There is something you always need to remember when dealing with problems where there is no air resistance involved and with a shotput no drag is an excellent assumption. The one constant of the problem is that the horizontal component of velocity remains constant throughout the flight of the projectile.
So if given an initial angle of inclination of flight, you can easily compute the horizontal velocity component (a constant) by trigonometry. Then you only need to compute the time of flight which is based on the vertical component of velocity, Vv. The vertical component of velocity will have the same magnitude going up as it does coming down for any respective height above release point. So the time of the flight is twice the time it takes to get to its maximum elevation or twice the time it takes to fall to the point of release achieving a vertical velocity Vv, the vertical component of the initial velocity. So if Vv is the vertical component of initial veloicity at some angle, the time of flight is t = 2*Vv/g, g being the gravitational acceleration. The distance the shotput travels is then distance = Vh * t, where Vh is the horizontal component of velocity based on the initial angle of inclination.
Another simple fact that you may already know. If you fire a bullet horizontally and at the same instant drop a bullet from the same elevation, both hit the ground at the same time.
Good luck with your test. The problem could always be stated in reverse order so what I might do is do a practice problem with some initial velocity at some angle. Determine the distance. Then using the distance, work the problem backwards to get either the initial velocity or angle of inclination. Needless to say, when going in reverse, you must come up with the same initial conditions.