# A triangle has sides A, B, and C. Sides A and B have lengths of 9 and 3, respectively. The angle between A and C is (pi)/3 and the angle between B and C is  (pi)/4. What is the area of the triangle?

Dec 27, 2015

$\frac{27 \left(1 + \sqrt{3}\right)}{4 \sqrt{2}}$

#### Explanation:

Since the interior angles of a triangle add up to $\pi$, the angle between $A$ and $B$, $\theta$, is

$\theta = \pi - \frac{\pi}{4} - \frac{\pi}{3} = \frac{5 \pi}{12}$.

Area of triangle is calculated using

$\frac{1}{2} \times A \times B \times \sin \left(\theta\right)$

$= \frac{1}{2} \times 9 \times 3 \times \sin \left(\frac{5 \pi}{12}\right)$

$= \frac{27 \left(1 + \sqrt{3}\right)}{4 \sqrt{2}}$

Note:

$\sin \left(\frac{5 \pi}{12}\right) = \sin \left(\pi - \frac{\pi}{4} - \frac{\pi}{3}\right)$

$= \sin \left(\frac{\pi}{4} + \frac{\pi}{3}\right)$

$= \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{3}\right) + \cos \left(\frac{\pi}{4}\right) \sin \left(\frac{\pi}{3}\right)$

$= \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{2}\right) + \left(\frac{1}{\sqrt{2}}\right) \left(\frac{\sqrt{3}}{2}\right)$

$= \frac{1 + \sqrt{3}}{2 \sqrt{2}}$