Difference between revisions of "2016 AMC 12B Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | We | + | We set up equations to find each angle. The larger angle will be represented as <math>x</math> and the larger angle will we represented as <math>y</math>, in degrees. This implies that |
<math>4x=5y</math> | <math>4x=5y</math> | ||
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since the larger the original angle, the smaller the complement. | since the larger the original angle, the smaller the complement. | ||
| − | We then find that <math>x=75</math> and <math>y=60</math>, | + | |
| + | We then find that <math>x=75</math> and <math>y=60</math>, and their sum is <math>\boxed{\textbf{(C)}\ 135}</math> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=3|num-a=5}} | {{AMC12 box|year=2016|ab=B|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:10, 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 set up equations to find each angle. The larger angle will be represented as 
and the larger angle will we represented as 
, in degrees. This implies that
and
since the larger the original angle, the smaller the complement.
We then find that 
and 
, and their sum is 
See Also
| 2016 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
