Question

If $sin\alpha =cos(90°-\beta )=\frac{1}{\sqrt{2}}$ and $0°\le \alpha ,\beta \le 90°$, then find $\alpha +\beta$.

• A. 90°
• B. 60°
• C. 45°
• D. 30°

Answer 1

The correct option is A 90°

Given, $sin\alpha =cos(90°-\beta )=\frac{1}{\sqrt{2}}$ and $0°\le \alpha ,\beta \le 90°$

$\Rightarrow sin\alpha =\frac{1}{\sqrt{2}}\text{and}cos(90°-\beta )=\frac{1}{\sqrt{2}}\Rightarrow \alpha =45°\text{and}90°-\beta =45°\Rightarrow \alpha =\beta =45°\Rightarrow \alpha +\beta =90°$