# A triangle has sides A, B, and C. Sides A and B have lengths of 2 and 4, respectively. The angle between A and C is (7pi)/24 and the angle between B and C is  (5pi)/8. What is the area of the triangle?

Aug 7, 2016

The area is $\setminus \sqrt{6} - \setminus \sqrt{2}$ square units, about $1.035$.

#### Explanation:

The area is one half the product of two sides times the sine of the angle between them.

Here we are given two sides but not the angle between them, we are given the other two angles instead. So first determine the missing angle by noting that the sum of all three angles is $\setminus \pi$ radians:

$\setminus \theta = \setminus \pi - \frac{7 \setminus \pi}{24} - \frac{5 \setminus \pi}{8} = \setminus \frac{\pi}{12}$.

Then the area of the triangle is

Area $= \left(\frac{1}{2}\right) \left(2\right) \left(4\right) \setminus \sin \left(\setminus \frac{\pi}{12}\right)$.

We have to compute $\setminus \sin \left(\setminus \frac{\pi}{12}\right)$. This can be done using the formula for the sine of a difference:

$\sin \left(\setminus \frac{\pi}{12}\right) = \setminus \sin \left(\textcolor{b l u e}{\setminus \frac{\pi}{4}} - \textcolor{g o l d}{\setminus \frac{\pi}{6}}\right)$
$= \setminus \sin \left(\textcolor{b l u e}{\setminus \frac{\pi}{4}}\right) \cos \left(\textcolor{g o l d}{\setminus \frac{\pi}{6}}\right) - \setminus \cos \left(\textcolor{b l u e}{\setminus \frac{\pi}{4}}\right) \sin \left(\textcolor{g o l d}{\setminus \frac{\pi}{6}}\right)$
$= \left(\frac{\setminus \sqrt{2}}{2}\right) \left(\frac{\setminus \sqrt{3}}{2}\right) - \left(\frac{\setminus \sqrt{2}}{2}\right) \left(\frac{1}{2}\right)$
$= \frac{\setminus \sqrt{6} - \setminus \sqrt{2}}{4}$.

Then the area is given by:

Area $= \left(\frac{1}{2}\right) \left(2\right) \left(4\right) \left(\frac{\setminus \sqrt{6} - \setminus \sqrt{2}}{4}\right)$
$= \setminus \sqrt{6} - \setminus \sqrt{2}$.