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, $\theta = \boxed{\textbf{(C)} 45^\circ}$

~Failure.net

Solution 4 (Vector Bash)

We can set up vectors $\vec{a} = <1,2>$ and $\vec{b} = <3,1>$ to represent the two lines. We know that $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \cos \theta$ . Plugging the vectors in gives us $\cos \theta = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$ . From this we get that $\theta = \boxed{\textbf{(C)} 45^\circ}$ .

~middletonkids

Solution 5 (Complex Numbers)

Let $Z_1 = 3 + i$ and $Z_2 = 1 + 2i$ $\begin{align*} Z_2 &= Z_1 \cdot re^{i\theta} \\ 1+2i&=(3+i) \cdot re^{i\theta} \\ 1+2i&=(3 + i) \cdot r(\cos\theta + i\sin\theta) \\ 1+2i&=3r\cos\theta - r\sin\theta + 3ri\sin\theta + ri\cos\theta \\ \end{align*}$

From this we have: $\begin{align} 1 &= 3r\cos\theta - r\sin\theta \\ 2 &= r\cos\theta + 3r\sin\theta \end{align}$

To solve this we must compute $r$ $\begin{align*} r &= \frac{|Z_2|}{|Z_1|} \\ &= \frac{\sqrt{5}}{\sqrt{10}} \\ &= \frac{\sqrt{2}}{2} \end{align*}$

Using elimination we have: $3\cdot(2) - (1)$ $\begin{align*} 5 &= 10r\sin\theta \\ \frac{1}{2r} &= \sin\theta \\ \frac{1}{2\frac{\sqrt{2}}{2}} &= \sin\theta \\ \frac{\sqrt{2}}{2} &= \sin\theta \\ \theta &= \boxed{\textbf{(C)} 45^\circ} \\ \end{align*}$

~luckuso

Solution 5.b (Complex Numbers - Simpler)

Let $Z_1 = 3 + i$ and $Z_2 = 1 + 2i$ $\begin{align*} Z_2 &= Z_1 \cdot re^{i\theta} \\ 1+2i &=(3+i) \cdot re^{i\theta} \\ \frac{1+2i}{3+i} &= re^{i\theta} \\ \frac{(1+2i)(3-i)}{(3+i)(3-i)} &= re^{i\theta} \\ \frac{3+2+6i-i}{(3+i)(3-i)} &= re^{i\theta} \\ tan(\theta) &= 5/5 \\ \theta &= \boxed{\textbf{(C)} 45^\circ} \\ \end{align*}$

~luckuso

Solution 6

The lines $y = 2x, y = \frac {1}{3}x$ , and $x = 3$ form a large right triangle and a small right triangle. Call the angle that is formed by the x-axis and the line $y = 2x$ $\alpha$ , and call the angle that is formed by the x-axis and the line $y = \frac {1}{3}x$ $\beta$ . We try to find $\sin (\alpha - \beta)$ first, and then try to see if any of the answer choices match up.

$\sin (\alpha - \beta)$ = $\sin \alpha$ $\cos \beta$ - $\sin \beta$ $\cos \alpha$ .

Using soh-cah-toa, we find that $\sin \alpha = \frac {2}{\sqrt 5}, \sin \beta = \frac {1}{\sqrt 10}, \cos \alpha = \frac {1}{\sqrt 5},$ and $\cos \beta = \frac {3}{\sqrt 10}$ .

Plugging it all in, we find that $\sin (\alpha - \beta) = \frac {5}{\sqrt {50}}$ , which is equivalent to $\frac {\sqrt 2}{2}$ . Since $\sin (\alpha - \beta) = \frac {\sqrt 2}{2}$ , we get that $\alpha - \beta = 45^{\circ}$ . Therefore, the answer is $\boxed {\textbf {(C)} 45^{\circ}}$ .

~Arcticturn

=Solution 7 (Cheese)

AMC 12 does not allow graph paper or protractor. https://amc-reg.maa.org/manual/AMC1012B.pdf

Using a makeshift ruler, draw an accurate to-scale diagram. You can do this by simply drawing the two lines such that they intersect at the origin. Then, measure the angle by eye or by folding paper to observe that they form a 45 degree angle. The answer is $\boxed{45^\circ}$ .

~InstallHelp_Hex

Video Solution (easy to digest) by Power Solve

https://youtu.be/YXIH3UbLqK8?si=Uh4K33tNEzBOXc_h&t=1406

Video Solution (Under 4 minutes)

https://youtu.be/PWBEUXTtUxI

~Education, the Study of Everything

Video Solution

https://youtu.be/kPcsTZpFzTY

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by Math4All999 (pretty easy)

https://youtu.be/sa2HHgMZjSg?feature=shared

2023 AMC 12A (Problems • Answer Key • Resources)
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