# A triangle has sides A, B, and C. Sides A and B have lengths of 5 and 4, respectively. The angle between A and C is (17pi)/24 and the angle between B and C is  (pi)24. What is the area of the triangle?

$5 \setminus \sqrt{2} \setminus \setminus {\textrm{u n i t}}^{2}$

#### Explanation:

We know that the sum of all interior angles of a triangle is always $\setminus \pi$ then the angle between sides A & B is given as

\pi-\text{angle between sides A & C}-\text{angle between sides B & C}

$= \setminus \pi - \frac{17 \setminus \pi}{24} - \frac{\setminus \pi}{24}$

$= \frac{6 \setminus \pi}{24}$

$= \setminus \frac{\pi}{4}$

hence, the area of given triangle having sides $A = 5$, $B = 4$ & their included angle $\setminus \angle C = \setminus \frac{\pi}{4}$ is given as follows

$\frac{1}{2} A B \setminus \sin \setminus \angle C$

$= \frac{1}{2} \left(5\right) \left(4\right) \setminus \sin \left(\setminus \frac{\pi}{4}\right)$

$= 10 \setminus \cdot \frac{1}{\setminus} \sqrt{2}$

$= 5 \setminus \sqrt{2} \setminus \setminus {\textrm{u n i t}}^{2}$