$\overline{AB}$ is a chord of $\odot(O, 10)$ such that $AB = 16$. Tangents at $A$ and $B$ to the circle intersect at $P$. Find the length of $PA$.

(40/3) Let $O$ be the center of the circle with radius $r = 10$. Let $M$ be the midpoint of chord $AB$. Since $AB = 16$,$AM = MB = 8$. In $\triangle OMA$,by the Pythagorean theorem,$OM = \sqrt{OA^2 - AM^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6$. Let $PA = x$. In $\triangle OAP$,$\angle OAP = 90^\circ$ (radius is perpendicular to the tangent). Let $\angle AOP = \theta$. Then $\triangle OMA \sim \triangle OAP$ by $AA$ similarity. Thus,$\frac{PA}{AM} = \frac{OA}{OM}$,which gives $\frac{x}{8} = \frac{10}{6}$. Solving for $x$,we get $x = \frac{80}{6} = \frac{40}{3}$.