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A particle has two velocities of equal magnitude inclined to each other at an angle θ. If one of them is halved, the angle between the other and the original resultant velocity is bisected by the new resultant velocity. Then θ is

1. 90°
2. 120°
3. 45°
4. 60°

Correct answer: 120°

Solution

Let both speeds = v at angle theta; the original resultant bisects theta. After halving one, new resultant R'=A+B/2 makes angle theta/4 with A, so tan(theta/4)=(sin theta)/(2+cos theta). Testing theta=120: RHS=(sqrt3/2)/(2-1/2)=0.577=tan30=tan(120/4). So theta=120 deg.

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