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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $A$.

Similar Questions

$\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}} = ?$

Evaluate the following:
$\sin 60^{\circ} \cos 30^{\circ} + \sin 30^{\circ} \cos 60^{\circ}$

If $\tan 2A = \cot(A - 18^{\circ})$,where $2A$ is an acute angle,find the value of $A$ (in $^{\circ}$).

If $\cot \theta = \frac{7}{8},$ evaluate:
$(i) \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$(ii) \cot^2 \theta$

Evaluate the following:
$\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module