Difference between revisions of "1994 AHSME Problems/Problem 29"
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<math> \textbf{(A)}\ \frac{1}{2}\csc{\frac{1}{4}} \qquad\textbf{(B)}\ 2\cos{\frac{1}{2}} \qquad\textbf{(C)}\ 4\sin{\frac{1}{2}} \qquad\textbf{(D)}\ \csc{\frac{1}{2}} \qquad\textbf{(E)}\ 2\sec{\frac{1}{2}} </math> | <math> \textbf{(A)}\ \frac{1}{2}\csc{\frac{1}{4}} \qquad\textbf{(B)}\ 2\cos{\frac{1}{2}} \qquad\textbf{(C)}\ 4\sin{\frac{1}{2}} \qquad\textbf{(D)}\ \csc{\frac{1}{2}} \qquad\textbf{(E)}\ 2\sec{\frac{1}{2}} </math> | ||
==Solution== | ==Solution== | ||
+ | First note that arc length equals <math>r\theta</math>, where <math>\theta</math> is the central angle in radians. Call the center of the circle <math>O</math>. Then <math>\angle{BOC} = 1</math> radian because the minor arc <math>BC</math> has length <math>r</math>. Since <math>ABC</math> is isosceles, <math>\angle{AOB} = \pi - \tfrac{1}{2}</math>. We use the Law of Cosines to find that <cmath>\frac{AB}{BC} = \frac{\sqrt{2r^2 - 2r^2\cos{(\pi - \frac{1}{2})}}}{\sqrt{2r^2 - 2r^2\cos1}} = \frac{\sqrt{1 + \cos{(\frac{1}{2})}}}{\sqrt{1 - \cos1}}.</cmath> | ||
+ | Using half-angle formulas, we have that this ratio simplifies to <cmath>\frac{\cos\frac{1}{4}}{\sin{\frac{1}{2}}} = \frac{\cos\frac{1}{4}}{\sqrt{1 - \cos^2{\frac{1}{2}}}} = \frac{\cos\frac{1}{4}}{\sqrt{(1 + \cos{\frac{1}{2}})(1 - \cos{\frac{1}{2}})}} = \frac{\cos{\frac{1}{4}}}{2\cos{\frac{1}{4}}\sin{\frac{1}{4}}}</cmath> <cmath>= \boxed{\frac{1}{2}\csc{\frac{1}{4}}.}</cmath> |
Revision as of 21:39, 16 February 2016
Problem
Points and on a circle of radius are situated so that , and the length of minor arc is . If angles are measured in radians, then
Solution
First note that arc length equals , where is the central angle in radians. Call the center of the circle . Then radian because the minor arc has length . Since is isosceles, . We use the Law of Cosines to find that Using half-angle formulas, we have that this ratio simplifies to