Question

Two corners of a triangle have angles of `(7 pi ) / 12` 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

Sum of the angles of a triangle `=pi`

Two angles are `(7pi)/12, pi/12`
Hence `3^(rd)`angle is `pi - ((7pi)/12 + pi/12) = (pi)/3`

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

To get the longest perimeter, length 2 must be opposite to angle `pi/12`

#:. 6/ sin(pi/12) = b/ sin((7pi)/12) = c / sin ((pi)/3)

`b = (6sin ((7pi)/12))/sin (pi/12) = 22.3923`

`c =( 6* sin(pi/3))/ sin (pi/12) = 20.0764`

Hence perimeter `= a + b + c = 6 + 22.3923 + 20.0764 = 48.4687`