$sin 24^{0} + sin 32^{0} + sin 204^{0} + sin 212^{0} =$

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Solution

The correct option is C $0$
Given,

$sin 24 + sin 32 + sin 204 + sin 212$

$= sin 24 + sin 32 + sin (180 + 24) + sin (180 + 32)$

$= sin 24 + sin 32 - sin 24 - sin 32$

$= 0$

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