Difference between revisions of "2023 AMC 12A Problems/Problem 11"
Arcticturn (talk | contribs) (→Solution 6) |
Arcticturn (talk | contribs) (→Solution 6) |
||
Line 74: | Line 74: | ||
==Solution 6 == | ==Solution 6 == | ||
− | The lines <math>y = 2x, y = \frac {1}{3}x</math>, and <math>x = 3</math> form a large right triangle and a small right triangle. Call the angle that is formed by the x-axis and the line <math>y = 2x </math> <math> \alpha</math>, and call the angle that is formed by the x-axis and the line <math>y = \frac {1}{3}x</math> <math>\beta</math>. We try to find <math>sin (\alpha - \beta)</math> first, and then try to see if any of the answer choices match up. <math>sin (\alpha - \beta)</math> = | + | The lines <math>y = 2x, y = \frac {1}{3}x</math>, and <math>x = 3</math> form a large right triangle and a small right triangle. Call the angle that is formed by the x-axis and the line <math>y = 2x </math> <math> \alpha</math>, and call the angle that is formed by the x-axis and the line <math>y = \frac {1}{3}x</math> <math>\beta</math>. We try to find <math>sin (\alpha - \beta)</math> first, and then try to see if any of the answer choices match up. <math>sin (\alpha - \beta)</math> = <math>sin \alpha</math> <math>cos \beta</math> - <math>sin \beta</math> <math>cos \alpha</math>. Using soh-cah-toa, we find that <math>sin \alpha = \frac {2}{\sqrt 5}, sin \beta = \frac {1}{\sqrt 10}, cos \alpha = \frac {1}{\sqrt 5}, </math> and <math>cos \beta = \frac {3}{\sqrt 10}</math>. Plugging it all in, we find that <math>sin (\alpha - \beta) = \frac {5}{\sqrt {50}}</math>, which is equivalent to <math>\frac {\sqrt 2}{2}</math>. Since <math>sin (\alpha - \beta) = \frac {\sqrt 2}{2}</math>, we get that <math>\alpha - \beta = 45^{\circ}</math>. Therefore, the answer is <math>\boxed {\textbf {(C)} 45^{\circ}}</math>. |
~Arcticturn | ~Arcticturn |
Revision as of 19:32, 11 November 2023
Contents
[hide]Problem
What is the degree measure of the acute angle formed by lines with slopes and ?
Solution 1
Remind that where is the angle between the slope and -axis. , . The angle formed by the two lines is . . Therefore, .
~plasta
Solution 2
We can take any two lines of this form, since the angle between them will always be the same. Let's take for the line with slope of 2 and for the line with slope of 1/3. Let's take 3 lattice points and create a triangle. Let's use , , and . The distance between the origin and is . The distance between the origin and is . The distance between and is . We notice that we have a triangle with 3 side lengths: , , and . This forms a 45-45-90 triangle, meaning that the angle is .
~lprado
Solution 3 (Law of Cosines)
Follow Solution 2 up until the lattice points section. Let's use , , and . The distance between the origin and is . The distance between the origin and is . The distance between and is . Using the Law of Cosines, we see the , where is the angle we are looking for.
Simplifying, we get .
.
.
.
Thus,
~Failure.net
Solution 4 (Vector Bash)
We can set up vectors and to represent the two lines. We know that . Plugging the vectors in gives us . From this we get that .
~middletonkids
Solution 5 (Complex Numbers)
Let and
From this we have:
To solve this we must compute
Using elimination we have:
~luckuso
Solution 6
The lines , and form a large right triangle and a small right triangle. Call the angle that is formed by the x-axis and the line , and call the angle that is formed by the x-axis and the line . We try to find first, and then try to see if any of the answer choices match up. = - . Using soh-cah-toa, we find that and . Plugging it all in, we find that , which is equivalent to . Since , we get that . Therefore, the answer is .
~Arcticturn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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.