JEE Main · 2006 · 30 Apr 2006 · Offline · Vector Algebra

Take the velocity that is left unchanged along the reference direction and let each original velocity have magnitude $u$ with angle $\theta$ between them. Because the two original velocities are equal in size, their resultant bisects the angle between them and so makes an angle $\frac{\theta }{2}$ with the reference direction. After the second velocity is halved to $\frac{u}{2}$, the new resultant makes an angle $\beta$ with the reference given by $\tan \beta =\frac{\frac{u}{2}\sin \theta }{u+\frac{u}{2}\cos \theta }=\frac{\sin \theta }{2+\cos \theta }$. The requirement that this new resultant bisects the angle between the unchanged velocity and the original resultant means $\beta =\frac{\theta }{4}$, so $\tan \frac{\theta }{4}=\frac{\sin \theta }{2+\cos \theta }$. Writing $\phi =\frac{\theta }{4}$ and clearing the fraction gives $2\sin \phi +\sin \phi \cos 4\phi =\cos \phi \sin 4\phi$, and since $\cos \phi \sin 4\phi -\sin \phi \cos 4\phi =\sin 3\phi$ this becomes $2\sin \phi =\sin 3\phi$. Using $\sin 3\phi =3\sin \phi -4{\sin }^3\phi$ reduces it to $4{\sin }^3\phi =\sin \phi$, hence ${\sin }^2\phi =\frac{1}{4}$ and $\sin \phi =\frac{1}{2}$. Thus $\phi =30^{\circ }$ and $\theta =120^{\circ }$.