1994 AMC 12 第 29 题

The minor arc $\overset{\frown}{BC}$ subtends a central angle of $1$ radian

Use the chord formula $2r\sin(\frac{\theta}{2})$ for both $BC$ and $AB$

The central angle subtending minor arc $\overset{\frown}{BC}$ is $\frac{r}{r}=1,$ so $BC=2r\sin\frac12.$ Since $AB=AC$ and $AB\gt r,$ point $A$ is the midpoint of the major arc $BC.$ The minor central angle from $A$ to $B$ is $\pi-\frac12,$ so $AB=2r$$\sin\left(\frac{\pi-\frac{1}{2}}{2}\right)=2r\cos\frac14.$ Therefore $\frac{AB}{BC}=\frac{\cos(\frac{1}{4})}{\sin(\frac{1}{2})}$$=\frac1{2\sin(\frac{1}{4})}=\frac12\csc\frac14.$ Thus the correct answer is A.