Difference between revisions of "2016 AMC 10B Problems/Problem 7"
(→Solution) |
(→Solution) |
||
| Line 17: | Line 17: | ||
Substituting this <math>x</math> value back into the first equation yields <math>y</math> <math>=</math> <math>75</math>, leaving <math>x</math> <math>+</math> <math>y</math> equal to <math>\boxed{\textbf{(C)}\ 135}</math>. | Substituting this <math>x</math> value back into the first equation yields <math>y</math> <math>=</math> <math>75</math>, leaving <math>x</math> <math>+</math> <math>y</math> equal to <math>\boxed{\textbf{(C)}\ 135}</math>. | ||
| − | Solution by akaashp11 | + | (Solution by akaashp11) |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=B|num-b=6|num-a=8}} | {{AMC10 box|year=2016|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:17, 21 February 2016
Problem
The ratio of the measures of two acute angles is 
, 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?
Solution
We can set up a system of equations where 
and 
are the two acute angles. WLOG, assume that 
in order for the complement of to be greater than the complement of . Therefore, 
and 
. Solving for in the first equation and substituting into the second equation yields
![\[\begin{split} 90 - x & = 2 (90 - 1.25x) \\ 1.5x & = 90 \\ x & = 60 \end{split}\]](http://latex.artofproblemsolving.com/3/7/6/37605e261d79ab5f53fd1c8add55b71c9584dbfc.png)
Substituting this value back into the first equation yields 
, leaving 
equal to 
.
(Solution by akaashp11)
See Also
| 2016 AMC 10B (Problems • Answer Key • Resources) | ||
| 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. 
