If $ \sin 3 \theta=\cos \left(\theta-6^{\circ}\right) $, where $ 3 \theta $ and $ \theta-6^{\circ} $ are acute angles, find the value of $ \theta $.
Given:
\( \sin 3 \theta=\cos \left(\theta-6^{\circ}\right) \), where \( 3 \theta \) and \( \theta-6^{\circ} \) are acute angles.
To do:
We have to find the value of \( \theta \).
Solution:
We know that,
$\cos\ (90^{\circ}- \theta) = sin\ \theta$
Therefore,
$\sin 3 \theta=\cos \left(\theta-6^{\circ}\right)$
$\cos (90^{\circ}- 3\theta)=cos \left(\theta-6^{\circ}\right)$
Comparing on both sides, we get,
$90^{\circ}- 3\theta=\theta-6^{\circ}$
$3\theta+\theta=90^{\circ}+6^{\circ}$
$4\theta=96^{\circ}$
$\theta=\frac{96^{\circ}}{4}$
$\theta=24^{\circ}$
The value of \( \theta \) is $24^{\circ}$.
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