# 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 5, what is the area of the triangle?

Jun 14, 2017

The area is $= 2.64 {u}^{2}$

#### Explanation:

The angle between $A$ and $B$ is

$\theta = \pi - \left(\frac{1}{12} \pi + \frac{1}{4} \pi\right) = \pi - \frac{4}{12} \pi = \frac{2}{3} \pi$

$\sin \theta = \sin \left(\frac{2}{3} \pi\right) = \frac{\sqrt{3}}{2}$

We apply the sine rule to find $= A$

$\frac{A}{\sin} \left(\frac{1}{12} \pi\right) = \frac{B}{\sin} \left(\frac{2}{3} \pi\right)$

$A = B \sin \frac{\frac{1}{12} \pi}{\sin} \left(\frac{2}{3} \pi\right)$

The area of the triangle is

$a = \frac{1}{2} A B \sin \left(\frac{1}{4} \pi\right)$

$a = \frac{1}{2} B \sin \frac{\frac{1}{12} \pi}{\sin} \left(\frac{2}{3} \pi\right) \cdot B \cdot \sin \left(\frac{1}{4} \pi\right)$

$= \frac{1}{2} \cdot 5 \cdot 5 \cdot \sin \left(\frac{1}{12} \pi\right) \cdot \sin \frac{\frac{\pi}{4}}{\sin} \left(\frac{2}{3} \pi\right)$

$= 2.64$

Jun 14, 2017

$2.64$ $u n i {t}^{2}$

#### Explanation:

Let say the angle, $a = \frac{\pi}{12} \mathmr{and} c = \frac{\pi}{4}$. Then angle between C and A, $b$ $= \pi - \frac{1}{4} \pi - \frac{1}{12} \pi = \frac{8}{12} \pi = \frac{2}{3} \pi$.

Area of triangle, $= \frac{1}{2} B C \sin a$
$= \frac{1}{2} \left(5\right) \left(C\right) \sin \left(\frac{\pi}{12}\right)$.$\to i$

use, $\frac{B}{\sin} b = \frac{C}{\sin} c$ to find C

$\frac{5}{\sin \left(\frac{2}{3} \pi\right)} = \frac{C}{\sin \left(\frac{1}{4} \pi\right)}$

$C = \frac{5}{\sin \left(\frac{2}{3} \pi\right)} \cdot \sin \left(\frac{1}{4} \pi\right) = 4.08$unit

Area of triangle, plug in $C$ in $\to i$
$= \frac{1}{2} \left(5\right) \left(4.08\right) \sin \left(\frac{\pi}{12}\right) = 2.64$ $u n i {t}^{2}$