HOW DO YOU SOLVE THIS

HOW DO YOU SOLVE THIS

Fuck you Veeky Forums you cuks post this shit and don't post how to solve

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duckware.com/tech/worldshardesteasygeometryproblem.html
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the answer
is 4

50 + a + b = 180
10 + 40 + b + a = 180
???
profit

>he can't solve a system of two equations with two unknowns.

I really hope you are underaged because if not you're basically a retard.

This guy has it.

>literally just wrote the same equation twice
That's not how simultaneous equations work friendo.

60 and 70... cyclic quadrilaterals.

a = [math]70^{\circ}[/math]
b = [math]60^{\circ}[/math]

Assuming that is a cyclic quadrilateral, there's no circle and it doesnt say that it is. (Either way though 60 and 70 is right so it obviously is one)

Why doesn't 50 and 80 work?

60+20+40+50 =! 180.

It's not obvious to me, care to explain?

Numerical computations say that a = 20 degrees. So any clever algebraic tricks had better agree with that.

Im no mathematican so pardon my informal solution:

1. Draw a line through intersection such that two identical new angles are formed (in addition to already present angle of 50 degrees). These angles are 130/2=65
2. Create a parallelogram from the triangle (a,b,50) such that the corners have the degrees 50, a+b, 50, a+b.
3. Mirror the triangle (65, 85, 30) in a similar fashion as with the parallelogram.
4. Blue angle is 180-5-15=160
5. We know the upmost angle 65 from symmetry.
6. The angle 10 is known from 10=180-65-90-15
7. Since a+b=130 and 360=(160-10-a)+10+40+a+b+30 = 360, a=10 and b=120

I've found a contradiction in your construction.

Look at the area in pic related.

Notice how the angle made of the angles of measure 65 and 15 on the top is an exterior angle of the triangle with angles 75, 40, 65.

A Theorem in Euclidean geometry is that the exterior of one angle of a triangle is equal to the sum of the two other interior angles of that triangle.

65 + 15 = 80
but
65 + 40 = 105

Also notice that the angles of 65, 15 and 75 should be complementary but 65 + 15 + 75 does not equal 180

So somewhere you fucked up. I cannot tell you where because without proper notation it is impossible for me to figure out exactly what you did to construct that but I can at least tell you that such a figure does not exist in the plane.

I meant to post this picture.

trace the image in geogebra and youll see those angles are totally made up

>posting your homeworks in Veeky Forums as challenge problems to get many replies.

It is not cyclic.

I think you mistake the blue angle you outlined for 65. It's actually 90 degrees, I just forgot to draw a square on it, see updated solution.png

so there is no triangle with an angle 65 and an angle 15.

It's a quadrilateral so the sum of the interior angles must be 360. That must mean a=60 degrees and b=70 degrees for there to be 360 degrees in total.

That only means a+b=130. Not enough information.

...

I was gonna say, 60 and 70 are correct but those angles are most definitely not drawn correctly.

What a waste of time. You can actually use geogebra to find the correct answer.

The center angles being equal on both sides means 40+a=30+b im not sure how but it do

after a bunch of law of sins calculations I get
b=118.72
a= 11.28

duckware.com/tech/worldshardesteasygeometryproblem.html

Fucking finally.

I got 19.6 and 110.4

Well go ahead because sure as hell didnt get it right.

The upshot of that link is that drawing a line parallel to the left edge gives you an isosceles triangle, so 30+a=80 => a=50 => b=80

How do you know the triangle is isosceles?

alright so we all know that a+b=130 and if you can't find any contradictions proving otherwise then every solution for a+b=130 works

lel. yeah he's a retard

a=0, b=130

5. What symmetry?

See pic on , the line splitting K and J is the same line as the hypotenuse of D. Since 65 is the lower angle of J, 65 must also be the upper angle of D.

100