13.7.9 Given the parabola $y^{2}=2 p x(p>0), A B$ is a chord passing through the focus $F$. If $A B$ makes an angle $\theta\left(0<\theta \leqslant \frac{\pi}{2}\right)$ with the $x$-axis, find $\angle A O B$.

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The original text has been translated into English while preserving the original formatting and line breaks.
In a polar coordinate system with $F$ as the pole and the positive x-axis as the polar axis, the equation of the parabola is $\rho = \frac{p}{1 - \cos \alpha}$ (where $\alpha$ is the polar angle).

Construct $AM \perp x$-axis, $BN \perp x$-axis, where $M$ and $N$ are the feet of the perpendiculars. Thus, $|FA| = \frac{p}{1 - \cos \theta}$, $|AM| = \frac{p \sin \theta}{1 - \cos \theta}$, $|FM| = \frac{p \cos \theta}{1 - \cos \theta}$, and $|OF| = \frac{p}{2}$,
then $|OM| = |OF| + |FM| = \frac{p(1 + \cos \theta)}{2(1 - \cos \theta)}$, $\tan \angle AOM =$
$$
\begin{array}{l}
\frac{|AM|}{|OM|} = \frac{2 \sin \theta}{1 + \cos \theta}, \text{ and } \\
|FB| = \frac{p}{1 + \cos \theta}, |BN| = \frac{p \sin \theta}{1 + \cos \theta}, |FN| = \frac{p \cos \theta}{1 + \cos \theta}, \\
|ON| = \frac{p}{2} - \frac{p \cos \theta}{1 + \cos \theta} = \frac{p(1 - \cos \theta)}{2(1 + \cos \theta)}, \tan \angle BON = \frac{2 \sin \theta}{1 - \cos \theta}
\end{array}
$$

Thus, $\tan \angle AOB = \tan (\angle AOM + \angle BON) = -\frac{4}{3 \sin \theta}$. Therefore, $\angle AOB = \pi - \arctan \left(\frac{1}{3 \sin \theta}\right)$.

When $0 < \theta < \frac{\pi}{2}$, the equation of $AB$ can be written as $y = \tan \theta \left(x - \frac{p}{2}\right)$, i.e., $x = \cot \theta \cdot y + \frac{p}{2}$. This result also holds for $\theta = \frac{\pi}{2}$.

Substituting it into the parabola equation yields $y^2 - 2p \cot \theta \cdot y - p^2 = 0$, solving which gives $y = \frac{p(\cos \theta \pm 1)}{\sin \theta}$. Thus, $y_A = \frac{p(\cos \theta + 1)}{\sin \theta}$, $y_B = \frac{p(\cos \theta - 1)}{\sin \theta}$. Therefore,
$$
\begin{array}{c}
x_A = \cot \theta \cdot \frac{p(\cos \theta + 1)}{\sin \theta} + \frac{p}{2} - \frac{p(1 + \cos \theta)^2}{2 \sin^2 \theta}, \\
x_B = \frac{p(\cos \theta - 1)^2}{2 \sin^2 \theta}.
\end{array}
$$

Construct perpendiculars from $A$ and $B$ to the $y$-axis, $AP$ and $BQ$, and connect $AO$ and $BO$. Then, $\tan \angle AOP = \frac{|AP|}{|OP|} = \frac{x_A}{y_A} = \frac{1 + \cos \theta}{2 \sin \theta}$, $\tan \angle BOQ = \frac{x_B}{-y_B} = \frac{1 - \cos \theta}{2 \sin \theta}$, so $\tan (\angle AOP + \angle BOQ) = \frac{4}{3 \sin \theta}$. Therefore, $\angle AOB = \pi - (\angle AOP + \angle BOQ) = \pi - \arctan \left(\frac{4}{3 \sin \theta}\right)$.