Question

Two opposite sides of a parallelogram each have a length of `16`. If one corner of the parallelogram has an angle of `(5 pi)/12` and the parallelogram's area is `64`, how long are the other two sides?

Answer 1

The other two sides each have length `4(sqrt6-sqrt2)`.

Explanation:

Since the area of a triangle can be written as `A_triangle=1/2ab sinC`, a parallelogram that is made up of two identical triangles has area `A=2times1/2ab sinC=ab sinC`.

If we know 3 of these variables, we can solve for the 4th.

`A=ab sinC`

`=>64=16bsin((5pi)/12)`

`=>(""^4cancel(64))/(""^1cancel(16)sin((5pi)/12))=b`

`=>4/((sqrt6+sqrt2)//4)=b`

`=>b=16/(sqrt6+sqrt2)color(brown)(*(sqrt6-sqrt2)/(sqrt6-sqrt2))`         (rationalize)

`=>b=(16(sqrt6-sqrt2))/(6-2)`

`=>b=(""^4cancel(16)(sqrt6-sqrt2))/(""^1cancel(4))`

`=>b=4(sqrt6-sqrt2)`

So the other two sides have length `b=4(sqrt6-sqrt2)`.

Bonus:

It doesn't matter if the angle that we know is the acute one or the obtuse one. In a parallelogram, any two adjacent angles are supplementary (they add to 180°, or `pi` radians), and we know that

`sintheta=sin(pi-theta)`.