Points $A,$ $B$ and $C$ on a circle of radius $r$ are situated so that $AB = AC, AB > r,$ and the length of minor arc $BC$ is $r$. If angles are measured in radians, then $AB/BC = ?$
A) $1/2 \csc(1/4)$
B) $2 \cos(1/2)$
C) $4 \sin(1/2)$
D) $\csc(1/2)$
E) $2 \sec(1/2)$
So, I got the lengths of $AB$ and $BC$ in terms of the radius using law of cosines.
$AB = \sqrt{2r^2(1+\cos(1/2))}$ and $BC = \sqrt{2r^2(1-\cos(1))}$.
I put them in a fraction and simplified using the double angle formula to get $AB/BC = 1/\sqrt{2(1+\cos(1/2))}$.
From then, I don't know what to do since all the answer choices don't have a square root.
I'd appreciate some help, thanks!
For your answer, $$ \frac 1 {\sqrt{2(1 + \cos (1/2))}} = \frac 1 {\sqrt {2 \cdot 2 \cos^2(1/4)}} = \frac 12 \sec (1/4), $$ so perhaps you made some mistake during computation.
Now I provide my solution.
Solution. $\blacktriangleleft$ Let the center of the circle be $O$. Since the length of the minor arc ${BC}$ is $r$, the angle $$\angle BOC = r/r = 1,$$ which means $\angle BAC = 1/2$. Now construct $AD \perp BC$ where $D$ is on $BC$. Since $AB=AC$, the line segment $AD$ bisect $\angle BAC$ and $BC$ simultaneously. Now in the right triangle $\triangle ABD$, we have $$\angle BAD = \angle BAC /2 =1/4,$$ hence $$BD/AB = \sin (\angle BAD) = \sin (1/4).$$ Therefore $$BC/AB = 2BD/AB = 2\sin (1/4),$$ or $AB/BC = \csc(1/4) /2$, which is exactly choice $(\mathrm A)$. $\blacktriangleright$