2016 AMC 10B Problems/Problem 7

Revision as of 13:16, 21 February 2016 by Akaashp11 (talk | contribs) (Solution)

Problem

The ratio of the measures of two acute angles is $5:4$, and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?

$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$


Solution

We can set up a system of equations where $x$ and $y$ are the two acute angles. WLOG, assume that $x$ $<$ $y$ in order for the complement of $x$ to be greater than the complement of $y$. Therefore, $5x$ $=$ $4y$ and $90$ $-$ $x$ $=$ $2$ $(90$ $-$ $y)$. Solving for $y$ in the first equation and substituting into the second equation yields \[\begin{split} 90 - x & = 2 (90 - 1.25x) \\ 1.5x & = 90 \\ x & = 60 \end{split}\]

Substituting this $x$ value back into the first equation yields $y$ $=$ $75$, leaving $x$ $+$ $y$ equal to $\boxed{\textbf{(C)}\ 135}$.

Solution by akaashp11.

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png