1994 AHSME Problems/Problem 9: Difference between revisions

Latest revision as of 16:33, 9 January 2021

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

Let $\angle A=x$ and $\angle B=y$ . From the first condition, we have $x=4y$ . From the second condition, we have $90-y=4(90-x).$ Substituting $x=4y$ into the previous equation and solving yields $\begin{align*}90-y=4(90-4y)&\implies 90-y=360-16y\\&\implies 15y=270\\&\implies y=\boxed{\textbf{(D) }18^\circ.}\end{align*}$

--Solution by TheMaskedMagician

1994 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AHSME Problems and Solutions

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 --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
+==See Also==
+{{AHSME box|year=1994|num-b=8|num-a=10}}
+{{MAA Notice}}

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1994 AHSME Problems/Problem 9: Difference between revisions

Latest revision as of 16:33, 9 January 2021

Problem

Solution

See Also