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An angle θ,0<θ<π2\theta, 0<\theta<\frac{\pi}{2}, which increases twice as fast as its sine, is

Solution

✅ Correct Option: 3

The condition means dθdt=2ddt(sin⁡θ)=2cos⁡θdθdt\frac{d\theta}{dt}=2\frac{d}{dt}(\sin\theta)=2\cos\theta\frac{d\theta}{dt}.

Cancelling dθdt\frac{d\theta}{dt} gives cos⁡θ=12\cos\theta=\frac{1}{2}, so θ=π3\theta=\frac{\pi}{3}.

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