Question

Two opposite sides of a parallelogram have lengths of `3`. If one corner of the parallelogram has an angle of `pi/12` and the parallelogram's area is `14`, how long are the other two sides?

Answer 1

Assuming a bit of basic Trigonometry...

Explanation:

Let x be the (common) length of each unknown side.
If b = 3 is the measure of the base of the parallelogram, let h be its vertical height.
The area of the parallelogram is `bh = 14`
Since b is known, we have `h = 14/3`.

From basic Trig, `sin(pi/12) = h/x`.

We may find the exact value of the sine by using either a half-angle or difference formula.
`sin(pi/12) = sin(pi/3 - pi/4) = sin(pi/3)cos(pi/4) - cos(pi/3)sin(pi/4)`
`= (sqrt6 - sqrt2)/4`.

So...
`(sqrt6 - sqrt2)/4 = h/x`
`x(sqrt6 - sqrt2) = 4h`
Substitute the value of h:
`x(sqrt6 - sqrt2) = 4(14/3)`
`x(sqrt6 - sqrt2) = 56/3`
Divide by the expression in parentheses:
`x = 56/(3(sqrt6 - sqrt2))`

If we require that the answer be rationalized:
`x = 56/(3(sqrt6 - sqrt2))*((sqrt6 + sqrt2)/(sqrt6 + sqrt2))`

`= 56(sqrt6 + sqrt2)/(3(4))`
`= (14(sqrt6 + sqrt2))/(3)`

NOTE: If you have the formula `A = ab sin(theta)`, you may use it to arrive at the same answer more rapidly.