# 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/6. If side B has a length of 17, what is the area of the triangle?

Dec 29, 2016

The area is 144.5.

#### Explanation:

let's first switch to standard notation. Sides are denotes a, b and c and the angle of the vertex opposite the sides are denotes A, B and C.

The question then becomes, what is the area of the triangle where $b = 17$, $A = \frac{\pi}{6}$ and $C = \frac{5 \pi}{12}$.

The best equation for the areas in $A r e a = \frac{1}{2} a b \sin C$. For this we need the length of side a. To calculate this we also, need angle B.

To calculate angle B we use the fact that $A + B + C = \pi$. We rewrite this as $B = \pi - A - C$

$B = \pi - \frac{\pi}{6} - \frac{5 \pi}{12} = \frac{5 \pi}{12}$

Using the sine rule:

$\frac{a}{\sin A} = \frac{b}{\sin B}$

This gives:

$a = \frac{b \sin A}{\sin B} = \frac{17 \sin \left(\frac{\pi}{6}\right)}{\sin \frac{5 \pi}{12}}$

So the areas is:

$A r e a = \frac{1}{2} \frac{b \sin A}{\sin B} = \frac{17 \sin \left(\frac{\pi}{6}\right)}{\sin \frac{5 \pi}{12}} 17 \sin \frac{5 \pi}{12} = {17}^{2} \sin \left(\frac{\pi}{6}\right)$

Now $\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

So:

$A r e a = {17}^{2} / 2 = \frac{289}{2} = 144.5$