A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/12, the angle between sides B and C is (2pi)/3, and the length of B is 20, what is the area of the triangle?

1 Answer

$100 \sqrt{6} \cdot \sin \frac{\pi}{12}$

Explanation:

We use ABC for points; and a,b,c for opposite sides.

angle between a and b = $\hat{C} = \frac{1}{12} \pi$

angle between b and c = $\hat{A} = \frac{2}{3} \pi$

$\hat{B} = \pi - \hat{C} - \hat{A} = \pi \left(1 - \frac{1}{12} - \frac{2}{3}\right) = \frac{1}{4} \pi$

Let $H \in A C$, such that $B H$ is perpendicular to $A C$.

$| B H | = h , | A H | = m$ and we want ${S}_{\Delta} = \frac{1}{2} \cdot 20 \cdot h$

$\tan \hat{A} = \frac{h}{m} R i g h t a r r o w m = \frac{h}{\tan} \hat{A}$

$\tan \hat{C} = \frac{h}{20 - m} R i g h t a r r o w 20 - m = \frac{h}{\tan} \hat{C}$

$20 - \frac{h}{\tan} \hat{A} = \frac{h}{\tan} \hat{C}$

$20 = h \left(\frac{1}{\tan} \hat{A} + \frac{1}{\tan} \hat{C}\right)$

$20 = h \left(\cos \frac{\hat{A}}{\sin} \hat{A} + \cos \frac{\hat{C}}{\sin} \hat{C}\right)$

$h = 20 \cdot \frac{\sin A \sin C}{\sin A \cos C + \sin C \cos A} = \frac{20}{\sin} \frac{3 \pi}{4} \cdot \sin \frac{2 \pi}{3} \sin \frac{\pi}{12}$

${S}_{\Delta} = 10 h = 200 \cdot \frac{2}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} \cdot \sin \frac{\pi}{12}$