check the picture below.
keeping in mind that the point of tangency for a radius line and a tangent is alway a right-angle, since the "red" chord is parallel to the "green" tangent line outside, then the chord is cutting the "green" radius there in two equal halves at a right-angle, as you see in the picture.
we know the chord is 10 units long, so 5 + 5, since is perpendicularity with the radius will also cut the chord in two equal halves.
anyhow, all that said, we end up with triangle you see on the right-hand-side, and then we can just use the pythagorean theorem.
[tex]\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2
\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
-------\\
c=2a\\
a=a\\
b=5
\end{cases}\implies (2a)^2=a^2+5^2[/tex]
[tex]\bf (2^2a^2)=a^2+25\implies 4a^2=a^2+25\implies 3a^2=25
\\\\\\
a^2=\cfrac{25}{3}\implies a=\sqrt{\cfrac{25}{3}}\implies a=\cfrac{\sqrt{25}}{\sqrt{3}}\implies a=\cfrac{5}{\sqrt{3}}
\\\\\\
\textit{and then we can \underline{rationalize the denominator} }
\\\\\\
a=\cfrac{5}{\sqrt{3}}\cdot \cfrac{\sqrt{3}}{\sqrt{3}}\implies a=\cfrac{5\sqrt{3}}{\sqrt{3^2}}\implies a=\cfrac{5\sqrt{3}}{3}[/tex]
