Question:
If $\sqrt{2} \sin \left(60^{\circ}-\alpha\right)=1$ then $\alpha=?$
(a) 15°
(b) 30°
(c) 45°
(d) 60°
Solution:
As we know that,
$\sin 45^{\circ}=\frac{1}{\sqrt{2}}$
Thus,
if $\sqrt{2} \sin \left(60^{\circ}-\alpha\right)=1$
$\Rightarrow \sin \left(60^{\circ}-\alpha\right)=\frac{1}{\sqrt{2}}$
$\Rightarrow 60^{\circ}-\alpha=45^{\circ}$
$\Rightarrow \alpha=60^{\circ}-45^{\circ}$
$\Rightarrow \alpha=15^{\circ}$
Hence, the correct option is (a).
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