Answer
(a) $50^\circ.$
(b) $150^\circ.$
Work Step by Step
(a) If the rotation is clockwise, then the projectile moves to the right in the $x$ direction. Also, if the inclination angle is $\theta$ at the initial point, then the direction of motion (which is the same as the direction of the velocity) has to rotate exactly for that angle $\theta$ clockwise to get to the completely horizontal orientation which is achieved at the highest point. So we conclude that the launching angle was $50^\circ$.
(b) If the rotation is counterclockwise, then the projectile moves to the left, opposite of the orientation of the $x$ axis. Reasoning the same way as in part $a$ we see that the launching angle had to be $30^\circ$ but with respect to the 'negative end' of the $x$ axis which means that it makes $180^\circ-30^\circ=150^\circ$ with the 'positive end' of the $x$ axis.
