Difference between revisions of "2023 AMC 12A Problems/Problem 11"

Revision as of 19:20, 9 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

@@ Line 1: / Line 1: @@
-did you really think the leaked problems would be here?
+==Problem==
-if we do this for every problem be like
+What is the degree measure of the acute angle formed by lines with slopes <math>2</math> and <math>\frac{1}{3}</math>?
+<math>\textbf{(A)} ~30\qquad\textbf{(B)} ~37.5\qquad\textbf{(C)} ~45\qquad\textbf{(D)} ~52.5\qquad\textbf{(E)} ~60</math>
+==Solution 1==
+Remind that <math>\text{slope}=\dfrac{\Delta y}{\Delta x}=\tan \theta</math> where <math>\theta</math> is the angle between the slope and <math>x</math>-axis. <math>k_1=2=\tan \alpha</math>, <math>k_2=\dfrac{1}{3}=\tan \beta</math>. The angle formed by the two lines is <math>\alpha-\beta</math>. <math>\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\dfrac{2-1/3}{1+2\cdot 1/3}=1</math>. Therefore, <math>\alpha-\beta=\boxed{\textbf{(C)} 45^\circ}</math>.
+~plasta

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Difference between revisions of "2023 AMC 12A Problems/Problem 11"

Revision as of 19:20, 9 November 2023

Problem

Solution 1