Two corners of a triangle have angles of (5 pi )/ 8 and ( pi ) / 6 . If one side of the triangle has a length of 5 , what is the longest possible perimeter of the triangle?

1 Answer

20.3264\ \text{unit

Explanation:

Let in \Delta ABC, \angle A={5\pi}/8, \angle B=\pi/6 hence

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

=\pi-{5\pi}/8-\pi/6

={5\pi}/24

For maximum perimeter of triangle , we must consider the given side of length 5 is smallest i.e. side b=5 is opposite to the smallest angle \angle B={\pi}/6

Now, using Sine rule in \Delta ABC as follows

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

\frac{a}{\sin ({5\pi}/8)}=\frac{5}{\sin (\pi/6)}=\frac{c}{\sin ({5\pi}/24)}

a=\frac{5\sin ({5\pi}/8)}{\sin (\pi/6)}

a=9.2388 &

c=\frac{5\sin ({5\pi}/24)}{\sin (\pi/6)}

c=6.0876

hence, the maximum possible perimeter of the \triangle ABC is given as

a+b+c

=9.2388+5+6.0876

=20.3264\ \text{unit