# A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/12, the angle between sides B and C is (3pi)/4, and the length of side B is 9, what is the area of the triangle?

Jun 27, 2016

Area of triangle is 14.777

#### Explanation:

Length of side b = 9

Angle between sides a and b = $\frac{\pi}{12}$ = $\angle C$

Angle between sides b and c = $\frac{3 \pi}{4}$ = $\angle A$

Sum of the angles of a $\triangle$ = $\pi$

Hence, angle between c and a = $\angle B$= $\pi - \left(\frac{\pi}{12} + \frac{9 \pi}{12}\right)$ = $\left(\frac{2 \pi}{12}\right)$ = $\frac{\pi}{6}$

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

we get $\frac{b}{\sin} \left(\frac{\pi}{6}\right) = \frac{a}{\sin} \left(\frac{3 \pi}{4}\right) = \frac{c}{\sin \left(\frac{\pi}{12}\right)}$

$\frac{9}{\frac{1}{2}} = \frac{a}{\sin} \left(\frac{3 \pi}{4}\right) = \frac{c}{\sin \left(\frac{\pi}{12}\right)}$

$a = 18 \cdot \sin \left(\frac{3 \pi}{4}\right)$ = $18 \cdot \frac{\sqrt{2}}{2}$ = $18 \cdot 0.707106781$ = $12.7279$

A =$\frac{1}{2} \cdot a \cdot b \cdot S \in C$ = $\frac{1}{2} \cdot 12.7279 \cdot 9 \cdot S \in \left(\frac{\pi}{12}\right)$

= $\frac{1}{2} \cdot 12.7279 \cdot 9 \cdot 2.58$ = 14.777