# 1993 AHSME Problems/Problem 23

## Problem

$[asy] draw(circle((0,0),10),black+linewidth(.75)); draw((-10,0)--(10,0),black+linewidth(.75)); draw((-10,0)--(9,sqrt(19)),black+linewidth(.75)); draw((-10,0)--(9,-sqrt(19)),black+linewidth(.75)); draw((2,0)--(9,sqrt(19)),black+linewidth(.75)); draw((2,0)--(9,-sqrt(19)),black+linewidth(.75)); MP("X",(2,0),N);MP("A",(-10,0),W);MP("D",(10,0),E);MP("B",(9,sqrt(19)),E);MP("C",(9,-sqrt(19)),E); [/asy]$

Points $A,B,C$ and $D$ are on a circle of diameter $1$, and $X$ is on diameter $\overline{AD}.$

If $BX=CX$ and $3\angle{BAC}=\angle{BXC}=36^\circ$, then $AX=$

$\text{(A) } cos(6^\circ)cos(12^\circ)sec(18^\circ)\quad\\ \text{(B) } cos(6^\circ)sin(12^\circ)csc(18^\circ)\quad\\ \text{(C) } cos(6^\circ)sin(12^\circ)sec(18^\circ)\quad\\ \text{(D) } sin(6^\circ)sin(12^\circ)csc(18^\circ)\quad\\ \text{(E) } sin(6^\circ)sin(12^\circ)sec(18^\circ)$

## Solution

We have all the angles we need, but most obviously, we see that right angle in triangle $ABD$.

Note also that angle $BAD$ is 6 degrees, so length $AB = cos(6)$ because the diameter, $AD$, is 1.

Now, we can concentrate on triangle $ABX$ (after all, now we can decipher all angles easily and use Law of Sines).

We get:

$AB /sin(\angle{AXB}) = AX / sin(\angle{ABX})$

That's equal to

$cos(6)/ sin(180-18) = AX / sin 12$

Therefore, our answer is equal to: $\fbox{B}$

Note that $sin(162) = sin(18)$, and don't accidentally put $\fbox{C}$ because you thought 1/sin was sec!