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

Let the velocities be $\overrightarrow{u_1}$ and $\overrightarrow{u_2}$ with $|\overrightarrow{u_1}|=|\overrightarrow{u_2}|=u$.The original resultant $\overrightarrow{R_1}$ makes an angle $\theta /2$ with each velocity.New velocities are $u$ and $u/2$. Let the new resultant $\overrightarrow{R_2}$ make an angle $\alpha$ with the velocity $u$.According to the question, the angle between the velocity $u$ and the original resultant ($\theta /2$) is bisected by $\overrightarrow{R_2}$.So, $\alpha =\frac{\theta /2}{2}=\theta /4$.Using the formula for the direction of resultant:
$$\tan \alpha =\frac{(u/2)\sin \theta }{u+(u/2)\cos \theta }=\frac{\sin \theta }{2+\cos \theta }$$
$$\tan \frac{\theta }{4}=\frac{2\sin (\theta /2)\cos (\theta /2)}{1+2{\cos }^2(\theta /2)}$$
Solving this or substituting options, for $\theta =120^{\circ }$:$\tan (30^{\circ })=\frac{\sin 120^{\circ }}{2+\cos 120^{\circ }}=\frac{\sqrt{3}/2}{2-1/2}=\frac{\sqrt{3}/2}{3/2}=\frac{1}{\sqrt{3}}$ This is true. Thus $\theta =120^{\circ }$.