# A triangle has sides A, B, and C. The angle between sides A and B is pi/3 and the angle between sides B and C is pi/12. If side B has a length of 18, what is the area of the triangle?

$37.592 \setminus \setminus {\textrm{u n i t}}^{2}$

#### Explanation:

Given in $\setminus \triangle A B C$ ,

$\setminus \angle A = \setminus \frac{\pi}{12}$

$\setminus \angle C = \setminus \frac{\pi}{3}$

$\setminus \therefore \setminus \angle B = \setminus \pi - \setminus \angle A - \setminus \angle C$

$= \setminus \pi - \setminus \frac{\pi}{12} - \setminus \frac{\pi}{3}$

$= \frac{7 \setminus \pi}{12}$

Now, using sine rule in $\setminus \triangle A B C$ as follows

$\setminus \frac{A}{\setminus \sin \setminus \angle A} = \setminus \frac{B}{\setminus \sin \setminus \angle B}$

$\setminus \frac{A}{\setminus \sin \left(\setminus \frac{\pi}{12}\right)} = \setminus \frac{18}{\setminus \sin \left(\frac{7 \setminus \pi}{12}\right)}$

$A = 18 \setminus \frac{\setminus \sin \left(\setminus \frac{\pi}{12}\right)}{\setminus \sin \left(\frac{7 \setminus \pi}{12}\right)}$

$A = 18 \left(2 - \setminus \sqrt{3}\right)$

Now, the area $\setminus \Delta$ of triangle $\setminus \triangle A B C$ with sides $A = 18 \left(2 - \setminus \sqrt{3}\right)$ & $18$ & included angle $\setminus \angle C = \setminus \frac{\pi}{3}$ is given as

$\setminus \Delta = \frac{1}{2} A B \setminus \sin \setminus \angle C$

$= \frac{1}{2} \left(18\right) \left(18 \left(2 - \setminus \sqrt{3}\right)\right) \setminus \sin \left(\setminus \frac{\pi}{3}\right)$

$= 162 \left(2 - \setminus \sqrt{3}\right) \left(\setminus \frac{\sqrt{3}}{2}\right)$

$= 81 \left(2 \setminus \sqrt{3} - 3\right)$

$= 37.592 \setminus \setminus {\textrm{u n i t}}^{2}$