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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Updated on: 10-Oct-2022

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