A hydrogen atom is in an orbital angular-momentum state $|l,m=l\rangle$. In the vector model, its angular momentum lies on a cone of half-angle 30° to the z-axis. Find l.
The vector-model magnitude is $|\mathbf L|=\sqrt{l(l+1)}\hbar$ and its z projection is $L_z=l\hbar$. Hence $\cos30^\circ=l/\sqrt{l(l+1)}$. Squaring gives $3/4=l/(l+1)$, so $3l+3=4l$ and l = 3. This describes the angular-momentum cone representation, not simultaneous sharp values of all vector components.
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