Given,
$AP=10cm$
$\angle APB={{60}^{\circ }}$
According to the figure
We know that,
A line drawn from centre to point from where external tangents are drawn, bisects the angle made by tangents at that particular point
$\angle OPB=\angle APO=\frac{1}{2}\times {{60}^{\circ }}={{30}^{\circ }}$
And, bisected perpendicularly the chord AB
$\therefore AB=2AM$
In $\vartriangle AMP$,
$\sin {{30}^{\circ }}=\frac{\text{Opposite}}{\text{Hypotenuse}}=\frac{AM}{AP}$
$AM=AP\sin {{30}^{\circ }}$
$\frac{AP}{2}=\frac{10}{2}$
$=5cm$ [As AB = 2AM]
So, $AP=2AM=10cm$
And,$AB=2AM=10cm$