# A triangle DeltaABC has sides a, b, and c. The angle between sides a and b is pi/12 and the angle between sides b and c is pi/12. If side b has a length of 12, what is the area of the triangle?

Jun 11, 2017

In $\Delta A B C$

$\angle C = \angle A = \frac{\pi}{12} \mathmr{and} \text{ side } b = 12$

So $\angle B = \pi - 2 \cdot \frac{\pi}{12} = \frac{5 \pi}{6}$

Now by rule of sine

$\frac{a}{\sin} A = \frac{b}{\sin} B$

Now area of the

$\Delta A B C = \frac{1}{2} a \times b \times \sin C$

$= \frac{1}{2} \frac{b}{\sin} B \times \sin A \times b \times \sin C$

$= {b}^{2} / 2 \times \frac{\sin A \sin C}{\sin} B$

$= {b}^{2} / 2 \times \frac{\sin \left(\frac{\pi}{12}\right) \sin \left(\frac{\pi}{12}\right)}{\sin} \left(\frac{5 \pi}{6}\right)$

$= {12}^{2} / 4 \times \frac{2 {\sin}^{2} \left(\frac{\pi}{12}\right)}{\sin} \left(\pi - \frac{\pi}{6}\right)$

$= 36 \times \frac{1 - \cos \left(\frac{\pi}{6}\right)}{\sin} \left(\frac{\pi}{6}\right)$

$= 36 \times \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}}$

$= 36 \left(2 - \sqrt{3}\right)$ squnit