1994 AHSME Problems/Problem 9: Difference between revisions
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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> | |||
--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}} | |||
Latest revision as of 16:33, 9 January 2021
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
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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