Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2016 AMC 10B Problems/Problem 7"

(→‎Solution 2)
(→‎Video Solution)
 
(2 intermediate revisions by 2 users not shown)
Line 6: Line 6:
  
  
−
==Solution==
+
==Solution 1==
  
 
We can set up a system of equations where <math>x</math> and <math>y</math> are the two acute angles. WLOG, assume that <math>x</math> <math><</math> <math>y</math> in order for the complement of <math>x</math> to be greater than the complement of <math>y</math>. Therefore, <math>5x</math> <math>=</math> <math>4y</math> and <math>90</math> <math>-</math> <math>x</math> <math>=</math> <math>2</math> <math>(90</math> <math>-</math> <math>y)</math>. Solving for <math>y</math> in the first equation and substituting into the second equation yields
 
We can set up a system of equations where <math>x</math> and <math>y</math> are the two acute angles. WLOG, assume that <math>x</math> <math><</math> <math>y</math> in order for the complement of <math>x</math> to be greater than the complement of <math>y</math>. Therefore, <math>5x</math> <math>=</math> <math>4y</math> and <math>90</math> <math>-</math> <math>x</math> <math>=</math> <math>2</math> <math>(90</math> <math>-</math> <math>y)</math>. Solving for <math>y</math> in the first equation and substituting into the second equation yields
Line 32: Line 32:
  
 
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>.
 +
 +
==Video Solution (CREATIVE THINKING)==
 +
https://youtu.be/4LsOG_HPYn8
 +
 +
~Education, the Study of Everything
 +
 +
 +
 +
 +
==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}}

Latest revision as of 11:46, 2 July 2023

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 1

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)

Solution 2

We let the measures be $5x$ and $4x$ giving us the ratio of $5:4$. We know $90-4x>90-5x$ since this inequality gives $x>0$, which is true since the measures of angles are never negative. We also know the bigger complement is twice the smaller, so

$90-4x=2(90-5x)$

$90-4x=180-10x$

$6x=90$

$x=15$

Therefore, the angles are $75$ and $60$, which sum to $\boxed{\textbf{(C)}\ 135}$.

Video Solution (CREATIVE THINKING)

https://youtu.be/4LsOG_HPYn8

~Education, the Study of Everything



Video Solution

https://youtu.be/x1rKDLjWNjg

~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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10B_Problems/Problem_7&oldid=195103"