The sum of sides $AB$ and $BC$ of triangle $ABC$ is 11, angle $B$ is $60^{\circ}$, and the radius of the inscribed circle is $\frac{2}{\sqrt{3}}$. It is also known that side $AB$ is greater than side $BC$. Find the height of the triangle dropped from vertex

A.

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In standard notation, $p-b=r \frac{\sqrt{3}}{2}=2.2$, $p=a+b+c=b+11$, from which $p=9$. $S=a c \frac{\sqrt{3}}{4}=r p=\frac{18}{\sqrt{3}}$, so $a c=24$. Since $a+c=11$, then $a=3, c=8$, and the desired height is $c \frac{\sqrt{3}}{2}=4 \sqrt{3}$.

Answer

$4 \sqrt{3}$.