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== | ||
| + | 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] | ||
Revision as of 18:55, 28 June 2014
Problem
If 
is four times 
, and the complement of is four times the complement of , then 
Solution
Let 
and 
. From the first condition, we have 
. From the second condition, we have ![\[90-y=4(90-x).\]](http://latex.artofproblemsolving.com/7/6/7/767d755dce7d34e9b99fd13f186c18ff190ff5dd.png)
Substituting into the previous equation and solving yields 
--Solution by TheMaskedMagician
