# A triangle has sides A,B, and C. If the angle between sides A and B is (pi)/12, the angle between sides B and C is pi/4, and the length of B is 16, what is the area of the triangle?

Apr 1, 2018

Area of the triangle is $27.05$ sq.unit.

#### Explanation:

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 $\angle a = \frac{\pi}{4} = \frac{180}{4} = {45}^{0} \therefore$

Angle between Sides $C \mathmr{and} A$ is $\angle b = 180 - \left(45 + 15\right) = {120}^{0}$

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=16 :. A/sina=B/sinb or

$\frac{A}{\sin} 45 = \frac{16}{\sin} 120 \therefore A = 16 \cdot \sin \frac{45}{\sin} 120 \approx 13.06 \left(2 \mathrm{dp}\right)$unit

Now we know sides $A \approx 13.06 , B = 16$ and their included angle

$\angle c = {15}^{0}$. Area of the triangle is ${A}_{t} = \frac{A \cdot B \cdot \sin c}{2}$

$\therefore {A}_{t} = \frac{13.06 \cdot 16 \cdot \sin 15}{2} \approx 27.05$ sq.unit

Area of the triangle is $27.05$ sq.unit [Ans]