# 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/12, and the length of B is 2, what is the area of the triangle?

Jan 9, 2016

$A r e a = 0.5773$ 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.
$\implies \angle A + \angle B + \angle C = \pi$
$\implies \frac{\pi}{12} + \angle B + \frac{7 \pi}{12} = \pi$
$\implies \angle B = \pi - \left(\frac{\pi}{12} + \frac{7 \pi}{12}\right) = \pi - \frac{8 \pi}{12} = \pi - \frac{2 \pi}{3} = \frac{\pi}{3}$
$\implies \angle B = \frac{\pi}{3}$

It is given that side $b = 2$.

Using Law of Sines

$\frac{S \in \angle B}{b} = \frac{\sin \angle C}{c}$

$\implies \frac{S \in \left(\frac{\pi}{3}\right)}{2} = \sin \frac{\frac{7 \pi}{12}}{c}$

$\implies \frac{0.86602}{2} = \frac{0.96592}{c}$

$\implies 0.43301 = \frac{0.96592}{c}$

$\implies c = \frac{0.96592}{0.43301}$

$\implies c = 2.2307$

Therefore, side $c = 2.2307$

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

$\implies A r e a = \frac{1}{2} \cdot 2 \cdot 2.2307 S \in \left(\frac{\pi}{12}\right) = 2.2307 \cdot 0.2588 = 0.5773$ square units
$\implies A r e a = 0.5773$ square units