper karma]

Why?

[storm portal]

That triangle's dimensions aren't accurate

A triangle like that can't be physically constructed

[trim breach]

Did you have specific problems in mind?

[storm portal]

The guy's name I can't take seriously lol

[dense island]

Hi can anyone pls help me with qn 1?

Im unsure of both parts

[gray jetty]

anyone?

[wintry tundra]

If you remember, for any function $a\cos{b(x-h)} + k$, amplitude = $|a|$ and period = $\frac{2\pi}{b}$

[somber coyote]

visual of Petter's ascendance

[upper karma]

to find the area ratio, we can first find the ratio of corresponding lengths and then square that

so we can compare the distance from the centroid of the triangle to the vertex

so to find big area / small area, we need to find (b/a)^2

a is the inradius b is the circumradius

so here we have a 30-60-90 triangle

b/a=2

(b/a)^2=4

the ratio of the big triangle's area to the small triangle's area is 4:1

[wintry tundra]

That's some brilliant math 👏

[upper karma]

this one is basically the same we can find the ratio of the radii and then square that

[dense island]

Hm ok thabks i will take time to read

[scarlet basin]

hey guys

ngl im pretty confused on how my teacher solved a couple of questions

any help is appreciated

here's the first one

this is a cylinder

the formula for volume on a cylider is pi(r^2)*h

idk how she did what she did to get the answer, and why it's incorrect as compared to what i did

what i did is recognize the 19.3 as the diameter, divide it by 2 to get the radius, then used 9.5 as the height, and plugged everything in to get 2779.25 m^3

[pure bronze]

Show your calculations, perhaps you just did something wrong..

[scarlet basin]

@pure bronze V = pi(9.65^2)*9.5 <-- that's what i plugged into the equation to get my answer below

V = 2779.25 m^3

But clearly she does something way different in the pic

[pure bronze]

She used the Pythagorean theorem to get the diameter, the diameter is x because this line segment x is shown to us by its intersection with both sides of the upper circle

To get the radius, divide by 2

[pure bronze]

The 19.3 is the hypotenuse in the right triangle

However, it’s not always the diameter, you assumed as if the hypotenuse was the diameter.

@scarlet basin got it?

[scarlet basin]

@pure bronze ive got classes rn but i just posted the question and what i did in advance, can we continue in a couple of hours or so?

[pure bronze]

Sure

[scarlet basin]

ill tag u when im finished

awesome, thx!!! ❤️

[pure bronze]

If I’m available though

[pure bronze]

Np bud

[daring nexus]

is there an identity like this?

(n,w,m, a_k are whole numbers)

i also dont know if this is trigonometry really

(geometry-trigonometry for that sake)

[daring nexus]

yes? no? maybe? who cares? :P

[wise pawn]

there might be, cause we can do something with powers of 2 in the denominator like that

[daring nexus]

👀

that would be awesome

[dire willow]

hi

<@&286206848099549185>

how do i solve this

[steady shoal]

From the figure you can see the radius of cylinder and sphere are same

The material used is only for the outer covering so use formula of csa of sphere and you’ll get your answer

[signal otter]

Can anyone help me figure out how to set my calculator up to solve the way it is asking me to?

Not sure how to graph over a certain interval

[wise pawn]

there's like a window button or something on the top row

you can set the x min, x max, y min and y max

[signal otter]

Is there any way to set the interval

[vapid stag]

did i do these right

[tulip grail]

Im lost

[silent plank]

consider outside angle theorem (for circle)

[tulip grail]

I havent learned that yet

[hard vale]

does anyone know this?

\

[silent plank]

yes

apply properties of cyclic quadrilaterals

[rugged sage]

so like

how to find reference angle of radian

as in radian being 3

not pi

[silent plank]

the same way you'd do it if your value had pi in it

[rugged sage]

3 - 2pi?

[silent plank]

no

consider that 3 is between pi/2 and pi

[rugged sage]

i dont understand this question

"Explain how translating the cosine graph can be used to justify the fact that for all theta: cos(theta - 90) = sin theta"

[pure bronze]

You can see that when you plot the graph of cosine theta you’d have an intersection point with the y axis that is 1, whereas the sine graph intersects the y axis at 0.

[rugged sage]

oh

i get it now

ty

[pure bronze]

np

[tiny snow]

@green forum You need to give a proper description of the construction if you want help. Can you copy the problem text exactly as stated?

[compact jasper]

I working on triangulation and want to check that point inside triangles (Triangle contains point).
https://github.com/mapbox/delaunator/blob/master/delaunator.png Looks something like that
I use brute force and search inside all triangles.
Is there is a better way to do it ?

[obsidian marsh]

Solve for v: sin(2v) = 0.65
I've found the first solution, 20.3, but I'm supposed to be able to find 4 different ones. Does anyone have any tips on where to start?

[molten mist]

does anyone know how to solve

i set the equation up for the top problem but i don’t know how to solve it unfortunately

[silent plank]

what equation did you set up

[molten mist]

i set the first problem up like this

but idk where to go from there

[silent plank]

dunno why you multiplied the left fraction by 25/25
it was pointless

[molten mist]

idek

[silent plank]

first isolate sin(x)
then use inverse trig to get x

[molten mist]

idk how tho that’s the problem i’m sorry

[silent plank]

which part?

[molten mist]

i just don’t know how to solve the problem

[silent plank]

first isolate sin(x)
then use inverse trig to get x

are you able to first isolate sin(x)

[molten mist]

dont you just keep it alone when you isolate it

[silent plank]

that's what isolate means

yes

[molten mist]

idk what inverse trig is tho

[silent plank]

inverse trig is stuff that you should definitely have already done in right triangle trig

$\arcsin, \arccos, \arctan$ or $\sin^{-1}, \cos^{-1}, \tan^{-1}$

[somber coyote]

ℝamonov

[molten mist]

yeah idk how to do that stuff i apologize

[silent plank]

then you're really lacking the prerequisites for this section if this is alien to you

[molten mist]

yes ik Lol

[silent plank]

are you saying you've never done problems like this where they ask you to find the size of angle x?

[molten mist]

yes we have

[silent plank]

...

really?

[molten mist]

yeah

[silent plank]

and how would you find x here?

[molten mist]

ok nvm i figured it out lol i made a small mistake

[silent plank]

because it really sounds like you've done inverse trig before

[molten mist]

i did i just don’t know the names of certain stuff

[upper karma]

inverse trigonometry 😮

[tiny snow]

@green forum Here's my best guess at the construction, but with arbitrary points A, B, C, D, and E (they can be moved freely in GeoGebra). The gray lines and circles are transporting the angle <BED to create an equal <DFC. The two blue circles are defined by the three points E, D, M and F, D, N respectively. P and Q are intersections of the blue and red circles, apart from D. The red dotted lines are tangents and their normals to the blue circles in D. The green lines are angle bisectors used to create the tangential circles M_1 and and M_2, where M_1 is tangential to AC and I_1, and M_2 to AB and I_2. The purple lines are the extensions of DE, DF, AB, and AC. https://www.geogebra.org/geometry/pgjxthj9

Given the construction, there are some good clues as to why A, D, Q, and P are colinear. Take particular note of how the angle is transported and where this construction coincide with the angle bisectors.

Since the transporation of an angle may be hard to distinguish, here's the transportation alone: https://www.geogebra.org/m/r5sdzakg

[night basin]

What is the radius of the smallest circle that encloses an equilateral triangle
And what is the radius of the largest circle that will fit inside a triangle
I know it relates to circumcenter i think🤔

[serene obsidian]

how do i solve mJI⌢

[pure bronze]

Could you explain in verbally?

[serene obsidian]

wdym

arch of mJI

i dont understand im a little lost onnhow to solve this

[pure bronze]

No like

What is given to us though

It’s kind of tangled all around the photo

For instance, are MJ and LK tangents?

[serene obsidian]

i dont know he didnt specify

but he said

if a line looks tangent on a question it is

[pure bronze]

Then tangents it is

Generally, the task is to find arc JI

Right?

[tawny cypress]

how would you find the length of CF

length = 5 | width = 1 cm | height = 4 cm

[pure bronze]

You use the Pythagorean theorem twice I believe

Not sure though

[upper karma]

Anyone can help

bru

[trim breach]

Do you know what types of lines those are?

[covert ridge]

I am a little confused on what the directions are referencing in this question

[stone sapphire]

part of my trig hw

how does the left turn into the right

[safe trail]

@stone sapphire multiply nominator and denomitor with sqrt(3) - 1

[stone sapphire]

oh yeaah i remember now

its like the same thing with imaginary numbers

whats it called again when u use that rule

[safe trail]

(a-b)*(a+b) = a^2 - b^2 stuff

yeah multiplying with compliment i think its similar

[unique condor]

İts not in english but you guys probably can understand its asking for shaded area

[steady forum]

Is abd isos triangle?

[unique condor]

Nope

Any ideas for solution ?

[steady forum]

Hm

[unique condor]

This question is on test named sin area formula by the way

İ would post it on stack exchange but my pc isnt here right now

[steady forum]

1/2 ab sin C right?

[unique condor]

Yes

[steady forum]

Wth

Im so confused 😭

[unique condor]

There is a similar problem

İts so messy hope you can understand

[warm elbow]

yo u still need help

[unique condor]

Yes

Goddamn i just want to study algebra but i have at least 1000 pages of geometry left

[buoyant ibex]

Does anyone know the answer to this? I got 15 but it says I’m wrong

[silent plank]

you're not applying the correct formula

first consider something like pythagoras

[rugged sage]

reference angle of sin195 is sin 15

correct

[wintry tundra]

definitely agree

@buoyant ibex use pythag formula

[leaden cloud]

How do I draw the triangle for sin -630

-630 and 90 are coterminals right but how do u draw a triangle for that

[livid moss]

Draw a plane, start from the positive x axis. Go 630 degrees clockwise, and you will end up in the same place as if you went 90 degrees counterclockwise.

[leaden cloud]

How would the triangle look like tho

[livid moss]

What do you mean?

[leaden cloud]

cuz it is 90 degrees

[livid moss]

There is no triangle

[leaden cloud]

o

lol ok thx

[livid moss]

The triangle definition for sin and cos and tan only works for angles strictly between 0 and 90

[leaden cloud]

o I didn’t know that

[livid moss]

Well, yeah, you can't draw a triangle with 2 90 degree angles

[tiny snow]

(on a sphere you can)

[stone skiff]

Oh didnt see this is chat, if anyone can help

[versed river]

is that meant to be y=3sin(2t)

[dark sparrow]

okay

[upper karma]

What do i do

[dark sparrow]

so, tell me:

have you ever done questions like this before

[upper karma]

No

[dark sparrow]

ok

if instead of the arc you were asked to find the length (ie circumference) of the whole circle

would you be able to do it?

[upper karma]

No

[dark sparrow]

so you don't know how to find the circumference of a circle?

[upper karma]

Wait

Isnt it

C=2pir

[dark sparrow]

there we go

yes, that's the circumference formula

now look back at your problem

your arc has a measure of 45°

[upper karma]

yh

[dark sparrow]

how many degrees make a full circle?

[upper karma]

360

[dark sparrow]

great, so this means that your arc is 45/360 of the circle.

or to put it simpler

your arc is 1/8 of the whole circle.

[upper karma]

I got

0.125

[dark sparrow]

you got 0.125, evidently by simply typing 1/8 into a calculator with little to no thought.

your arc is 1/8 of the circumference.

[upper karma]

Huh

[dark sparrow]

why do you think i was asking you about the circumference? just for fun?

[upper karma]

No lol

[dark sparrow]

there are two quantities that matter here:

• the circumference of the circle, bc you know how to find it (as you just confirmed)
• the length of the arc, bc you're asked for it

and i just told you, in as clear language as i could manage, the relationship between the two.

[upper karma]

So its 0.125?

[dark sparrow]

"it"?

no, the length of your arc is not 0.125.

[upper karma]

Ok

So what do i do next

[dark sparrow]

okay, so i guess this needs to be spelled out explicitly.

find the circumference of the circle first.

as an intermediate step.

the circumference will not by itself be the answer to your problem. but it'll enable you to find the answer.

[upper karma]

erm

[dark sparrow]

and don't give me a decimal this time.

[upper karma]

Eo 64