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

(Solution 2: Trial and Error)
(Solution 1)
Line 11: Line 11:
 
<cmath>\alpha+\beta=\tan^{-1}(\frac{4}{3})</cmath>
 
<cmath>\alpha+\beta=\tan^{-1}(\frac{4}{3})</cmath>
 
<cmath>\alpha+\beta=\frac{\pi}{2}-\alpha</cmath>
 
<cmath>\alpha+\beta=\frac{\pi}{2}-\alpha</cmath>
<cmath>\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}</cmath>
+
<cmath>\beta=\fbox{(C) \frac{\pi}{2} -2\alpha}</cmath>
 
~lptoggled
 
~lptoggled
  

Revision as of 21:42, 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 the question, \[\tan\alpha=\frac{3}{4}, \space \tan\beta=\frac{7}{24}\] \[\tan(\alpha+\beta)= \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\] \[\tan(\alpha+\beta)= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}}\] \[\tan(\alpha+\beta)=\frac{4}{3}\] \[\alpha+\beta=\tan^{-1}(\frac{4}{3})\] \[\alpha+\beta=\frac{\pi}{2}-\alpha\]

\[\beta=\fbox{(C) \frac{\pi}{2} -2\alpha}\] (Error compiling LaTeX. Unknown error_msg)

~lptoggled

Solution 2: Trial and Error

Another approach to solving this problem is trial and error, comparing the sine of the answer choices with $\sin\beta = \frac{7}{25}$. Starting with the easiest sine to compute from the answer choices (option choice D). We get: \[\sin{(\frac{\alpha}{2})} = \sqrt{\frac{1 - \cos{\alpha}}{2}}\] \[= \sqrt{\frac{1 - \frac{4}{5}}{2}}\] \[= \sqrt{\frac{1}{10}}\] \[\neq \frac{7}{25}\]

The next easiest sine to compute is option choice C. \[\sin{(\frac{\pi}{2} - 2\alpha)} = \sin{(\frac{\pi}{2})}\cos{(2\alpha)}\] \[=\cos{2\alpha}\] \[=\cos^2{\alpha} - \sin^2{\alpha}\] \[=\frac{16}{25} - \frac{9}{25}\] \[=\frac{7}{25}\]

Since $\sin(\frac{\pi}{2} - 2\alpha)$ is equal to $\sin\beta$, option choice C is the correct answer. ~amshah

Solution 3:

sin(2B) = 24 /25 = 2 * 12 / 25 = 2 * 3 / 5 * 4 / 5 = 2 * sin(A) * cos(A) = Sin(2A) = Cos( 90 - 2A)

choice C is the correct answer ~luckuso

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