Question
If $\sin 3A = \cos 6A,$ then $\angle A = ?$
✓
Answer
$\sin 3 A=\cos 6 A \quad \ldots . .[\text { Given }]$
$\therefore \sin 3 A=\sin \left(90^{\circ}-6 A\right) \quad \ldots . .\left[\cos \theta=\sin \left(90^{\circ}-\theta\right)\right]$
$\therefore 3 A=90^{\circ}-6 A$
$\therefore 3 A+6 A=90^{\circ}$
$\therefore 9 A=90^{\circ}$
$\therefore A=\frac{90^{\circ}}{9}$
$\therefore A=10^{\circ}$
$\therefore \sin 3 A=\sin \left(90^{\circ}-6 A\right) \quad \ldots . .\left[\cos \theta=\sin \left(90^{\circ}-\theta\right)\right]$
$\therefore 3 A=90^{\circ}-6 A$
$\therefore 3 A+6 A=90^{\circ}$
$\therefore 9 A=90^{\circ}$
$\therefore A=\frac{90^{\circ}}{9}$
$\therefore A=10^{\circ}$
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