1980 AMC 12 第 4 题

Question

在所附图形中，

CDE

是等边三角形，

ABCD

和

DEFG

是正方形。

\angle GDA

的度数是

In the adjoining figure,

CDE

is an equilateral triangle and

ABCD

and

DEFG

are squares. The measure of

\angle GDA

is

90^\circ

105^\circ

120^\circ

135^\circ

150^\circ

找出点

D

周围三个已知角

Identify the three known angles at

D

利用点

D

周围一周的角度为

360^\circ

Use the full

360^\circ

angle around point

D

在点

D

处，两个正方形的角各为

90^\circ

，等边三角形的角为

60^\circ

。因此

\begin{aligned} \angle GDA &=360^\circ-90^\circ\\ &\quad-60^\circ-90^\circ\\ &=120^\circ \end{aligned}\text{。}

因此，正确答案是 C。

At

D,

the two square angles are

90^\circ

each and the equilateral-triangle angle is

60^\circ.

Hence

\begin{aligned} \angle GDA &=360^\circ-90^\circ\\ &\quad-60^\circ-90^\circ\\ &=120^\circ. \end{aligned}

Therefore, the correct answer is C.

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