If sin 3A = cos (A - 26^{circ}),where 3A is a | vedclass

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(A) Given: $\sin 3A = \cos (A - 26^{\circ})$We know that $\sin 	heta = \cos (90^{\circ} - 	heta)$.Therefore,$\cos (90^{\circ} - 3A) = \cos (A - 26^{

Vedclass · Accepted Answer

$29$

Vedclass · Answer

(A) Given: $\sin 3A = \cos (A - 26^{\circ})$We know that $\sin 	heta = \cos (90^{\circ} - 	heta)$.Therefore,$\cos (90^{\circ} - 3A) = \cos (A - 26^{\circ})$.Comparing the angles,we get: $90^{\circ} - 3A = A - 26^{\circ}$.Rearranging the terms: $90^{\circ} + 26^{\circ} = 3A + A$.$116^{\circ} = 4A$.$A = \frac{116^{\circ}}{4} = 29^{\circ}$.

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

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