# A triangle has two corners with angles of  pi / 2  and  ( pi )/ 8 . If one side of the triangle has a length of 13 , what is the largest possible area of the triangle?

Apr 16, 2018

The area is $\frac{169}{2 \left(\sqrt{2} - 1\right)} \approx 204$

#### Explanation:

If one of the angles of the triangle is $\frac{\pi}{2}$, it is a right triangle. The measure of the third angle of the triangle must be $\frac{\pi}{2} - \frac{\pi}{8} = \frac{3 \pi}{8}$. The shortest side of this right triangle will be the leg that is opposite the angle measuring $\frac{\pi}{8}$. Let $x$= the length of the longer leg. The area, $A$, of the triangle will be

$A = \frac{13}{2} x$

From trigonmetry

$x = \frac{13}{\tan} \left(\frac{\pi}{8}\right)$

So

$A = {13}^{2} / \left(2 \tan \left(\frac{\pi}{8}\right)\right) = \frac{169}{2 \left(\sqrt{2} - 1\right)} \approx 204$