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

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<math> \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} </math>
 
<math> \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} </math>
 
==Solution==
 
==Solution==
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Let <math>\angle A=x</math> and <math>\angle B=y</math>. From the first condition, we have <math>x=4y</math>. From the second condition, we have <cmath>90-y=4(90-x).</cmath> Substituting <math>x=4y</math> into the previous equation and solving yields <cmath>\begin{align*}90-y=4(90-4y)&\implies 90-y=360-16y\\&\implies 15y=270\\&\implies y=\boxed{\textbf{(D) }18^\circ.}\end{align*}</cmath>
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--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
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==See Also==
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{{AHSME box|year=1994|num-b=8|num-a=10}}
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{{MAA Notice}}

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

See Also

1994 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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