# 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 15, what is the area of the triangle?

Nov 30, 2017

Area of the triangle is $41.18$ sq.unit.

#### Explanation:

Angle between Sides $A \mathmr{and} B$ is $\angle c = \frac{\pi}{6} = \frac{180}{6} = {30}^{0}$

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

Angle between Sides $C \mathmr{and} A$ is $\angle b = 180 - \left(30 + 15\right) = {135}^{0}$

The sine rule states if $A , B \mathmr{and} C$ are the lengths of the sides

and opposite angles are $a , b \mathmr{and} c$ in a triangle, then:

A/sina = B/sinb=C/sinc ; B=15 :. A/sina=B/sinb or

$\frac{A}{\sin} 15 = \frac{15}{\sin} 135 \therefore A = 15 \cdot \sin \frac{15}{\sin} 135 \approx 5.49 \left(2 \mathrm{dp}\right)$unit

Now we know sides $A = 5.49 , B = 15$ and their included angle

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

$\therefore {A}_{t} = \frac{5.49 \cdot 15 \cdot \sin 30}{2} \approx 41.18 \left(2 \mathrm{dp}\right)$ sq.unit

Area of the triangle is $41.18$ sq.unit [Ans]