Question

Two corners of a triangle have angles of `(5 pi )/ 12` and `( pi ) / 12`. If one side of the triangle has a length of `9`, what is the longest possible perimeter of the triangle?

Answer 1

`P=9(3+sqrt3+sqrt6+sqrt2)approx77.36`.

Explanation:

In `triangleABC`, let `A=(5pi)/12,B=pi/12`. Then

`C=pi-A-B`
`C=(12pi)/12-(5pi)/12-pi/12`
`C=(6pi)/12=pi/2`.

In all triangles, the shortest side is always opposite the shortest angle. Maximizing the perimeter means putting the largest value we know (9) in the smallest position possible (opposite `angleB`). Meaning for the perimeter of `triangleABC` to be maximized, `b=9`.

Using the law of sines, we have

`sinA/a=sinB/b=sinC/c`

Solving for `a`, we get:

`a=(bsinA)/sinB=(9sin((5pi)/12))/sin(pi/12)=(9(sqrt6+sqrt2)//4)/((sqrt6-sqrt2)//4)=...=9(2+sqrt3)`

Similarly, solving for `c` yields

`c=(bsinC)/sinB=(9sin(pi/2))/(sin(pi/12))=(9(1))/((sqrt6-sqrt2)//4)=...=9(sqrt6+sqrt2)`

The perimeter `P` of `triangleABC` is the sum of all three sides:

`P=color(orange)a+color(blue)b+color(green)c`
`P=color(orange)(9(2+sqrt3))+color(blue)9+color(green)(9(sqrt6+sqrt2))`
`P=9(2+sqrt3+1+sqrt6+sqrt2)`
`P=9(3+sqrt3+sqrt6+sqrt2)approx77.36`