The chord of a circle is $10\text{cm}$ in length and at a distance of $\frac{10}{\sqrt{3}}\text{cm}$ from the centre. Calculate the angle subtended by the chord at the centre of the circle.

Let $AB$ be the chord of length $10\text{cm}$. Let $O$ be the centre of the circle.The perpendicular from the centre to the chord bisects the chord. Let this perpendicular meet $AB$ at $M$.Thus, $AM=MB=\frac{10}{2}=5\text{cm}$.The distance from the centre is $OM=\frac{10}{\sqrt{3}}\text{cm}$.Let $\angle AOM=\theta$. Then the total angle subtended at the centre is $\angle AOB=2\theta$.In right-angled triangle $△OAM$:
$$\tan \theta =\frac{AM}{OM}=\frac{5}{\frac{10}{\sqrt{3}}}=\frac{5\sqrt{3}}{10}=\frac{\sqrt{3}}{2}$$
$$\theta ={\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$$
Therefore, the total angle subtended at the centre is:
$$2\theta =2{\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$$