A triangle has two corners with angles of # (3 pi ) / 4 # and # ( pi )/ 12 #. If one side of the triangle has a length of #11 #, what is the largest possible area of the triangle?

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Answer 1 · 2017-07-21T10:20:27

The area is #=82.6u^2#

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The angles ot the triangle are

#hatA=3/4pi#

#hatB=1/12pi#

#hatC=pi-(3/4pi+1/12pi)=pi-10/12pi=1/6pi#

The side of length #11# is opposite the smallest angle in the triangle

The smallest angle is #=hatB#

So,

#b=11#

We apply the sine rule to the triangle

#a/sin hatA=b/sin hatB#

#a/sin(3/4pi)=11/sin(1/12pi)#

#a=11*sin(3/4pi)/sin(1/12pi)=30.05#

The area of the triangle is

#area =1/2ab sin hatC=1/2*30.05*11*sin(1/6pi)#

#=82.6u^2#