# A triangle has sides A, B, and C. Sides A and B have lengths of 10 and 8, respectively. The angle between A and C is (11pi)/24 and the angle between B and C is  (3pi)/8. What is the area of the triangle?

Feb 5, 2016

${\text{Area" = 20 " units}}^{2}$

#### Explanation:

Let $\alpha$ be the angle opposite to the side $A$, $\beta$ be the angle opposite to the side $B$ and $\gamma$ be the angle opposite to the side $C$.

Thus, you have:

$A = 10$, $B = 8$, $\beta = \frac{11 \pi}{24}$ and $\alpha = \frac{3 \pi}{8}$.

Let's find out the length of the third angle first.

As the sum of all three angles in the triangle must be ${180}^{\circ} = \pi$, you know that

$\gamma = \pi - \alpha - \beta = \pi - \frac{11 \pi}{24} - \frac{3 \pi}{8} = \frac{\pi}{6}$

Now you have the sides $A$ and $B$ and $\gamma$, the angle between those two sides. With this information, you can use the formula

$\text{Area} = \frac{1}{2} A \cdot B \cdot \sin \gamma$

$= \frac{1}{2} \cdot 10 \cdot 8 \cdot \sin \left(\frac{\pi}{6}\right)$

$= 5 \cdot 8 \cdot \frac{1}{2}$

$= 20 {\text{ units}}^{2}$