A triangle has sides A,B, and C. If the angle between sides A and B is $pi/6$ , the angle between sides B and C is $pi/12$ , and the length of B is 8, what is the area of the triangle?

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Answer 1 · 2017-11-14T11:23:58

Area of the triangle is $5.86$ sq.unit.

Angle between Sides $A and B$ is $/_c= pi/6=180/6=30^0$

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

Angle between Sides $C and A$ is $/_b= 180-(30+15)=135^0$

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:

$A/sina = B/sinb=C/sinc ; B=8 :. A/sina=B/sinb$ or

$A/sin15=8/sin135 :. A = 8* sin15/sin135 ~~ 2.93(2dp)$ unit

Now we know sides $A=2.93 , B=8$ and their included angle

$/_c = 30^0$ . Area of the triangle is $A_t=(A*B*sinc)/2$

$:.A_t=(2.93*8*sin30)/2 ~~ 5.86$ sq.unit

Area of the triangle is $5.86$ sq.unit [Ans]

A triangle has sides A,B, and C. If the angle between sides A and B is pi/6pi/6, the angle between sides B and C is pi/12pi/12, and the length of B is 8, what is the area of the triangle?