A triangle has sides A, B, and C. If the angle between sides A and B is #(pi)/3#, the angle between sides B and C is #(5pi)/12#, and the length of B is 2, what is the area of the triangle?

1 Answer
Jun 10, 2016

Area of the triangle is #2.36(2dp)# sq units

Explanation:

The angle between sides A and B is #/_C=pi/3 = 180/3 = 60^0#
The angle between sides B and C is #/_A= (5pi)/12 = (5*180)/12= 75^0#
The angle between sides C and A is #/_B=180-(60+75) = 45^0#
#B=2# (given) Applying sine law we get #A/sinA=B/sinB or A=2*sin75/sin45=2.73# Now #A=2.73 ; B=2# and their included angle #/_C=60^0# So area of the triangle is #a*b*sinC/2= 2.73*2*sin60/2=2.36(2dp)#sq units [Ans}