geometry-and-trigonometry

[blissful spire]

Right?

[thin locust]

yes

[blissful spire]

Cause the circles area would be pi r squared

Which is 25pi

Ok Tysm

[thin locust]

ye

[upper karma]

Area of the Circle= pi r^2= pi*5^2 which is approx. 78.53982 (round it accordingly)

[blissful spire]

I think it’s just asking for the exact answer so I can keep it in pi form

[upper karma]

alr!

[bronze sand]

in a rectangle, the base is twice as long as the height. calculate the area of the rectangle if the diagonal is √215 cm.

Can someone tell me how to solve this question??

[nocturne remnant]

represent the base and height in terms of the same variable

[wind coral]

Here I drew you a model

Use Pythagoras to determine x

[upper karma]

what do they mean by this

[dry atlas]

Trig is used all over the place where angles are involved, not necessarily only when triangles are involved

[upper karma]

ah thanks

[limpid onyx]

Prove CU x CV=DU x DV where U and V are the circumcenters of AOC and BOD respectively

[cold chasm]

is there some sort of list for theorems abt circles

[wind coral]

https://thirdspacelearning.com/gcse-maths/geometry-and-measure/circle-theorems/ found this. It might help

[cold chasm]

i was told there was 14

i can send a list but i need help proving them

[wind coral]

Oh ok, I can't help prove them sty

[cold chasm]

its ok

[visual ermine]

Guys so like if I'm evaluating functions and like it gets to cos(-1) like then what

Cus tbh my trig is not great and like I just keep getting messy decimals and no good values or anything

Unless I'm being really dumb and it's pi but I'm pretty sure that's not how trig works

[south pawn]

could somebody help

<@&286206848099549185>

[foggy parcel]

How far u have done with this ques?

[wicked narwhal]

@south pawn

[dull seal]

Use the cosine rule

[rough stratus]

hi

@rough stratus hi

[river blaze]

Find X?

[river blaze]

1610sin(67°)

[lapis wind]

help pls

[pallid vine]

Start off with y= tan x
You can see that the graph is translated pi/4 to the right first, so we now have y = tan (x - pi/4).
From here, we check for at least one point:
E.g. when x = pi/2, the y-coordinate should be 3. As tan (pi/2 - pi/4) = 1, we need to multiply this equation by three to stretch it parallel to the y-axis, and get the desired equation y = 3 tan (x-pi/4)

[upper karma]

"Prove that if a plane is perpendicular to two secant planes, then it is
perpendicular to the intersection of these planes."

help pls

[floral shale]

You can try doing the cross product of the planes' normal vectors first

Not entirely sure but that's something you could try

[upper karma]

hello. im not sure if i can post this question because its in calculus. but basically im trying to solve a question that deals with arctan and the limit as "a" is approaching -infinity of arctan(a). should i look at the tan(x)'s domain to justify the behavior of arctan?

[pallid vine]

Seems like a reasonable step, but just be careful to justify, since tan(x) has the domain over all the reals, whereas the domain of tan(x) is restricted to define arctan(x) as a function, such that it's range is (-pi/2, pi/2).
Since we're in geometry chat, you could also consider, if you had a right-angled triangle, and one of the acute angles has fixed adjacent side, but an opposite side which has its length approaching infinity...
Seems kind of wishy washy, but it seems like as if the angle gets closer to pi/2.

[upper karma]

thank you for explaining. that's what i was worried about, getting confused with the domain. ok just so im clear completely, tan(x) is only restricted to define arctan(x) from -pi/2 to +pi/2, so the limit would only exist between -pi/2 and +pi/2?

[pallid vine]

That would probably be fine, so long as you keep domains/ranges clear for each function, and don't get mixed up between the two :)

[upper karma]

ok great thanks for helping

[granite sparrow]

Hi

I need some help

1. Prove that
   (Sinx + Cosx )² + (Sinx - Cosx )²x = 2

[fallow sable]

Can someone explain the operations on deriving half angle identifies from the Euler formula

my tutor showed me the formula for if I forgot my identities

I kinda forgot how he even started it 🥲

I’m in precalc 2 btw if that’s necessary information

[verbal jacinth]

What is the given informations

[nocturne remnant]

There are at least 9 things named “Euler’s formula”

[spiral mesa]

You may want to look at this too

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

[warm canopy]

the things i sent are all the info i have

[upper karma]

I think the sum is 180 degrees

You just have a star right ?

a random pentagram with nothing specified

lemme try

Lemme upload pic

@warm canopy

[warm canopy]

thank you!!

[upper karma]

no problem

Glad to help

Now I disappear into the shadows once more ...

[nocturne remnant]

...you literally asked them if it was asking for angle sum and it wasnt tho

tho honestly that's the only reasonable thing that would be asked

[warm canopy]

wait which angle is which and where did the A’ B’ C’ D’ E’ come from are those the interior angles of the polygons

[upper karma]

Yes

I wrote it in a quick way

[fallow sable]

I meant this one

[half eagle]

how do i solve this?

[grave pond]

Doesn't your textbook explain how to find the distance between points in a coordinate system?

[half eagle]

it does

i think'

but i dont have it on me rn

soo

its a formula

but i didnt memorise it

[grave pond]

You might have to rederive it, then. Draw a hoizontal line through one of the points and a vertical line though the other. Together with the line connecting your two points, they form a triangle with one right angle. Apply the most famous theorem about right triangles you remember.

... or you could also use what Daht posted and lose the learning opportunity.

[flat owl]

My bad

[timid thistle]

is there things like

uh

like you have $\sin \theta = \frac{a}{b}$ and then $\sin^2 \theta = \frac{a^2}{b^2}$, is there such a thing as eg. $\arcsin^2$?

[grave pond]

There's not a particular name for it, no.

In some situations it can be useful to know that (by trigonometric identities about squared sines) we do have $$\sin^{-1}\sqrt x = \frac{cos^{-1}(1-2x)}{2}$$

[somber coyote]

Troposphere

[timber cargo]

You could use the pythagorean theorem or the distance formula

[smoky jetty]

if i drew this correctly, line AB is hypotenuse right?

[pliant pumice]

Hello! I have a question about proofs. I learned that proofs are usually written using table of statements and reasons, and every small step should have a reason (name of some property, not just explanation). I find this way very long and exhausting. Are there other ways to write proofs and solutions in geometry?

[dark sparrow]

you learned that 2-col proofs were the "usual" format?

cause yeah thats bureaucratic as shit and tiring to stick to

usually a porof is just plain-english prose maybe interspersed with some calculation

[pliant pumice]

So usual "non-bureaucratic" proof is just explanations in words and some calculations when needed?

[dark sparrow]

yes, essentially.

[silent plank]

two col is excessive in the need to explicitly state "given" and trivial theorems

which yeh, can be exhausting, there's no real need to have all that

[grave pond]

Yes.

I think "two-column" proofs could make some sense if they were taught as a scratchwork technique for making sure you have all your ducks in a row before you start writing out the human-readable proof. But they're likely harmful if they become a homework objective in and of themselves.

[pliant pumice]

Thanks!

[dire mortar]

yo hello can anyone help with my homework? trig

[upper karma]

@everyone does anyone know about pythagoras theorem?

[grave pond]

Nobody knows about Pythagoras' theorem; the last surviving description of it was lost when the Great Library of Alexandria was sacked.

[upper karma]

I think you're talking about arithmetic, Pythagoras theorem is actually described in the Bible

[timber cargo]

Okay

[upper karma]

how do I know if my foci is parallel to the y-axis or the x-axis?

im wondering if its related to the semi-major, but im not quite certain

[grave pond]

The foci of an allipse always lie on the major axis.

[devout warren]

is there any reasoning behind this being ex. sin(v+pi) and not sin(v ±pi)?

[upper karma]

ok, where a is greatest, that's where the focal is?

since semimajor = the a component

[tender token]

So I'm assuming you're asking about the focal point of an ellipse, which is one of the two points that define the major axis of the ellipse.

In an ellipse, the distance between the center and each of the two foci is denoted by c, and the distance between the center and each vertex (i.e., the endpoints of the major axis) is denoted by a. The semi-major axis of the ellipse is then given by a, and the semi-minor axis is given by b, where b = sqrt(a^2 - c^2).

So to answer your question, no, the location of the focal point(s) is not determined solely by the value of a. The location of the focal point(s) depends on the values of both a and c, as well as the position and orientation of the ellipse.

However, it is true that the value of a is related to the length of the major axis of the ellipse, and therefore to the distance between the two vertices. Specifically, the length of the major axis is given by 2a.

[upper karma]

i was asking more so about the foci's relationship to the y-axis versus the x-axis. because sometimes you need to use different equations for the same problem

@tender token

is it just the semimajor that decides?

sorry if im asking the same thing after you explained but i dont really understand ellipses

[tender token]

What would the points of my own geometrical algebra equation be? I used such 2 + 2 = 2x (squared (2)) which both sides of a square is 2 it converts into a cube which is entered into cube = 2x (squared (2))?

Summary: Square (both sides are 2) = 2x² = Cube 2x²

[median igloo]

My whole class strugglin need som help

[foggy parcel]

26 should be 22.83 the value of DE

26 no. Ques ans: 22.83

@median igloo

[dark sparrow]

!nosols

[lime crown]

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

[snow tide]

"find the length of the unknown sides of this right trapezoid, and find the area of the trapezoid"

im struggling with this problem

[austere light]

not sure what i did wrong here, deltamath says im wrong

nevermind i figured it out

[abstract dagger]

👌

[copper lark]

I need help

Can u do this question

Help

Ignore the writing around it can you help with that question tho

[lofty tangle]

do you still need help

[lofty tangle]

for the bonus (29) use Heron's formula to solve ((1/2)(a+b+c))=s then sqrt(s)(s-a)(s-b)(s-c)

[lofty tangle]

for #25 use law of sines to find angle z, then use angle sum theorem to find angle y

[grave pond]

I've drawn a few more lines -- red is horizontal and green is parallel to the inclined plane so the angle between red and green is the angle of the ramp.

Now red is perpendicular to black, and green is perpendicular to dashed blue.

[smoky jetty]

the only way I could think of rn, is by finding the opposide side to angle 25 (you'd need a calculator), then it should be easy.

[grand heron]

you can figure out the sides easy with the trigonometric rules of right angled triangles. Is thr problem figuring out the area of a triangle?

[graceful quest]

b

[timid thistle]

what can you do to that angle that gives you some insight to the sides?

[grand heron]

with these trigonometric rules you can figure out the sides if the triangle

[smoky jetty]

you may also use law sine for the oblique trianglge

[timid thistle]

indeed

[foggy parcel]

Ups here B & C are the centre of the two circles

Since radius 2
AB = BC = AC = 2
so according to the tringle being equilateral
angleABC = angleACB = angleBAC = 60degrees

so [ABC circular cut] = theta/360 * pier^2
= 60/360 pie * 2^2
= 1/6 * pie * 4
= 2pie/3

and [tringleABC] = rootover3 a^2/4
= rootover3 * 4/4
= rootover 3

so [black region under ABC circular cut] = 2pie/3 - rootover 3
= 2pie - 3rootover3/3

So the other one is also 2pie - 3rootover3/3

And the down black region can be divided with the tringle where it creates the same situation like the upper one

So there should be 2 other black region under circular cut with same area

So all black regions under circular cut are 4
[all black regions under circular cut] = 4{(2pie - 3rootover3)/3}
= 8pie - 12rootover3/3

And the other same tringle (with same area) should also be added as the down black region except the 2 down black regions under circular cut

So the actual [black region] = 8pie - 12rootover3/3 +root3
= 8pie -12rootover3 + 3rootover3/3
= 8pie - 9rootover3/3
Hence,
8pie - 9rootover3 / 3 = apie - b rootover c/d

a = 8
b = 9
c = 3
d = 3

and gcd{8,9,3,3} = 1

              So     ( a + b + c + d)   =  8 + 9 + 3 +3 
                                                    = 23
                                                     {Ans?)

Am I right or not?

<@&286206848099549185>

[foggy parcel]

THis ques

[dense galleon]

Isn’t ABC triangle equilateral

[foggy parcel]

Yea

[foggy parcel]

Well I have added a solution but not sure if this is correct or not plz help me <@&286206848099549185>

[chrome timber]

Hey anyone here studies Euclidean geometry

[grand heron]

from all of the people that are in this server there is a very high chance that someone is or did

[reef frigate]

obc is 50 because ob bisects ac. so obc is 50 so u know bac. ob = oa so z = 50 - bac.

[split grail]

aob = 180 - oba - oab = 180 - 2*50 = 80;

bac + bca + abc = 180 ==> bac = (180 - abc )/2 ==> bac = 40

[upper karma]

hello. just a question, is the definition of horizontal tranverse axis means that the parabola opens upwards and down?

[graceful idol]

can anyone help me with this one, i thought the answer was 82 but i guess not

[upper karma]

OMG I'M LEARNING THIS, TOO

SOO FRIKIN HARD+EASY LMAO

just use this formula... m<1=(m of arc AB + m of arc CD)
(These are just random letters, you can do it accordingly for your problem)

he left oml

[upper karma]

76= (82 + x)/2
76= 41 + x/2
x=70 {x: arc BC}

[upper karma]

hello guys, i have a question. Idk if this is basic or advanced math too so i ask here

i saw this guys calculate circumcircle of a given triangle's coordinates with something called D. I have no idea what that is and how do this work

you may confuse by coding, but at the end he give result of the triangle circumcircle coordinate is:
(X / D, Y / D)

[grand cargo]

Hi, can someone help me with this one? We're supposed to find the area for the ones that are shaded.

[upper karma]

assume 2 rectangle's edges is x and y, you calculate "7" and "5" based on x and y

and reversed the formular

[grand cargo]

Something like this?

[upper karma]

yeah

[grand cargo]

Ah

So x= 7 and y=5?

Wait nvm I don't think it's like that 💀

[upper karma]

i dont really remember any formula from high school, but there would be something to do with the angle

if you have 1 angle and 2 edge, you can calculate the other edge

[grand cargo]

Huh

[upper karma]

you apply that formula to both triangle, you will have an equation

solve that equation you will have the answer

[grand cargo]

does that only applies for right triangles or not?

[upper karma]

no, obs not

[grand cargo]

The angle is unknown in the problem however

[upper karma]

that is why we need the equation

if you notice closer, sum of that 2 angle is 180

[grand cargo]

Wait which one

[upper karma]

well, seem like everything is obvious when youre asian haha

the one in middle

[grand cargo]

💀

Huh

Still trying to figure it out

[upper karma]

you see it now?

what is 360 - 90 - 90?

[grand cargo]

180

Ok yeah I see it now

[grand cargo]

Wait how do we find the 2 angles though, they clearly don't make a 90°

And how do we find the area after that 💀

[nocturne remnant]

You meant cosine rule

just use it on each triangle and hope that stuff cancel

A solution without trigonometry probably exists but is not very easy to find

[grand cargo]

e

I was thinking about using that but the angle is unknown

[nocturne remnant]

The angles are supplementary; their cosines sum to 0

[grand cargo]

Huh

[nocturne remnant]

Fyi I was talking about this

[grand cargo]

Ah

[worldly tundra]

im confused an what nEZ does

[exotic yarrow]

It means $n$ is an integer, $\mathbb{Z}$ represents the set of integers

[somber coyote]

Civil Service Pigeon

[exotic yarrow]

It’s because you need to add integer multiples of the period

Cause otherwise, you could say stuff like $\sin(\pi/2)=\sin(\pi)$, which is clearly false.

[somber coyote]

Civil Service Pigeon

[worldly tundra]

ty!

[exotic yarrow]

Pythagorean theorem

[astral viper]

Yes

Ok

What’s m

=

[exotic yarrow]

Probably the unit metres?

[timber cargo]

Such a tall building with only 3 floors

[worldly tundra]

what i did so far was made 3x=a and then said 10(2sinacosa)=0

but im not sure where to go from there

[gaunt path]

How do I find x

[timber cargo]

What do you know about trigonometric functions?

[gaunt path]

Ik sin cosine and tangent and law of sine and law of cosine but I forgot how to implement the law of sine and cosine

law of sine right

since its an enclosed angle

[timber cargo]

For this problem it's more convenient to use tangent

You are given a side length that is opposite of the angle and adjacent to the angle

So, you'll have to use tan^-1 (30/8) to get the angle

[gaunt path]

ohhhhh

ok thank you that makes sense

I forgot abt the inverse ones

I was trying to write 8tan(x)=30

so 75.07 deg?

[timber cargo]

If they ask you to round to the nearest hundreds, then yes

[blissful spire]

Is there a clear formula to calculate volume?

Sry it was just unclear for me

Nvm sry got it

[timber cargo]

The formula varies by the shape

[shy sun]

Not really sure how to explain this. I have a line to a point and a 3d vector from that point with an X and Y rotation. How can I find the angle between the line and the vector?

[gilded topaz]

Can someone please tell me if 13 is right

I think I did it wrong but idk how else to do it😭

guys?

pls

[nocturne remnant]

i think arc YVU stands for the arc that starts from Y and goes counterclockwise around the circle until meeting U

[lethal fable]

how to do this

😭

please help

[hexed shale]

are you trying to find the hypotnuse

[lethal fable]

yes

[timber cargo]

Use your trigonometric function rules

[lethal fable]

😭

What the formula I dont have my notes

[neat plank]

sin 31 = 16/x

[proven badger]

Need help with these questions (last three especially) Thanks!!

[agile oyster]

area of triangle = ((ab/2)*sin(C)) where a and b are two side lengths and C is the angle between the sides. the answer is now just plug to calculator

[smoky jetty]

You may also refer to Heron's formula as 3 sides are also given

use calculator for convenience

[dire snow]

How do I prove that in a triangle the semi perimeter is always bigger than a side?

[silent plank]

a late?

[dire snow]

A side?

I think you call it side

[silent plank]

what have you tried?

[dire snow]

This

Maybe I have to subtract ab and bc in every part

[void charm]

its the semi perimeter
semi perimeter=(a+b+c)/2

[dire snow]

Yeah I mean that

[silent plank]

use lowercase notation instead of segment notation,
it'd be much easier to read

[dire snow]

?

[void charm]

whats the equation?

[dire snow]

Wdym

It's here

[void charm]

brb

[dire snow]

.

Oki

[silent plank]

instead of triangle inequality, you can use something like
a <= b<= c

wlog

[dire snow]

Uhm

No I don't know i know only this

Btw

[silent plank]

its just basically sorting the lengths in order for convenience

[dire snow]

Is it fine if I subtract two egual sides in every part?

[dire snow]

Oh never heard

But I can try it

Then I have to do another thing

It asks me another thing

[silent plank]

and apply that to get more info about
s = (a+b+c)/2

[dire snow]

Did I do it well?

[void charm]

backk

[silent plank]

well your inequalities are just saying that double the perimeter is greater than the perimeter

which is trivially true

[dire snow]

Yeah but

[void charm]

can u tell me the equation?

[dire snow]

What equation pt2

Xd

[void charm]

lmao

i mean the question!

that u wanna solve

[dire snow]

.

[void charm]

let me think ;-;

The sum of any two sides in a triangle is greater than the third side.

Let a b and c be the length of the three sides

So we have (a+b) >c

Adding c to both sides we bave

a+b+c > 2c

This implies (a+b+c)/2 > c

The LHS is semi- perimeter of a triangle.

So semi- perimeter of a triangle is greater than any side of the triangle.

[dire snow]

Lmao you are a computer to write this in a second

[void charm]

its basically like

[silent plank]

consider given
a <= b
then
if a = k
you can conclude that
b >= k

[void charm]

b+c>a

or like

a+b+c>2a

that is

[dire snow]

Okk

Thx now I get it

[void charm]

(a+b+c)/2>a

then

[silent plank]

did you want greater than at least one side, or the longest side

[void charm]

a+b+c/2>b

and

[dire snow]

Then I have to find how the semi perimeter is more long that the height

I try

[void charm]

(a+b+c)/2 >c

did u get it

;-;

[dire snow]

Yeah

Thx

[void charm]

yeh

np :)

what grade are u in rn?

[dire snow]

I try to do the other part

[dire snow]

Uhmm

[void charm]

?

[dire snow]

Let me check

[void charm]

huh?

i asked what grade like 9th grade 10th grade and stuff ;-;

[dire snow]

I'm Italian we don't have grade lmao

[void charm]

oh

i didnt know

what year?

[dire snow]

Year 11 for Uk

[void charm]

oh

okay

[dire snow]

I looked only my age lol

[void charm]

so like how old are u

wait i have this equation

[dire snow]

15

[void charm]

ohk

do u know the formula of finding CSA of a cylinder

wait i forgot-

[dire snow]

What is csa

[void charm]

curved surface area?

LSA and TSA

um?

[dire snow]

Oh

[void charm]

yeh

[dire snow]

We did the last year

Perimeter *height?

[void charm]

yehh im 14 rn so yeh

wrong

[dire snow]

What is a curved surface area lol

No we didn't do it

[void charm]

2 pie r h

[dire snow]

Didn't do it yeah

[void charm]

lateral surface area?

total surface area?

u didnt do it?

[dire snow]

For lateral you mean all the area without counting the bases right?

[void charm]

yeh

tsa is with bases

[dire snow]

Oh k I did it

[void charm]

yehh

lol

do u remember statistics?

ig u do have it now also

[dire snow]

Oh Lmao I wrote perimetef

I mean

Area*height

[void charm]

yeh ;-;

[dire snow]

Yes

[void charm]

mean,median and tsuff

stuff*

[dire snow]

Not probability bcs I can't do nothing

[dire snow]

Ok this yeah

[void charm]

yeh that kinda complicated but yeh

[dire snow]

Btw I can't do the other part

[void charm]

which one?

[dire snow]

How do I prove that in a triangle the semi perimeter is always bigger than the height ?

[void charm]

uh wait

[dire snow]

I try it too

[void charm]

i thought u understood?

[dire snow]

Oh got it

[void charm]

;-;

[dire snow]

It's the height now

[void charm]

ohh

the height

[dire snow]

I let you see if my think is good

Wait

[void charm]

idk that ;-; let me try

yeah idk

ask someone from ur class

if they know it

[void charm]

i seeeeee

noicee

[dire snow]

Yeah lesgo

[void charm]

goodjob

[dire snow]

I have mainly to start thinking on what it asks

I have another exercise I will see if I have problems

[void charm]

okay

whens ur finals?

[dire snow]

Thx for the help again

[void charm]

ur final exams?

[void charm]

happy to help

[dire snow]

You mean to graduate to university?

[void charm]

yeh

[dire snow]

I have to do another 4 years

[void charm]

ohhh okay

cool

[dire snow]

So basically I go to university at 18 years

[void charm]

yeah i got it

ur 15 now

so yeah

3 or 4

more yrs

[dire snow]

Yup

[void charm]

are u good at bio?

if u are explai types of animal tissues

[dire snow]

Maybe I think we have done like a melting pot of science in the last years ahaha

[void charm]

uh

[dire snow]

Ok Lmao I didn't do tissues

[void charm]

bruh

tell me more about mole concept in chem

[dire snow]

Oh wait

I've thought you were referring of animals in general 💀

[void charm]

is it me or is it u?

being dumb rn

[dire snow]

Me that I don't read

Lmao

[void charm]

help!

im so bad in bio

fr

i lost like 4 marks

forgot

we shouldnt curse

;-;

and the assertion and reason ong 💀

[dire snow]

We've only did muscular and nervous but in general we didn't do biology. I can help you more in astronomy and chemistry

[void charm]

yeh chem

[dire snow]

We've only studied the human body

[void charm]

oh

okay

and the compounds atomic mass

[dire snow]

Lmao at 14 you've already did these things? 💀

[void charm]

i have just memorised first 20 elements in the periodic table

[void charm]

uh yeahhh

[dire snow]

Ok we have studied like 60

[void charm]

yeh

[dire snow]

We'll doing these things the next year

[void charm]

u fr rn?

what the freak

then why do i have these

[dire snow]

Lmao

[void charm]

i finished it

and then suspensions and colloids

[dire snow]

It's because the schools in the world aren't the same

[void charm]

homogeneous and heterogeneous mixtures!

those are the easiest

[dire snow]

Yeah we are here

[void charm]

yehh

so tell me what do u understand by it

homogeneous has like solutions

and heterogeneous is of 2 types

colloids and suspensions

have u studied about saturated ,un saturated and super saturated solutions?

[dire snow]

I'm writing what I know about them lmao

[dire snow]

Nop

Satured is like when you can't dissolve anything else

[void charm]

do u want me to give u a brief about it

so u can answer confidently in ur class when the q comes!

[dire snow]

Lmao we are i the geometry part btw

[void charm]

oh okay

so u want me to or not?

[dire snow]

Yeah but maybe not here

I didn't see it

[void charm]

what?

[dire snow]

That we are in this channel

[void charm]

oh yeah

so where?

[dire snow]

Where do you want but I will have to study astronomy and chimics for tomorrow test

[void charm]

okay study i dont wanna disturb ima do maths ig

or science

[foggy parcel]

Good luck for your exam bud!

[dire snow]

No problem

You can say it and I will see when I can

[dire snow]

only 60 pages

[foggy parcel]

🙂

[dire snow]

Ok goodbye and thanks everyone

[foggy parcel]

Then why r u doin maths go study that topics xD

[dire snow]

I have homework for tomorrow

[void charm]

okay ig

[dire snow]

I have even to do Olympics of Italian then 💀

[void charm]

byee all the best for tmrw

[dire snow]

At least it's for Monday (the Olympics)

[dreamy cave]

cos(x-pi) = cos(pi-x)
sin(x-pi) = -sin(pi-x)

this is correct?

[dark sparrow]

yes

[dreamy cave]

thanks ♥️

[sudden marlin]

Can someone help me with this?

[timber cargo]

You know that a circle is 360 degrees

and 98 degrees is 40.62 in^2 in area

so make a proportion

[sudden marlin]

Ah, I get it, Thx

[upper karma]

can i get help on this (both perimeter and area)

[wind coral]

firstly label all of the side lengths to make this easier

[upper karma]

ok

[wind coral]

after you do so, find the perimeter by adding all of the side lengths together

[upper karma]

are these correct?

[mighty sparrow]

Yes

[cedar grotto]

can the unit circle be used if the hypotenuse of a triangle is not 1

[mighty sparrow]

No because the unit circle has a radius of 1

[cedar grotto]

so whats the point of it then

if it can only be used with a triangle that has a hypotenuse of one

[mighty sparrow]

Because the hypotenus = sqtrt(cos(x)²+sin(x)²)

You have a triangle defined by sinus and cosinus

[cedar grotto]

what does that mean

you can use it with a traingle that has a hpyotenuse of anything?

[mighty sparrow]

I don't understand

[cedar grotto]

i dont understand what you are saying

i think my stupidity has confused you and me

[mighty sparrow]

If you take a point on the unit circle it is defined by a cosinus and sinus which bot have a value between -1 and 1 and if you use the pythagoras theorem with cos(x) and sin(x) you get the radius of the circle

So if you have a cosine you can get it's related sinuses

Thanks to the pythagoras theorem

[cedar grotto]

what is a cosinus

[mighty sparrow]

Cos(x)

[cedar grotto]

x is theta right

[mighty sparrow]

Yes the angle

[cedar grotto]

ok so let me get this correct because i have a very basic understanding of the unit circle

the unit circle is used to find the x and y of a triangle when you have the angle right?

like sin is y and cos is x

[mighty sparrow]

Yes

[cedar grotto]

so i just dont understand how it can be very useful if you can only use it to find the x and y values of a triangle where the hypotenuse is 1

unless im wrong about that

[mighty sparrow]

With a cosine you can find a sinus and vice-versa because the hypotenus is always 1 it's easy

[cedar grotto]

ok but im still not understanding

[mighty sparrow]

What are you not understanding?

[cedar grotto]

this

like what if the hypotenuse of my triangle is 34

[mighty sparrow]

Then you don't use the unit circle

[cedar grotto]

but then why even use the unit circle

[mighty sparrow]

Because in trigonometry you have a circle and a triangle that are alike

[cedar grotto]

so what im not understanding about this is the 1 being the hypotenuse

[mighty sparrow]

I guess

[pale snow]

It has nothing to do with that. The unit circle has a radius of 1 unit because it is easier to work that way but you can really have ANY MEASURE OF RADIUS. Since, in reality, there the concept of RADIANS is at stake (which has to do with the relationship of a radius and the perimeter of the circumference). That said, those RADIANS are used to measure an angle (counterclockwise, from the x axis).

That angle determines a point on the circumference (in any of the four quadrants) and from there the cosine (abscissa of the point on the circumference) and the sine (the ordinate of the point) are defined.

[cedar grotto]

i know why it is 1 because its the UNIT circle and 1 radian but im not getting why its that useful if you can only use it for a triangle with 1 as the hypotenuse

[cedar grotto]

so can the unit circle be altered to be able to use 34 as the hypotenuse

[mighty sparrow]

I don't think so

[cedar grotto]

but he said " but you can really have ANY MEASURE OF RADIUS. "

[limpid condor]

does e=mc2?

[mighty sparrow]

He said that it's easier to use 1 as a radius but you could tweak it to have any radius

[cedar grotto]

yeah so why did u say i dont think so when the hypotenuse is the radius

and how would you tweak it to use and radius and therefore any hypotenuse

[mighty sparrow]

Because i personaly didn't learn any purporse for a unit circle to have a different radius than 1

[mighty sparrow]

Multiply cos(x) by the new radius as well as sin(x) i guess

[timber cargo]

What are you talking about

[cedar grotto]

is that all you would have to do?

[pale snow]

Yes, but you would have to change the measures of the angles (there is a formula, I think it is x/r) but just understanding the unit circle is enough. If you want, to understand better, looks for "the Euler function" that function associated to each real number an arc on the unit circle. From there, it can be expanded.

[upper karma]

are these correct

[silent plank]

perimeters are wrong

[sand matrix]

It is valid for all triangles. Remember that sinx and cosx are periodic functions which means they repeat values on a certain interval.

Think about a simple value like sin(30*), this evaluates to some ratio which is 1/2. Now let's use your aforementioned example of a hypotenuse of 37. By using the unit circle we know that a value of 30 degrees or pi/6 radians evaluates to 1/2 and this can be applied to any triangle.

Let's say that your triangle has sin(x) = 37/74, we know that x = pi/6 works because of the unit circle.

I hope this makes more sense @cedar grotto

[frail ridge]

This might be a dumb question, but how do you draw a triangle given it's name (∆abc for example) ?

And is ∆abc and ∆cab the same triangle or is it different?

[silent plank]

usually capitals should be used for points

triangle ABC, is just a triangle with points,vertices A,B,C

you can have your A,B,C clockwise or anticlockwise, doesn't matter

[frail ridge]

But it's still in some form of order

Right?

[silent plank]

well for a triangle it doesn't matter, it'll be in order no matter which way you arrange them

[frail ridge]

So ∆abc has points a, b, and c in either cw or ccw

But it can't be a, c, b

Right?

[silent plank]

it can

well for a triangle it doesn't matter, it'll be in order no matter which way you arrange them

try as much as you can to screw it up

[frail ridge]

Now my second question

Do different names results to different triangles?

Given that both names share the same points

[silent plank]

wdym

[frail ridge]

Like I said before

[frail ridge]

This

[silent plank]

triangles ABC, ACB, BAC, BCA, CAB, CBA are all names for the same triangle

order doesn't matter

[frail ridge]

Okay then

[silent plank]

if you're referencing just a lone triangle

if you're doing stuff with similarity/congruence then
you'd order the names based on corresponding points

[frail ridge]

That's what I'm doing

Finding the corresponding sides and angles

[silent plank]

you should always start with the original problem

[frail ridge]

My assignment is that given two triangles that are Cogurrent, enumerate all the corresponding sides and angles

[silent plank]

you can use whatever order you want for the first triangle
but for the name of the second triangle
the corresponding points will need to be in the same order

[frail ridge]

Ah I get it

So for ∆abc and ∆efg

Angle a and e are corresponding

Right?

[silent plank]

yes,

assuming they have been for some reason been represented with lowercase letters

[frail ridge]

But what about the sides? Seems tricky to do

[silent plank]

sides between the corresponding angles would be corresponding

[lavish portal]

Anyone does have idea 🌚

[nocturne remnant]

(somewhat) famous problem https://www.cut-the-knot.org/triangle/80-80-20/IndexToLong.shtml

[frail ridge]

What does two capital letters together means?

No line above it, something like AB

[nocturne remnant]

The length of the line segment

[frail ridge]

How is it used?

[nocturne remnant]

Wdym how it is used

AB means the length of the line segment joining A and B. Like i just said

[frail ridge]

So I can say AB = BC if seg AB is equal in length as BC?

[nocturne remnant]

Yes

[frail ridge]

What is the difference between AB and AB with a line over it?

[nocturne remnant]

Some people use AB with a line over it to emphasise that they’re talking about the line(/line segment) itself rather than its length

Some people don’t put a line over it for any occasion

[frail ridge]

Okay then

Thanks

Sorry

One last question

What does <ABC mean?

[nocturne remnant]

The angle ABC

..have you not done any geometry at all????

[frail ridge]

You could say that

[nocturne remnant]

Oh

[frail ridge]

What does abc mean here?

[nocturne remnant]

You have a long way ahead of you then xd

[frail ridge]

Just learning triangle congruence

[nocturne remnant]

The angle ABC specifically means “the angle formed by the arms AB and BC”

[frail ridge]

Is <AB also valid?

[nocturne remnant]

No

[frail ridge]

Good to know

[lavish portal]

why ∆ACE = 10°

[nocturne remnant]

Congruence

As they proved in the previous line

[lavish portal]

Oh I see 😁

[vast trail]

is this a soild geometry channel ?

[dark sparrow]

you can post solid geometry here

[vast trail]

what is the vol of a parallelpiped if he gave me the 3 points in 3d directions ?

[dark sparrow]

who's "he"...?

can you show the problem maybe

[fossil lava]

Guys I want to clarify something, if we have tan(nx) it's period is π/n right

[native jay]

Next time, google it

But yeah you're right

[fossil lava]

I don't know how to put them into words

[native jay]

type in "period of tan nx"

[cobalt bone]

help pls

[grave pond]

Note that cos(pi/6 - x) = sin(pi/3 + x). That gives a pretty nice set-up for using a sum-to-product formula.

[lilac otter]

Do you mind telling me how you came up with this result please? I see that cos(pi/6-x)=sin(pi/3 - x/2) so I don't know where the mistake lies.

[grave pond]

It just cos(a) = sin(pi/2 - a), and then simplify pi/2 - (pi/6- x) = pi/3 + x.

[lilac otter]

oh ok thank you very much

[upper karma]

I need help with a test, I’m struggling

[rough timber]

how did they get ybar = 215

for the triangle

you would think it would be 45/3 = 15, no?

[grave pond]

Which triangle where?

[upper karma]

someone help

[hollow turtle]

i’m pretty sure this is really simple but i got it wrong:

a pyramid with a square base of a side length 8m has a height of 3m find the length of a sloping edge to once decimal place

[silent plank]

show your attemp

[smoky jetty]

you may find the value of the diagonal using Pythagorean theorem

where base^2 + height^2 = hypotenuse^2

since the diagonal cuts the rectangle into two congruent right triangles, then u can use the inverse of sin function (or the other trigo funcs) to find the angle (use calculator for convenience)

And refer to the theorems of parallel lines cut by a transversal so that you need not solve for the other right triangle's angle measures. Take note of the vertical angle theorem

[astral bison]

help ploz

[smoky jetty]

for the first part, I would say that "also a blop" is like a bigger set that the set of "something is a blip" is enclosed in. Though, I am not completely sure.

For the second part, u may do the Pythagorean theorem i.e sec(theta)= a/n

[thin locust]

is this correct?

[thin locust]

contrapositive since normal = contrapositive (in truth values)

[smoky jetty]

seems like it

[upper karma]

after i subtract an area from another area, what is that red area called in english?

im a bit new to english math so i dont know those term

[dark sparrow]

dunno of a special word for that. is there a term for it in your language?

[glad sphinx]

remainder?

I don't know what to say

you can also say rectangular donut but I don't think that's a real word

[valid comet]

Usually we just call it the shaded region

[upper karma]

ah thats the word, thanks

imma die googling by myself to find that keyword

[pliant pumice]

Hello! My question is about triangle congruency. SSS - side/side/side rule is that if all 3 sides of 2 triangles are equal, they are congruent. The reason for this is showed on the picture. We can see that triangles AB1C and AB2C are absolutely same, but how do we prove that?

[pallid vine]

Knowing that AB1 and AB2 are radii of the left circle, CB1 and CB2 are radii of the right circle, each pair of those sides have the exact same length. From there, as you said, it will be SSS congruency as they also share the side AC

[deft tiger]

hie

can anyone help me

to prove : if a triangle has two angles equal to one another than the sides subtending the equal angles will also be equal to one another.

can anyone??

[pliant pumice]

But we are proving SSS! How can we prove SSS is true using SSS?

[pallid vine]

Ah I see... sorry for misunderstanding your question. I'd imagine it's something to do with the angles in the triangles, and the properties of isosceles triangles, since it's a super symmetrical figure, and the fact that it uses circles.

[astral bison]

ty

[river galleon]

It feels like this should be easy. But i'm stumped

[pliant pumice]

I will try to use it, thanks for the idea!

[bleak blade]

sin area

[timber cargo]

Multiply the two side lengths by the sin of 114 and divide the whole thing by 2

Let me rephrase that

$absin(c)/2$

[somber coyote]

Daniel S.

[timber cargo]

@river galleon A and B being your side lengths, and sin(114 degrees)

[river galleon]

that was very helpful thank you

[timber cargo]

Is that a genuine expression or... Because it sounded sarcastic

[river galleon]

not sarcasm

i do have a follow up question

what if you only have one side and two angles

for example 61 degrees 80 degrees and 17ft

to again find the area

[timber cargo]

Let me draw it out for you

[timber cargo]

Take this as an example

[river galleon]

right. how did u acquire the height values

from 114

[timber cargo]

Are you asking, how to get 17 km

[river galleon]

from your example how you got 60 and 54

[timber cargo]

Drawing a perpendicular line that bisects the two side lengths

But notice that I added 30 degrees as the angle

[river galleon]

oh 30 60 90

[timber cargo]

Wait, I screwed up the triangle

[river galleon]

XD

[timber cargo]

But, do you get the idea/

[river galleon]

so when I find all the angles i can just use the same formula from before?

[timber cargo]

If the two side lengths are adjacent to the angle, then yes

[river galleon]

alright works for me yea.

thank you for your help

[timber cargo]

✅

[upper karma]

A ship sails on a bearing of 156°T until it is 45 km south of its starting point. How far
east is it, correct to two decimal places?

help

[grave pond]

What does the T mean?

[wind herald]

can someone dm I have somethings I need help with

[timber cargo]

You know sending the problem here works too

In fact you'll get more people to help you

[grave pond]

Many helpers are motivated at least in part by the fact that they can be seen to be helpful and/or knowledgeable. If you want a closed DM conversation that motivation disappears.

[floral shale]

Shhh don't expose us

[subtle palm]

lol

[neat plank]

does anyone know how to solve or could point me in the right direction of finding the local min and max of 3/(4+2sinx-cosx) without differentiation?

[silent plank]

harmonic trig identity

[upper karma]

can someone explain to me wtf just happened

[thin locust]

given a convex quadrilateral ABCD where ABC + ADC = 180°, is the quadrilateral always cyclic

im not sure if
cyclic quadrilateral -> ABC+ADC=180°
or
cyclic quadrilateral <-> ABC+ADC=180°

[upper karma]

hm

well think of it this way

can you imagine a quadrilateral where opposite angles add up to 180 and it isn't cyclic

[thin locust]

no

i tried drawing one

and im 99% that it isnt possible

but i may be wrong

[upper karma]

"The converse of this theorem is also true, which states that if opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic."

I think this answers your question

[thin locust]

ah ok

ty

[upper karma]

supplementary angles are angles adding up to 180

[thin locust]

yes thank you

[upper karma]

np

[thin locust]

could this be proven by dividing the quadrilateral into 4 right triangles

and then wait

let me try it

nvm

but thank you

[latent compass]

I can’t find the answer anywhere on google so 😭

[thin locust]

1.since ABC = ADC = 90, ABC+ADC=180 -> cyclic

2.since ABC=90, then AOC has to be 180 -> A, O, C are colinear -> diameter

this is correct right?

[blissful spire]

I just started learning about tangents and stuff and I’m confused how to solve for x here

[floral shale]

Start with identifying what tan(49) would be in terms of the triangle parts

Remember that tangent is opposite/adjacent

[blissful spire]

Ok

Wait what is opposite/adjacent?

That would mean it would be tan(49)=14/x?

If I did it correctly I got x=16.1056

How do I solve for an angle using tangents when I have all the sides and a 90 degree angle alr?

Like this

[trim lion]

😭😭

[upper karma]

help

[trim lion]

bru

rhombus

[scarlet shell]

Could anyone dm me a good trigonometry book recommendations? 😄

[timber cargo]

Pythagorean theorem to find the height

Then you base times height your way to victory

[timber cargo]

Use your knowledge of angles and trig functions to solve this problem

[upper karma]

bro i dont have knowledge of everything i literally flunked school

[upper karma]

ok find dis angle using trig and then u can find the base of that triangle on the left with more trig functions and then do the same with the other triangle and then use base times height divided by 2

then just add all of it up

[sage bison]

send the full question ty

[viscid shoal]

Hi, can I get a help or tips about these problems?

What is the area of quadrilateral ABCD given 2 sides and radius of inscribed circle? AD= 5cm, BC = 4cm, radius of circle is 2 cm

And how to find the length of a radius of circumscribed circle around a regular hexagon, if I know the length of the radius of inscribed in it circle (it is 6√3 cm)?

[smoky jetty]

also has something to do with special triangle 45-45-90 caSE

[silent plank]

for geo, always make a diagram

[glad sphinx]

art of problem solving precalculus

[chilly zinc]

Set up your equation as normal with SOHCAHTOA, but this time, theta is your unknown variable. You'll have to solve for theta using an inverse tangent

[timber cargo]

What do you know about trigonometric function

I.e. tan, sin, cos

[crisp valley]

Can you just give answers

Nevermind i threw the paper away

[timber cargo]

What do you think this is? Google, Chat gpt, Twitter?

[crisp valley]

@timber cargo Bro be fr

[smoky jetty]

we're here to guide you on getting it.

[golden nymph]

Could someone point out my error here? The answer should be 0 if you use the 1+tan^2(5) identity. But shouldn't this work too? I'm sure it's a stupid mistake, but I'm not seeing it.

[exotic yarrow]

This should be a negative sign

[golden nymph]

Why?

[exotic yarrow]

b/c $\sin^2 5^{\circ}-1=-(1-\sin^2 5^{\circ})=-\cos^2 5^{\circ}$

[prime musk]

Cos square theta plus sine square theta is one

[somber coyote]

Civil Service Pigeon

[golden nymph]

🤣 I knew it was something stupid. Thank you!

[deep vapor]

good afternoon , i don't know ,what i did wrong

[leaden seal]

Angle C and E are flipped

[arctic breach]

Determine all values of k for which the points A=(1,2), B=(11,12), and C=(k,6) form the vertices of a right-angled triangle.
I tried this using the pythagorean theorem by calculating AB^2, AC^2, and BC^2 and using the Pythagorean theorem but i'm not getting the right answer

[nocturne remnant]

maybe you could show us your working

[arctic breach]

AB^2 = (11-1)^2 + (12-2)^2 = 200
AC^2 = (k-1)^2 + (6-2)^2 = k^2-2k+17
BC^2 = (k-11)^2 + (6-12)^2 = k^2 - 22k + 157

[leaden seal]

Isn’t it all the points on a circle

[nocturne remnant]

after that what did you do

[arctic breach]

AB^2 + AC^2 = BC^2
simplifying this you get k = -3

if you sub in the values

the answer is supposed to be 1 and 11 i believe

[nocturne remnant]

so you assumed A is the right angle

I think that's the problem

[arctic breach]

i did the other ones too

and none of them were 1 or 11

[nocturne remnant]

hmm

probably some algebra mistake then

[nocturne remnant]

...hey, where did the 12 come from?

it's supposed to be 2 i think

[upper karma]

I NEED HELPPP

[leaden seal]

This is every point for a right triangle using those 2 points

I think

[nocturne remnant]

is that the answer given? because i think there should be 4 answers

[leaden seal]

Why are there 2 answers?

The y value is equal to 6

[upper karma]

Ion know y this is wrong... what's my mistake?

[arctic breach]

yeah it only says 1 and 11

[nocturne remnant]

hmm whatttttttt

i think the answers are k = 1, 3, 9, 11

[arctic breach]

how do you get that?

[nocturne remnant]

for k = 3 and 9 <ACB is 90 deg

for k = 1 <CAB is 90 deg

for k = 11 <ABC is 90 deg

[arctic breach]

but how did you find those numbers?

[nocturne remnant]

for k = 1 and 11 should be obvious

the other ones i used desmos

[arctic breach]

could you explain it? im not really sure

[nocturne remnant]

well line AB is horizontal right

so if k = 1 then AC is vertical and CAB = 90

[arctic breach]

isnt AB on an angle?

[nocturne remnant]

?

[arctic breach]

if its going from (1,2) to (11,12)

[nocturne remnant]

oh

your problem then

[arctic breach]

oh yes sorry i wrote the wrong number

[nocturne remnant]

is that true or is your original message true because you made a typo

[arctic breach]

i meant 11,12

[arctic breach]

sorry, my mistake

[nocturne remnant]

hmm

in that case it's even weirder than k = 1 and 11 is the given answer

[maiden kelp]

?

[arctic breach]

sorry its 1 and 13 thats the answer 😭

i screwed up the numbers

Determine all values of k for which the points A=(1,2), B=(11,12), and C=(k,6) form the vertices of a right-angled triangle.

the answer is 1 and 13

[nocturne remnant]

huh i got -1, 13 and others

[floral shale]

Knowing that $C = 2\pi r \implies r = \frac C{2\pi}$, we can say that:

$$\begin{align*}
A &= \pi r^2 \
&= \pi\parens{\frac C{2\pi}}^2 \
&= \frac{C^2}{4\pi}
\end{align*}$$

You can plug in $C$ directly. It's probably an intermediate rounding error that caused you to goof up because your work is fine.

[somber coyote]

Umbraleviathan
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[floral shale]

Yup

Although it only wants points on the circle where y = 6

Which has two answers

Either you use projection or pythag

Although your circle uses (11,2) instead of (11,12)

[deft helm]

Please help. (This is probably a easy geometry question, but I don't know how to do it)

[timber cargo]

The two triangles are similar, meaning their side lengths are in proportion

24 is proportional to 8 as 10 is to x

so

24-8
10-x

solve for x by cross multiplying

or you can just do this intuitively

and see that it’s 10/3

^ This should help you with analyzing future problems

[deft helm]

Oh sorry I was afk

And thanks for the answer

[timber cargo]

But do you understand why

[deft helm]

I think that if the shape of both of the triangles is similar, it is proportional.

[smoky jetty]

the bigger one was essentially just scaled up to a factor of 3

[scarlet shell]

Thanks

[vernal parrot]

can someone help me with my problem (help -12)

[sick hemlock]

I got 436.9

You probably shouldn't have rounded the value of r at the start

Just kept it as it is, square it and multiply with pie

[vast vapor]

C^2/4pi^2

don't forget

[timber cargo]

It's 436.944776539

You rounded your r, don't do that

[dusk marten]

do vectors belong here or somehwere else

[valid comet]

#linear-algebra probably

[static magnet]

can someone please explain this to me

because 36.8*2

and 180- the sum = 106

which is wrong

and hwo they got the equation at the bottom

complete ss

or how do they prove tan^2 x = 0.5

I dont' understand that part

[viral merlin]

why does a conic section always touch the cone in one point?

<@&268886789983436800>

[floral shale]

Aww what did I miss

[viral merlin]

porn

[floral shale]

Ah yes

[viral merlin]

very weird porn

[floral shale]

My favorite math topic

Porn

(Parabolic horns)

[cold chasm]

what are tangent circles

[valid comet]

circles that touch each other at exactly one point

[maiden kelp]

wht

[timber cargo]

Set (3x+19) equal to (4x+14)

Solve for x

y+(3x+19) are supplementary

they add up to 180 degrees

If you know what x is, you’ll know what y is

[maiden kelp]

so 5?

then for y we plug the x into y+(3x+19)?

x=5 y=24?

[grave pond]

If x is 5, what is then 3x+19?

[maiden kelp]

oop i ment 34

lol

anyways thx for your help

lol it was wrong

[grave pond]

Yes, but does 34 + 24 make 180?

[maiden kelp]

no

[tender rain]

Is anyone good with factors ?

[grave pond]

So x=5, y=24 is not a solution.

[cunning acorn]

can someone help me with this please

[cunning pumice]

could someone teach me how to identify the side lengths

i'm not sure why it's so difficult for me

i feel like it should be easy

[gilded lion]

Find RQ and Triangle of PRQ

me and my whole class has trouble with this 😂

ifa nyone wanna help

[atomic moat]

should PR is = 3.2^1/2

?

[foggy parcel]

u shouyld extand the line PQ

and create a perpeniducular from R to that line

then u can use the cosine

to identity the whole adjacent line

[smoky jetty]

Since it's an SAS case, u may refer to law of cosine where p^2 (or segment RQ^2) = q^2 + r^2 - 2qr cos(45deg)

[smoky jetty]

yea i think so as well

[foggy parcel]

like this

extand the PQ

and draw a perpendicualr

[gilded lion]

tysm

god bless

[foggy parcel]

my plesesure

have u got what u have to do

?

U can use cosine to get the value of PQ'

then u can get QQ' and same way u can get the valuye of RQ' using sine

[floral storm]

hey,,

what are your POVs about existence of perfect cuboid

[smoky jetty]

u mean 10cos45deg correct?

or im confused, lol

[foggy parcel]

cos 15deggres = PQ'/PR

in the triangle PRQ'

Pr = 10

[smoky jetty]

oh wait

[foggy parcel]

SO from here u can get the value of PQ'

[smoky jetty]

I essentially just got the value of PQ'

[foggy parcel]

Okay good

[smoky jetty]

since it's a right triangle, i just did 10cos45deg.. is this correct?

[foggy parcel]

ah u should just use cos15

wait

is the angle 45 or 15

[smoky jetty]

for triangle PRQ'

[smoky jetty]

45

[foggy parcel]

the angle is 15 right ? or 45?

Oh lol

[gilded lion]

45*

[foggy parcel]

Oh then u can just use cos 45

[gilded lion]