A triangle has two corners with angles of # (3 pi ) / 4 # and # ( pi )/ 6 #. If one side of the triangle has a length of #1 #, what is the largest possible area of the triangle?

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Answer 1 · 2017-07-17T14:27:17

The area is #=0.68u^2#

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The angles are

#hatA=1/6pi#

#hatB=3/4pi#

#hatC=pi-(3/4pi+pi/6)=pi-11/12pi=1/12pi#

To have the greatest area, the length #=1# is opposite the angle #hatC#

So, #c=1#

We apply the sine rule to the triangle

#a/sin hatA=c/sin hatC#

#a/sin(1/6pi)=1/sin(1/12pi)#

#a=sin(1/6pi)/sin(1/12pi)=1.93#

The area of the triangle is

#=1/2*a*c*sin hatB#

#=1/2*1*1.93*sin(3/4pi)#

#=0.68u^2#