# A triangle has sides A, B, and C. The angle between sides A and B is (2pi)/3 and the angle between sides B and C is pi/12. If side B has a length of 6, what is the area of the triangle?

$A r e a = 5.70576$ square units

#### Explanation:

To compute for the Area by using the given, there are several ways to do it.
I will present 2 solutions.
1st solution: $A r e a = \frac{1}{2} \cdot b \cdot h$
Compute height $h$ first. The altitude from angle B to side b:

Given angle $A = \frac{\pi}{12}$ and angle $C = \frac{2 \pi}{3}$ and side $b = 6$

$h = \frac{b}{\cot A + \cot C}$

$h = \frac{6}{\cot \left(\frac{\pi}{12}\right) + \cot \left(\frac{2 \pi}{3}\right)}$

$h = 1.90192$

Compute Area:

$A r e a = \frac{1}{2} \cdot b \cdot h = \left(\frac{1}{2}\right) \left(6\right) \left(1.90192\right)$

$A r e a = 5.70576$ square units

2nd solution:
If there are 2 sides and an included angle then, the area is determined.
Compute Angle $B$ then apply sine law to compute side $c$
So that , sides $b$, $c$, and angle $A$ area available.

Compute angle $B$:

$B = \pi - A - C = \pi - \frac{\pi}{12} - \frac{2 \pi}{3} = \frac{\pi}{4}$

Compute side $c$ using Sine Law:

$c = \frac{b \cdot \sin C}{\sin} B = \frac{6 \cdot \sin \left(\frac{2 \pi}{3}\right)}{\sin} \left(\frac{\pi}{4}\right)$

$c = 7.34847$

Compute the $A r e a$:

$A r e a = \frac{1}{2} \cdot b \cdot c \cdot \sin A = \frac{1}{2} \left(6\right) \left(7.34847\right) \sin \left(\frac{\pi}{12}\right)$

$A r e a = 5.70576$ square units

Have a nice day!!! from the Philippines..