smc

First we extend the line with $A$ and the line with $B$ so that they both meet the line with $C$, forming an equilateral triangle. Let the vertices of this triangle be $D$, $E$, and $F$. We know it is equilateral because of the angle of $60^{\circ }$ shown, and because the tangent lines $\overline{EF}$ and $\overline{DE}$ are congruent. We can see, because $A$, $B$, and $C$ are points of tangency, that circle $ABC$ is inscribed in $△DEF$. The height of an equilateral triangle is exactly $3$ times the radius of a circle inscribed in it. Let the height of $△DEF$ be $h$. We can see that the radius of the circle equals $\frac{3}{16}$. Thus $$h=\frac{3\cdot 3}{16}=\frac{9}{16}.$$ Subtracting $\frac{1}{2}$ from $h$ gives us $$x=h-\frac{1}{2}=\frac{1}{16},$$ so our answer is $\boxed{\frac{1}{16}"}$.