How smart is /b?

how did u figure out the two angles were 50 and 80? I feel stupid right now

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oh shit never mind I'm retarded this won't work

I dont get that part either.

360° is a full circle

Corona made me do it.

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How did you get the 65 and 45?

how did you continue from here?
>pic related?

bullshit

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This is wrong because the angles of the top triangle add to 280

Geometric rules to remember:

1) Triangle = 180 degrees total
2) Quadrangle = 360 degrees total
3) At an intersection of two lines, the opposite vertices are equal angles

agreed with germanon here that this is fishy
checked for x=30 on geogebra though

This is where I got stuck. I think there may be no solution

Umm, you mean 40, not 140

50 because of a reflection, and the straight line is 180 minus 50. Then split in two. Then reflection of 65 is also 65.

as said x=30 is the solution but I could only verify it

Oh shit typo.

I dont know. I think there should be a solution. If you vary the angle x then the top left corner would also change its position and the angles in the left triangle would change.
So intuitively I would assume there is only one angle x which fulfills that constellation. But I cant figure out how to calculate it.

How do you know the line is split evenly?
Would measuring these lines do anything?

The rhombus still adds up to 390 degrees so it's wrong

No where in the op image does it imply that the line where you "split in two" actually splits the angle into two equal angles.

My bad I used a calculator this time.

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69
the sexy number
get it
XD

I got cocky and did it in my head, but the reflection rule is correct.

do your own homework

Impossible to tell, some numbers given are wrong.

OK faggots, for all of you who "did it in their head", you most likely found the right answer but you have by no means the right way of getting here, which means you're all pre school faggots who guess the answer to your math.

I admit I found x=30° and I admit I couldn't justify it myself. A justification exists here: geogebra.org/m/wq2WM564 and is mostly valid, but requires a bit more than just angles
So no, the reflection rule is not correct.

original problem is en.wikipedia.org/wiki/Langley’s_Adventitious_Angles and you're all faggots

If you split the left triangle into a single three sided triangle, you end up with 170 deg. You're a retard, x=30, this is easymode shit boiz

You can find the length of both sides of the top 70 degree angle by by solving all the triangles up

So what's the answer? Not 35. Reflection is true,. Go hold this picture up to a mirror.

Your maths is bad and you should feel bad

PITA but doable

i said the answer was x=30 you fucking nigger lover cocksucker faggot. You can't find it "just" with reflection though, or you're inferring something but your ridiculously small brain can't comprehend the point of a justification to a solution.

your answer is like "dude it's 30 just trust me", go get an actual education

Hi /sci/ fag here answers 30

its not solvable given the current information

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problem not scale do you want head pats trad

All I have is this if it's solvable idk, it's been a long time.

140=x+(140-x)

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Draw {\displaystyle BG}BG at {\displaystyle 20^{\circ }}20^{\circ } to {\displaystyle BC}BC intersecting {\displaystyle AC}AC at {\displaystyle G}G and draw {\displaystyle FG.}{\displaystyle FG.} (See figure on the lower right.)
Since {\displaystyle \angle {BCG}=80^{\circ }}{\displaystyle \angle {BCG}=80^{\circ }} and {\displaystyle \angle {CBG}=20^{\circ }}{\displaystyle \angle {CBG}=20^{\circ }} then {\displaystyle \angle {BGC}=80^{\circ }}{\displaystyle \angle {BGC}=80^{\circ }} and triangle {\displaystyle BCG}{\displaystyle BCG} is isosceles with {\displaystyle BC=BG.}{\displaystyle BC=BG.}
Since {\displaystyle \angle {BCF}=50^{\circ }}{\displaystyle \angle {BCF}=50^{\circ }} and {\displaystyle \angle {CBF}=80^{\circ }}{\displaystyle \angle {CBF}=80^{\circ }} then {\displaystyle \angle {BFC}=50^{\circ }}{\displaystyle \angle {BFC}=50^{\circ }} and triangle {\displaystyle BCF}{\displaystyle BCF} is isosceles with {\displaystyle BC=BF.}{\displaystyle BC=BF.}
Since {\displaystyle \angle {FBG}=60^{\circ }}{\displaystyle \angle {FBG}=60^{\circ }} and {\displaystyle BF=BG}{\displaystyle BF=BG} then triangle {\displaystyle BGF}{\displaystyle BGF} is equilateral.
Since {\displaystyle \angle {BGE}=100^{\circ }}{\displaystyle \angle {BGE}=100^{\circ }} and {\displaystyle \angle {GBE}=40^{\circ }}{\displaystyle \angle {GBE}=40^{\circ }} then {\displaystyle \angle {GEB}=40^{\circ }}{\displaystyle \angle {GEB}=40^{\circ }} and triangle {\displaystyle BGE}{\displaystyle BGE} is isosceles with {\displaystyle GB=GE.}{\displaystyle GB=GE.}
Therefore all the red lines in the figure are equal.
Since {\displaystyle GE=GF}{\displaystyle GE=GF} triangle {\displaystyle EFG}{\displaystyle EFG} is isosceles with {\displaystyle \angle {GEF}=70^{\circ }.}{\displaystyle \angle {GEF}=70^{\circ }.}
Therefore {\displaystyle \angle {BEF}=30^{\circ }.}{\displaystyle \angle {BEF}=30^{\circ }.}

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[spoiler]I'm an engineer and got stuck on the circle angles[/spoiler]

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not exact because I didn't want to rememeber fractions, but it's real close

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A bunch of scholars you are. I’m impressed, good job gentlemen.

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also you could get the exact result by only computing all the sines at the end but that would've been a lot to write so I calculated them after each step and rounded to 4 digits.

I got this. Haven't done this in a decade so

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I didn't round in the following post except for the result:
Btw: x=28.496

use the internets
calculator.net/triangle-calculator.html

wrong argumentation, see . if you change x and y then the top left edge would move and the angles in the left triangle could not be 50, 110 and 20 anymore.

False

My own solution to the Adventitious Quandrangle was always just straight linear algebra. Introduce some coordinates and calculate the intersection of lines. Messy as fuck. But it works. I have heard that there are more elegant solutions. Would you care to enlighten us, OP?

You're not retarded. We've all been there with this problem. It's the obvious solution, but it doesn't work because of that bizarre linear dependence.

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