If the angles A, B,C of a triangle are in arithmetic progression such that $$\sin(2A + B) = 1/2$$ then $$\sin(B + 2C)$$ is equal to

If the angles A, B,C of a triangle are in arithmetic progression such

that $$\sin(2A + B) = 1/2$$ then $$\sin(B + 2C)$$ is equal to

The sum of all angles in a triangle = 180$$^{\circ\ }$$

Angles A, B and C are in AP.

Thus, 2B = A + C

B + 2B = 180

B = 60

Also we are given that $$\sin(2A + B) = 1/2$$

This is only possible if 2A+B = 30 degree or 150 degree

2A+B cannot be equal to 30 as B = 60 and A cannot be negative.

Therefore, 2A+B = 150

Since, B = 60

A = 45 and C = 75

Therefore, $$\sin(B + 2C)$$ =

B + 2C = 60+150 = 210

$$\sin(210^{\circ\ })$$ = -1/2