# 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 (13pi)/24 and the angle between B and C is  (pi)24. What is the area of the triangle?

Apr 18, 2018

Since triangle angles add to $\pi$ we can figure out the angle between the given sides and the area formula gives
$A = \setminus \frac{1}{2} a b \sin C = 10 \left(\sqrt{2} + \sqrt{6}\right)$.

#### Explanation:

It helps if we all stick to the convention of small letter sides $a , b , c$ and capital letter opposing vertices $A , B , C$. Let's do that here.

The area of a triangle is $A = \frac{1}{2} a b \sin C$ where $C$ is the angle between $a$ and $b$.

We have $B = \setminus \frac{13 \setminus \pi}{24}$ and (guessing it's a typo in the question) $A = \setminus \frac{\pi}{24}$.

Since triangle angles add up to ${180}^{\setminus} \circ$ aka $\setminus \pi$ we get

$C = \setminus \pi - \setminus \frac{\pi}{24} - \frac{13 \pi}{24} = \setminus \frac{10 \pi}{24} = \setminus \frac{5 \pi}{12}$

$\setminus \frac{5 \pi}{12}$ is ${75}^{\setminus} \circ .$ We get its sine with the sum angle formula:

$\sin {75}^{\circ} = \sin \left(30 + 45\right) = \sin 30 \cos 45 + \cos 30 \sin 45$

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

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

So our area is

$A = \setminus \frac{1}{2} a b \sin C = \setminus \frac{1}{2} \left(10\right) \left(8\right) \setminus \frac{1}{4} \left(\sqrt{2} + \sqrt{6}\right)$

$A = 10 \left(\sqrt{2} + \sqrt{6}\right)$

Take the exact answer with a grain of salt because it's not clear we guessed correctly what the asker meant by the angle between $B$ and $C$.