# A triangle has sides A,B, and C. If the angle between sides A and B is (pi)/3, the angle between sides B and C is pi/6, and the length of B is 9, what is the area of the triangle?

Dec 28, 2015

17.537

#### Explanation:

As the sum of all the angles of a triangle is ${180}^{0}$ or $\pi$ radians
So, from your sum it can be seen that the third angle between A and C $= \pi - \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{2}$
Thus making the side B, which is opposite to the ${90}^{0}$ angle hypotenuse of the triangle.
So, the area of the triangle would be $\frac{1}{2} \cdot b a s e \cdot h e i g h t = \frac{1}{2} \cdot A \cdot C$
Given that side $B = 9$, we can find the other two side by triangle geometry
$A = B \cdot \cos \left(\frac{\pi}{3}\right) = 9 \cdot \cos \left(\frac{\pi}{3}\right) = 4.5$
$C = B \cdot \sin \left(\frac{\pi}{6}\right) = 9 \cdot \sin \left(\frac{\pi}{6}\right) = 7.794$
Hence, $A r e a = \frac{1}{2} \cdot A \cdot C = \frac{1}{2} \cdot 4.5 \cdot 7.794 = 17.537$