If sin 3A=cos (a-26∘), where 3A is an acute angle. find the value of A.

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

A
$32^{\circ}$
B
$30^{\circ}$
C
$29^{\circ}$
D
$24^{\circ}$

Solution:

Given,

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

, where 3A is an acute angle.
Use the identity,

\cos (90^{\circ} - θ) = \sin A

Simplify the given expression.

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

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

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

\Rightarrow 4 A = 116^{\circ}

\Rightarrow A = 29^{\circ}

Therefore, the value of A is

29^{\circ}

.
Hence, option 3 is correct.

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