# A triangle has sides A, B, and C. The angle between sides A and B is pi/12 and the angle between sides B and C is pi/12. If side B has a length of 17, what is the area of the triangle?

Sep 30, 2017

The area of the triangle is $19.36$ sq.unit [Ans]

#### Explanation:

The angle between sides $A \mathmr{and} B$ is $\angle c = \frac{\pi}{12} = \frac{180}{12} = {15}^{0}$

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

Angle between sides $C \mathmr{and} A$ is /_b=(180-(15+15)=150^0

The Law of Sines (or Sine Rule) is:$\frac{A}{\sin} a = \frac{B}{\sin} b = \frac{C}{\sin} c$

$B = 17 \therefore \frac{A}{\sin} a = \frac{B}{\sin} b \mathmr{and} \frac{A}{\sin} 15 = \frac{17}{\sin} 150$ or

$A = 17 \cdot \sin \frac{15}{\sin} 150 \approx 8.8$ So the sides $A \mathmr{and} B$ and their

included angle are $A = 8.8 , B = 17 \mathmr{and} \angle c = {15}^{0}$

The area of triangle is ${A}_{t} = \frac{A \cdot B \cdot \sin c}{2} = \frac{8.8 \cdot 17 \cdot \sin 15}{2}$ or

${A}_{t} \approx 19.36$ sq.unit [Ans]