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?

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Answer 1 · 2017-12-08T09:44:56

Longest possible perimeter = 17.1915

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#