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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
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