I have a isosceles trapezoid question. The top line is 13" long, the sides are 60" long, and the bottom is 27" long. I do not know what is the equation or formula is to determine the acute angle between the bottom and sides or top and sides. Can you please let me know what the equation or math formula is?
geometry - Reiny, Wednesday, March 9, 2011 at 10:46pm
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.
geometry - tchrwill, Thursday, March 10, 2011 at 10:19am
The lower angle derives from arcos[(27-13)/2]/60 = 83.3º.
The upper angle is therefore 96.7º.