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 `2`, what is the longest possible perimeter of the triangle?

Answer 1

Longest possible perimeter = 17.1915

Explanation:

Sum of the angles of a triangle `=pi`

Two angles are `(5pi)/12, pi/12`
Hence `3^(rd)`angle is `pi - ((5pi)/12 + pi/12) = (pi)/2`

We know`a/sin a = b/sin b = c/sin c`

To get the longest perimeter, length 2 must be opposite to angle `pi/24`

`:. 2/ sin(pi/12) = b/ sin((5pi)/12) = c / sin ((pi)/2)`

`b = (2 sin((5pi)/12))/sin (pi/12) = 7.4641`

`c =( 2* sin((pi)/2))/ sin (pi/12) = 7.7274`

Hence perimeter `= a + b + c = 2 + 7.4641 + 7.7274 = 17.1915`