# A triangle has sides A, B, and C. The angle between sides A and B is (7pi)/12 and the angle between sides B and C is pi/12. If side B has a length of 2, what is the area of the triangle?

Dec 17, 2017

Area of the triangle is $0.58$ sq.unit.

#### Explanation:

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

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 $\angle b = 180 - \left(105 + 15\right) = {60}^{0}$

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=2 :. A/sina=B/sinb or

$\frac{A}{\sin} 15 = \frac{2}{\sin} 60 \therefore A = 2 \cdot \sin \frac{15}{\sin} 60 \approx 0.60 \left(2 \mathrm{dp}\right)$unit

Now we know sides $A = 0.60 , B = 2$ and their included angle

$\angle c = {105}^{0}$. Area of the triangle is ${A}_{t} = \frac{A \cdot B \cdot \sin c}{2}$

$\therefore {A}_{t} = \frac{0.6 \cdot 2 \cdot \sin 105}{2} \approx 0.58$ sq.unit [Ans]