Since the angle for novice competition measures half the steepness of the angle for the high level competition, and [latex]\tan \theta =\frac{5}{3}[/latex] for high competition, we can find [latex]\cos \theta[/latex] from the right triangle and the Pythagorean theorem so that we can use the half-angle identities.

[latex]\begin{gathered}{3}^{2}+{5}^{2}=34 \\ c=\sqrt{34} \end{gathered}[/latex]

We see that [latex]\cos \theta =\frac{3}{\sqrt{34}}=\frac{3\sqrt{34}}{34}[/latex]. We can use the half-angle formula for tangent: [latex]\tan \frac{\theta }{2}=\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}[/latex]. Since [latex]\tan \theta[/latex] is in the first quadrant, so is [latex]\tan \frac{\theta }{2}[/latex]. Thus,

[latex]\begin{align} \tan \frac{\theta }{2}&=\sqrt{\frac{1-\frac{3\sqrt{34}}{34}}{1+\frac{3\sqrt{34}}{34}}} \\ &=\sqrt{\frac{\frac{34 - 3\sqrt{34}}{34}}{\frac{34+3\sqrt{34}}{34}}} \\ &=\sqrt{\frac{34 - 3\sqrt{34}}{34+3\sqrt{34}}} \\ &\approx 0.57\end{align}[/latex]

We can take the inverse tangent to find the angle: [latex]{\tan }^{-1}\left(0.57\right)\approx {29.7}^{\circ }[/latex]. So the angle of the ramp for novice competition is [latex]\approx {29.7}^{\circ }[/latex].