Question

Two opposite sides of a parallelogram each have a length of `9`. If one corner of the parallelogram has an angle of `(3pi)/8` and the parallelogram's area is `72`, how long are the other two sides?

Answer 1

Approximately `8.66`.

Explanation:

Height on the side of unknown length `b` shall equal to
`h_b=a*sin theta`
where `a` is the length of the side given in the question, and `theta` is the angle between the two adjacent sides. This relationship is true since the height here essentially creates a right triangle where the known side `BC` is the hypotenuse and the height `CH` the side opposite to the corner `hat B= theta`.

The question states that `a=9` and `theta = (3 pi)/8`. Thus
`h_b=9*sin((3pi)/(8))=8.31`
(rounded, you might need your calculator when evaluating `sin((3pi)/8)=sin(67.5^"o")`.)

From the area parallelogram area formula `A=b*h` where `h` is the height of the corresponding base `b`,
`b=A/h_b=72/8.31=8.66`

Thus the length of the side in question equals `8.66`.