# A triangle has sides A, B, and C. If the angle between sides A and B is (5pi)/8, the angle between sides B and C is (pi)/3, and the length of B is 16, what is the area of the triangle?

Nov 9, 2017

$784.42$ square units (2 d.p.)

#### Explanation: $A = \frac{1}{2} a b \sin C$
this formula can be used when working with $2$ sides and $1$ angle - $a$ and $b$ are the two sides that are not opposite the angle we are using.

however, there is only $1$ side given, so we need to find a second side - either $a$ or $c$.

first, find the angle between $a$ and $c$:

${\Sigma}_{\angle} \left(\triangle\right) = \pi$

angle between $a$ and $c = \pi - \left(\frac{5 \pi}{8}\right) - \left(\frac{\pi}{3}\right)$

$= \frac{\pi}{24}$

sine rule: $\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C$

$\frac{16}{\sin} \left(\frac{\pi}{24}\right) = \frac{a}{\sin} \left(\frac{\pi}{3}\right)$

$16 \sin \left(\frac{\pi}{3}\right) = a \sin \left(\frac{\pi}{24}\right)$

$16 \times 0.86603 = 0.13053 a$

$a = \frac{16 \times 0.86603}{0.13053} =$

$a = 106.16 , b = 16$

angle opposite $c = \frac{5 \pi}{8}$

$\text{Area} = \frac{1}{2} \left(106.16 \cdot 16\right) \sin \frac{5 \pi}{8}$

$= 849.028 \times 0.9239$

$\text{Area} = 784.42$ square units (2 d.p.)