Question : If $\cos (3 \alpha)=\sin \left(\alpha-22^{\circ}\right)$, where $3 \alpha<90^{\circ}$, then what is the value of $\alpha$?
Option 1: 29°
Option 2: 26°
Option 3: 27°
Option 4: 28°
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Correct Answer: 28°
Solution :
Given that $\cos (3 \alpha)=\sin \left(\alpha-22^{\circ}\right)$ and $3 \alpha<90^{\circ}$,
$⇒\cos (3 \alpha)=\cos \left\{90^{\circ}-(\alpha-22^{\circ}\right)\}$ [$\because \cos (90^{\circ} - \theta) = \sin \theta$]
$⇒3 \alpha = 90^{\circ}-\alpha+22^{\circ}$
$⇒4 \alpha = 112^{\circ}$
$⇒\alpha = 28^{\circ}$
Hence, the correct answer is 28°.
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