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

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Answer 1 · 2017-12-11T05:29:02

Largest possible area of the triangle is 144.1742

Given are the two angles #(7pi)/12# and #pi/8# and the length 1

The remaining angle:

#= pi - ((7pi)/12) + pi/8) = (7pi)/24#

I am assuming that length AB (1) is opposite the smallest angle.

Using the ASA

Area#=(c^2*sin(A)*sin(B))/(2*sin(C)#

Area#=( 12^2*sin((7pi)/24)*sin((7pi)/12))/(2*sin(pi/8))#

Area#=144.1742#