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2024 AMC 12A Problems/Problem 10

Revision as of 19:57, 8 November 2024 by Dondee123 (talk | contribs) (→‎Solution 1)

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, \[\tan\alpha=\frac{3}{4}, \space \tan\beta=\frac{7}{24}\] \[\tan(\alpha+\beta)= \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\] \[\tan(\alpha+\beta)= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}}\] \[\tan(\alpha+\beta)=\frac{4}{3}\] \[\alpha+\beta=\tan^{-1}(\frac{4}{3})\] \[\alpha+\beta=\frac{\pi}{2}-\alpha\] \[\beta=\boxed{(C) \frac{\pi}{2} -2\alpha}\] ~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

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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