# A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/8, the angle between sides B and C is (pi)/2, and the length of B is 3, what is the area of the triangle?

Jun 14, 2018

Area of the triangle is $1.86$ sq.unit.

#### Explanation:

Angle between sides $A \mathmr{and} B$ is $\angle c = \frac{\pi}{8} = \frac{180}{8} = {22.5}^{0}$

Angle between sides $B \mathmr{and} C$ is $\angle a = \frac{\pi}{2} = \frac{180}{2} = {90}^{0} \therefore$

Angle between sides $C \mathmr{and} A$ is

$\angle b = 180 - \left(22.5 + 90\right) = {67.5}^{0}$ 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, $\frac{A}{\sin} a = \frac{B}{\sin} b = \frac{C}{\sin} c$

$B = 3 \therefore \frac{A}{\sin} a = \frac{B}{\sin} b \therefore \frac{A}{\sin} 90 = \frac{3}{\sin} 67.5$

$\therefore A = 3 \cdot \left(\sin \frac{90}{\sin} 67.5\right) \approx 3.25 \left(2 \mathrm{dp}\right)$

Now we know sides $A \approx 3.25 , B = 3$ and their included angle

$\angle c = {22.5}^{0}$. Area of the triangle is ${A}_{t} = \frac{A \cdot B \cdot \sin c}{2}$

$\therefore {A}_{t} = \frac{3.25 \cdot 3 \cdot \sin 22.5}{2} \approx 1.86$ sq.unit

Area of the triangle is $1.86$ sq.unit [Ans]