How smart are you, Sup Forums?

How smart are you, Sup Forums?

22

30

60 sry

35

x= filthy jew

...

It can be any number between 0 and 90
As long as the angle to the right is 90-x

Lol retard. Look at the angles bottom left. 50 degree angle is smaller than 40

>implying the angles on the drawing are relevant in any way

it's the numbers that count, you fucking autist

Smart enough to ignore this problem and move on.

I got to here, and I do not believe we have enough information to continue.

Though it does ask the question: HOW did you get 50 degrees and 60 degrees as the angles of the bottom triangle?

Can anyone explain the step after solving the angles 70, 40 and 50? I have an exam soon and stuff like this is gonna pop up.

It can be 60 or 50
60 would work for you
But 50 works for

Gcse maths m8. Are you 12?

educated guess, since is correct in saying that there's not enough information to continue purely mathematically, or if there is, then it's some retarded theorem that you learn once in high school and never use again in your entire life

This

as the guy who got 50, this

fucking 8th graders

How to get 60

This, you cannot proceed after this

but you do know that x + y (other unkown angle) = 180 - 70

meaning x+y = 110

so x could be 50
and y could be 60

or x could be 45
and y could be 65. whatever you want.

There are 8 triangles, faggots

You could've spent the same amount of energy just answering the fucking question. But instead, you chose to be a prick.

Are you 12?

No.

>found it

prove him wrong then, faggot

Should be 55 unless I'm retarded.

No. For refference we also do taylor polynomials, but stuff like this is there for kicks and giggles, to test 'cleverness' i guess.

did you make 60 = y and did a SoE?

Oh shit, I've not thought about doing that. Looks like it checks out, too.

the fuck is SoE

35, but there's no way to know

lol there are other triangles to consider

It can't be 60. If it were then the quadrilateral would be cyclic, yet the top chord is subtended by a 40 and 50. Impossible!

on the bottom left side, how did you get 55 (the angle under x)

This makes no sense.

He's basically saying that because the opposite angle is 110, that x and the angle under it are both exactly half of 110. They are clearly different angles

The problem is 100% unsolveable

Meme magic lads.
X can literally be any number you want it to be and all the triangles work out under the one constraint that the other unknown angle is 110-X.
55 is just the nicest number.

It's blatantly clear that the whole figure is completely defined by the four angles given, meaning that x can have one and only one value. If you think x can have several values, try drawing two different versions of the figure with two different values of x with the four given angles remaining the same. It can't be done.

This. OP bamboozled us and is a filthy rotting nigger

Can law of sines or cosines be used to continue?

i got this

...

Wrong. See

why did you decide 130=55+75 and not 70+60 ?

You ARE retarded you dumb cuck

You have decided without proof that the angles adding up to 110 would be 55 and 55 and not any other number.

Yes they can. It boils down to (sin(30)sin(60))/(sin(40)sin(20)) = sin(x)/sin(110-x)

Which I just calculated is x = 30 degrees.

what sides are you using?

I... what?

Which law are you using?

also, had the wrong equation its sin(110-x)/sin(x)

35

Call the top a, the bottom b. Then by sine law

asin(60)/sin(40) = bsin(110-x)/sin(30)

and

asin(50)/sin(50) = bsin(x)/sin(30)

but you don't know the lengths of a or b

wtf is this law?

Oops, bsin(x)/sin(20) in the last equation. Which rearranges to b = asin(20)/sin(x). Plug into first equation, cancels a.

Sine law.

Don't need to, they'll cancel out in the end.

image not drawn to scale, so "size" of angles means nothing. I am a high school math teacher with a BS in pure mathematics.
>tl;dr
>I wasted my fucking education

30 degrees.

I also got a bachelors in pure math. also a masters and phd. Uni prof.

Hope you're trolling. Four of the angles have their sizes explicitly given, this has nothing to do with the image being drawn to scale.

actually, it looks like process of elimination.
With the other side, where the 50/130 lines were made, the non-x, unknown angle was determined to be 55, and the only angle that could fit 55 and 70 in a triangle would be 55.

regardless, in high school, we teach the students to disregard what the picture looks like. We even use pictures that not mathematically possible to drive home the idea of solving for unknowns using only geometric properties

All 4 angles in the quadrilateral equal 360.

Now you can solve.

the post I was replying to suggested it was "impossible" to draw the image with angles of different measures than user was trying to prove. my point was that it doesnt matter what the angles "look" like

I'm not the guy from before. Just making friends!

30+80 also makes sense

The post you were replying to made no mention of what angles "look" like. It said simply that you can't construct two versions of this figure with the four given angles remaining the same but with different values for x. That's true.

lol, I tip my hat to you math prof. It was a dream of mine to be that skilled. I am a huge math nerd! Luckily, I am an IB math teacher, so at least I get to teach some basic calculus

>to test 'cleverness' i guess.
Sounds like you're screwed!

Good job, OP. You got us where we deserve to be ;)

isn't sine law:

a sin (B) = b sin (A)

how are you getting sin (60) / sin (40)?

not very

Thought I'd make a picture detailing the use of the sine law.

Which rearranges to b/sin(B) = a/sin(A) which was how I learned it way back in the day.

Well done, not a retard

Yeah, well done on disproving the retards who claimed this was unsolvable or that x have have any number of values.

Yeah I figured this one was probably a troll.
I'm retarded for even attempting to solve it.

thanks for this

why is it that when I plug the equation into wolfram alpha, it does not give me a solution for x?

(sin(30)sin(60))/(sin(40)sin(20)) = sin(x)/sin(110-x)

Typo all day. Heres the proper one. Just forgot a sin(40) in a couple places when I typed it up.

Not clue as to why it won't, but I got out a calculator to verify the 30 degrees. I don't use alpha.

Thank you.

holy fuck check my image

holy fuck reported nigger

Don't forget that any shape has a maximum of 360*

wrong.

the answer is 30

Don't forget that you're a retard.

fuck x. he can solve his own goddamn problems for once.

I believe you're mistaken.