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 "2023 AMC 12A Problems/Problem 11"
(→‎Solution 3 (Law of Cosines))
Line 21: Line 21:
  
 
Simplifying, we get <math>-60 = -2\times(\sqrt{20}) \times(\sqrt{90}) \times\cos(\theta)</math>.
 
Simplifying, we get <math>-60 = -2\times(\sqrt{20}) \times(\sqrt{90}) \times\cos(\theta)</math>.
−
<math>30 =  sqrt{1800} \times\cos(\theta)</math>.
+
 
 +
<math>30 =  \sqrt{1800} \times\cos(\theta)</math>.
  
 
<math>30 =  30\sqrt{2} \times\cos(\theta)</math>.
 
<math>30 =  30\sqrt{2} \times\cos(\theta)</math>.
Line 27: Line 28:
 
<math>\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}= \cos(\theta)</math>.
 
<math>\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}= \cos(\theta)</math>.
  
−
Thus, <math>\theta = \boxed{\textbf{C} 45}</math>
+
Thus, <math>\theta = \boxed{45}</math>
 
~Failure.net
 
~Failure.net
 +
 
==See also==
 
==See also==
 
{{AMC12 box|year=2023|ab=A|num-b=10|num-a=12}}
 
{{AMC12 box|year=2023|ab=A|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:13, 10 November 2023

Problem

What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$?

$\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60$

Solution 1

Remind that $\text{slope}=\dfrac{\Delta y}{\Delta x}=\tan \theta$ where $\theta$ is the angle between the slope and $x$-axis. $k_1=2=\tan \alpha$, $k_2=\dfrac{1}{3}=\tan \beta$. The angle formed by the two lines is $\alpha-\beta$. $\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\dfrac{2-1/3}{1+2\cdot 1/3}=1$. Therefore, $\alpha-\beta=\boxed{\textbf{(C)} 45^\circ}$.

~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 $y=2x$ for the line with slope of 2 and $y=\frac{1}{3}x$ for the line with slope of 1/3. Let's take 3 lattice points and create a triangle. Let's use $(0,0)$, $(1,2)$, and $(3,1)$. The distance between the origin and $(1,2)$ is $\sqrt{5}$. The distance between the origin and $(3,1)$ is $\sqrt{10}$. The distance between $(1,2)$ and $(3,1)$ is $\sqrt{5}$. We notice that we have a triangle with 3 side lengths: $\sqrt{5}$, $\sqrt{5}$, and $\sqrt{10}$. This forms a 45-45-90 triangle, meaning that the angle is $\boxed{45^\circ}$.

~lprado

Solution 3 (Law of Cosines)

Follow Solution 2 up until the lattice points section. Let's use $(0,0)$, $(2,4)$, and $(9,3)$. The distance between the origin and $(2,4)$ is $\sqrt{20}$. The distance between the origin and $(9,3)$ is $\sqrt{90}$. The distance between $(2,4)$ and $(9,3)$ is $\sqrt{50}$. Using the Law of Cosines, we see the $50 = 90 + 20 - 2\times\sqrt{20}$ $\times\sqrt{90}$ $\times\cos(\theta)$, where $\theta$ is the angle we are looking for.

Simplifying, we get $-60 = -2\times(\sqrt{20}) \times(\sqrt{90}) \times\cos(\theta)$.

$30 = \sqrt{1800} \times\cos(\theta)$.

$30 = 30\sqrt{2} \times\cos(\theta)$.

$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}= \cos(\theta)$.

Thus, $\theta = \boxed{45}$ ~Failure.net

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_11&oldid=201709"