# A triangle has sides A, B, and C. The angle between sides A and B is pi/4 and the angle between sides B and C is pi/12. If side B has a length of 13, what is the area of the triangle?

Nov 12, 2017

The area of the triangle is $17.86$ sq.unit.

#### Explanation:

Angle between Sides $A \mathmr{and} B$ is $\angle c = \frac{\pi}{4} = \frac{180}{4} = {45}^{0}$

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 $\angle b = 180 - \left(45 + 15\right) = {120}^{0}$

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=13 :. A/sina=B/sinb or

$\frac{A}{\sin} 15 = \frac{13}{\sin} 120 \mathmr{and} A = 13 \cdot \sin \frac{15}{\sin} 120 \approx 3.89 \left(2 \mathrm{dp}\right)$ unit.

Now we know $A = 3.89 , B = 13$ and their included angle

$\angle c = {45}^{0}$. The area of triangle is ${A}_{t} = \frac{A \cdot B \cdot \sin c}{2}$

$\therefore {A}_{t} = \frac{3.89 \cdot 13 \cdot \sin 45}{2} \approx 17.86 \left(2 \mathrm{dp}\right)$ sq.unit.

The area of the triangle is $17.86 \left(2 \mathrm{dp}\right)$ sq.unit. [Ans]