$PQ$ is a chord of length $8 \, cm$ of a circle of radius $5 \, cm$. The tangents at $P$ and $Q$ intersect at a point $T$ (see figure). Find the length $TP$.

(D) Join $OT$. Let it intersect $PQ$ at the point $R$. Then $\triangle TPQ$ is isosceles and $TO$ is the angle bisector of $\angle PTQ$. So,$OT \perp PQ$ and therefore,$OT$ bisects $PQ$ which gives $PR = RQ = 4 \, cm$.
Also,$OR = \sqrt{OP^2 - PR^2} = \sqrt{5^2 - 4^2} \, cm = \sqrt{25 - 16} \, cm = \sqrt{9} \, cm = 3 \, cm$.
Now,$\angle TPR + \angle RPO = 90^{\circ}$ and $\angle TPR + \angle PTR = 90^{\circ}$ (since $\triangle TRP$ is a right triangle).
So,$\angle RPO = \angle PTR$.
Therefore,right triangle $TRP$ is similar to the right triangle $PRO$ by $AA$ similarity.
This gives $\frac{TP}{PO} = \frac{RP}{RO}$,i.e.,$\frac{TP}{5} = \frac{4}{3}$ or $TP = \frac{20}{3} \, cm$.