Question

一艘船从 $A$ 到 $B$ ，沿直线航行 $10$ 英里，转过一个介于 $45^\circ$ 和 $60^\circ$ ，之间的角，然后再航行 $20$ 英里到达 $C$ 。令 $AC$ 以英里为单位。下列哪个区间包含 $AC^2$ ？

A ship sails $10$ miles in a straight line from $A$ to $B,$ turns through an angle between $45^\circ$ and $60^\circ,$ and then sails another $20$ miles to $C.$ Let $AC$ be measured in miles. Which of the following intervals contains $AC^2?$

$[400, 500]$

$[500, 600]$

$[600, 700]$

$[700, 800]$

$[800, 900]$

答案：D

知识点：余弦定理极限情形界定

难度评级：1770

解答：

由余弦定理， $\begin{aligned} AC^2 &= 10^2 + 20^2 \\ &\quad {}- 2\cdot 10\cdot 20\cos(\angle ABC) \\ &= 500 - 400\cos(\angle ABC). \end{aligned}$

船转过的角介于 $45^\circ$ 和 $60^\circ$ ，所以内角 $\angle ABC$ 介于 $120^\circ$ 和 $135^\circ$ 。

因为 $\cos 120^\circ = -\dfrac{1}{2}$ 且 $\cos 135^\circ = -\dfrac{\sqrt{2}}{2}$ ， $\begin{aligned} 700 &= 500 + 200 \\ &\le AC^2 \le 500 + 200\sqrt{2} \\ &\lt 800. \end{aligned}$

所以 $AC^2$ 位于 $[700, 800]$ 中。

因此，正确答案是 D。

By the Law of Cosines, $\begin{aligned} AC^2 &= 10^2 + 20^2 \\ &\quad {}- 2\cdot 10\cdot 20\cos(\angle ABC) \\ &= 500 - 400\cos(\angle ABC). \end{aligned}$

The ship turns through an angle between $45^\circ$ and $60^\circ,$ so the interior angle $\angle ABC$ lies between $120^\circ$ and $135^\circ.$

Since $\cos 120^\circ = -\dfrac{1}{2}$ and $\cos 135^\circ = -\dfrac{\sqrt{2}}{2},$ $\begin{aligned} 700 &= 500 + 200 \\ &\le AC^2 \le 500 + 200\sqrt{2} \\ &\lt 800. \end{aligned}$

So $AC^2$ lies in $[700, 800].$

Thus, the correct answer is D.

2009 AMC 12A 第 13 题

答案：D

难度评级：1770

解答：

其他年份的第 13 题