A triangle has sides A, B, and C. The angle between sides A and B is pi/6 and the angle between sides B and C is pi/12. If side B has a length of 18, what is the area of the triangle?

Aug 12, 2017

Area of the triangle is $29.65$ sq.unit.

Explanation:

The angle between sides A and B is $\angle c = \frac{\pi}{6} = \frac{180}{6} = {30}^{0}$

The angle between sides B and C is $\angle a = \frac{\pi}{12} = \frac{180}{12} = {15}^{0}$

The angle between sides C and A is

$\angle b = 180 - \left(30 + 15\right) = {135}^{0} , B = 18$ Applying sine law we get

$\frac{A}{\sin} a = \frac{B}{\sin} b \mathmr{and} A = B \cdot \sin \frac{a}{\sin} b = 18 \cdot \sin \frac{15}{\sin} 135 \approx 6.59$

Now we have side $A = 6.59 , B = 18$ and their included angle

$\angle c = {30}^{0}$ . Area of the triangle is ${A}_{t} = \frac{A \cdot B \cdot \sin c}{2}$ or

${A}_{t} = \frac{6.59 \cdot 18 \cdot \sin 30}{2} \approx 29.65$ sq.unit [Ans]