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?

Feb 2, 2016

${\text{Area" = 15 sqrt(3) " units}}^{2}$

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

${\text{Area" = A * C * sin((7pi)/18) = 5 * 6 * sin((7pi)/18) = 30 sin ((7pi)/18) " units}}^{2}$

Now, let's evaluate $\sin \left(\frac{7 \pi}{18}\right)$.

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

$\sin \left(x - y\right) = \sin x \cos y - \cos x \sin y$,

we can express $\sin \left(\frac{7 \pi}{18}\right)$ as follows:

$\sin \left(\frac{7 \pi}{18}\right) = \sin \left(\frac{9 \pi}{18} - \frac{2 \pi}{18}\right) = \sin \left(\frac{\pi}{2} - \frac{\pi}{6}\right)$

$= \sin \left(\frac{\pi}{2}\right) \cos \left(\frac{\pi}{6}\right) - \cos \left(\frac{\pi}{2}\right) \sin \left(\frac{\pi}{6}\right)$

$= 1 \cdot \frac{\sqrt{3}}{2} - 0 \cdot \frac{1}{2}$

$= \frac{\sqrt{3}}{2}$

Thus, we have

${\text{Area" = 30 sin ((7pi)/18) = 30 * sqrt(3) /2 = 15 sqrt(3) " units}}^{2}$