A triangle has sides A, B, and C. The angle between sides A and B is $(7pi)/12$ 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 · 2017-12-17T14:36:10

Area of the triangle is $0.58$ sq.unit.

Angle between Sides $A and B$ is $/_c= (7pi)/12=105^0$

Angle between Sides $B and C$ is $/_a= pi/12=180/12=15^0 :.$

Angle between Sides $C and A$ is $/_b= 180-(105+15)=60^0$

The sine rule states if $A, B and C$ are the lengths of the sides

and opposite angles are $a, b and c$ in a triangle, then:

$A/sina = B/sinb=C/sinc ; B=2 :. A/sina=B/sinb$ or

$A/sin15=2/sin60 :. A = 2* sin15/sin60 ~~ 0.60(2dp)$ unit

Now we know sides $A=0.60 , B=2$ and their included angle

$/_c = 105^0$ . Area of the triangle is $A_t=(A*B*sinc)/2$

$:.A_t=(0.6*2*sin105)/2 ~~ 0.58$ sq.unit [Ans]

A triangle has sides A, B, and C. The angle between sides A and B is (7pi)/12(7pi)/12 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?