2012 AMC 10A 第 4 题

Question

已知

\angle ABC = 24^\circ

，

\angle ABD = 20^\circ

。

\angle CBD

的最小可能度数是多少？

Let

\angle ABC = 24^\circ

and

\angle ABD = 20^\circ .

What is the smallest possible degree measure for

\angle CBD?

0

2

4

6

12

两个角共用射线

BA

。

Both angles share ray

BA

当两个给定角尽可能重叠时，剩下的角最小。

The smallest angle occurs when the two given angles overlap as much as possible

注意两个角都共用射线

AB

。为了使所求角最小，让

D

位于

A

和

C

之间。

这样就有

\begin{aligned} \angle CBD &= \angle ABC - \angle ABD \\ &= 4^{\circ} \end{aligned}\text{。}

所以正确答案是 C。

Note that both of the angles share the ray

AB.

To minimize the desired degree, we want

D

to be between

A

and

C.

This would make

\begin{aligned} \angle CBD &= \angle ABC - \angle ABD \\ &= 4^{\circ}. \end{aligned}

Thus, C is the correct answer.

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