A triangle has sides A,B, and C. If the angle between sides A and B is $(5pi)/12$ , the angle between sides B and C is $pi/4$ , and the length of B is 14, what is the area of the triangle?

Question

1601 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-18T12:34:38

Area of the triangle is $77.29$ sq.unit.

Angle between Sides $A and B$ is

$/_c= (5 pi)/12=(5*180)/12=75^0$

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

Angle between Sides $C and A$ is $/_b= 180-(75+45)=60^0$

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:

$A/sinA = B/sinb=C/sinc ; B=14 :. B/sinb=C/sinc$ or

$14/sin 60=C/sin75 or C= 14* (sin 75/sin 60) ~~ 15.61 (2dp)$

Now we know sides $B=14 , C ~~15.61$ and their included angle

$/_a = 45^0$ . Area of the triangle is $A_t=(B*C*sin a)/2$

$:.A_t=(14*15.61*sin 45)/2 ~~ 77.29 (2 dp)$ sq.unit

Area of the triangle is $77.29$ sq.unit [Ans]

A triangle has sides A,B, and C. If the angle between sides A and B is (5pi)/12(5pi)/12, the angle between sides B and C is pi/4pi/4, and the length of B is 14, what is the area of the triangle?