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

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Question

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

Vedclass · Accepted Answer

$29$

Vedclass · Answer

(A) Given equation: $\sin 3A = \cos(A - 26^{\circ})$We know that $\sin 	heta = \cos(90^{\circ} - 	heta)$.Therefore,$\sin 3A = \cos(90^{\circ} - 3A)$.Substituting this into the given equation:$\cos(90^{\circ} - 3A) = \cos(A - 26^{\circ})$Since the angles are acute,we can equate them:$90^{\circ} - 3A = A - 26^{\circ}$Rearranging the terms to solve for $A$:$90^{\circ} + 26^{\circ} = A + 3A$$116^{\circ} = 4A$$A = \frac{116^{\circ}}{4} = 29^{\circ}$Thus,the value of $A$ is $29^{\circ}$.

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

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