# A triangle has sides A, B, and C. The angle between sides A and B is pi/6 and the angle between sides B and C is pi/12. If side B has a length of 3, what is the area of the triangle?

Jan 8, 2016

$A r e a = 0.8235$ square units.

#### Explanation:

First of all let me denote the sides with small letters $a$, $b$ and $c$.
Let me name the angle between side $a$ and $b$ by $\angle C$, angle between side $b$ and $c$ by $\angle A$ and angle between side $c$ and $a$ by $\angle B$.

Note:- the sign $\angle$ is read as "angle".
We are given with $\angle C$ and $\angle A$. We can calculate $\angle B$ by using the fact that the sum of any triangles' interior angels is $\pi$ radian.
$\implies \angle A + \angle B + \angle C = \pi$
$\implies \frac{\pi}{12} + \angle B + \frac{\pi}{6} = \pi$
$\implies \angle B = \pi - \left(\frac{\pi}{6} + \frac{\pi}{12}\right) = \pi - \frac{3 \pi}{12} = \pi - \frac{\pi}{4} = \frac{3 \pi}{4}$
$\implies \angle B = \frac{3 \pi}{4}$

It is given that side $b = 3.$

Using Law of Sines
$\frac{S \in \angle B}{b} = \frac{\sin \angle C}{c}$
$\implies \frac{S \in \left(\frac{3 \pi}{4}\right)}{3} = \sin \frac{\frac{\pi}{6}}{c}$

$\implies \frac{\frac{1}{\sqrt{2}}}{3} = \frac{\frac{1}{2}}{c}$

$\implies \frac{\sqrt{2}}{6} = \frac{1}{2 c}$

$\implies c = \frac{6}{2 \sqrt{2}}$

$\implies c = \frac{3}{\sqrt{2}}$

Therefore, side $c = \frac{3}{\sqrt{2}}$

Area is also given by
$A r e a = \frac{1}{2} b c S \in \angle A$

$\implies A r e a = \frac{1}{2} \cdot 3 \cdot \frac{3}{\sqrt{2}} S \in \left(\frac{\pi}{12}\right) = \frac{9}{2 \sqrt{2}} \cdot 0.2588 = 0.8235$ square units
$\implies A r e a = 0.8235$ square units