# 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 4, what is the area of the triangle?

Mar 16, 2017

Area=1.73$u n i t {s}^{2}$

#### Explanation:

First you can figure out the last angle of the triangle which is $\frac{2 \pi}{3}$ (if it helps better the angles are $15 , 45 , \mathmr{and} 120$ degrees.)
then you can use the law of sines to figure out another side.
$\sin \frac{\frac{2 \pi}{3}}{4} = \sin \frac{\frac{\pi}{4}}{x}$
$x = \frac{4 \sin \left(\frac{\pi}{4}\right)}{\sin} \left(\frac{2 \pi}{3}\right) = 3.27$
then do that again to find the last side
$\sin \frac{\frac{2 \pi}{3}}{4} = \sin \frac{\frac{\pi}{12}}{x}$
$x = \frac{4 \sin \left(\frac{\pi}{12}\right)}{\sin} \left(\frac{2 \pi}{3}\right) = 1.20$
then you can use Heron's formula.
First find S
$S = \frac{a + b + c}{2} = 4.24$
then substitute
$A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$
$A = \sqrt{4.24 \left(4.24 - 1.20\right) \left(4.24 - 4\right) \left(4.24 - 3.27\right)} = 1.73$