Difference between revisions of "2016 AMC 12B Problems/Problem 4"
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<math>\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270</math> | <math>\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270</math> | ||
| − | ==Solution== | + | ==Solution 1== |
| − | We set up equations to find each angle. The larger angle will be represented as <math>x</math> and the | + | By: dragonfly |
| + | |||
| + | We set up equations to find each angle. The larger angle will be represented as <math>x</math> and the smaller angle will be represented as <math>y</math>, in degrees. This implies that | ||
<math>4x=5y</math> | <math>4x=5y</math> | ||
| Line 17: | Line 19: | ||
We then find that <math>x=75</math> and <math>y=60</math>, and their sum is <math>\boxed{\textbf{(C)}\ 135}</math> | We then find that <math>x=75</math> and <math>y=60</math>, and their sum is <math>\boxed{\textbf{(C)}\ 135}</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | We can visualize the problem like so: | ||
| + | <asy> | ||
| + | path b = brace((0,10),(90,10),5); | ||
| + | draw(b); | ||
| + | label("$90^\circ$",b,N); | ||
| + | |||
| + | draw("$5x$",(0,0)--(75,0),N); | ||
| + | draw((75,2.5)--(75,-2.5)); | ||
| + | draw("$1y$",(75,0)--(90,0),N); | ||
| + | |||
| + | draw("$4x$",(0,-10)--(60,-10),S); | ||
| + | draw((60,-7.5)--(60,-12.5)); | ||
| + | draw("$2y$",(60,-10)--(90,-10),S); | ||
| + | |||
| + | draw((0,5)--(0,-15)); | ||
| + | draw((90,5)--(90,-15)); | ||
| + | </asy> | ||
| + | <cmath>5x+1y = 90^\circ = 4x+2y</cmath> | ||
| + | Moving like terms to the same side gets <math>x = y</math>, and substituting this back gets <math>6x = 90^\circ \implies x = \frac{90^\circ}{6} = 15^\circ</math>, so the sum of the degree measures is <math>5x + 4x = 9x = 9(15) = \boxed{\textbf{(C)}\ 135}</math>. ~[[User:emerald_block|emerald_block]] | ||
==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}} | ||
Latest revision as of 19:49, 6 November 2021
Contents
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 1
By: dragonfly
We set up equations to find each angle. The larger angle will be represented as 
and the smaller angle will be 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 
Solution 2
We can visualize the problem like so:
![[asy] path b = brace((0,10),(90,10),5); draw(b); label("$90^\circ$",b,N); draw("$5x$",(0,0)--(75,0),N); draw((75,2.5)--(75,-2.5)); draw("$1y$",(75,0)--(90,0),N); draw("$4x$",(0,-10)--(60,-10),S); draw((60,-7.5)--(60,-12.5)); draw("$2y$",(60,-10)--(90,-10),S); draw((0,5)--(0,-15)); draw((90,5)--(90,-15)); [/asy]](http://latex.artofproblemsolving.com/3/c/3/3c363938cdab3ccee9d3ec538a6815acf71fd045.png)
![\[5x+1y = 90^\circ = 4x+2y\]](http://latex.artofproblemsolving.com/0/1/9/019e74a0011e4998cf849f4168982b02d5f39a7f.png)
Moving like terms to the same side gets 
, and substituting this back gets 
, so the sum of the degree measures is 
. ~emerald_block
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. 
