geometry-and-trigonometry

[obtuse quiver]

i guess thats stating the obvious

but the key is practice

[heavy solstice]

help i dont know what to do im freaking out 😿

[hard moon]

it doesn’t appear to have exactly one solution

[north kindle]

I’m having trouble with this problem, Im trying something with a line from A to the perpendicular bisector of CB but it doesn’t seem to be working

I think I have that EFA + BAC = 180

[grave pond]

That would require EF to be horizontal..

It should be EFA + 2BAC = 180­°.

[north kindle]

Why exactly

[grave pond]

It you draw a horizontal line through F, then its angle with FE and FC are both the same as angle BAC.

[north kindle]

How do you know that

[grave pond]

Because it's parallel to BEDA.

[north kindle]

How do you know that

[grave pond]

Because that's what I deined my line to be.
Angle FEA = angle BAC because isosceles triangle.
Angle KFE = angle FEA because parallel lines
Angle KFC = angle BAC becaluse parallel lines.

So angle EFC is two times angle BAC.

[north kindle]

Damn

I must’ve gotten something wrong in my algebra somewhere

[grave pond]

And then angle ECF = angle EFC, again because isosceles triangle ...

[upper karma]

I thought tangent lines were all the same lenght at intersections

but it has coome to my attention i am critically wrong

I am off about 2 when I multiply 23.2 (One whole triangle side lenght) by 3

[grave pond]

You're probably assuming that the three points of tangency are 120° apart on the circle, but that cannot be true.

[upper karma]

how do i approach this then

I have 3 points, they are all tangent (Questions states: "If line appears tangent, then they are tangent"

Im going to label the image for easier explanation'

i forgor the C 💀

[grave pond]

I don't know what it is you're computing, but you can find all the side lengths using this:

[hard moon]

instead of perpendiculars, label the measure of any angle “x” then start finding all the other angles in terms of x

[maiden brook]

this is a great problem

i love aops

[north kindle]

It is

I got it btw

[maiden brook]

good job!

[north kindle]

Why is it necessarily true that they share an altitude?

[dark sparrow]

from B onto AC

[north kindle]

Yes

Why is that necessarily both of those triangles altitudes

[dark sparrow]

i mean we are considering AX and CX as their bases arent we

[north kindle]

Yes

[dark sparrow]

so like

those two come from the same straight line

so of course the altitude dropped from B will be geometrically the same in both cases

not merely equal length but like the segment is one and the same for both

[north kindle]

Ok but like, why is that segment the altitude

[dark sparrow]

what's your defn of an altitude of a triangle

it ought to be something equivalent to "a segment from one of the vertices to a point on the straight line containing its opposite side, and perpendicular to said line"

[north kindle]

Yeah

[dark sparrow]

yeah os

so*

i think you're overthinking it

we are literally just dropping a perpendicular from the same point (B) onto the same straight line (AC) in both cases

[north kindle]

How is BX perpendicular to AC

Or I guess BD

[dark sparrow]

nobody said BX or BD are perp to AC

[north kindle]

Looks about right

What about the other one though, how have we shown they are the same

[dark sparrow]

wym??

this red segment is the literal altitude for both triangles

[north kindle]

How?

[north kindle]

Oh wait I see I think

[dark sparrow]

if you drop a perp from B onto the line containing XC, to construct the altitude of triangle BXC, you get BH

if you drop a perp from B onto the line containing AX, to construct the altitude of triangle BAX, you still get BH

it's literally the same one

[north kindle]

Ok that makes sense

Thanks

[upper karma]

I need Help with option C,D

[dark sparrow]

holy shit they missed some important punctuation in there

• ABCD is a tetrahedron.
• Face BCD is equilateral, with side length 12 units.
• H is the foot of the perpendicular from A onto BCD, and it lies inside the face.
• The area of CDH is 3 times that of BCH.
• The area of DBH is 2 times that of BCH.
• M is the midpoint of side BD.

[upper karma]

I understood the question

It is clear?

What to do ?

[dark sparrow]

well i needed to write it out like this to make any sense of it.

cause the run-on sentence going on there is insane

anyway... do you have a diagram of this thing?

[upper karma]

Nope

They didn't give any

[dark sparrow]

i'm not asking if they give one

i am asking whether you have even thought to make one

[upper karma]

Oh yes yes

[dark sparrow]

"do you have [thing]" is not, in general, the same question as "does the problem give you [thing]".

[upper karma]

My bad , I am sorry

By any chance, is the foot of A on plane BCD THR CENTroid ?

[grave pond]

I think Ann wants to see your diagram.

[dark sparrow]

^

[upper karma]

Oh

I do realize the fact that BCD does not look like an equilateral triangle

[dark sparrow]

ok let me remind you once again

hidden edges are drawn with dotted lines

[upper karma]

I do remember that,but drawing it this way make me to visualize it in a easy way

[dark sparrow]

it's actually way easier if you do it my way

that way you actually get a sense of the picture being 3D lol

[upper karma]

Okay I will do it from now on

[dark sparrow]

also i don't see H anywhere

[upper karma]

H won't be usefully for options C AND D

[dark sparrow]

i would not be so sure

the position of H relative to B, C and D seems relevant!

especially given A is directly above it

[upper karma]

OHHH

I forgot that point

1 min

[dark sparrow]

i also recommend you make a subdiagram showing only face BCD

[upper karma]

Okay

[dark sparrow]

cause the area ratios between CDH, BCH and BDH should be more easily shown on that

[upper karma]

(I know how my x looks like)

[dark sparrow]

æ is an interesting choice of letter for area

anyway i am pretty sure the distances of H to each side can just be computed directly.

knowing BC=CD=DB=12 and these area ratios

[upper karma]

Yes

[dark sparrow]

do that

[upper karma]

Yrs

I already got it...

Option C and D

That is what i need help with

[dark sparrow]

drop the perpendiculars from H onto each side of the triangle

give names to each of the three points that are their bases

and send a photo of the resulting diagram here

mark their lengths as well

[upper karma]

👍

[dark sparrow]

ok

[upper karma]

CHF are not collinear

[dark sparrow]

ok so you want the distance from A to CM...

and the logical first step to that is to find the distance from H to CM

[upper karma]

Okay i will try to

Yay

Got it

The perpendicular distance is 2

2

So now we can use pythogorean theorem

Thanks

[obtuse quiver]

solve $cos(3x-15^{\circ}) = \frac{\sqrt{3}}{2} for, 0 \leq x < 180$

[dark sparrow]

did you mean \leq?

[obtuse quiver]

oh yeah

shit

[somber coyote]

uh

[obtuse quiver]

I got the reference angle and everything

[dark sparrow]

solve $\cos(3x-15^{\circ})=\frac{\sqrt{3}}{2}$ for $0\leq x\leq 180^{\circ}$

[somber coyote]

|Ann⟩

[obtuse quiver]

but normally the quadrants are like 90 180 270 and 360 right in a range of 0 to 360

i can't wrap my head around the changed ranges

[west radish]

15

[obsidian harness]

you should explain if you want to answer

cause wolfram can do it for them ofc

[dark sparrow]

maybe you should try solving for all real values of x first

[west radish]

As do cosinverse on the other side it will equal 30 then add 15 which will equal 45 and then divide by 3

Which equals 15

[obtuse quiver]

yeah i know that

but i'm trying to know what to do when the range is changed

[west radish]

K

[obsidian harness]

But the way to think of it is:
0 <= 3x <= 540
-15 <= 3x - 15 <= 525

There you go, that's the domain you need to serach for

And x = 15 is only one solution ofc

[dark sparrow]

this gives you only one solution and loses all the rest.

[obtuse quiver]

i got x=15, 115, and 135

[obsidian harness]

,calc 3*135-15

[somber coyote]

Result:

390

[obsidian harness]

keep searching

you're not at the upper limit yet

Let me doublecheck

[west radish]

180 minus answer

Or 90 plus answer

As quadranta

Quadrants matter

[obsidian harness]

Ah yes you are

Yeah see what happens when you try to find the next highest solution

[obsidian harness]

Correct btw

[obtuse quiver]

all i wanna know is what to do when the range is changed, at school they drew something on the quadrants

[obsidian harness]

@obtuse quiver

[dark sparrow]

imo the best way to do this is to solve for x in R and then filter to your range

instead of trying to memorize a thousand procedures for what to do when the range is in this or that form

[obtuse quiver]

how'd i do that

[west radish]

If the range is changed like to 360 then minus the ansr with 360 or with 270 or with 180

According to what the range is

[dark sparrow]

cos(t) = sqrt(3)/2 <=> t = ±pi/3 + 2pi*n, where n ranges over all integers

or in degrees ±60 + 360n

3x - 15 = ±60 + 360n and solve for x as normal

[obtuse quiver]

ohh

now i got it

thanks

[obsidian harness]

np

[dark sparrow]

augh

sorry. 30, not 60

typo on my part.

[obtuse quiver]

all good, thanks ann

btw is this correct $cos\theta = \frac{2}{3} , -180\leq x <180 \newline reference angle=48.1^{\circ} \newline \theta=-131.9,131.9$

[somber coyote]

uh

[dark sparrow]

,calc cos(48.1 deg)

[somber coyote]

Result:

0.66783255547105

[dark sparrow]

as long as you recognize that 48.1° is an approximation, yes.

though.

er wait

[obtuse quiver]

still getting used to it lol

[dark sparrow]

no

why ±131.9 and not ±48.1?

cos(131.9°) is **-**2/3 ish.

[obtuse quiver]

one sec let me show you this

i drew this quadrant to do that problem

instead of 0 and 360 i put in -180 and 180

because the range was changed

and then -180+48.1 and 180-48.1

[dark sparrow]

... hold on

these should both be on the left.

[obtuse quiver]

idk cos^-1(2/3) comes out as 48.1

[obtuse quiver]

wait what

[dark sparrow]

yes but you should have used those values as is, not taken their supplements.

angles are always measured from the positive x axis. the positive x axis is always 0, or an integer multiple of 360.

-180 to 180 still sweeps out a full circle exactly once but it doesn't start on the positive x axis

[obtuse quiver]

ah

[dark sparrow]

it starts a half turn across from that

[obtuse quiver]

so i should've done this

[dark sparrow]

upside down

[obtuse quiver]

is this good

[dark sparrow]

worse

[obtuse quiver]

shit

[dark sparrow]

the ±180 are in the right places but the 0 and 90 are both misplaced and unnecessary.

[obtuse quiver]

alright so ignoring the 0 and 90

how would i proceed

i have the reference as 48.1

[dark sparrow]

... you would get x = ±48.1 but the whole construction of your diagram makes it nigh impossible to justify that cleanly.

[obtuse quiver]

could you tell me how to do it from scratch

if it doesn't take up too much of ur time

[dark sparrow]

afraid it will

[obtuse quiver]

ah no problem then

[rapid rivet]

I was following a video online on how to graph cos and sin equations. I was a bit confused why they replaced the 2 in the formula solving for period 2pie/b as 3pie/(1/3) i was thinking because the amplitude is three that’s why but i’m still a bit confused if there is like a rule for it

[dark sparrow]

can you link the video

also $\frac{3\pi}{1/3}$ is not even equal to $4\pi$ lmao

[somber coyote]

|Ann⟩

[rapid rivet]

https://youtu.be/Vw-RwPBWS8g?si=F1P9ICZpMdoSeOUd at 4:25

[dark sparrow]

huh

wow ok he's just plain wrong there

[rapid rivet]

oh dang

[dark sparrow]

a comment from the video to that effect

[rapid rivet]

OH

okay ty i was so confused

[dark sparrow]

yeah

the period should be 2pi/(1/3) = 6pi

[rapid rivet]

watching on my school account so the comments were turned off

okay thats what i thought thank you!

[steady mason]

is the range and domain of a cubic function always infinity unless said otherwise

[dark sparrow]

it's not "infinity"

it's all real numbers.

or (-∞, +∞),

or just R.

but yes.

and the same goes for any polynomial function of odd degree.

[static hearth]

if sin(alpha) is equal to -4/5 in quad 4, does this mean that sin(theta) also has the same ratio of -4/5? what is the difference between them, exactly?

[steady mason]

.close

oh

right

dumb me

[dark sparrow]

... alpha and theta are two different letters which probably denote different angles

thats the difference

you cannot tell anything about sin(θ) from only the info you wrote here

[obtuse quiver]

,rccw

[obtuse quiver]

Any help on the trig question

[amber hollow]

any advice? i did P = 2 so B = pi, and midpoint = 7+1/2 = 4, and amplitude = 7-1/2 = 2, and since it starts at 0 im assuming its a flipped cos function

[dark sparrow]

!nogpt

[lime crown]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[dark sparrow]

@static hearth sin, cos and tan still denote the same trig functions no matter which letter you write for their input. but if a problem involves two different angles, one of them called α and the other θ, then you obviously have to distinguish between values of trig functions at alpha and those at theta

like idk how to explain it except to repeat the same shit 17000 times

only to have you fire off this fucking GPT shit

[static hearth]

this diagram^ is what was confusing me. but now I know that beta, theta, and potentially gamma just represent angles in the context of any given problem.

[dark sparrow]

the problem will tell you all the letters it uses.

they are greek but they are just letters

so yeah

[static hearth]

for instance, here I believed that on A) I had to calculate the Beta value in the context of sina = -4/5

but you just go over on the right side and find beta in the context of tanB = 4/1 right?

ew wtf. <@&268886789983436800>

good. it’s gone

[dark sparrow]

yes obviously

to find shit for alpha you work with alpha

to find shit for beta you work with beta

[static hearth]

so for this, which is just a reference I put on my wall, why would sin(a) = sin(a)cos(B) ?

why wouldn’t sin(a) just equal sin(a)

bahhh nevermind.

[elder anchor]

I do not see sin(a) = sin(a)cos(B) in your notes

[amber hollow]

also stuck on this one

[obtuse quiver]

Yeah this was wrong it will be just 48.2

[west tundra]

can i get an answer and explanation for this problem, I know to take inverse and use inverse of tan but I get -0.174 and 2.967 but mathaway says its wrong

[zealous olive]

to plug it into your calculator turn cot into 1/tan and solve for tan then use inverse tan on your calculator

and this problem has infinite solutions

[clear owl]

Why can’t I understand this? I’m stupid that’s what

Idiot here trying to actually be good at trig, what a fool I am

[ruby flame]

x = 10cos(60) i think

which is 5

[clear owl]

Well, I check and yeah, it’s 5

My apologies, it’s just that my math anxiety tends to get the best of me

Grade 11 math hasn’t been that kind to me

Anyway to find another method to get the same result as in ink? (52.23)

Yeah

I tend to worry a lot because I sometimes feel like my teacher does not get that I AM trying to understand, but everything feels so cramped and doesn’t have enough space to breath, I do well in all my other classes, but math annoys me the most

Like the problem above as I do not know HOW anybody could find a second method to the question without the use of a second side for the pythagorean theorem

[trail tendon]

you can use sin of one angle or cos of the other angle

and you can find the other angle because all the angles add up to 180

oh wait

it probably wants you to do pythagorean theorem

oh

i forgot you just said without pythagorean theorem

lol

theres actually a lot of ways you could do it

also it doesn't say not to use pythagorean theorem...?

[clear owl]

The paper was made weirdly

[trail tendon]

yeah

[clear owl]

Gotta love grade 11 math for uni (Hell is real)

/j

[trail tendon]

i bet it wanted you to use pythagorean theorem

[clear owl]

Yup and that was supposed to be the Correct Answer BUT, the side didn’t have any other values

But it’s okay

I am focusing on other stuff

[trail tendon]

wdym?

[clear owl]

All the stuff from the previous quizzes

I dk why this was graded weirdly

I’m sure I showed all my steps

Isn’t cot literally cos/sin? Therefore making if x/r / y/r

[trail tendon]

i think it may have wanted to you multiply by the reciprocal

idk

[clear owl]

And this is weirder

I tried to solve it as best as I could using the Pythagorean identity

[trail tendon]

ngl i have no idea what you did

[ruby flame]

use difference of squares

[trail tendon]

i think hes supposed to prove it better than that...

maybe not 💀

i guess they might have just wanted him to multiply it out like (1-sinx)(1+sinx) = 1 - sin^2(x)

but thats not really a proof

i thought it wants him to prove cos^2(x) = 1-sin^2(x) after distributing

but maybe not

[clear owl]

Yeah

I tried to do it using this

That wasn’t my best idea

[trail tendon]

yeah, if you can assume that then thats the way

but i don't think you distributed correctly

that might be the problem

[clear owl]

But I now know

[trail tendon]

i'm not entirely sure what you did

lol

[clear owl]

How to do it

[trail tendon]

yeh

[clear owl]

I just had to times them together

[trail tendon]

yeah

[clear owl]

And then use that identity

[trail tendon]

yeah

[clear owl]

And cross the sinses? Out

[ruby flame]

but why is difference of squares not a good proof

[clear owl]

For the cos^2x

[ruby flame]

you would use the pythagorean identity

[trail tendon]

because cos^2(x) = 1 - sin^2(x) is a trig identity, it says "prove the trig identity"
if it wants you to prove cos^2(x) = (1-sin(x))(1+sin(x)) simply because it equals another identity, thats not really a good proof :l

[jagged wyvern]

Could someone help me with this? At first I tried using surface area of a sphere using 4pi(r)^2, but it marked it wrong, then i found out if it was half a sphere it would be 2pi(r)^2

Both are wrong?

[ruby flame]

you need to account for the circle base

[trail tendon]

is that exactly half a circle?

sphere*

[jagged wyvern]

Im not sure

Thats what I found on google

[trail tendon]

i dont know exactly what a hemisphere is

[ruby flame]

hemisphere is half a sphere

[trail tendon]

assuming it is exactly half a sphere

ok

[ruby flame]

from google

the problem is you didn't account for the base

[clear owl]

Did you use the radius or the diameter

?

[jagged wyvern]

radius

[jagged wyvern]

What do i do with the base?

[ruby flame]

add an extra pi r^2

[trail tendon]

huh?

[jagged wyvern]

So what should the formula look like

[ruby flame]

$$3\pi r^2$$

[jagged wyvern]

If sphere is 4pi(r)^2

[somber coyote]

jasoney

[trail tendon]

what??

[ruby flame]

because when you cut it in half there is a circle at the bottom

[trail tendon]

isn't the volume of a sphere like (4pir^2)/3?

oh

mb

[ruby flame]

surface area

[trail tendon]

surface area

ahhh i see

[jagged wyvern]

So this should be used

Hm alr

[clear owl]

Then it should be 226.19 2pi (6) ^ 2, 2pi x 36 or 6.283 x 36 = 226.188 or 266.189

Make use of bed as

Bedmas

[jagged wyvern]

339.29

Uh could someone confirm? It doesnt let me check no more

If im rounding to nearest hundreths

[trail tendon]

uh

[ruby flame]

i got 339.29

[trail tendon]

how?

[ruby flame]

3pi(6^2) = (3)(36)pi = 108pi = 339.29

[trail tendon]

4pir^2 is the surface area of entire circle tho

so 2pir^2 should be half circle...

i missing sum?

[ruby flame]

but when you cut it in half you add a circle on the bottom

to close it

[jagged wyvern]

I tried using 2pir^2 it was wrong

[trail tendon]

oh

oh gotcha

i see

loll

yea u right

[jagged wyvern]

Can someone explain how i got this wrong? I solved for the radius using the given circumference. I kept the radius in fraction form for better approximation

29.5/(2pi)

V = 4/3pi(29.5/2pi)^3
= 433.53 if rounded to nearest hundreth

[jagged wyvern]

@trail tendon

[trail tendon]

that is....very weird

uhh

[hard moon]

That’s not even wrong

yeah idk

[trail tendon]

yeah

idk

XD

ur program is just dumb maybe 💀

[hard moon]

it may be displaying the correct answer after all tries have been attempted

[dark sparrow]

,calc 4/3 * pi * (29.5/(2 * pi))^3

[somber coyote]

Result:

433.52590365166

[jagged wyvern]

Bruh

[jagged wyvern]

Apparently volume is 434.19

[hard moon]

oh we got it wrong, then

[jagged wyvern]

🤷‍♂️

[dark sparrow]

??

ok wait

,calc 4/3 * 3.14 * (29.5/(2 * 3.14))^3

[somber coyote]

Result:

433.96579644881

[dark sparrow]

no, can't be that

it smells like some kind of rounding fuckup tho

[faint pasture]

They might have done pi = 22/7 instead of 3.14 maybe?

[dark sparrow]

,calc 4/3 * 22/7 * (29.5/(44/7))^3

[somber coyote]

Result:

433.17712637741

[faint pasture]

Oh lmao

[slate frost]

can someone please help me with this?

its permutation

but idk how to find the favorable outcomes

[faint pasture]

Well you set aside the two books as said by the question, then you just have to choose 2 out of the 3 books remaining

[dark sparrow]

#discrete-math is a better place for this

[faint pasture]

Id guess #competition-math would be better but okay

[slate frost]

if i set two books aside wont i have 4 left?

[faint pasture]

Nvm yea

Choose two out of the 4

[slate frost]

i used permutation to find total outcomes (360)

prbability = fav outcomes/total outcomes

[faint pasture]

,calc 654*3

[somber coyote]

Result:

360

[faint pasture]

Yea

How many ways can you select 2 from 4?

[slate frost]

my teacher said the fav outcomes was 12 in his notes but idk how he got 12 for fav outcomes

[slate frost]

let me see

oh its 12

[faint pasture]

Yep

[slate frost]

so i have to use the formula twice

once for total outcomes and once for fav

[faint pasture]

Yep

[slate frost]

thanks

[gaunt iron]

<@&286206848099549185>

[hard moon]

Try showing it’s a parallelogram; you’re almost there, then use the fact that opposite angles of a parallelogram are congruent

Alternatively, prove two triangles containing those angles are congruent

[foggy gust]

In a parallelogram, the acute angle is 45 . Find the area of a parallelogram
if its diagonals are 6 cm and 8 cm.

can someone help me with this task please

i dont quite understand it

[obsidian harness]

Do you know the formula A = 1/2 ab sin C for a triangle?

[foggy gust]

yes

[obsidian harness]

Right then from the picture above, there are 4 smaller triangles

And the centre bisects both diagonals

So we have 2 * (1/2 * 3 * 4 sin 45) + 2 * (1/2 * 3 * 4 sin(180 - 45))

[foggy gust]

but that gives us approximately 11.27

but the correct answer is 7 as i was told but how do we get 7

[obsidian harness]

Oh wait I misinterpreted the question

https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2Fcalcworkshop.com%2Fwp-content%2Fuploads%2Fconsecutive-angles-parallelogram.png&f=1&nofb=1&ipt=74a4926a27cd379ba3deba5dc068c3843e15a578a2e584bfb9862e1edfe7d2f3&ipo=images

So the question must mean that n = 45 right

[foggy gust]

yes

[obsidian harness]

https://www.desmos.com/calculator/lhdlr0yvia

You could try using coordinate geometry

Then we must have (a + h)^2 + h^2 = 8^2, (a - h)^2 + h^2 = 6^2 if a is the base and h is the height

So (a + h)^2 - (a - h)^2 = 4ah = 28

[upper karma]

Status 1

Correct channel,right?

[obsidian harness]

yes but JEE questions are hell so

Btw, use Approach0 to search for answers to competition problems

https://approach0.xyz/search/?q=OR content%3A%24\cos x%2B\cos y%2B\cos z%3D1%24%2C OR content%3A%24\sin x%2B\sin y%2B\sin z%3D\frac{1}{\sqrt{2}}%24&p=1

[upper karma]

I will be back after the problem is precalc is done

[obsidian harness]

ok

The first link in Approach0 tells you that tan y = sqrt(2)/2

So options A and C are incorrect

in fact it's B

[upper karma]

Yes,the correct answer is B

[upper karma]

These are just the mock test questions

The original one's are of a different level

[obsidian harness]

don't bring me more hell lmao

[upper karma]

Thr paper comes with solutions but , in most cases they dont make any sense

That is why I ask them here

[obsidian harness]

RIP

but no worries ig

[upper karma]

Got this one though

Thank you

[obsidian harness]

np

[rotund stump]

can someone explain how this works and how this is able to rotate a absolute value function

[faint pasture]

Are you sure that is all the context?

[upper karma]

https://media.discordapp.net/attachments/1191957938756124774/1234676884269301791/image.png?ex=66460934&is=6644b7b4&hm=400852cabf4d55c6419e8c0206d66dd1720da11405ea2bedcef0ac17080a32e6&=&format=webp&quality=lossless

If area of ​​the triangular region AQP is 50
What is the area of ​​square ABCD

5(PB)=12(PQ)

[proud nest]

Its saying this is correct, but in the wrong form

oh it was 1/sqrt11

[round gale]

a and b was easy but for c I tried setting 2sin2x=root2 which gave pi/8 for quadrant 1 (which was correct) I did the same for quad 2 I got 7pi/8 and that wasn't correct The answer key said the answer is x=-pi/8,x=9pi/8, x=3pi/8, and 11pi/8. Can someone explain?

[faint pasture]

Do you know the formula for 2cos(a)cos(b)?

[upper karma]

help

i forgot to study so uh help

i remember first do some triangle?

Then get midpoint of each line?

[hard moon]

if you construct a perpendicular bisector of segment AB, then every point on that perpendicular bisector will be same distance from A as is to B

Likewise, if you construct a perpendicular bisector of BC, then every point on that perpendicular bisector will be the same distance from B as is to C

How can you use that to find a point (the boat) which is the same distance away from all three points A, B, and C?

[upper karma]

😭

[clear owl]

Any clues on how to find the measurement of D to G?

Would I have to find the “height” of the second triangle?

[wicked island]

how can I prove that a>b?

[trail tendon]

maybe you could do it using law of cosines?

[wicked island]

good idea, but I'm not sure I can use it already, considering the book didn't prove it yet

[hard moon]

hinge theorem

[trail tendon]

😔

[wicked island]

I'll search about it, thanks

[maiden brook]

cos 100 < cos 80

that’s all u need

cause cos 100 is negative

[grave pond]

If you don't have any trigonometry: Drop a perpendicular from A towards BC, meeting it in P, such that the triangles ABP and ACP are right. Then use the fact that BP > PC and Pythagoras.

[daring compass]

do i ask geometry questions here instead of in the help section?

[grave pond]

Each is accepted, just please don't do both.

[sudden shell]

I don't know how to do both of the problems

[buoyant kraken]

I believe that for the first one it is angles KIR and KER because they are both opposite of the diameter.

for the second one I believe it is FHG and FYG for the same reasons

[rotund stump]

is it possible to rotate this or no

[haughty tangle]

Somebody help please

[primal jasper]

What part do you need help with?

[haughty tangle]

Like on how to start. I'm so stuck

[primal jasper]

Alright so lets do the simple parts first, so when you have a line interesting a line?

[haughty tangle]

They meet at a single point right?

[primal jasper]

Yes but it’s also split into four parts right?

[haughty tangle]

Oh yeah

[primal jasper]

If you’re thinking of all of those parts as the angles surrounding that one point, they’d all add up to 360

[static hearth]

I just failed my trig class. anyone have an online resource/mega test that I can study so that next semester I pass for sure?

[haughty tangle]

Right because of the full rotation

[primal jasper]

Yup! You can also see it as two half circles

[haughty tangle]

If not sure if I’m correct but the half circle isn’t it 180 degrees

[primal jasper]

Exactly

[haughty tangle]

:0

[primal jasper]

If you have one part of that 180, you can subtract that part from 180 to find the other part.

[haughty tangle]

Nice 👍

[primal jasper]

Then we also see that the next angle is the same as the one opposite of it. This is always the case

Ones right next to the angle are ones that will “complete” the angle making it 180. And the ones opposite of it are just the same as the angle

[haughty tangle]

oh yeah the supplementary angles

i think

[primal jasper]

Yup!

[haughty tangle]

phew

quick question on the 25 degress its called a vertical angle right

so is 40

and 155?

[primal jasper]

Vertical angles is the rule that says opposite angles are always congruent (the-same)

[haughty tangle]

oh yeah their congruent

[primal jasper]

Do you know what the || means in a||b

Okay discord doesn’t like me

[haughty tangle]

in ab?

[primal jasper]

But the do you know what this means |

When in between the a and b

[haughty tangle]

oh

In geometry its a notation where for example a is parallel to line b just like c||d

so its just saying the parallele line are in the same plane that never intersect

[trail tendon]

| |

[primal jasper]

Mhm!

[trail tendon]

oh

i see

[haughty tangle]

yeah

that why they put it in these line ||

so i tried one on my own and idk if its correct so lemme send it

cause 130+50=180

[primal jasper]

Yup perfect!

[haughty tangle]

Yay 😭

I’ve only done these two so far

This one Cuase it’s the same thing like the first one

So I have measures 1 and 3

and 2 because I noticed 1 and 2 are the same degree 130

[primal jasper]

Mhm and why is that?

[haughty tangle]

becuase their on the same line

i think it has a name

im not sure tho

[primal jasper]

Yes, but not quite. It’s because the two lines intersecting that line are parallel

[haughty tangle]

ooh

[primal jasper]

If they weren’t parallel or if not stated they are, you can’t assume that.

[haughty tangle]

wait

is it corresponding?

[haughty tangle]

i see

[primal jasper]

Yup!

[haughty tangle]

oh so thats what i saw

[primal jasper]

The 130 to the 130 & 50 to the 50 are corresponding angles

[haughty tangle]

thats pretty simples

[primal jasper]

So what would be the relationship between 1 & 5

[haughty tangle]

Corresponding?

[primal jasper]

Exactly

[haughty tangle]

o

[primal jasper]

What about 1 to 8

[haughty tangle]

hnm

idk if this is the name

but

same side extrior

[zealous olive]

supplementary

[haughty tangle]

yeah ss exterior is supplementary aswell

[primal jasper]

Exactly, which means they are supplementy

[haughty tangle]

lets go

[primal jasper]

What about 4 to 5

[haughty tangle]

same side interior

which i also think is supplementary

[primal jasper]

Mhm!

And then 4 to 6

[haughty tangle]

alt interior

or alternate

[primal jasper]

Which means the angles are?

[haughty tangle]

congruent

[primal jasper]

Nice

[haughty tangle]

thank god i thought it was supplementary for a second

[primal jasper]

You should be able to complete your assignment pretty easily with that

[haughty tangle]

Ok lemme try

can i send it on here to make sure im correct?

when im done

[primal jasper]

Mhm

[haughty tangle]

👍

[haughty tangle]

This was actually easy 😨

Now I just hope I didn’t mess up

sorry for bad lighting and quality and messy hand writing 😭

[primal jasper]

You got it good job!

[haughty tangle]

Yessss Thank you for you help!!! I just needed a refresher ❤️

[primal jasper]

Ofc ofc!

[autumn pelican]

I really feel stupid right now

I can't figure out how to do (sin(x)+cos(x)) ^2

In my brain I think it'll be sin^2(x) + 2sin(x)cos(x) + cos^2(x) but then a math app tells me the answer is sin(2x) + 1

[zealous olive]

those are the same things

sin^2(x) + cos^2(x) = 1
and
2sinxcosx = sin2x

[autumn pelican]

I think I am lost, why are those the same

[zealous olive]

trig identities

you probably haven’t learned them yet

[autumn pelican]

I'm in the process of learning them but this is my first day so far

Oohh.. so for Pythagorean identities I can add the sin^2 + cos^2 to equal 1, but 2sin(x)cos(x) still confuses me

Nevermind I found it, I forgot about double angle identities

[surreal escarp]

How would I get x without cancelling it both

[dark sparrow]

(x+8)^2 is not equal to x^2 + 64.

@surreal escarp

[surreal escarp]

how so?

[dark sparrow]

dumb example: if x were equal to 2, then
(x+8)^2 would be (2+8)^2 = 10^2 = 100
but x^2 + 64 would be 2^2 + 64 = 4 + 64 = 68

either recall how to expand (a+b)^2, or expand (x+8)(x+8) the long way.

[surreal escarp]

i see

so 256+x^2 = x^2+16x+64 instead?

[dark sparrow]

yes of course

in general (a+b)^n is not equal to a^n+b^n. powers don't distribute over addition like that.

[surreal escarp]

so what should i do now

[dark sparrow]

how would you solve the equation

x^2 + 256 = x^2 + 16x + 64

if it was given to you like that to begin with?

what first step would you take

[shrewd haven]

Get the X’s by themselves?

[surreal escarp]

square root the x

[dark sparrow]

the question was not addressed to you.

[surreal escarp]

or minus the 256 on both sides

[shrewd haven]

Sorry

[dark sparrow]

if you mean to square root both sides, that's supremely unhelpful.

why not just subtract x^2 from both sides?

[surreal escarp]

oo

i got x=12

[gaunt iron]

https://cdn.discordapp.com/attachments/1172366291269587075/1240512082701586502/image.png?ex=6646d468&is=664582e8&hm=88691c343bff3047cfdf75d6ef1fc0511f61891b40198ec00e9a94544225a52a&

I need help with this

[surreal escarp]

that seems right when i plug it in in the reg pythagoreon theorem equation, tysm

[dark sparrow]

yup there you have it

[gaunt iron]

@dark sparrow do yk how to do my problem ive been stuck on it for a while plz

[dark sparrow]

i am not going to even touch it

[upper karma]

Does anyone know how to do this?

😭

[dark sparrow]

there are people on this server who know how to do this, yes.

@upper karma what is troubling you here?

[upper karma]

I don't know where to start

[dark sparrow]

ok so you have a circular sheet of metal and you are cutting a hexagon out of it.

everything inside the circle but outside the hexagon is the scraps that they ask you about.

do you understand this? y/n
(i am not done yet, but i am bringing you up to the same page as myself)

@upper karma

[upper karma]

yes

[dark sparrow]

ok

so the scrap area equals the area of the circle minus the area of the hexagon.

do you see how to continue?

[autumn pelican]

If I am prompted cot = 2/3 and sinx > 0 and then asked

sin2x

Would my answer using the double angle identity be

sin2x = ~1.2569

or

sin2x = 12/13

I can't remember if I replace sin(x)cos(x) with that or if I compute it. Forgive me I have been working for 15 hrs straight and I am tired

[dark sparrow]

how did you get 1.2569

also go take a fucking nap for gods sake

15 hours straight is like a million times longer than you should be working

[autumn pelican]

I have a quiz due in 2 hours that I haven't had time to study for 😭

[autumn pelican]

2sin(3/sqrt13)cos(2/sqrt13)

[dark sparrow]

what. why are you evaluating the sine of 3/sqrt(13) radians.

???

[trail tendon]

people should only be working like 5 seconds per work time 👍

[dark sparrow]

you got yourself royally confused.

[trail tendon]

😭

[autumn pelican]

I have lost the barrier between knowing what tf radians and degree mode does atm

[dark sparrow]

in this instance my use of "million" is not to be taken literally, but also 1/1,000,000 of 15 hours is less than that.

[trail tendon]

oh yeah its less than one second

[autumn pelican]

Was I right to still compute the sin and cos but meant to do this in degree mode?

[trail tendon]

half a second

💀

[dark sparrow]

it actually doesn't matter what mode you use bc this is a non calculator question

[autumn pelican]

Oh, so I wasn't meant to compute the sin and cos

so 12/13 was correct?

[dark sparrow]

...

i mean you weren't meant to try to get x itself.

[autumn pelican]

I.. am even more cdonfused now, let me just take a picture of the question

From my fading mental capacity only thing I could deduce is to use double angle identity

[dark sparrow]

give me a couple minutes to clock in to work

[autumn pelican]

Sure, I'll just write what I did rq

What I did was

find hypot = sqrt13

sin2x = 2sin(3/sqrt13)cos(2/sqrt13) = ~1.2569
or
sin2x = 2(3/sqrt13)(2/sqrt13) = 12/13

I finally discovered I was dumb and not meant to compute it, the 2nd option was right (Dear God I hope)

[dark sparrow]

yes, it is.

it's sin(x) = 3/sqrt(13), not the sine of 3/sqrt(13)

[autumn pelican]

It only gets worse from here

[dark sparrow]

95% of it getting worse is because you have overworked yourself

[autumn pelican]

Oh 100% but if I don't I get a lovely zero

All I have left is another of these but with half angle and then a power reduction and something I don't even know

Could someone quickly confirm for me if the U in half angle identities for.. say sin(u/2) is equal to y/c?

[spiral lodge]

What are y and c?

[autumn pelican]

opposite = y and hypotenuse = c

[spiral lodge]

And what is u?

Without any picture it's impossible to answer

[autumn pelican]

T

u as in the half identity u (I think)

[spiral lodge]

Yes, this is correct

[autumn pelican]

Oh okay great, ty

[spiral lodge]

U is just a name for the angle, you can use whatever letter you like

[autumn pelican]

ah okay

Ok I am swinging back full circle, if I don't have radians to easily get cosu do I replace cosu with -4/5 if my opposite is -4 and hypot is 5?
to make it

+- sqrt( 1-(-4/5) / 2 )

Coincidentally my teachers lecture has the exact same problem on the quiz, so nvm I can figure this out

[upper karma]

Status 1

[dark sparrow]

cos ec typesetting fail

[upper karma]

?

[dark sparrow]

they tried to type cosec as in cosecant, but whatever software they used for typesetting took the first three letters as meaning cos

[upper karma]

Oh

[dark sparrow]

it is an inconsequential but noticeable typesetting defect

csc anyway...

(i don't know how to do this problem without headache)

[upper karma]

This is what the solution says(Screen shot was not allowed so I had to write it down)

[dark sparrow]

you mean 2^(r-1) and not 2^(n-1) in the sum?

[upper karma]

Something to remember or can it be derived in a easy way ?

[upper karma]

mb

[dark sparrow]

no idea how they would derive it

it looks annoying

[upper karma]

Okay

[nocturne remnant]

Surely you just induct or sth

[upper karma]

Induct?

Induction .

?

Anyway

Guessed it

Status 1

[autumn pelican]

Is this all the question is asking me to do?

Sorry let me flip that

[upper karma]

,rotate

[autumn pelican]

That works, ty

[dark sparrow]

keep going

expand this

you will get a cos^2(2x)

apply that identity again

[autumn pelican]

Got it, thanks I'll do that rq

Why isn't it cos^2(4x)?

Oh wait, nvm

[upper karma]

[autumn pelican]

Is the answer

3 + 3cos(2x)
/
4

[dark sparrow]

never write fractions like this again.

but also definitely not.

[autumn pelican]

sorry lmao I thought it was going to look better

Yeah I figured 😦

[dark sparrow]

show your work so i can tell you where you screwed up.

[autumn pelican]

,rotate

[dark sparrow]

cos^2(2x) = (1+cos(2*2x))/2.

(1+cos(2x))/2 would be cos^2(1x) and not cos^2(2x).

[autumn pelican]

oh okay so the 2 * goes inside the cos not on the outside of the parenthesis

[dark sparrow]

..... yes obviously since you are working with cos^2(2x) and not with 2cos^2(x)...

[autumn pelican]

So right after I convert cos^2(2x) it should look like

( 1 + 2cos(2x) + ((1 + cos(4x)/2) ) / 4
```?

[upper karma]

Any idea, Ann ?

[dark sparrow]

yes, and now clean that up

[dark sparrow]

no

[upper karma]

For the first time

[dark sparrow]

it's more like i don't want to touch that problem.

[upper karma]

Oh

[autumn pelican]

It's all for naught, I can't finish all these questions in time 😦

[obsidian harness]

Wait I don't get what the question wants you to do

You can write cos^4 x in terms of a constant, cos 2x, and cos 4x

[faint pasture]

that is exactly what it wants you do i guess

[upper karma]

@obsidian harness

[obsidian harness]

which one?

[upper karma]

Both

[autumn pelican]

I CAN SLEEEPPP I submitted it a whopping 4 minutes before I would have gotten 0 points. Fuck college

[upper karma]

South?

[obsidian harness]

cba

[upper karma]

Cba?

[obsidian harness]

can't be assed

[obsidian harness]

pls ask in #calculus

or in one of the help channels, see #❓how-to-get-help to know where to find them

[upper karma]

Okay

In which category does complex numbers come under ?

[dark sparrow]

#precalculus but honestly just go wherever in pre-uni.

[upper karma]

Okay thank you

[gilded egret]

hey guys

any1 good at like cosine proofs?

[lime dune]

!da2a

[lime crown]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[zealous olive]

what’s the question

[gilded egret]

so uh

triangle with A 38degrees, B 55degrees, C 87degrees and a 14m, c and b are unknown and they want me to find out what c is

[lime dune]

that’s not “cosine proof”???

law of sines instakills it

[gilded egret]

oh i already got proofs down i meant like

well i would love some help understanding it

but

this

sorry for crude drawings

[gilded egret]

this one 🙏

[lime dune]

there is nothing particularly deep going on here

just plug into law of sines and solve for c

a/sin(A)=b/sin(B)=c/sin(C)

[upper karma]

Nvm sorry oops

[bronze wind]

can someone give me the step by step process of solving for the surface area of this figure? Im stumped and I have gotten like 5 different answers

(I just need the process, I know that this answer here is correct because I checked the answer key to the assignment)

[silent plank]

can you show your most recent attempt

[bronze wind]

I erased it

I just need to know how to use the formula

because I currently have no idea how to plug the numbers in

[silent plank]

area of each face (there are 5)

doable with basic area of rectangles and triangles

then add them all up

the 30in side length you calculated is correct btw

[bronze wind]

do you have like the actual formula im supposed to use for a polyhedron

and do you know how I would plug it in for this particular scenario

[silent plank]

general formula would be
add up area of each face/surface

[bronze wind]

ok

[silent plank]

there are formulae for certain solids,
but that's where they come from, so better off remembering that instead
instead of a different formula for each one

[bronze wind]

yeah its weird because our instructor had us remember a specific formula for a polyhedron and it makes little sense

[obsidian harness]

ew

memorisation

[silent plank]

priority is to remember where stuff comes from

if you rely fully on memorising formula and don't know where they come from
and you for some reason forget or misremember, you're screwed

[bronze wind]

I wish this is how we did it

unfortunately public education thinks that everything is needed to be memorized

[sharp ruin]

Heyy
Can somebody help me derive the formula for the excenter of a triangle?
I know it would require me to use external angle bisector theorem but I can't figure out how can I do that?

[ebon swallow]

can someone help me with geometry and solving for volume

[pale spire]

Given the positions of the Moon, Alice, and Bob, find the length of the shortest path that starts at one of the players, touches (or crosses) the edge or the interior of the Moon, and ends at the position of the other player.

[sudden shell]

Can someone explain this to me someone told me the answer was 80 but I need to understand how it’s 80

[trail tendon]

assuming that dot is the center...

[sudden shell]

oh Thank you for the help know I understand it

[quick stump]

ok so basically angle F is a right angle because the diameter is the hypotenuse, and the measure of the arc is twice the measure angle GFP, which means it is 80 degrees

[halcyon sand]

Let MA and MB be tangents to (O). MCD is the secant line. Draw BK and BL perpendicular to AC and AD at K and L respectively. KL intersects CD at I. Prove that: IK = IL

can someone help me?

[sharp ruin]

Can somebody help?
I'm really stuck on this for 2 days.

[dusky roost]

for the 4 square root of 2, is it just asking for the square root of 2 multiplied by 4?

[halcyon sand]

wdym by that

[dusky roost]

the top left thing

[halcyon sand]

wut abt it

[dusky roost]

what does it mean

im so confused

wtf does 4 squareroot 2 mean

[halcyon sand]

it means 4 multiply wit sqrt2

[dusky roost]

oh ok

thanks gang

am i cooked for my math final

[halcyon sand]

yes

wut grade rr u in

no offense

[dark sparrow]

it's not asking for that as such.

but $4\sqrt{2}$ does \textit{mean} $4 \times \sqrt{2}$, yes.

[somber coyote]

|Ann⟩

[ripe prairie]

wait

so

2 legs on a 45 45 triangle are 4 root 2

and A^2 + B^2 = C^2

so 4 root 2 squared+ 4 root 2 squared would equal c^2

but the ^2 cancels out the root

so its 4X2+4X2= C^2

which means c^2 equals 16

but its not

the hypontenuse is 8

so why is it 8

[north kindle]

This is not correct

Also don’t use x as a multiplication symbol

[ripe prairie]

ik

i dont see a dot on my keyboard

[dark sparrow]

use *

[ripe prairie]

alr

[dark sparrow]

or \* to stop discord eating it as italics

[ripe prairie]

*

[dark sparrow]

also use sqrt() for square root

[ripe prairie]

k

[dark sparrow]

also $(4 \sqrt{2})^2 \neq 4 \cdot 2$.

[somber coyote]

|Ann⟩

[ripe prairie]

how come

[dark sparrow]

$(a \cdot b)^2 \neq a \cdot (b^2)$.

[somber coyote]

|Ann⟩

[ripe prairie]

ohhhh

cuz the 2 squared is outside

alr

[sharp ruin]

Is somebody here actually available now?
Please refer to my question above

[sharp ruin]

This one

[ripe prairie]

someone help my boy

[sharp ruin]

Not gonna lie
I thought I'd get help here

But I have been ignored for more than 24 hrs straight

[north kindle]

No one owes you their time

[sharp ruin]

Yeah they don't of course

I just said I thought someone would help me
Sorry if it sounded rude

[north kindle]

Also that’s a fairly non trivial question so the number of people willing and able to help is lower than some other questions

I’d recommend going to a dedicated help channel, those are normally easier to get help in from my experience

[sharp ruin]

Can you send some links?
I don't really know any

Even this channel was recommended to me by a friend of mine

[north kindle]

#❓how-to-get-help

[sharp ruin]

Ohh here

Thanks for letting me know

[dark sparrow]

not only is it a nontrivial question

it's also very unclear as stated

you didn't tell us what data you are given about your triangle and what output you expect (coordinates of the excenter, maybe?)

[sharp ruin]

Yes I need the coordinates of the excenter

The given can be summed up in this figure
I've posted the same in the help section though

Do you have any idea though? @dark sparrow

[dark sparrow]

i don't really care about this enough to work anything out, so no.

[trail tendon]

ppl on here either aren't that good at that kind of geometry or don't care, apparently lol

[sharp ruin]

Whatever man
If I get something on the help section, well nice
If I get something here, well that's nice too
If I don't, I'll eventually figure out by myself after sometime

[trail tendon]

for the record im more of the first one 😂

[sharp ruin]

Well I'm also new to geometry though

So maybe count me in with you too 😆

[trail tendon]

that doesn't look like "new to geometry" 😭

like i aint even know what an excenter is

[dark sparrow]

the center of a circle tangent to one side and the extensions of the other two.

[sharp ruin]

Maybe man
I'm an Asian
And you know how it is here 😁

[sharp ruin]

That is kinda the consequence of what it primarily is

[trail tendon]

when i think of geometry i think of like basic finding degrees and arc lengths of circles n stuff

the hard part is like all the weird things liek excenter

[dark sparrow]

??

[trail tendon]

what is tangent, to one side of what, and the extensions of the other two what

is this a triangle?

[sharp ruin]

It's the intersection point of the internal angle bisector of one of the angles and the external angle bisectors of the other 2

[dark sparrow]

that's your definition of it?

[dark sparrow]

you don't know what the word "tangent" means? as in a line tangent to a circle?

but yes excircles and excenters are defined for a triangle

[sharp ruin]

That's the definition on wiki, that's what I was told by my teacher

[trail tendon]

i thought you were saying the center was tangent to the circle, i don't understand english 💀

[dark sparrow]

what is your native language

[trail tendon]

english

i'm just bad lol

[dark sparrow]

... what other languages do you speak

[trail tendon]

i would say the only language im fluent at is english

i speak a little spanish ig

the center of a circle, tangent to one side
to me, another way of saying this
the center of a circle is tangent to one side

the center of a circle is a noun, tangent to one side is i guess an adjective phrase of some sort? idk grammar but the sentence doesnt make sense to me

;-;

[dark sparrow]

the center of [a circle which is tangent to [one side and the extensions of the other two]]

apologies for not making my sentences unmisinterpretable

[sharp ruin]

After thinking a lot, I've reached the result

It's really satisfying at the end. Not gonna lie

[amber hollow]

how am i supposed to tell when im supposed to use the angle i get and when im supposed to subtract it from 180

[dark sparrow]

how exactly did you get these angles?

[amber hollow]

cosine rule into sine rules

i mean law

[dark sparrow]

... why not just use the cosine law 3 times lol

you've got all 3 sides

[amber hollow]

cause the sine one is easier

[dark sparrow]

if you're using a calculator it aint much of a difference

but anyway

[amber hollow]

easier to make mistakes with the cosine one

[dark sparrow]

with the sine law you run into the issue that sin(180°-x) = sin(x),
so you have to know if the angle you're calculating is acute or obtuse

since sin^-1 will only ever give you the acute ones

[amber hollow]

oh

[dark sparrow]

and that, if you so desire, can be accomplished by comparing a^2 + b^2 with c^2 (where a and b are the sides adjacent to the angle in question and c is the 3rd side)

if a^2+b^2 were equal to c^2 the angle would obviously be right, bc pythagoras

if a^2 + b^2 < c^2 the angle is obtuse, and if a^2 + b^2 > c^2 the angle is acute.

[dusky roost]

https://cdn.discordapp.com/attachments/1136489990528970932/1240876003866316820/copy_CCEB1945-7E3F-43A8-9BE9-8C49DF8F406C.mov?ex=66482755&is=6646d5d5&hm=1e2cd90940a53a295d0e0012636d4d41fbcb991c4f1dc0050deabeebb1d0b2b8&

oops wrong photo

[dark sparrow]

very wrong photo.

[dusky roost]

what the skibidi

[dark sparrow]

but then again, this is about as difficult as just applying the cosine law to begin with.

since cosine law would have you calculate (a^2 + b^2 - c^2)/(2ab) anyway.

[amber hollow]

so long story short, use the cosine rule if i have 3 sides unless i can clearly tell if the angles are obtuse or acute at least

[dark sparrow]

nothing wrong with just always using the cosine rule if you have 3 sides.