A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/3π3, the angle between sides B and C is (5pi)/125π12, 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)2.36(2dp) sq units

Explanation:

The angle between sides A and B is /_C=pi/3 = 180/3 = 60^0C=π3=1803=600
The angle between sides B and C is /_A= (5pi)/12 = (5*180)/12= 75^0A=5π12=518012=750
The angle between sides C and A is /_B=180-(60+75) = 45^0B=180(60+75)=450
B=2B=2 (given) Applying sine law we get A/sinA=B/sinB or A=2*sin75/sin45=2.73AsinA=BsinBorA=2sin75sin45=2.73 Now A=2.73 ; B=2A=2.73;B=2 and their included angle /_C=60^0C=600 So area of the triangle is a*b*sinC/2= 2.73*2*sin60/2=2.36(2dp)absinC2=2.732sin602=2.36(2dp)sq units [Ans}