(B) Given,$\sin \alpha = \frac{1}{2} = \sin 30^{\circ}$ $\left[\because \sin 30^{\circ} = \frac{1}{2}\right]$$\Rightarrow \alpha = 30^{\circ}$And,$\co

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(B) Given,$\sin \alpha = \frac{1}{2} = \sin 30^{\circ}$ $\left[\because \sin 30^{\circ} = \frac{1}{2}\right]$$\Rightarrow \alpha = 30^{\circ}$And,$\cos \beta = \frac{1}{2} = \cos 60^{\circ}$ $\left[\because \cos 60^{\circ} = \frac{1}{2}\right]$$\Rightarrow \beta = 60^{\circ}$Therefore,$\alpha + \beta = 30^{\circ} + 60^{\circ} = 90^{\circ}$