# A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/12 and the angle between sides B and C is pi/12. If side B has a length of 63, what is the area of the triangle?

Aug 13, 2016

$= 496.125$

#### Explanation:

Clearly this is a right-angled triangle since Angle between sides$A$ and $C$ is $= \pi - \left(\frac{5 \pi}{12} + \frac{\pi}{12}\right)$
$= \pi - \frac{\pi}{2}$
$= \frac{\pi}{2}$
In this right angled triangle side$A$ is height and side $C$ is base and side $B = 63$ is hypotenuse
$A = h e i g h t = 63 \sin \left(\frac{\pi}{12}\right)$ and
$C = b a s e = 63 \cos \left(\frac{\pi}{12}\right)$
Therefore
Area of the triangle
$= \frac{1}{2} \times h e i g h t \times b a s e$
$= \frac{1}{2} \times 63 \sin \left(\frac{\pi}{12}\right) \times 63 \cos \left(\frac{\pi}{12}\right)$
$= \frac{1}{2} \times 63 \left(0.2588\right) \times 63 \left(0.966\right)$
$= 496.125$