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AHSME 1995 Problem 18

    Two rays with common endpoint OO forms a 30∘30^\circ angle. Point AA lies on one ray, point BB on the other ray, and AB=1AB = 1. The maximum possible length of OBOB is (A) 1(B) 1+32(C) 3(D) 2(E) 43\mathrm{(A) \ 1 } \mathrm{(B) \ \frac {1 + \sqrt {3}}{\sqrt 2} } \mathrm{(C) \ \sqrt{3} } \mathrm{(D) \ 2 } \mathrm{(E) \ \frac{4}{\sqrt{3}} }

    Dataset text NuminaMath-1.5. Quoted with source.

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