Given the parabola $y^{2}=12 x$, we draw a chord through its focus, which forms a $60^{\circ}$ angle with the $x$-axis. We need to determine the length of this chord and the angle enclosed by the tangents drawn at the endpoints of this chord.
If the coordinates of point $A$ of the chord $AB$ are $x_{1}$ and $y_{1}$, and the coordinates of point $B$ are $x_{2}$ and $y_{2}$, then (since the coordinates of the focus $F$ are 3 and 0) the equation of the chord $AB$ is:

$$
y=(x-3) \tan 60^{\circ}
$$

or

$$
y^{2}=3(x-3)^{2}
$$

If we combine this equation with the given equation of the parabola, we obtain the abscissas of points $A$ and $B$;

$$
3(x-3)^{2}=12 x
$$

from which

$$
x_{1}=9, x_{2}=1
$$

and thus

$$
AB=\frac{x_{1}-x_{2}}{\sin 30^{\circ}}=16
$$

The tangent at $A$ bisects the angle $LAF$ and the tangent at $B$ bisects the angle $ABK$ (denote by $L$ and $K$ the points where the lines parallel to the abscissa axis from $A$ and $B$ intersect the ordinate axis); therefore,

$$
\angle OAF + \angle OBF = \frac{1}{2}(\angle LAF + \angle FBK) = 90^{\circ}
$$

and thus

$$
\angle AOB = 90^{\circ}
$$

(Mihály Devecis.)

The problem was also solved by: Bojedain F., Goldziher K., Groffits G., Juvancz I., Koós A., Krausz B., Lukhaub Gy., Probst E., Prohászka J., Sasvári G., Spitzer S.