The trajectory of a projectile is described by $y = 2x - 0.01x^{2}$, where x is the horizontal displacement and y is the vertical displacement from the point of projection.

The angle of projection is the angle the path makes with the horizontal at the launch point, so it is the gradient of the trajectory at $x = 0$.

$\dfrac{dy}{dx} = 2 - 0.02x$, and at $x = 0$ this is 2.

So $\tan\theta = 2$ and $\theta = \tan^{-1}2 = 63.4^{\circ}$, that is $63^{\circ}$ to the nearest degree.

The same result comes from comparing $y = 2x - 0.01x^{2}$ with the standard form $y = x\tan\theta - \dfrac{gx^{2}}{2V^{2}\cos^{2}\theta}$, in which the coefficient of x is $\tan\theta$.