# 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 18, what is the area of the triangle?

Dec 31, 2015

The area of the triangle is $\frac{81 \cdot \sqrt{3}}{2}$ or aproximately 70.148

#### Explanation:

Let's call "c" the angle ($= \frac{\pi}{3}$) between sides A and B, "a" the angle ($= \frac{\pi}{6}$) between sides B and C, and "b" the angle between sides A and C.

For it is a triangle, $a + b + c = \pi$
Then b=pi-c-a=pi-pi/3-pi/6=(6pi-2pi-pi)/6=(3pi) /6=pi/2
(Because of $b = \frac{\pi}{2}$, the triangle is a right one. The side B is the hypotenuse)

In a right triangle it is valid the following statement:
$\sin x = \left(\text{opposed cathetus")/("hypotenuse}\right)$

Applying the aforementioned expression
$\sin a = \frac{A}{B}$ => $A = B \cdot \sin a = 18 \cdot \sin \left(\frac{\pi}{6}\right) = 18 \cdot \frac{1}{2} = 9$
$\sin c = \frac{C}{B}$ => $C = B \cdot \sin c = 18 \cdot \sin \left(\frac{\pi}{3}\right) = 18 \cdot \frac{\sqrt{3}}{2} = 9 \cdot \sqrt{3}$

Now, to the formula of the triangles area:
${S}_{\triangle} = \frac{b a s e \cdot {h}_{\triangle}}{2}$
(Since the angle between sides A and C is $\frac{\pi}{2}$, it's convenient to use one of this side as the base and the other one as the height of the triangle)

Then
${S}_{\triangle} = \frac{A \cdot C}{2} = : \frac{9 \cdot 9 \cdot \sqrt{3}}{2} = \frac{81 \cdot \sqrt{3}}{2}$