CUET Preparation Today
If $\sqrt{2} \sin(60^\circ - \alpha) = 1, 0^\circ < \alpha < 90^\circ$, then $\alpha$ is equal to: |
$45^\circ$ $30^\circ$ $15^\circ$ $60^\circ$ |
$15^\circ$ |
√2 sin (60º - α) = 1 sin (60º - α) = \(\frac{1}{√2 }\) { we know, sin 45º = \(\frac{1}{√2 }\) } So, α = 45º |
