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Difference between revisions of "1994 AHSME Problems/Problem 9"

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Revision as of 20:26, 27 June 2014

Problem

If $\angle A$ is four times $\angle B$, and the complement of $\angle B$ is four times the complement of $\angle A$, then $\angle B=$

$\textbf{(A)}\ 10^{\circ} \qquad\textbf{(B)}\ 12^{\circ} \qquad\textbf{(C)}\ 15^{\circ} \qquad\textbf{(D)}\ 18^{\circ} \qquad\textbf{(E)}\ 22.5^{\circ}$

Solution

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