Posted by Steven on Wednesday, March 9, 2011 at 10:00pm.
draw in a diagonal, let its length be x
Since you have parallel lines, alternate angles are equal, let that angle be Ø
we can use the cosine law twice
1. 60^2 = x^2 + 13^2 - 2(x)(13)cosØ
cosØ = (x^2 + 169 - 3600)/(26x)
2. 60^2 = x^2 + 27^2 - 2(x)(27)cosØ
cosØ = (x^2 + 27^2 - 3600)/(54x)
then (x^2 - 3431)/(26x) = (x^2 - 2871)/(54x)
(x^2 - 3431)/(26) = (x^2 - 2871)/(54) , after multiplying both sides by x
54x^2 - 185274 = 26x^2 - 74646
28x^2 = 110628
x^2 = 3951
x = 62.857
put that back into one of the first two equations, to get cosØ, and then Ø
then use the Sine Law to get a 2nd angle in one of the triangles.
check my arithmetic.
The lower angle derives from arcos[(27-13)/2]/60 = 83.3º.
The upper angle is therefore 96.7º.
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