Question

Alice 和 Bob 相距 $10$ 英里。某天，Alice 从家中向正北方看到一架飞机。与此同时，Bob 从家中向正西方看到同一架飞机。Alice 看到的仰角为 $30^\circ$ ，Bob 看到的仰角为 $60^\circ$ 。下列哪一项最接近飞机的高度（英里）？

Alice and Bob live $10$ miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is $30^\circ$ from Alice's position and $60^\circ$ from Bob's position. Which of the following is closest to the airplane's altitude, in miles?

$3.5$

$4$

$4.5$

$5$

$5.5$

答案：E

知识点：特殊直角三角形勾股定理立体几何

难度评级：1630

解答：

设飞机位于点 $C$ ，其正下方的地面点为 $D$ ，高度为 $h$ 。三角形 $ACD$ 和 $BCD$ 都是 $30$ - $60$ - $90$ 直角三角形，所以 $AD=\sqrt3\,h$ ， $BD=\dfrac{h}{\sqrt3}$ 。Alice 向北看而 Bob 向西看，因此 $\angle ADB=90^\circ$ ，从而 $AD^2+BD^2=AB^2=100$ 。于是 $3h^2+\dfrac{h^2}{3}=\dfrac{10h^2}{3}=100$ ，解得 $h=\sqrt{30}\approx5.48$ ，最接近 $5.5$ 。

所以正确答案是 E。

Let the airplane be at $C,$ directly above point $D$ on the ground at altitude $h.$ Triangles $ACD$ and $BCD$ are $30$ - $60$ - $90$ right triangles, so $AD=\sqrt3\,h$ and $BD=\dfrac{h}{\sqrt3}.$ Since Alice looks north and Bob looks west, $\angle ADB=90^\circ,$ so $AD^2+BD^2=AB^2=100.$ Then $3h^2+\dfrac{h^2}{3}=\dfrac{10h^2}{3}=100,$ giving $h=\sqrt{30}\approx5.48,$ closest to $5.5.$

Thus, the correct answer is E.

2016 AMC 12B 第 13 题

答案：E

难度评级：1630

解答：

其他年份的第 13 题