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Difference between revisions of "1995 AHSME Problems/Problem 18"

(New page: ==Problem== Two rays with common endpoint <math>O</math> forms a <math>30^\circ</math> angle. Point <math>A</math> lies on one ray, point <math>B</math> on the other ray, and <math>AB = 1<...)
(No difference)

Revision as of 11:33, 7 July 2008

Problem

Two rays with common endpoint $O$ forms a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB = 1$. The maximum possible length of $OB$ is

$\mathrm{(A) \ 1 } \qquad \mathrm{(B) \ \frac {1 + \sqrt {3}}{\sqrt 2} } \qquad \mathrm{(C) \ \sqrt{3} } \qquad \mathrm{(D) \ 2 } \qquad \mathrm{(E) \ \frac{4}{\sqrt{3}} }$

Solution

Triangle $OAB$ has the property that $\angle AOB=30^{\circ}$ and $AB=1$. Form the Law of Sines, $\frac{\sin{OAB}}{OB}=\frac{1}{2}$. Thus $2\sin{OAB}=OB$. The maximum of $\sin{OAB}$ is 1, so the maximum of $OB$ is $2\Rightarrow \mathrm{(D)}$.

See also

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