# 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/12, and the length of B is 3, what is the area of the triangle?

Feb 6, 2016

Area $= 1.356 {\text{ units}}^{2}$ to 3 decimal places

#### Explanation:

$\textcolor{b l u e}{\text{Method}}$

Find h using the sin rule. Then use h to determine the area.

$\textcolor{b l u e}{\text{Solution}}$

Target is to be able to apply $\frac{C}{\sin \left(c\right)} = \frac{B}{\sin \left(b\right)}$

To do this we need to find $\angle c b a$

The sum of the internal angles in a triangle is ${180}^{o}$

$\implies \angle c b a = \pi - \frac{5 \pi}{8} - \frac{\pi}{12} = \frac{7 \pi}{24} \text{ } \left({52.5}^{o}\right)$

Thus we have

$\frac{C}{\sin \left(\frac{5 \pi}{8}\right)} = \frac{3}{\sin \left(\frac{7 \pi}{24}\right)}$

$C = 3 \times \sin \frac{\frac{5 \pi}{8}}{\sin \left(\frac{7 \pi}{24}\right)}$

$h = C \sin \left(\frac{\pi}{12}\right)$

$h = 3 \times \sin \frac{\frac{5 \pi}{8}}{\sin \left(\frac{7 \pi}{24}\right)} \times \sin \left(\frac{\pi}{12}\right)$

$A r e a = \frac{B}{2} \times h \text{ "=" } \frac{9}{2} \times \sin \frac{\frac{5 \pi}{8}}{\sin \left(\frac{7 \pi}{24}\right)} \times \sin \left(\frac{\pi}{12}\right)$

$\frac{9}{2} \times \frac{\sin \left({112.5}^{o}\right)}{\sin \left({52.5}^{o}\right)} \times \sin \left({15}^{o}\right)$

$= 1.356 {\text{ units}}^{2}$