Question

Two corners of a triangle have angles of `(3 pi ) / 8` and `pi / 12`. If one side of the triangle has a length of `6`, what is the longest possible perimeter of the triangle?

Answer 1

Largest possible perimeter of the triangle is **50.4015#

Explanation:

Sum of the angles of a triangle `=pi`

Two angles are `(3pi)/8, pi/12`
Hence `3^(rd)`angle is `pi - ((3pi)/8 + pi/12) = (13pi)/24`

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`

`:. 6/ sin(pi/12) = b/ sin((3pi)/8) = c / sin ((13pi)/24)`

`b = (6 sin((3pi)/8))/sin (pi/12) = 21.4176`

`c =( 6* sin((13pi)/24))/ sin (pi/12) = 22.9839`

Hence perimeter `= a + b + c = 6 + 21.4176 + 22.9839 = 50.4015`