# A triangle has sides A,B, and C. If the angle between sides A and B is (7pi)/12, the angle between sides B and C is pi/4, and the length of B is 7, what is the area of the triangle?

Jun 19, 2017

The area of the triangle is $= 33.5 {u}^{2}$

#### Explanation:

The angle between $A$ and $C$ is

$= \pi - \left(\frac{7}{12} \pi + \frac{1}{4} \pi\right)$

$= \pi - \frac{10}{12} \pi$

$= \frac{2}{12} \pi = \frac{1}{6} \pi$

$B = 7$

We apply the sine rule

$\frac{A}{\sin} \left(\frac{1}{4} \pi\right) = \frac{B}{\sin} \left(\frac{1}{6} \pi\right)$

$A = \frac{B \sin \left(\frac{1}{4} \pi\right)}{\sin} \left(\frac{1}{6} \pi\right)$

The area of the triangle is

$= \frac{1}{2} \cdot A \cdot B \cdot \sin \left(\frac{7}{12} \pi\right)$

$= \frac{1}{2} \cdot B \cdot \frac{B \sin \left(\frac{1}{4} \pi\right)}{\sin} \left(\frac{1}{6} \pi\right) \cdot \sin \left(\frac{7}{12} \pi\right)$

$= \frac{1}{2} \cdot {B}^{2} \cdot \frac{\sin \left(\frac{1}{4} \pi\right)}{\sin} \left(\frac{1}{6} \pi\right) \cdot \sin \left(\frac{7}{12} \pi\right)$

$= \frac{1}{2} \cdot 49 \cdot 1.366$

$= 33.5$