Question

A parallelogram has a side of length 40 and a diagonal of length 75. If the angle between these two is 37 degrees, how do you find the length of the other side of the parallelogram?

Answer 1

Call ABCD the parallelogram. AB the small base , and DC the large base. In the triangle ABD, we have AB = 40, BD = 75 and B = 37 deg -> cos 37 = 0.80.

`b^2 = 40^2 + 75^2 - 2.(40)(75)(0.80) =`

= 1600 + 5625 - 4800 = 2425 -> `Side AD = b = 49.20`

In the triangle ABD, we have now: B = 37 -> sin B = 0.60

`sin A/(75) = (0.60)/(49.24)` ->`sin A = (75(0.6))/49.24` = 0.91

sin A = 0.91 -> A = 66.05 (rejected because < 90 deg)
and `x = (180 -66.05) = 113.95` (accepted).
In the parallelogram, angle A = angle B = 113.95
In the triangle BCD, we have: side BD = 75, side BC = 49.20 and angle B = (113.95 - 37) = 76.95 -> cos 76.95 = 0.226.

`Side DC = x^2 = 75^2 + (49.2)^2 - 2(75)(49.2)(0.226) = 6,380.76`

Large base DC = x = 79.88