AMC12 2024 A

AMC12 2024 A · Q10

AMC12 2024 A · Q10. It mainly tests Manipulating equations, Triangles (properties).

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 $\beta$?

设$\alpha$是一个$3{-}4{-}5$直角三角形中最小角的弧度测度。设$\beta$是一个$7{-}24{-}25$直角三角形中最小角的弧度测度。用$\alpha$表示，$\beta$是多少？

(A) \frac{\alpha}{3} \frac{\alpha}{3}

(B) \alpha - \frac{\pi}{8} \alpha - \frac{\pi}{8}

(D) \frac{\alpha}{2} \frac{\alpha}{2}

(E) \pi - 4\alpha\qquad \pi - 4\alpha\qquad

Answer

Correct choice: (C)

正确答案：（C）

Solution

We are given that \[\tan\alpha=\frac{3}{4} \text{ and } \tan\beta=\frac{7}{24}.\] We have \begin{align*} \tan(\alpha+\beta) &= \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \\ &= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}} \\ &= \frac{4}{3}. \end{align*} It follows that \begin{align*} \alpha+\beta&=\tan^{-1}\left(\frac{4}{3}\right) \\ &=\frac{\pi}{2}-\tan^{-1}\left(\frac{3}{4}\right) \\ &=\frac{\pi}{2}-\alpha. \end{align*} Therefore, the answer is \[\beta=\boxed{\textbf{(C) }\frac{\pi}{2}-2\alpha}.\]

给定\[\tan\alpha=\frac{3}{4} \text{ and } \tan\beta=\frac{7}{24}.\] 我们有 \begin{align*} \tan(\alpha+\beta) &= \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \\ &= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}} \\ &= \frac{4}{3}. \end{align*} 由此 \begin{align*} \alpha+\beta&=\tan^{-1}\left(\frac{4}{3}\right) \\ &=\frac{\pi}{2}-\tan^{-1}\left(\frac{3}{4}\right) \\ &=\frac{\pi}{2}-\alpha. \end{align*} 因此，答案是 \[\beta=\boxed{\textbf{(C) }\frac{\pi}{2}-2\alpha}.\]

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