How smart are you, Yea Forums?

nvm, I'm stupid

also meant 40, but still wrong

This

You can solve all the other angles...

Attached: t4.png (348x467, 57K)

Is it smarter to worry about this? Or if you should be in a compromise with 10 of millions of people?

Or that you've built your life around 3$ gasoline?

wrong!

give me a moment to recreate my drawing in paint

Thats pretty much how far you can go

60° according to the law of vertical angles.

you're an idiot and the answer is 30 deg

Does not apply there though

if anyone's curious

Attached: 4chan.jpg (325x386, 32K)

pretty sure its 40 degrees
- my crippled GCSE brain

Ok explanation:
1. calculate both red angles, notice ABD has 2 equal angles therefore is an isosceles triangle (noted by marking sides with a).
2. Assume point E so that angle is 80°, creating another issceles triangle, calculate both green angles.
3.1 Notice triangle EBD is at least isosceles, because calculating angles creates 3 60° it's even even-sided.
3.2 using the 40° red and green we have triangle EBC also isosceles, therefore triangle EBD is also isosceles forcing angles ECD and EDC to be equal resulting in them being (180-40) / 2 = 70° and x = 70°-40° = 30°

Attached: solved for x.jpg (593x733, 89K)

small correction in the last line of the explanation, it should be:
therefore triangle EDC is also isosceles ...

Pretty sure x is 60.

Angle above x is 40. Opposite angle is 110-x, and the middle angle is 70.

Rule of quadrilaterals equal 360:
(40+x)+(50+y)=200
90+x+y=200
110=x+y

Rule of triangles (equaling 180)
y=110-x

Substitute and solve?
40+x=y+50
40+x=(110-x)+50
2x=120
X=60

I think?

and another correction, this time for step 3.1, it's not even-sided it's equilateral

Made with protractor image and GIMP,
almost certain that the answer is 30, recreated as accurately as possible. I'm bad at maths ;)

Attached: Capture.png (550x469, 58K)

I don't know how you drew it, but just wanted to mention that the numbers may not accurately represent the visual angles?

I wrote this out, but I don't understand why it's wrong. Could someone point out where my logic is flawed? The answer itself doesn't seem right, but the process seems correct? But I feel like the process is flawed.

I feel like it's because I am operating under the assumption that there will be an equal divide between the left and right bottom corners.

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it's not 40+x = y+50
it's
40+x + 50+y = 200 (as you stated above)
but if you use the rule of triangles to substitute here you'll get 0 = 0 in the end.

I used the protractor to draw all the angles i had available, so all the angles are almost pixel accurate.

Attached: Capture.png (485x388, 10K)

Originally the lines were longer, but where they crossed eachother I cut them down. I'm pretty sure 30 is the only possible answer

Right, thank you. This was a case of staring at a problem too long and wanting it to work and missing the obvious.

Is this not the best way to solve this problem? Just assign a length to a side and then use cosine to solve?

(0;110)

This is how far I know, I'm sure you can calculate the other angles from knowing that 2 things are the same, or that you can build a thing around it.

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Why the FUCK haven't you joined this server yet?
discordapp.com\invite\v3HXgkt
Copy to your browser... EZ...NO-RULES, no yannies
GO GO GO

-j86

You can just work it out visually, using the top horizontal line at the top as reference, if you draw each line longer than it needs to be you can just trim them once they all connect, this gives a solution of 30, which i believe is the only possible answer.

The accuracy would increase with picture resolution, but as you would expect a clean answer its relatively unnecessary imo

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Nice, man! Me convinced!

Hmm

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slightly improved my explanation and all in 1 image that's it for today
x is still 30°

Attached: solved for x better.jpg (1774x744, 280K)

someone failed their exams :?

SUCCESS!! I figured it out! Lots of math involved. The answer is 28.5 degrees

I'll put the maths up in a second.

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Wrong. You just need to make use of equilateral triangle and parallel line theorems.

Ok this is better than mine. My math degree is getting rusty

I assigned a numerical value to the hypotenuse of green triangle : 100m

Equations used:

(Sin(a)) / A = (Sin(b)) / B

A^2 = B^2 + C^2 - (2BC)(Cos(a))

The gist of this was to solve each side the whole way down, so that you can finally solve for x.

Green triangle:

B = [A (Sin(b)) ] / (Sin(a))

Red Triangle

Side B = Side D

Repeat earlier step to find Side E/H

Do this same shit to solve for values for the yellow triangle.

For the black triangle with x, you need to solve for Side I before solving for x. To do this you need to use the second equation.

"I" ends up equaling 68.4039
Put "I" into x = Sin^-1 [ (XSin(70)) / H ]

x = 28.49580

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Attached: 7209424337.jpg (1014x1670, 116K)

Ñigga had Yea Forums do his homework

>Reply
5/7 smart

The photo is showing up as a girl...but that's not the photo I uploaded. I also can't seem to repost it?

either way! Free girl.

Haha, if so, it's a great question.