In the figure, $ A B $ is a chord of length $ 16 \mathrm{~cm} $ of a circle of radius $ 10 \mathrm{~cm} $. The tangents at $ A $ and $ B $ intersect at a point $ P $. Find the length of $ P A $."

Given:

In the figure, \( A B \) is a chord of length \( 16 \mathrm{~cm} \) of a circle of radius \( 10 \mathrm{~cm} \). The tangents at \( A \) and \( B \) intersect at a point \( P \). 

To do:

We have to find the length of \( P A \).

Solution:

Join $OA$.

$OA = 10\ cm$

$OL$ is perpendicular to $AB$.

$AL=LB= \frac{16}{2}=8\ cm$  (AL and LB are the radii of the circle)

In right angled triangle $OLA$,

By Pythagoras theorem,

$OA^2=OL^2 +LA^2$ 

$OL^2=OA^2-LA^2$

$OL^2=(10)^2-8^2$

$=100-64$

$=36$

$\Rightarrow OL=6\ cm$

$\tan\ AOL=\frac{AL}{OL}$

$=\frac{8}{6}$

$=\frac{4}{3}$

From $\triangle OAP$,

$\tan\ AOL=\frac{PA}{OA}$

$\frac{4}{3}=\frac{PA}{10}$

$PA=\frac{10\times4}{3}$

$PA=\frac{40}{3}\ cm$

The length of PA is $\frac{40}{3}\ cm$.