A triangle has sides A,B, and C. If the angle between sides A and B is $(pi)/12$ , the angle between sides B and C is $pi/4$ , and the length of B is 16, what is the area of the triangle?

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Answer 1 · 2018-04-01T10:58:02

Area of the triangle is $27.05$ sq.unit.

Angle between Sides $A and B$ is $/_c= pi/12=180/12=15^0$

Angle between Sides $B and C$ is $/_a= pi/4=180/4=45^0 :.$

Angle between Sides $C and A$ is $/_b= 180-(45+15)=120^0$

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:

$A/sina = B/sinb=C/sinc ; B=16 :. A/sina=B/sinb$ or

$A/sin45=16/sin120 :. A = 16* sin45/sin120 ~~ 13.06(2dp)$ unit

Now we know sides $A~~13.06 , B=16$ and their included angle

$/_c = 15^0$ . Area of the triangle is $A_t=(A*B*sinc)/2$

$:.A_t=(13.06*16*sin15)/2 ~~ 27.05$ sq.unit

Area of the triangle is $27.05$ sq.unit [Ans]

A triangle has sides A,B, and C. If the angle between sides A and B is (pi)/12(pi)/12, the angle between sides B and C is pi/4pi/4, and the length of B is 16, what is the area of the triangle?