1. **Problem statement:** Given that $\sin \alpha \neq \frac{1}{2}$ and $\cos \beta = \frac{1}{2}$, find the value of $\alpha + \beta$. 2. **Recall the values of cosine:** $\cos \beta = \frac{1}{2}$ implies $\beta = 60^\circ$ or $\beta = 300^\circ$ (since cosine is positive in the first and fourth quadrants). 3. **Check the sine condition:** Since $\sin \alpha \neq \frac{1}{2}$, $\alpha$ cannot be $30^\circ$ or $150^\circ$ (the angles where sine equals $\frac{1}{2}$). 4. **Calculate $\alpha + \beta$ for possible $\beta$ values:** - If $\beta = 60^\circ$, then $\alpha + \beta = \alpha + 60^\circ$. - If $\beta = 300^\circ$, then $\alpha + \beta = \alpha + 300^\circ$. 5. **From the options given:** - a) $0^\circ$ - b) $90^\circ$ - c) $30^\circ$ - d) $60^\circ$ 6. Since $\sin \alpha \neq \frac{1}{2}$, $\alpha$ is not $30^\circ$ or $150^\circ$. The only way to get one of the options for $\alpha + \beta$ is if $\alpha + \beta = 90^\circ$. 7. Therefore, if $\beta = 60^\circ$, then $\alpha = 30^\circ$ (which is not allowed), so discard. 8. If $\beta = 300^\circ$, then $\alpha + 300^\circ = 90^\circ$ implies $\alpha = -210^\circ$ (which is equivalent to $150^\circ$ modulo $360^\circ$), but $\sin 150^\circ = \frac{1}{2}$, which is not allowed. 9. The only consistent value from the options is $\alpha + \beta = 60^\circ$. **Final answer:** $\boxed{60^\circ}$