# A triangle has sides A, B, and C. If the angle between sides A and B is (7pi)/12, the angle between sides B and C is pi/3, and the length of B is 1, what is the area of the triangle?

Jan 23, 2018

Area of the triangle is $1.62$ sq.unit.

#### Explanation:

Angle between Sides $A \mathmr{and} B$ is $\angle c = \frac{7 \pi}{12} = \frac{7 \cdot 180}{12} = {105}^{0}$

Angle between Sides $B \mathmr{and} C$ is $\angle a = \frac{\pi}{3} = \frac{180}{3} = {60}^{0} \therefore$

Angle between Sides $C \mathmr{and} A$ is $\angle b = 180 - \left(105 + 60\right) = {15}^{0}$

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=1 :. B/sinb=C/sinc or

$\frac{1}{\sin} 15 = \frac{C}{\sin} 105 \mathmr{and} C = 1 \cdot \left(\sin \frac{105}{\sin} 15\right) \approx 3.73 \left(2 \mathrm{dp}\right)$

Now we know sides $B = 1 , C = 3.73$ and their included angle

$\angle a = {60}^{0}$. Area of the triangle is ${A}_{t} = \frac{B \cdot C \cdot \sin a}{2}$

$\therefore {A}_{t} = \frac{1 \cdot 3.73 \cdot \sin 60}{2} \approx 1.62 \left(2 \mathrm{dp}\right)$ sq.unit.

Area of the triangle is $1.62$ sq.unit [Ans]