AMC12 2023 A

AMC12 2023 A · Q15

AMC12 2023 A · Q15. It mainly tests Coordinate geometry, Trigonometry (basic).

Usain is walking for exercise by zigzagging across a $100$-meter by $30$-meter rectangular field, beginning at point $A$ and ending on the segment $\overline{BC}$. He wants to increase the distance walked by zigzagging as shown in the figure below $(APQRS)$. What angle $\theta = \angle PAB=\angle QPC=\angle RQB=\cdots$ will produce a length that is $120$ meters? (This figure is not drawn to scale. Do not assume that he zigzag path has exactly four segments as shown; there could be more or fewer.)

Usain 通过在 $100$ 米长 $30$ 米宽的矩形场地内之字形行走锻炼，从点 $A$ 开始，结束在线段 $\overline{BC}$ 上。他想通过如图所示（$APQRS$）之字形路径增加行走距离。什么角度 $\theta = \angle PAB=\angle QPC=\angle RQB=\cdots$ 会使路径长度为 $120$ 米？（图未按比例绘制，不要假设之字形路径恰好有四段；可能更多或更少。）

(A) \arccos\frac{5}{6} \arccos\frac{5}{6}

(B) \arccos\frac{4}{5} \arccos\frac{4}{5}

(D) \arcsin\frac{4}{5} \arcsin\frac{4}{5}

(E) \arcsin\frac{5}{6} \arcsin\frac{5}{6}

Answer

Correct choice: (A)

正确答案：（A）

Solution

By "unfolding" $APQRS$ into a straight line, we get a right angled triangle $ABS'$. It follows that \begin{align*} \cos(\theta)&=\frac{100}{120} \\ \theta&=\boxed{\textbf{(A) } \arccos\left(\frac{5}{6}\right)}. \end{align*}

通过将 $APQRS$ "展开"成直线，得到直角三角形 $ABS'$。因此 $\cos(\theta)=\frac{100}{120}$，$\theta=\boxed{\textbf{(A) } \arccos\left(\frac{5}{6}\right)}$。

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