AMC10 2023 A

AMC10 2023 A · Q13

AMC10 2023 A · Q13. It mainly tests Triangles (properties), Trigonometry (basic).

Abdul and Chiang are standing $48$ feet apart in a field. Bharat is standing in the same field as far from Abdul as possible so that the angle formed by his lines of sight to Abdul and Chiang measures $60^\circ$. What is the square of the distance (in feet) between Abdul and Bharat?

Abdul和Chiang在田野中相距$48$英尺。Bharat站在同一田野中，尽可能远离Abdul，同时他看向Abdul和Chiang的视线形成的夹角为$60^\circ$。Abdul和Bharat之间的距离平方（英尺）是多少？

(A) 1728 1728

(B) 2601 2601

(D) 4608 4608

(E) 6912 6912

Answer

Correct choice: (C)

正确答案：（C）

Solution

Error creating thumbnail: Unable to save thumbnail to destination Let $\theta=\angle ACB$ and $x=\overline{AB}$. According to the Law of Sines, we know that $\dfrac{\sin\theta}x=\dfrac{\sin60^\circ}{48}=\dfrac{\sqrt3}{96}$. Rearranging, we get that $x=\dfrac{\sin\theta}{\frac{\sqrt3}{96}}=32\sqrt3\sin\theta$ where $x$ is a function of $\theta$. We want to maximize $x$. We know that the maximum value of $\sin\theta=1$, so this yields $x=32\sqrt3\implies x^2=\boxed{\textbf{(C) }3072.}$ A quick check verifies that $\theta=90^\circ$ indeed works.

设$\theta=\angle ACB$，$x=\overline{AB}$。根据正弦定律，$\dfrac{\sin\theta}x=\dfrac{\sin60^\circ}{48}=\dfrac{\sqrt3}{96}$。重新整理，$x=\dfrac{\sin\theta}{\frac{\sqrt3}{96}}=32\sqrt3\sin\theta$，其中$x$是$\theta$的函数。我们要最大化$x$。 $\sin\theta$的最大值为1，因此$x=32\sqrt3\implies x^2=\boxed{\textbf{(C) }3072}$。快速验证$\theta=90^\circ$确实可行。

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