# A triangle has sides A, B, and C. The angle between sides A and B is (3pi)/4 and the angle between sides B and C is pi/12. If side B has a length of 8, what is the area of the triangle?

May 4, 2018

Area of the triangle is $11.71$ sq.unit.

#### Explanation:

Angle between Sides $A \mathmr{and} B$ is $\angle c = \frac{3 \pi}{4} = \frac{3 \cdot 180}{4} = {135}^{0}$

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 $\angle b = 180 - \left(135 + 15\right) = {30}^{0}$

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/sin a = B/sin b=C/sin c ; B=8 :. B/sin b=C/sin c or

$\frac{8}{\sin} 30 = \frac{C}{\sin} 135 \mathmr{and} C = 8 \cdot \left(\sin \frac{135}{\sin} 30\right) \approx 11.31 \left(2 \mathrm{dp}\right)$

Now we know, sides $B = 8 , C \approx 11.31$ and their included angle

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

$\therefore {A}_{t} = \frac{8 \cdot 11.31 \cdot \sin 15}{2} \approx 11.71$ sq.unit

Area of the triangle is $11.71$ sq.unit [Ans]