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

Question and options

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

Correct answer: A - $\frac{-1}{2}$

AP angles: B=60, A+C=120. 2A+B = 180+A-C, so sin(2A+B)=sin(C-A)=1/2. B+2C = 180+C-A, so sin(B+2C) = -sin(C-A) = -1/2.