A triangle has sides A, B, and C. The angle between sides A and B is $pi/6$ and the angle between sides B and C is $pi/12$ . If side B has a length of 15, what is the area of the triangle?

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Answer 1 · 2017-11-30T11:34:34

Area of the triangle is $41.18$ 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:

$A/sina = B/sinb=C/sinc ; B=15 :. A/sina=B/sinb$ or

$A/sin15=15/sin135 :. A = 15* sin15/sin135 ~~ 5.49(2dp)$ unit

Now we know sides $A=5.49 , B=15$ and their included angle

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

$:.A_t=(5.49*15*sin30)/2 ~~ 41.18 (2dp)$ sq.unit

Area of the triangle is $41.18$ sq.unit [Ans]

A triangle has sides A, B, and C. The angle between sides A and B is pi/6pi/6 and the angle between sides B and C is pi/12pi/12. If side B has a length of 15, what is the area of the triangle?