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**Hint:**We are given three people standing on the circumference of a circle. We shall represent every person with a separate point on the circle to properly analyze their location and the relationship between the given distances between every point with the radius of the circle. Then, we will use the basic properties of geometry in order to get the final distance.

**Complete step by step solution:**

Let point A represent Anjali, point B represent Sonia and point C represent Surjit. Also, let O be the center of the circle.

Given that $AB=BC=8m$ and the radius of the circle is a 8m.

$\Rightarrow OA=OC=OB=8m$ ………………….. (1)

Thus, we see that $\Delta OAC$ is an equilateral triangle because all the sides of the triangle (OA, OB and AB) are equal. This implies that every angle of this triangle would be of ${{60}^{\circ }}$ each.

$\angle AOD={{60}^{\circ }}$ ……………….. (2)

We know that any line from the center intersecting the chord of the circle always bisects the chord perpendicularly. Thus, OD perpendicularly bisects AC.

$\Rightarrow AD=DC$ and $\angle ODA={{90}^{\circ }}$. ……………….. (3)

Now, in the right-angled triangle, $\Delta ODA$, we shall use some trigonometric values.

We know that $\sin \theta =\dfrac{P}{H}$

Where, $P=$ perpendicular and $H=$ hypotenuse

$\Rightarrow \sin AOD=\dfrac{AD}{AO}$

Substituting values from (1) and (2), we get

$\Rightarrow \sin {{60}^{\circ }}=\dfrac{AD}{8}$

Since, $\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$ ,

$\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{AD}{8}$

$\Rightarrow AD=4\sqrt{3}$

Now, we have $AC=AD+DC$

From (3), we have

$\Rightarrow AC=2AD$

Putting the values of AD, we get

\[\Rightarrow AC=2\left( 4\sqrt{3} \right)\]

$\therefore AC=8\sqrt{3}$

**Therefore, the distance between Anjali and Surjit is $8\sqrt{3}m$.**

**Note:**In order to solve geometrical problems where trigonometric concepts are involved, we must memorize the values of the angles of trigonometric functions. Also, the answer would have remained the same if we would have solved using $\Delta ODC$ instead of $\Delta ODA$.

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