1995 AHSME Problems/Problem 18

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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 $2\Rightarrow \mathrm{(D)}$ .

1995 AHSME (Problems)
Preceded by Problem 17	Followed by Problem 19
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1995 AHSME Problems/Problem 18

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