2007 AMC 10B 第 4 题

Question

点

O

是

\triangle ABC

外接圆的圆心，如图，

\angle BOC = 120^\circ

，且

\angle AOB = 140^\circ

。

\angle ABC

的度数是多少？

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?

35

40

45

50

60

所有半径相等，所以三角形

AOB, BOC, COA

都是等腰三角形。

Each radius is equal, so each of the triangles

AOB, BOC, COA

is isosceles

\angle ABC=\angle ABO+\angle OBC

，两部分都是等腰三角形的底角。

\angle ABC=\angle ABO+\angle OBC,

where each part is a base angle of an isosceles triangle

因为

OA=OB=OC

，三角形

AOB, BOC

和

COA

都是等腰三角形。底角给出

\angle ABO=\dfrac{180^\circ-140^\circ}{2}=20^\circ

，

\angle OBC=\dfrac{180^\circ-120^\circ}{2}=30^\circ

。

因此

\angle ABC=20^\circ+30^\circ=50^\circ

。

所以正确答案是 D。

Since

OA=OB=OC,

triangles

AOB, BOC,

and

COA

are isosceles. The base angles give

\angle ABO=\dfrac{180^\circ-140^\circ}{2}=20^\circ

and

\angle OBC=\dfrac{180^\circ-120^\circ}{2}=30^\circ.

Therefore

\angle ABC=20^\circ+30^\circ=50^\circ.

Thus, the correct answer is D.

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