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

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<cmath>\alpha+\beta=\frac{\pi}{2}-\alpha</cmath>
 
<cmath>\alpha+\beta=\frac{\pi}{2}-\alpha</cmath>
 
<math></math>\beta=\fbox{(C) <math>\frac{\pi}{2} -2\alpha</math>}<math></math>
 
<math></math>\beta=\fbox{(C) <math>\frac{\pi}{2} -2\alpha</math>}<math></math>
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~lptoggled
  
 
==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 18:47, 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) ~lptoggled

See also

2024 AMC 12A (Problems • Answer Key • Resources)
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

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