# Two corners of a triangle have angles of  (3 pi )/ 8  and  ( pi ) / 2 . If one side of the triangle has a length of  4 , what is the longest possible perimeter of the triangle?

$8 + 4 \setminus \sqrt{2} + 4 \setminus \sqrt{4 + 2 \setminus \sqrt{2}}$

#### Explanation:

Let in $\setminus \Delta A B C$, $\setminus \angle A = \frac{3 \setminus \pi}{8}$, $\setminus \angle B = \setminus \frac{\pi}{2}$ hence

$\setminus \angle C = \setminus \pi - \setminus \angle A - \setminus \angle B$

$= \setminus \pi - \frac{3 \setminus \pi}{8} - \setminus \frac{\pi}{2}$

$= \frac{\setminus \pi}{8}$

For maximum perimeter of triangle , we must consider the given side of length $4$ is smallest i.e. side $c = 4$ is opposite to the smallest angle $\setminus \angle C = \setminus \frac{\pi}{8}$

Now, using Sine rule in $\setminus \Delta A B C$ as follows

$\setminus \frac{a}{\setminus \sin A} = \setminus \frac{b}{\setminus \sin B} = \setminus \frac{c}{\setminus \sin C}$

$\setminus \frac{a}{\setminus \sin \left(\frac{3 \setminus \pi}{8}\right)} = \setminus \frac{b}{\setminus \sin \left(\setminus \frac{\pi}{2}\right)} = \setminus \frac{4}{\setminus \sin \left(\frac{\setminus \pi}{8}\right)}$

$a = \setminus \frac{4 \setminus \sin \left(\frac{3 \setminus \pi}{8}\right)}{\setminus \sin \left(\setminus \frac{\pi}{8}\right)}$

$a = 4 \left(\setminus \sqrt{2} + 1\right)$ &

$b = \setminus \frac{4 \setminus \sin \left(\frac{\setminus \pi}{2}\right)}{\setminus \sin \left(\setminus \frac{\pi}{8}\right)}$

$b = 4 \setminus \sqrt{4 + 2 \setminus \sqrt{2}}$

hence, the maximum possible perimeter of the $\setminus \triangle A B C$ is given as

$a + b + c$

$= 4 \left(\setminus \sqrt{2} + 1\right) + 4 \setminus \sqrt{4 + 2 \setminus \sqrt{2}} + 4$

$= 8 + 4 \setminus \sqrt{2} + 4 \setminus \sqrt{4 + 2 \setminus \sqrt{2}}$