Question

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?

Answer 1

`8+4\sqrt2+4\sqrt{4+2\sqrt2}`

Explanation:

Let in `\Delta ABC`, `\angle A={3\pi}/8`, `\angle B=\pi/2` hence

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

`=\pi-{3\pi}/8-\pi/2`

`={\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 `\angle C=\pi/8`

Now, using Sine rule in `\Delta ABC` as follows

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

`\frac{a}{\sin ({3\pi}/8)}=\frac{b}{\sin (\pi/2)}=\frac{4}{\sin ({\pi}/8)}`

`a=\frac{4\sin ({3\pi}/8)}{\sin (\pi/8)}`

`a=4(\sqrt2+1)` &

`b=\frac{4\sin ({\pi}/2)}{\sin (\pi/8)}`

`b=4\sqrt{4+2\sqrt2}`

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

`a+b+c`

`=4(\sqrt2+1)+4\sqrt{4+2\sqrt2}+4`

`=8+4\sqrt2+4\sqrt{4+2\sqrt2}`