Question

Two corners of a triangle have angles of `pi / 12` and `pi / 3`. If one side of the triangle has a length of `6`, what is the longest possible perimeter of the triangle?

Answer 1

`18+9\sqrt2+6\sqrt3+3\sqrt6`

Explanation:

Let in `\Delta ABC`, `\angle A=\pi/12`, `\angle B=\pi/3` hence

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

`=\pi-\pi/12-\pi/3`

`={7\pi}/12`

For maximum perimeter of triangle , we must consider the given side of length `6` is smallest i.e. side `a=6` is opposite to the smallest angle `\angle A=\pi/12`

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

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

`\frac{6}{\sin (\pi/12)}=\frac{b}{\sin (\pi/3)}=\frac{c}{\sin ({7\pi}/12)}`

`b=\frac{6\sin (\pi/3)}{\sin (\pi/12)}`

`b=9\sqrt2+3\sqrt6` &

`c=\frac{6\sin ({7\pi}/12)}{\sin (\pi/12)}`

`c=12+6\sqrt3`

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

`a+b+c`

`=6+9\sqrt2+3\sqrt6+12+6\sqrt3`

`=18+9\sqrt2+6\sqrt3+3\sqrt6`