# 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 3, what is the area of the triangle?

Dec 27, 2017

Area of the triangle is $1.65$ sq.unit.

#### Explanation:

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

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/sina = B/sinb=C/sinc ; B=3 :. A/sina=B/sinb or

$\frac{A}{\sin} 15 = \frac{3}{\sin} 30 \mathmr{and} A = 3 \cdot \left(\sin \frac{15}{\sin} 30\right) \approx 1.55 \left(2 \mathrm{dp}\right)$

Now we know sides $A \approx 1.55 , B = 3$ and their included angle

$\angle c = {135}^{0}$. Area of the triangle is ${A}_{t} = \frac{A \cdot B \cdot \sin c}{2}$

$\therefore {A}_{t} = \frac{1.55 \cdot 3 \cdot \sin 135}{2} \approx 1.65$ sq.unit

Area of the triangle is $1.65$ sq.unit [Ans]