If sin 3A = cos (A − 26°), where 3A is an acute angles, find the value of A.
Question:

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

Solution:

We are given 3A is an acute angle

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

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

$\Rightarrow \sin 3 A=\sin \left(116^{\circ}-A\right)$

$\Rightarrow 3 A=116^{\circ}-A$

 

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

$\Rightarrow A=29^{\circ}$

Hence the correct answer is $29^{\circ}$

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