Answer: \(\beta=69.0^\circ\) approximately.

Let \(v\) be the speed of the particle at the point \(B\) where it loses contact with the sphere.

First use energy from the highest point \(A\) to the ground. The particle starts with speed \(u\), and hits the ground with speed \(3u\), after falling a vertical distance \(2a\). Hence

\(\frac12mu^2+mg(2a)=\frac12m(3u)^2.\)

So

\(u^2+4ag=9u^2, \qquad u^2=\frac12ag.\)

Now use energy from \(A\) to \(B\). The vertical drop from \(A\) to \(B\) is \(a(1-\cos\theta)\), so

\(v^2=u^2+2ag(1-\cos\theta).\)

At the point where contact is lost, the normal reaction is zero. Resolving towards the centre of the sphere gives

\(mg\cos\theta=\frac{mv^2}{a}.\)

Therefore

\(v^2=ag\cos\theta.\)

Substitute \(u^2=\frac12ag\) into the energy equation:

\(ag\cos\theta=\frac12ag+2ag(1-\cos\theta).\)

Divide by \(ag\):

\(\cos\theta=\frac12+2-2\cos\theta.\)

Hence

\(3\cos\theta=\frac52, \qquad \cos\theta=\frac56.\)

So

\(v^2=ag\cos\theta=\frac56ag, \qquad v=\sqrt{\frac56ag}.\)

After the particle leaves the sphere, its horizontal component of velocity remains constant. At \(B\), this horizontal component is

\(v\cos\theta.\)

When the particle strikes the ground, its speed is \(3u\), and the angle with the horizontal is \(\beta\). Therefore

\(3u\cos\beta=v\cos\theta.\)

Thus

\(\cos\beta=\frac{v\cos\theta}{3u}.\)

Substitute \(v=\sqrt{\frac56ag}\), \(\cos\theta=\frac56\), and \(u=\sqrt{\frac12ag}\):

\(\cos\beta =\frac{\sqrt{\frac56ag}\left(\frac56\right)}{3\sqrt{\frac12ag}} =\frac{5\sqrt{15}}{54}.\)

Therefore

\(\beta=\cos^{-1}\left(\frac{5\sqrt{15}}{54}\right) \approx 69.0^\circ.\)

Hence

\(\beta=69.0^\circ.\)