# A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/6, the angle between sides B and C is (5pi)/12, and the length of B is 17, what is the area of the triangle?

Jun 4, 2016

It turns out to be simply $\frac{289}{4} = 72.25$ square units.

#### Explanation:

The three angles of the triangle add up to $\setminus \pi$ radians, so the angle between $A$ and $C$ is $\setminus \pi - \left(\setminus \frac{\pi}{6}\right) - \left(\frac{5 \setminus \pi}{12}\right) = \frac{5 \setminus \pi}{12}$ radians.

Then the angles on both ends of side $C$ are congruent and thus the triangle is isosceles, with side $A$ congruent with side $B$. So side $A$ measures $17$ units along with side $B$. Now the area of the triangle is half the product of the two sides $A$ and $B$ times the sine of the angle between them:

$\sin \left(\setminus \frac{\pi}{6}\right) = \frac{1}{2}$

Area = $\left(\frac{1}{2}\right) \times \left(17\right) \times \left(17\right) \times \left(\frac{1}{2}\right) = \frac{289}{4}$ square units.