Question

A parallelogram has sides A, B, C, and D. Sides A and B have a length of `5` and sides C and D have a length of `6`. If the angle between sides A and C is `(7 pi)/18`, what is the area of the parallelogram?

Answer 1

`"Area" = 15 sqrt(3) " units"^2`

Explanation:

You have the lengths of all the sides and the angle between the sides `A` and `C`.

Thus, you can compute the area of the parallelogram with the formula

`"Area" = A * C * sin((7pi)/18) = 5 * 6 * sin((7pi)/18) = 30 sin ((7pi)/18) " units"^2`

Now, let's evaluate `sin((7pi)/18)`.

Remember the table of `sin` and `cos` values:

#{: ("angle (deg)", color(white)(xx) 0^@ color(white)(xxx) 30^@ color(white)(xxx) 45^@ color(white)(xxx) 60^@ color(white)(xxx) 90^@), ("angle (rad)", color(white)(xx) 0 color(white)(xxxx) pi/6 color(white)(xxxx) pi/4 color(white)(xxxxi) pi/3 color(white)(xxxx) pi/2), (,), (" "sin, color(white)(xx) 0 color(white)(xxxx) 1/2 color(white)(xxxiii) sqrt(2)/2 color(white)(xxxx) sqrt(3)/2 color(white)(xxxiii) 1), (" "cos, color(white)(xx) 1 color(white)(xxxiii) sqrt(3)/2 color(white)(xxxii) sqrt(2)/2 color(white)(xxxx) 1/2 color(white)(xxxxi) 0) :}#

Using the formula

`sin(x-y) = sin x cos y - cos x sin y`,

we can express `sin((7 pi)/18)` as follows:

`sin((7pi)/18) = sin((9pi)/18 - (2pi)/18) = sin(pi/2 - pi/6)`

`= sin(pi/2) cos(pi/6) - cos(pi/2) sin(pi/6)`

`= 1 * sqrt(3)/2 - 0 * 1/2`

`= sqrt(3)/2`

Thus, we have

`"Area" = 30 sin ((7pi)/18) = 30 * sqrt(3) /2 = 15 sqrt(3) " units"^2`