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If $\sin (120^{0} - α) = \sin (120^{0} - β)$ and $0 < α; β < π$ then all the values of $α, β$ are

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$α = β$

$α + β = \frac{π}{3}$

$α = β or α + β = \frac{π}{3}$

$α = β = 0$

answer is C.

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Detailed Solution

$\sin (120^{0} - α) = \sin (120^{0} - β) 120^{0} - α = 120^{0} - β (or) 120^{0} - α = 180^{0} - (120^{0} - β) α = β 120 - α = 60^{0} + β \Rightarrow α + β = 60^{0}$

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