# A triangle has sides A,B, and C. If the angle between sides A and B is (5pi)/8, the angle between sides B and C is pi/4, and the length of B is 19, what is the area of the triangle?

Jan 14, 2016

Area $= \frac{1}{2} \cdot 19 \cdot 32.44 \approx 308.13$

#### Explanation:

The area of a triangle $= \frac{1}{b} \cdot h$ where $b =$base and $h =$height

In this case $\tan \left(\frac{5 \pi}{8}\right) = \frac{h}{x}$ and $\tan \left(\frac{\pi}{4}\right) = \frac{h}{y}$ where

$x + y = B = 19$
$y = 19 - x$

So $x \tan \left(\frac{5 \pi}{8}\right) = y \tan \left(\frac{\pi}{4}\right)$
$x \tan \left(\frac{5 \pi}{8}\right) = \left(19 - x\right) \tan \left(\frac{\pi}{4}\right)$
$x \left(\tan \left(\frac{5 \pi}{8}\right) + \tan \left(\frac{\pi}{4}\right)\right) = 19 \tan \left(\frac{\pi}{4}\right)$

$\therefore x = 19 \tan \frac{\frac{\pi}{4}}{\tan \left(\frac{5 \pi}{8}\right) + \tan \left(\frac{\pi}{4}\right)}$

$h = \left(19 \tan \frac{\frac{\pi}{4}}{\tan \left(\frac{5 \pi}{8}\right) + \tan \left(\frac{\pi}{4}\right)}\right) \tan \left(\frac{5 \pi}{8}\right)$
$\approx \frac{19 \cdot 1 \cdot 2.414}{2.414 + 1}$
$\approx 32.44$

Area $= \frac{1}{2} \cdot 19 \cdot 32.44 \approx 308.13$