Question

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?

Answer 1

`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`