# A triangle has sides A, B, and C. The angle between sides A and B is (pi)/3 and the angle between sides B and C is pi/6. If side B has a length of 11, what is the area of the triangle?

Nov 19, 2016

${\text{Area}}_{\triangle} = \frac{121 \sqrt{3}}{8}$ (using a calculator: $\approx 26.197$)

#### Explanation:

Since $\angle \left(A : B\right) = \frac{\pi}{3}$ and $\angle \left(B : C\right) = \frac{\pi}{6}$

$\rightarrow \angle \left(A : C\right) = \pi - \left(\frac{\pi}{3} + \frac{\pi}{6}\right) = \frac{\pi}{2}$
and the triangle is a right angled triangle:

$\frac{A}{B} = \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow A = \frac{1}{2} \times B = \frac{11}{2}$

$\frac{C}{B} = \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow C = \frac{\sqrt{3}}{2} \times B = \left(11 \frac{\sqrt{3}}{2}\right)$

$\text{Area"_triangle = 1/2* "base" * "height} = \frac{1}{2} \times \frac{11}{2} \times \frac{11 \sqrt{3}}{2} = \frac{121 \sqrt{3}}{8}$