1954 AMC 12 Problem 46

The marked $\frac{3}{8}$-inch measure is the circle’s diameter

With radius $\frac{3}{16},$ the center is $\frac{3}{8}$ inch above the $60^\circ$ vertex because $\sin30^\circ=\frac{1}{2}$

The radius is $\frac{3}{16}$ inch. The center lies on the angle bisector. The perpendicular radius to either sloping side and the segment from the vertex to the center form a right triangle with a $30^\circ$ angle, so the center is $$\frac{\frac{3}{16}}{\sin30^\circ}=\frac38$$ inch above the vertex. Hence the top tangent is $\frac{3}{8}+\frac{3}{16}=\frac{9}{16}$ inch above the vertex. Since the ledge is $\frac{1}{2}$ inch above the vertex, $$x=\frac9{16}-\frac12=\frac1{16}\text{ in.}$$

Thus, the correct answer is E.