A) \[60{}^\circ \]
B) \[90{}^\circ \]
C) \[30{}^\circ \]
D) \[45{}^\circ \]
Correct Answer: B
Solution :
Given, \[\sin \,\alpha =\frac{1}{2}\] and \[\cos \beta =\frac{1}{2}\] \[\Rightarrow \,\,\sin \alpha =\sin 30{}^\circ \] \[\left[ \because \,\sin 30{}^\circ =\frac{1}{2} \right]\] And \[\cos \beta =\cos 60{}^\circ \]\[\left[ \because \,\cos 60{}^\circ =\frac{1}{2} \right]\] \[\Rightarrow \,\,\,\,\alpha =30{}^\circ \] and \[\beta =60{}^\circ \] \[\therefore \,\,\alpha +\beta =30{}^\circ +60{}^\circ =90{}^\circ \]
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