Answer
The launch angle is:
$$\theta=30^\circ.$$
Work Step by Step
When the projectile is launched, its direction of motion, i.e. the direction of velocity is inclined for $\theta$ above the horizontal, while at the landing point its direction of motion is inclined for the same $\theta$ but below the horizontal. This means that, in total, rotating clockwise, the direction rotates for $2\theta$. Thus we have
$$60^\circ=2\theta\Rightarrow \theta=30^\circ.$$
