AMC10 2002 B

AMC10 2002 B · Q24

AMC10 2002 B · Q24. It mainly tests Circle theorems, Trigonometry (basic).

Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius 20 feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point 10 vertical feet above the bottom?

摩天轮上的乘客在垂直平面内沿圆周运动。该摩天轮半径为 20 英尺，每分钟匀速转一圈。乘客从轮底到轮底上方 10 英尺垂直高度处需要多少秒？

(A) 5 5

(B) 6 6

(D) 10 10

(E) 15 15

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) In the figure, the center of the wheel is at $O$, and the rider travels from $A$ to $B$. Since $AC = 10$ and $OB = OA = 20$, the point $C$ is the midpoint of $\overline{OA}$. In the right $\triangle OCB$, we have $OC$ half of the length of the hypotenuse $OB$, so $m\angle COB = 60^\circ$. Since the wheel turns through an angle of $360^\circ$ in 60 seconds, the time required to turn through an angle of $60^\circ$ is $$ 60\left(\frac{60}{360}\right)=10 \text{ seconds.} $$

（D）在图中，车轮的圆心在 $O$，骑手从 $A$ 移动到 $B$。由于 $AC = 10$ 且 $OB = OA = 20$，点 $C$ 是线段 $\overline{OA}$ 的中点。在直角三角形 $\triangle OCB$ 中，$OC$ 是斜边 $OB$ 长度的一半，因此 $m\angle COB = 60^\circ$。由于车轮在 60 秒内转过 $360^\circ$，转过 $60^\circ$ 所需的时间为 $$ 60\left(\frac{60}{360}\right)=10 \text{秒。} $$

Topics