1993 AMC 12 第 23 题

Because $BX=CX,$ the diameter through $X$ bisects $\angle BXC$

Find $AB$ using right triangle $ABD,$ then apply the Law of Sines in $\triangle ABX$

Since $BX=CX,$ the line $AD$ bisects $\angle BXC,$ and symmetry also gives $\angle BAD=6^\circ.$ Because $AD=1$ is a diameter, $\angle ABD=90^\circ,$ so $AB=\cos6^\circ.$ Also $$\begin{gathered} \angle AXB=180^\circ-18^\circ=162^\circ, \\ \angle ABX=180^\circ-162^\circ\\ {}-6^\circ=12^\circ. \end{gathered}$$ The Law of Sines in $\triangle ABX$ gives $$\begin{aligned} AX&=\frac{AB\sin12^\circ}{\sin162^\circ}\\ &=\cos6^\circ\sin12^\circ\\ &\qquad{}\csc18^\circ. \end{aligned}$$ Thus the correct answer is B.