Question

Two corners of a triangle have angles of `(7 pi )/ 12` and `pi / 8`. If one side of the triangle has a length of `4`, what is the longest possible perimeter of the triangle?

Answer 1

`4(1+sin({7π}/12)/sin(π/8)+sin({7π}/24)/sin(π/8))`

Explanation:

The three angles are `{7pi}/12`, `pi/8` and `pi - {7pi}/12-pi/8={7pi}/24`. The sine law for triangles tells us that the sides must be in the ratio of the sines of these angles.
For the perimeter of the triangle to be the largest possible, the given side must be the smallest of the sides - i.e. the side opposite the smallest angle. The length of the other two sides must then be
`4 xx sin({7pi}/12)/sin(pi/8) and 4 xx sin({7pi}/24)/sin(pi/8)` respectively. The perimeter is thus
`4+4 xx sin({7pi}/12)/sin(pi/8) + 4 xx sin({7pi}/24)/sin(pi/8)`