Difference between revisions of "2024 AMC 12A Problems/Problem 10"

(Created page with "==Problem== Let <math>\alpha</math> be the radian measure of the smallest angle in a <math>3{-}4{-}5</math> right triangle. Let <math>\beta</math> be the radian measure of the...")
 
(Problem)
Line 3: Line 3:
  
 
<math>\textbf{(A) }\frac{\alpha}{3}\qquad \textbf{(B) }\alpha - \frac{\pi}{8}\qquad \textbf{(C) }\frac{\pi}{2} - 2\alpha \qquad \textbf{(D) }\frac{\alpha}{2}\qquad \textbf{(E) }\pi - 4\alpha\qquad</math>
 
<math>\textbf{(A) }\frac{\alpha}{3}\qquad \textbf{(B) }\alpha - \frac{\pi}{8}\qquad \textbf{(C) }\frac{\pi}{2} - 2\alpha \qquad \textbf{(D) }\frac{\alpha}{2}\qquad \textbf{(E) }\pi - 4\alpha\qquad</math>
 +
==Solution 1==
 +
From question,
 +
<cmath>tan\alpha=\frac{3}{4}, \space tan\beta=\frac{7}{24}</cmath>
 +
<cmath>tan(\alpha+\beta)= \frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}</cmath>
 +
<cmath>tan(\alpha+\beta)= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}}</cmath>
 +
<cmath>tan(\alpha+\beta)=\frac{4}{3}</cmath>
 +
<cmath>\alpha+\beta=tan^{-1}(\frac{4}{3})</cmath>
 +
<cmath>\alpha+\beta=\frac{\pi}{2}-\alpha</cmath>
 +
<math></math>\beta=\fbox{(C) <math>\frac{\pi}{2} -2\alpha</math>}<math></math>
  
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=A|num-b=9|num-a=11}}
 
{{AMC12 box|year=2024|ab=A|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:08, 8 November 2024

Problem

Let $\alpha$ be the radian measure of the smallest angle in a $3{-}4{-}5$ right triangle. Let $\beta$ be the radian measure of the smallest angle in a $7{-}24{-}25$ right triangle. In terms of $\alpha$, what is $\beta$?

$\textbf{(A) }\frac{\alpha}{3}\qquad \textbf{(B) }\alpha - \frac{\pi}{8}\qquad \textbf{(C) }\frac{\pi}{2} - 2\alpha \qquad \textbf{(D) }\frac{\alpha}{2}\qquad \textbf{(E) }\pi - 4\alpha\qquad$

Solution 1

From question, \[tan\alpha=\frac{3}{4}, \space tan\beta=\frac{7}{24}\] \[tan(\alpha+\beta)= \frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}\] \[tan(\alpha+\beta)= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}}\] \[tan(\alpha+\beta)=\frac{4}{3}\] \[\alpha+\beta=tan^{-1}(\frac{4}{3})\] \[\alpha+\beta=\frac{\pi}{2}-\alpha\] $$ (Error compiling LaTeX. Unknown error_msg)\beta=\fbox{(C) $\frac{\pi}{2} -2\alpha$}$$ (Error compiling LaTeX. Unknown error_msg)

See also

2024 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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