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 3, what is the area of the triangle?

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Answer 1 · 2017-12-27T11:22:12

Area of the triangle is $1.65$ sq.unit.

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

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:

$A/sina = B/sinb=C/sinc ; B=3 :. A/sina=B/sinb$ or

$A/sin15=3/sin30 or A= 3* (sin15/sin30) ~~ 1.55 (2dp)$

Now we know sides $A~~1.55 , B=3$ and their included angle

$/_c = 135^0$ . Area of the triangle is $A_t=(A*B*sinc)/2$

$:.A_t=(1.55*3*sin135)/2 ~~ 1.65$ sq.unit

Area of the triangle is $1.65$ 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 3, what is the area of the triangle?