# 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 (5pi)/12, and the length of B is 2, what is the area of the triangle?

Area of the triangle is $2.36 \left(2 \mathrm{dp}\right)$ sq units
The angle between sides A and B is $\angle C = \frac{\pi}{3} = \frac{180}{3} = {60}^{0}$
The angle between sides B and C is $\angle A = \frac{5 \pi}{12} = \frac{5 \cdot 180}{12} = {75}^{0}$
The angle between sides C and A is $\angle B = 180 - \left(60 + 75\right) = {45}^{0}$
$B = 2$ (given) Applying sine law we get $\frac{A}{\sin} A = \frac{B}{\sin} B \mathmr{and} A = 2 \cdot \sin \frac{75}{\sin} 45 = 2.73$ Now A=2.73 ; B=2 and their included angle $\angle C = {60}^{0}$ So area of the triangle is $a \cdot b \cdot \sin \frac{C}{2} = 2.73 \cdot 2 \cdot \sin \frac{60}{2} = 2.36 \left(2 \mathrm{dp}\right)$sq units [Ans}