How smart are you, Sup Forums?

How smart are you, Sup Forums?

it can't be solved faggot

answer fgt.

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smart enough to lean back and enjoy the show

Its really easy

7

is this one of the crappy ones where you have to add an extra line? them ones are impossible to solve

x=x

>thread

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How did you get the 100?

the red numbers and the blue 40 are clear to me. but how did you get the 100 and x?

120

Hopefully this clears things up.

>autism intensifies

60 and 70 give a same angle, kurt angle.

God I'm stupid. But I am also drunk as fuck.

Yep this, not the best drawing put you get the point

How do u know x is 30 and the kther sngle is 100. As i see it it could also be 29 and 101 for example.

That doesnt prove that the green one is 40 at all

My thoughts exactly

ok, now you are trolling, that 100 on the right and the lower 40 make no sense at all.

Smart enough to know I suck at math

>Error or true
watches at $ 0?

Ok but you do understand how we get that 40 deg which is highlighted, yes?

>How smart are you, Sup Forums?
Infinite number of solutions.

Angle arcs of the same color indicate same angles.

Dumbass

Only the green angles there are known to be 40. The other angles are unknown.

thx u

You get 40 by adding all angles in that triangle together. The sum of the angles in a triangle should equal 180 degrees.

130+10=140
180-140=40

i agree with green is know, the rest isn't

Go back to high school, dropout. This is why you got a 2 on the AP calc test and then cried yourself to sleep.

Prove that there are infinite solutions then, retard

no fucking shit, it still does not determine how he got dat 40 starrrrsssss (lel) on the other congruent without using the 180 rule.

X could be any measure depending on the length of the bottom-most edge.

Oh shit, my bad. You guys are right, I messed up on that. I don't think there's enough info given to find x.

Changing the length of the bottom edge would change at least one of the four angles whose values are given, which you obviously can't do. Try again, retard.

meh

20°

My mistake. I see that now

Is there a way to find it without graphing it on a computer?

allright bitches


there's supposed to be a 180 rule (turn around) were the sum of angles equals half a sphere (your nutsack), ergo 50-130 etc

someone explain to me how he got 40 without the sum available. now fgt.

not really

At this point I pretty much believe that the lower left edge isn't fixed by the given angles and that x just as to fullfill 0>x>130°, and would change any of the given angles.

argh typo.
* and it wouldn't change any of the given angles

Of course. If there wasn't, the computer couldn't find it either.

Still wrong, though

Still keeping up?

20

How do you know that BCE and ACD are line segments?

I made a mistake and thought the 2 angles indicated followed the alternate interior angle rule. That's how I got the second set of 40 degree angles.

how can x and the angle beside it be 30? they're obviously different just by looking at it

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>so much autism in this thread

This continues to assume that the middle vertex is the intersection of two line segments. Tell me how given the original information you can prove that this is so?

Creating a double system with 2 unknown variables solves it. On phone so can't sketch.

>20

Ok but it's ultra long:
We know that in a triangle, all angle makes a total of 180°. That means:
- The angle ACB is 180-(60+70) = 50
- The angle ADB is 180-(60+80) = 40 -> The angle BCD is 180-(10+40) = 130
- The angle AEB is 180-(70+80)=30 -> The angle ACE is 180-(20+30)=130
- The angle ECD is part of a full circle so 360 -(130+130+50)=50
- The angle AEC is 180-(20+130)=30
But then we are blocked: The only way to calculate x is to know ADE but I found no other way to calculate it than having the angle x so go ahead and continue if you feel it, I've named the angles as in my text

Correct

sad thing is, it actually is 20 the scale is just messed up

I'll ask again... what PROOF from the original drawing do you have that ACD and BCE are line segments? What if there exists an Angle ACD and a quadrilateral ABDC? How do you KNOW this isn't true other than saying it just looks like it...

The way the crossed lines are drawn, it would be estimated that both lines are straight. I would ask OP but he's too busy sucking dicks. And if it isn't, I am working on this possibility

and what proof do you have of OP sucking dicks?

OP is always a faggot, that's a known unspoken rule of Sup Forums

OP here requesting cocks. and an explanation, it's for my kfc going shit turded from the ass 6 year old black son.

obama AFFIRMATIVE ACTION NEEDS YOU.

see?

You need to use the law of sines to solve it.

>180 = 50 + x + y
>130 = x + y
>x = 130 - y

>180 = 50 + (130 - y) + y
>130 =130 - y + y
>130 + y = 130 + y
>0 = 0
>x = x

and so on and so forth.

But if you measure the side lengths, there must be a true answer. Unless the angles don't jive with real measurements.

Draw a rectangle with the top line being your start point and you'll easily be able to work out x

X=10