Solve this
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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