Hey Veeky Forums.
This problem was filed under one of those "you should be able to do this" gimmicks. Unlike the norm where the problem has no solution, this one seems different.
The challenge is to find 'a' in the figure without using trig. Can anybody illustrate a solution, or show that you cannot solve this using classical geometry alone?
Other urls found in this thread:
duckware.com
twitter.com
My intuition tells me that you need to extend the figure to something like what's in the image.
Well, some reasoning and a little trial and error will show that a and b can equal 70 and 60. There's no profoundly rigorous proof that I know, but basic algebra carried me through it.
Triangles equal 180 and quadrilaterals equal 360, so 70+10+60+20+40+30+a+b=360, which simplifies down to a+b=130. 50+a+b=180, so a+b=130.
Assuming the figure and angles aren't to scale and can be ignored, a and b can be anything that sums to 130. 60 and 70 seem the most logical, since they match up to the other angles.
Like I said, not a rigorous way of solving the problem, and I may not have the "correct" answer, but it works.
You have two equations and two unknowns here.
First the sum of the angles of a quadrilateral and the sum of the angle of a triangle.
Both those sums individually will include a+b.
Solve from there.
Do not be tricked, this is not a trivial problem.
>a and b can be anything that sums to 330
This is not true. Counter example, a=0 and b=130. The problem is scale invariant, but the normalized lengths of the sides are fully constrained by the angles given.
The equations are not linearly independent.
360=(60+20)+(70+10)+(40+b)+(30+a) => 130=a+b
180=(10)+(40+b)+(a) => 130=a+b
180=(20)+(b)+(30+a) => 130=a+b
Believe me, I've tried.
Original thread was posted in /g/ under post
an answer won`t du?
All corners need to add to 360. Because the two corners add up to 80, the two right corners should add up to 100. A=70, B=60. 70+60+50=180. QED
I know I've seen something similar before...
duckware.com
You are pretty much solving problem 1.
This is an isosceles trapezoid. The two base angles are both 80. This means:
40 + b = 30 + a
10 + b = a
10 + b + b = 130
b = 60
a = 70
It's not a trapezoid.
It is. But I wish doing math was as simple as telling people what's right and wrong and not bothering with an explanation.
See
Not him but for a quadrilateral to be a trapezium at least one pair of sides have to be parallel.
I took my freshman geometry textbook to tell you this ffs what am I doing.
Anyways, looking at OP's pic there are no parallel sides, something you can confirm by extending the lines that contain those segments.
Nice try though, I almost thought you were correct.
0 is recused anyway as the angles are assumed to exist, or the shape would not be consistent.
130=a+b (a,b 0) is the correct solution.
Linear independence is the wrong way to look at it. All of the equations reducing to 130=a+b means your diagram is consistent. Consider that all of the triangles in the diagram have either an angle of 50, and two angles summing to 130 or vice versa.
There can be no more exact solution than 130=a+b, as the only information given are angles and the shape being quadrilateral. Line lengths are not given, so any line can be moved or stretched, making any pair of angles in the range 0
this
traced over the image in geogebra and the angles listed are not accurate
Any pics of what the image would look like if the angles provided were drawn correctly?
Oh god. Analytical geometry was the only mathematics I hated in high school. I'm done.
Apparently:
a = 20
b = 110
>this
is wrong.
See for solution. (thanks btw)
The angles fully constrain the geometry of the problem such that there is only one solution. If you disagree with me, draw two geometrically accurate solution.
I'll save you the time and draw the correct solution. The angles and labels correspond with each other in this figure.
a=30, b=100
a=100 b=30. Both solutions work.
Draw the figure and I'll believe you.
Draw it? We aren't retarded here on sci, they are variables, just picture it in your head. Nothing described this as being to scale..
The figure here is geometrically accurate. Check the angles.
A picture isn't proof, though, so go here for that If you can't produce a figure even approximately accurate to your proposed 'solutions', you should rethink your claim.
The fact that the angles (60 and 70 degrees) at the bottom aren't symmetrical locks the system in a single configuration.