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.