You can draw triangle \(PAS\) with \(DS\) perpendicular(the altitude) to \(AP\). You know \(PS = 10\). Let's name the midpoint of \(PS\) the letter \(F\). \(PF = 5, PA = 13, \) so \(AF = 12\). Now, cos \(PAS = \frac{AD}{AS}\). We already know \(AS = 13,\) so we just need to solve for \(AD\). To find \(AD\), we first need to solve for \(DS\), which is one of the altitudes of triangle \(PAS\). We know that the area of a triangle is \(\frac{1}{2}bh\), so we can make an equation. \(\frac{1}{2}\times10\times12=\frac{1}{2}\times13\times DS\). From this, we get \(DS = \frac{120}{13}\). Now, we can use Pythagorean Theorum to find \(AD\). \(12^2-(\frac{120}{13})^2=AD^2\). After calculating this, we can get \(AD = \frac{12\sqrt{69}}{13}\). Now, lets plug this into our fraction. We can get \(\frac{\frac{12\sqrt{69}}{13}}{13}\) which simplifies to \(\frac{12\sqrt{69}}{169}\).

- Daisy