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Pre-Calculus Help!!!!?
I had this problem and it kinda confused me. I don't know if I'm overthinking it or what but I just can't figure it out.
The diagonals of a parallelogram are 25 and 18 and intersect to form an angle of 113.5 degrees. Find the lengths of each side of the parallelogram.
1 Answer
- ?Lv 77 years agoFavorite Answer
Let a be the length of the longer sides of the parallelogram and b be the length of the shorter sides.
The diagonals with the sides of the parallelogram, form 4 triangles.
The diagonals intersect at their midpoints
Two triangles have leg lengths of 12.5, 9, a, and apex angles 113.5
Two triangles have leg lengths of 12.5, 9, b, and apex angles 180-113.5 = 66.5
Now, use the law of cosines to calculate the unknown lengths, a and b.
a² = 12.5² + 9² - 2(12.5)(9)cos(113.5) = 326.969
a = 18.08
b² = 12.5² + 9² - 2(12.5)(9)cos(66.5) = 147.531
b = 12.15
The lengths of the sides of the parallelogram are 18.08 and 12.15