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If $ \sin \theta=\cos \left(\theta-45^{\circ}\right) $, where $ \theta $ and $ \theta-45^{\circ} $ are acute angles, find the degree measure of $ \theta $.
Given:
\( \sin \theta=\cos \left(\theta-45^{\circ}\right) \), where \( \theta \) and \( \theta-45^{\circ} \) are acute angles.
To do:
We have to find the degree measure of \( \theta \).
Solution:
We know that,
$cos\ (90^{\circ}- \theta) = sin\ \theta$
Therefore,
$\sin \theta=\cos \left(\theta-45^{\circ}\right)$
$\Rightarrow cos\ (90^{\circ}- \theta)=\cos \left(\theta-45^{\circ}\right)$
Comparing on both sides, we get,
$90^{\circ}- \theta=\theta-45^{\circ}$
$\theta+\theta=90^{\circ}+45^{\circ}$
$2\theta=135^{\circ}$
$\theta=\frac{135^{\circ}}{2}$
$\theta=67\frac{1}{2}^{\circ}$
The degree measure of \( \theta \) is $67\frac{1}{2}^{\circ}$.
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