# A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/6, the angle between sides B and C is (5pi)/12, and the length of B is 2, what is the area of the triangle?

Jan 3, 2016

$A r e a = 1.93184$ 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" $\angle A$ 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.
$\implies \angle A + \angle B + \angle C = \pi$
$\implies \frac{\pi}{6} + \angle B + \frac{5 \pi}{12} = \pi$
$\implies \angle B = \pi - \frac{7 \pi}{12} = \frac{5 \pi}{12}$
$\implies \angle B = \frac{5 \pi}{12}$

It is given that side $b = 2.$

Using Law of Sines
$\frac{S \in \angle B}{b} = \frac{\sin \angle C}{c}$
$\implies \frac{S \in \left(\frac{5 \pi}{12}\right)}{2} = \sin \frac{\frac{5 \pi}{12}}{c}$
$\implies \frac{1}{2} = \frac{1}{c}$
$\implies c = 2$

Therefore, side $c = 2$

Area is also given by
$A r e a = \frac{1}{2} b c S \in \angle A = \frac{1}{2} \cdot 2 \cdot 2 S \in \left(\frac{7 \pi}{12}\right) = 2 \cdot 0.96592 = 1.93184$square units
$\implies A r e a = 1.93184$ square units