If $$\sqrt{2} \sin(60^\circ - \alpha) = 1, 0^\circ < \alpha < 90^\circ$$, then $$\alpha$$ is equal to:

As per the given question,

$$\sqrt(2) \sin(60^\circ - \alpha)= 1 $$ and $$0^\circ < \alpha < 90^\circ$$

$$\Rightarrow \sin(60^\circ - \alpha) = \dfrac{1}{\sqrt(2)}$$ 

$$\Rightarrow \sin(60^\circ - \alpha) = \sin(45^\circ) $$  (  put the value )

Taking the inverse of both side with $$\sin$$

$$\Rightarrow (60^\circ - \alpha) = 45^\circ $$ 

$$\Rightarrow \alpha = 60^\circ - 45^\circ $$

$$\alpha = 15^\circ $$ Ans