Problem:
Points A,B,C and D are on a circle of diameter 1, and X is on diameter AD. If BX=CX and 3∠BAC=∠BXC=36∘, then AX=
Answer Choices:
A. cos6∘cos12∘sec18∘
B. cos6∘sin12∘csc18∘
C. cos6∘sin12∘sec18∘
D. sin6∘sin12∘csc18∘
E. sin6∘sin12∘sec18∘
Solution:
The center of the circle is not X since 2∠BAC=∠BXC. Thus AD bisects ∠BXC and ∠BAC. (Why?) Since ∠ABD=90∘ and AD=1, it follows that AB=ADcos∠BAD=cos21∠BAC=cos6∘. Note that ∠AXB=162∘ and ∠ABX=12∘. By the Law of Sines,
sin∠AXBAB=sin∠ABXAX so sin162∘cos6∘=sin12∘AX.
Since sin162∘=sin18∘, we have
AX=sin18∘cos6∘sin12∘=cos6∘sin12∘csc18∘.
