A particle has two velocities of equal magnitude 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 two initial velocities be $$ \vec{v}_1 $$ and $$ \vec{v}_2 $$ with equal magnitude $$ v $$, inclined at an angle $$ \theta $$. The original resultant velocity is given by:

$$ \vec{R} = \vec{v}_1 + \vec{v}_2 $$

Since the two component vectors have equal magnitudes, their resultant $$ \vec{R} $$ perfectly bisects the angle $$ \theta $$ between them. Therefore, the angle between the vector $$ \vec{v}_2 $$ and the original resultant $$ \vec{R} $$ is:

$$ \alpha = \frac{\theta}{2} $$

One of the velocities is halved. Let $$ \vec{v}_1 $$ be halved to $$ \frac{v}{2} $$ while $$ \vec{v}_2 $$ remains $$ v $$. The new resultant velocity is:

$$ \vec{R}' = \frac{1}{2}\vec{v}_1 + \vec{v}_2 $$

The new resultant $$ \vec{R}' $$ bisects the angle between the unchanged velocity $$ \vec{v}_2 $$ and the original resultant $$ \vec{R} $$. Since that angle is $$ \frac{\theta}{2} $$, the angle $$ \alpha' $$ between $$ \vec{v}_2 $$ and the new resultant $$ \vec{R}' $$ is:

$$ \alpha' = \frac{1}{2} \left(\frac{\theta}{2}\right) = \frac{\theta}{4} $$

The angle $$ \alpha' $$ that a resultant makes with one of its components $$ \vec{v}_2 $$ is given by the standard vector direction formula:

$$ \tan \alpha' = \frac{\left(\frac{v}{2}\right) \sin \theta}{v + \left(\frac{v}{2}\right) \cos \theta} $$

Substitute $$ \alpha' = \frac{\theta}{4} $$ and simplify the fraction by canceling out $$ v $$:

$$ \tan\left(\frac{\theta}{4}\right) = \frac{\sin \theta}{2 + \cos \theta} $$

To make the equation easier to solve, let $$ \phi = \frac{\theta}{4} $$, which implies $$ \theta = 4\phi $$. Substituting this into the expression yields:

$$ \tan \phi = \frac{\sin 4\phi}{2 + \cos 4\phi} $$

Express $$ \tan \phi $$ as $$ \frac{\sin \phi}{\cos \phi} $$ and cross-multiply:

$$ \frac{\sin \phi}{\cos \phi} = \frac{\sin 4\phi}{2 + \cos 4\phi} $$

$$ \sin \phi (2 + \cos 4\phi) = \cos \phi \sin 4\phi $$

$$ 2 \sin \phi + \sin \phi \cos 4\phi = \cos \phi \sin 4\phi $$

$$ 2 \sin \phi = \cos \phi \sin 4\phi - \sin \phi \cos 4\phi $$

Using the trigonometric identity $$ \sin(A - B) = \sin A \cos B - \cos A \sin B $$:

$$ 2 \sin \phi = \sin(4\phi - \phi) $$

$$ 2 \sin \phi = \sin 3\phi $$

Expand $$ \sin 3\phi $$ using the triple-angle identity $$ \sin 3\phi = 3\sin \phi - 4\sin^3 \phi $$:

$$ 2 \sin \phi = 3 \sin \phi - 4 \sin^3 \phi $$

$$ 4 \sin^3 \phi - \sin \phi = 0 $$

$$ \sin \phi (4 \sin^2 \phi - 1) = 0 $$

Since $$ \theta > 0 $$, $$ \phi \neq 0 $$, which means $$ \sin \phi \neq 0 $$. Therefore:

$$ 4 \sin^2 \phi - 1 = 0 $$

$$ \sin^2 \phi = \frac{1}{4} $$

$$ \sin \phi = \frac{1}{2} $$

For an acute angle $$ \phi $$:

$$ \phi = 30^{\circ} $$

Now, solve back for $$ \theta $$:

$$ \theta = 4\phi = 4 \times 30^{\circ} = 120^{\circ} $$

Final Answer:

The angle $$ \theta $$ between the two original velocities of equal magnitude is 120 degrees or $$ \frac{2\pi}{3} $$ radians.