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?

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Answer 1 · 2017-12-05T14:50:24

Largest possible perimeter of the triangle is **50.4015#

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#