A triangle has sides A, B, and C. The angle between sides A and B is $(5pi)/12$ and the angle between sides B and C is $pi/12$ . If side B has a length of 7, what is the area of the triangle?

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Answer 1 · 2018-01-16T11:19:20

Area of the triangle is $** 6.12$ sq.unit.**

Angle between Sides $A and B$ is $/_c= (5pi)/12=75^0$

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

Angle between Sides $C and A$ is $/_b= 180-(75+15)=90^0$

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:

$A/sina = B/sinb=C/sinc ; B=7 :. A/sina=B/sinb$ or

$A/sin15=7/sin90 :. A = 7* sin15/sin90 ~~ 1.81(2dp)$ unit.

Now we know sides $A=1.81 , B=7$ and their included angle

$/_c = 75^0$ . Area of the triangle is $A_t=(A*B*sinc)/2$

$:.A_t=(1.81*7*sin75)/2 ~~ 6.12$ sq.unit

Area of the triangle is $6.12$ sq.unit [Ans]

A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/12(5pi)/12 and the angle between sides B and C is pi/12pi/12. If side B has a length of 7, what is the area of the triangle?