AMC10 2025 B

AMC10 2025 B · Q5

AMC10 2025 B · Q5. It mainly tests Triangles (properties), Circle theorems.

In $\triangle ABC$, $AB = 10$, $AC = 18$, and $\angle B = 130^\circ$. Let $O$ be the center of the circle containing points $A, B, C$. What is the degree measure of $\angle CAO$?

在$\triangle ABC$中，$AB = 10$，$AC = 18$，且$\angle B = 130^\circ$。设$O$为包含点$A, B, C$的圆的圆心。求$\angle CAO$的度数。

(A) \ 20^\circ \ 20^\circ

(B) \ 30^\circ \ 30^\circ

(D) \ 50^\circ \ 50^\circ

(E) \ 60^\circ \ 60^\circ

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let us extend $\overline{AO}$ to meet $\odot{O}$ at $D$. Since $m\angle{ABC}=130$°, by the Inscribed Angle Theorem, the measure of major arc $\widehat{AC}=260$°. $\overline{AD}$ is a diameter of $\odot{O}$, so the measure of top arc $\widehat{AD}=180$, so $m\widehat{CD}=260-180=80$° and $m\angle{CAO}=\boxed{\textbf{(C) }40\textbf{°}}$.

令$\overline{AO}$延长交圆$\odot{O}$于点$D$。因为$\angle ABC=130^\circ$，根据圆周角定理，弧$\widehat{AC}$的度数为$260^\circ$。$\overline{AD}$是$\odot{O}$的直径，因此弧$\widehat{AD}$的度数为$180^\circ$，所以弧$\widehat{CD} = 260^\circ - 180^\circ = 80^\circ$，故$\angle CAO = \boxed{\textbf{(C) }40^\circ}$。

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