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April 18, 2014

April 18, 2014

Posted by **Steven** on Wednesday, March 9, 2011 at 10:00pm.

- geometry -
**Reiny**, Wednesday, March 9, 2011 at 10:46pmdraw 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.

- geometry -
**tchrwill**, Thursday, March 10, 2011 at 10:19amThe lower angle derives from arcos[(27-13)/2]/60 = 83.3º.

The upper angle is therefore 96.7º.

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