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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