PQ is a chord of length $8\text{ cm}$ of a circle of radius $5\text{ cm}$. The tangents at $P$ and $Q$ intersect at a point $T$. Find $TP$. 

Join the centre $O$ to $P,Q,$ and $T$, and let $OT$ meet chord $PQ$ at $R$. Because tangents from $T$ are equal, triangle $TPQ$ is isosceles, so $TR$ is perpendicular to $PQ$ and bisects it; hence $PR = RQ = 4\text{ cm}$. In right triangle $OPR$, $OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} = 3\text{ cm}$. Now triangles $TRP$ and $PRO$ are similar: each is right-angled, and $\angle PTR = \angle RPO$. Therefore $\dfrac{TP}{PO} = \dfrac{PR}{RO} = \dfrac{4}{3}$, so $TP = 5\cdot \dfrac{4}{3} = \dfrac{20}{3}\text{ cm}$. The other options come from ignoring either the bisection of the chord or the similarity step.