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

Revision as of 17:55, 28 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

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

@@ Line 4: / Line 4: @@
 <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==
+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>
+--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]

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

Revision as of 17:55, 28 June 2014

Problem

Solution