# A triangle has sides A,B, and C. If the angle between sides A and B is (5pi)/12, the angle between sides B and C is pi/4, and the length of B is 14, what is the area of the triangle?

May 18, 2018

Area of the triangle is $77.29$ sq.unit.

#### Explanation:

Angle between Sides $A \mathmr{and} B$ is

$\angle c = \frac{5 \pi}{12} = \frac{5 \cdot 180}{12} = {75}^{0}$

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 $\angle b = 180 - \left(75 + 45\right) = {60}^{0}$

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=14 :. B/sinb=C/sinc or

$\frac{14}{\sin} 60 = \frac{C}{\sin} 75 \mathmr{and} C = 14 \cdot \left(\sin \frac{75}{\sin} 60\right) \approx 15.61 \left(2 \mathrm{dp}\right)$

Now we know sides $B = 14 , C \approx 15.61$ and their included angle

$\angle a = {45}^{0}$. Area of the triangle is ${A}_{t} = \frac{B \cdot C \cdot \sin a}{2}$

$\therefore {A}_{t} = \frac{14 \cdot 15.61 \cdot \sin 45}{2} \approx 77.29 \left(2 \mathrm{dp}\right)$ sq.unit

Area of the triangle is $77.29$ sq.unit [Ans]