A particle of mass m is placed on a fixed, smooth sphere of radius...

Let $\theta$ be the angle made by the radius vector of the particle with the vertical when it leaves the sphere.

At the point of leaving the sphere, the normal reaction $N = 0$. The centripetal force is provided by the component of gravity:

$mg \cos \theta = \dfrac{mv^2}{R} \Rightarrow v^2 = Rg \cos \theta$

By conservation of mechanical energy from the initial position $\theta_0 = 30^{{}^{{}^{\circ}}}$ to $\theta$:

$mgR(\cos 30^{{}^{{}^{\circ}}} - \cos \theta) = \dfrac{1}{2}mv^2$

Substituting $v^2 = Rg \cos \theta$:

$mgR(\cos 30^{{}^{{}^{\circ}}} - \cos \theta) = \dfrac{1}{2}mgR \cos \theta$

$\cos 30^{{}^{{}^{\circ}}} = \dfrac{3}{2} \cos \theta$

$\cos \theta = \dfrac{2}{3} \cos 30^{{}^{{}^{\circ}}} = \dfrac{2}{3} \times \dfrac{\sqrt{3}}{2} = \dfrac{1}{\sqrt{3}}$

$\theta = \cos^{-1}\left(\dfrac{1}{\sqrt{3}}\right) \approx 54.74^{{}^{{}^{\circ}}}$

The angle described by the particle is $\Delta \theta = \theta - 30^{{}^{{}^{\circ}}} = 54.74^{{}^{{}^{\circ}}} - 30^{{}^{{}^{\circ}}} = 24.74^{{}^{{}^{\circ}}}$.

Converting to radians:

$\Delta \theta = 24.74^{{}^{{}^{\circ}}} \times \dfrac{\pi}{180^{{}^{{}^{\circ}}}} \approx 0.43 \text{ rad}$

The distance travelled by the particle is $s = R \Delta \theta \approx 0.43R$.