If $ \sin 3 A=\cos \left(A-26^{\circ}\right) $, where $ 3 A $ is an acute angle, find the value of $ A $.

Given:

\( \sin 3 A=\cos \left(A-26^{\circ}\right) \), where \( 3 A \) is an acute angle.

To do:

We have to find the value of \( A \).

Solution:  

We know that,

$cos\ (90^{\circ}- \theta) = sin\ \theta$

This implies,

$\sin 3 A=\cos\ (90^{\circ}- 3A)$ 

Therefore,

$\sin 3 A=\cos \left(A-26^{\circ}\right)$

$\Rightarrow \cos\ (90^{\circ}- 3A)=\cos \left(A-26^{\circ}\right)$

Comparing on both sides, we get,

$90^{\circ}- 3A=A-26^{\circ}$

$3A+A=90^{\circ}+26^{\circ}$

$4A=116^{\circ}$

$A=\frac{116^{\circ}}{4}$

$A=29^{\circ}$

Therefore, the value of $A$ is $29^{\circ}$.   

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Updated on: 10-Oct-2022

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