$\overline{AB}$ is a chord of $\odot(O, 13)$ such that $AB = 24$. Tangents at $A$ and $B$ to the circle intersect at $P$. Find $PA$.

(D) Let $R$ be the intersection of $OP$ and $AB$. Since $PA$ and $PB$ are tangents from an external point $P$,$PA = PB$ and $\triangle OAP \cong \triangle OBP$. Thus,$OP$ is the perpendicular bisector of chord $AB$.
Given $AB = 24$,so $AR = RB = 12$.
In right-angled $\triangle ORA$,by Pythagoras theorem:
$OR^2 = OA^2 - AR^2 = 13^2 - 12^2 = 169 - 144 = 25$.
So,$OR = 5$.
In $\triangle OAP$,$\angle OAP = 90^\circ$ (tangent is perpendicular to radius at point of contact). $AR$ is the altitude to the hypotenuse $OP$.
By property of right triangles,$OA^2 = OR \cdot OP$.
$13^2 = 5 \cdot OP \implies 169 = 5 \cdot OP \implies OP = \frac{169}{5} = 33.8$.
In right-angled $\triangle OAP$,$PA^2 = OP^2 - OA^2 = (33.8)^2 - 13^2 = 1142.44 - 169 = 973.44$.
$PA = \sqrt{973.44} = 31.2 = \frac{156}{5}$.