# A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/4, the angle between sides B and C is (5pi)/12, and the length of B is 4, what is the area of the triangle?

$A r e a = \frac{4}{3} \cdot \left(3 + \sqrt{3}\right)$

$A r e a = 6.3094$ square units

#### Explanation:

Try drawing the triangle
Angle $A = \frac{5 \pi}{12}$ and angle $C = \frac{\pi}{4}$ and side $b = 4$
side $b$ is the base of the triangle. There 's a need to solve for the height $h$ from angle B to side b to compute the area.

$h \cot A + h \cot C = b$
h=b/(cot A+cot C)=4/(cot ((5pi)/12)+cot (pi/4)
From double angle formulas:

$\cot \left(\frac{5 \pi}{12}\right) = \cos \frac{\frac{\pi}{4} + \frac{\pi}{6}}{\sin} \left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \frac{\cos \left(\frac{\pi}{4}\right) \cdot \cos \left(\frac{\pi}{6}\right) - \sin \left(\frac{\pi}{4}\right) \cdot \sin \left(\frac{\pi}{6}\right)}{\sin \left(\frac{\pi}{4}\right) \cdot \cos \left(\frac{\pi}{6}\right) + \cos \left(\frac{\pi}{4}\right) \cdot \sin \left(\frac{\pi}{6}\right)}$

$\cot \left(\frac{5 \pi}{12}\right) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$
Solve $h$ now
$h = \frac{4}{\cot \left(\frac{5 \pi}{12}\right) + \cot \left(\frac{\pi}{4}\right)} = \frac{4}{\frac{\sqrt{3} - 1}{\sqrt{3} + 1} + 1}$
$h = \frac{2}{3} \cdot \left(3 + \sqrt{3}\right)$
Solve $A r e a = \frac{1}{2} \cdot b \cdot h$
$A r e a = \frac{1}{2} \cdot 4 \cdot \frac{2}{3} \cdot \left(3 + \sqrt{3}\right)$

$A r e a = \frac{4}{3} \cdot \left(3 + \sqrt{3}\right)$

$A r e a = 6.3094$ square units

Have a nice day !!! from the Philippines .