# A triangle has sides A,B, and C. If the angle between sides A and B is (3pi)/8, the angle between sides B and C is pi/12, and the length of B is 5, what is the area of the triangle?

Apr 15, 2016

Hence area of triangle is $3.015$ units

#### Explanation:

The length of side $B$ is $5$ and angle opposite this side is angle between sides $A$ and $C$, which is not given. But as other two angles are $\frac{3 \pi}{8}$ and $\frac{\pi}{12}$, this angle would be

$\pi - \frac{3 \pi}{8} - \frac{\pi}{12} = \frac{24 \pi - 9 \pi - 2 \pi}{24} = \frac{13 \pi}{24}$

Now for using sine formula for area of triangle given by $\frac{1}{2} \times a b \times \sin \theta$, we need one more side. Let us choose the side $C$, which is opposite the angle $\frac{3 \pi}{8}$.

Now using sine rule we have

$\frac{5}{\sin} \left(\frac{13 \pi}{24}\right) = \frac{C}{\sin} \left(\frac{3 \pi}{8}\right)$ or $C = 5 \times \sin \frac{\frac{3 \pi}{8}}{\sin} \left(\frac{13 \pi}{24}\right)$

Hence area of triangle is $\frac{1}{2} \times 5 \times 5 \times \sin \frac{\frac{3 \pi}{8}}{\sin} \left(\frac{13 \pi}{24}\right) \times \sin \left(\frac{\pi}{12}\right)$

= $\frac{25}{2} \times \frac{0.9239 \times 0.2588}{0.9914} = 3.015$ units