AMC12 2016 B

AMC12 2016 B · Q13

AMC12 2016 B · Q13. It mainly tests Triangles (properties), Pythagorean theorem.

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?

爱丽丝和鲍勃相距 10 英里。一天，爱丽丝从家中向正北方向看，看到一架飞机。与此同时，鲍勃从家中向正西方向看，也看到了同一架飞机。从爱丽丝的位置看，飞机的仰角是 $30^\circ$；从鲍勃的位置看，飞机的仰角是 $60^\circ$。下列哪个选项最接近飞机的高度（单位：英里）？

(A) 3.5 3.5

(B) 4 4

(D) 5 5

(E) 5.5 5.5

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let Alice, Bob, and the airplane be located at points $A$, $B$, and $C$, respectively. Let $D$ be the point on the ground directly beneath the airplane, and let $h$ be the airplane’s altitude, in miles. Then $\triangle ACD$ and $\triangle BCD$ are $30$–$60$–$90^\circ$ right triangles with right angles at $D$, so $AD=\sqrt{3}h$ and $BD=\dfrac{h}{\sqrt{3}}$. Then by the Pythagorean Theorem applied to the right triangle on the ground, \[ 100=AB^2=AD^2+BD^2=(\sqrt{3}h)^2+\left(\dfrac{h}{\sqrt{3}}\right)^2=\dfrac{10h^2}{3}. \] Thus $h=\sqrt{30}$, and the closest of the given choices is $5.5$.

答案（E）：设 Alice、Bob 和飞机分别位于点 $A$、$B$、$C$。设 $D$ 为飞机正下方地面上的点，$h$ 为飞机的高度（单位：英里）。则 $\triangle ACD$ 与 $\triangle BCD$ 都是以 $D$ 为直角的 $30$–$60$–$90^\circ$ 直角三角形，所以 $AD=\sqrt{3}h$，$BD=\dfrac{h}{\sqrt{3}}$。对地面上的直角三角形应用勾股定理， \[ 100=AB^2=AD^2+BD^2=(\sqrt{3}h)^2+\left(\dfrac{h}{\sqrt{3}}\right)^2=\dfrac{10h^2}{3}. \] 因此 $h=\sqrt{30}$，在给出的选项中最接近的是 $5.5$。

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