Two corners of a triangle have angles of # (7 pi )/ 12 # and # (3 pi ) / 8 #. If one side of the triangle has a length of # 15 #, what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-12-05T13:54:14

Largest possible perimeter 232.1754

Given two angles are #(7pi)/12, (3pi)/8#
Third angle #= (pi - ((7pi)/12 - (3pi)/8) = pi/24#

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

To get the longest perimeter, length 15 must be opposite to angle #pi/24#

#:. 15/ sin(pi/24) = b/ sin((7pi)/12) = c / sin ((3pi)/8)#

#b = (15 sin((7pi)/12))/sin (pi/24) = 111.0037#

#c =( 15 sin((3pi)/8))/ sin (pi/24) = 106.1717#

Hence perimeter #= a + b + c = 5 + 111.0037 + 106.1717 = 232.1754#