Difference between revisions of "2016 AMC 10B Problems/Problem 7"
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Therefore, the angles are <math>75</math> and <math>60</math>, which sum to <math>\boxed{\textbf{(C)}\ 135}</math>. | Therefore, the angles are <math>75</math> and <math>60</math>, which sum to <math>\boxed{\textbf{(C)}\ 135}</math>. | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/x1rKDLjWNjg | ||
+ | |||
+ | ~savannahsolver | ||
==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 15:32, 7 July 2020
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
Substituting this value back into the first equation yields , leaving equal to .
(Solution by akaashp11)
Solution 2
We let the measures be and giving us the ratio of . We know since this inequality gives , which is true since the measures of angles are never negative. We also know the bigger complement is twice the smaller, so
Therefore, the angles are and , which sum to .
Video Solution
~savannahsolver
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 |
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