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

$area=1/2ab sin hatC$

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

$=0.63u^2$

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