A triangle has sides A, B, and C. The angle between sides A and B is $(3pi)/4$ and the angle between sides B and C is $pi/12$ . If side B has a length of 8, what is the area of the triangle?

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Answer 1 · 2018-05-04T12:23:11

Area of the triangle is $11.71$ sq.unit.

Angle between Sides $A and B$ is $/_c= (3pi)/4=(3*180)/4=135^0$

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

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

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:

$A/sin a = B/sin b=C/sin c ; B=8 :. B/sin b=C/sin c$ or

$8/sin 30=C /sin 135 or C= 8* (sin 135/sin 30) ~~ 11.31 (2 dp)$

Now we know, sides $B=8,C~~ 11.31$ and their included angle

$/_a = 15^0$ . Area of the triangle is $A_t=(B*C*sin a)/2$

$:.A_t=(8*11.31*sin 15)/2 ~~ 11.71$ sq.unit

Area of the triangle is $11.71$ sq.unit [Ans]

A triangle has sides A, B, and C. The angle between sides A and B is (3pi)/4(3pi)/4 and the angle between sides B and C is pi/12pi/12. If side B has a length of 8, what is the area of the triangle?