Two corners of a triangle have angles of # (7 pi )/ 12 # and # pi / 4 #. 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 · 2017-07-26T07:53:00

The perimeter is #=65.2u#

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The angles are

#hatA=7/12pi#

#hatB=1/4pi#

#hatC=pi-(7/12pi+1/4pi)=pi-(7/12pi+3/12pi)=2/12pi=1/6pi#

To have the longest perimeter, the side #=15# must be opposite the smallest angle

Therefore,

#c=15#

Applying the sine rule to the triangle

#a/sin hatA=b/sin hatB=c/sin hatC#

#a/sin(7/12pi)=b/sin(1/4pi)=15/sin(1/6pi)=30#

So,

#a=30sin(7/12pi)=29#

#b=30sin(1/4pi)=21.2#

The perimeter is

#P=15+29+21.2=65.2u#