AMC10 2007 B

AMC10 2007 B · Q4

AMC10 2007 B · Q4. It mainly tests Angle chasing, Circle theorems.

The point $O$ is the center of the circle circumscribed about $\triangle ABC$, with $\angle BOC = 120^\circ$ and $\angle AOB = 140^\circ$, as shown. What is the degree measure of $\angle ABC$?

点 $O$ 是 $\triangle ABC$ 的外接圆圆心，已知 $\angle BOC = 120^\circ$，$\angle AOB = 140^\circ$，如图所示。求 $\angle ABC$ 的度数？

(A) 35 35

(B) 40 40

(D) 50 50

(E) 60 60

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Since $OA=OB=OC$, triangles $AOB$, $BOC$, and $COA$ are all isosceles. Hence $$ \angle ABC=\angle ABO+\angle OBC=\frac{180^\circ-140^\circ}{2}+\frac{180^\circ-120^\circ}{2}=50^\circ. $$

答案（D）：由于 $OA=OB=OC$，三角形 $AOB$、$BOC$ 和 $COA$ 都是等腰三角形。因此 $$ \angle ABC=\angle ABO+\angle OBC=\frac{180^\circ-140^\circ}{2}+\frac{180^\circ-120^\circ}{2}=50^\circ。 $$

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