A triangle has sides A, B, and C. The angle between sides A and B is #(2pi)/3# and the angle between sides B and C is #pi/12#. If side B has a length of 2, what is the area of the triangle?

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Answer 1 · 2017-07-24T14:26:37

The area of the triangle is #=0.63u^2#

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The angles are

#hatC=2/3pi#

#hatA=1/12pi#

Therefore,

#hatB=pi-(2/3pi+1/12pi)=pi-(8/12pi+1/12pi)=3/12pi=1/4pi#

The side #b=2#

We apply the sine rule to the triangle #DeltaABC#

#a/sin hatA=b/sin hatB#

#a/sin(1/12pi)=2/sin(1/4pi)#

#a=2sin(1/12pi)/sin(1/4pi)=0.73#

The area of the triangle is

#area=1/2ab sin hatC#

#=1/2*2*0.73*sin(2/3pi)#

#=0.63u^2#