If $\sqrt{2} \sin(60^\circ - \alpha) = 1, 0^\circ < \alpha < 90^\circ$, then $\alpha$ is equal to:
Answer & explanation
Correct answer: option 3
√2 sin (60º - α) = 1
sin (60º - α) = \(\frac{1}{√2 }\)
{ we know, sin 45º = \(\frac{1}{√2 }\) }
So, α = 45º