9. 如图,分别用α,β,γ标注∠BOC,∠BOE,∠COD.

答案:
解:标注如下:
∠BOC 标注为 α;
∠BOE 标注为 β;
∠COD 标注为 γ。
∠BOC 标注为 α;
∠BOE 标注为 β;
∠COD 标注为 γ。
10. 如图,用阴影部分表示∠AOP的内部.

答案:
11. 计算(结果用度、分、秒表示):
(1)25°36′+55°45′;
(2)134°14′-46°38′;
(3)28°18′×5;
(4)78°35′÷5.
(1)25°36′+55°45′;
(2)134°14′-46°38′;
(3)28°18′×5;
(4)78°35′÷5.
答案:
(1)解:25°36′+55°45′
=(25°+55°)+(36′+45′)
=80°+81′
=80°+1°21′
=81°21′
(2)解:134°14′-46°38′
=133°74′-46°38′
=(133°-46°)+(74′-38′)
=87°+36′
=87°36′
(3)解:28°18′×5
=28°×5+18′×5
=140°+90′
=140°+1°30′
=141°30′
(4)解:78°35′÷5
=78°÷5+35′÷5
=15°+3°÷5+7′
=15°+180′÷5+7′
=15°+36′+7′
=15°43′
(1)解:25°36′+55°45′
=(25°+55°)+(36′+45′)
=80°+81′
=80°+1°21′
=81°21′
(2)解:134°14′-46°38′
=133°74′-46°38′
=(133°-46°)+(74′-38′)
=87°+36′
=87°36′
(3)解:28°18′×5
=28°×5+18′×5
=140°+90′
=140°+1°30′
=141°30′
(4)解:78°35′÷5
=78°÷5+35′÷5
=15°+3°÷5+7′
=15°+180′÷5+7′
=15°+36′+7′
=15°43′
12. 如图,分别说出∠OAB,∠APQ,∠OPB的顶点和边.

答案:
解析:本题考查角的顶点和边的识别。
答案:∠OAB的顶点是点A,边是AO和AB;∠APQ的顶点是点P,边是PA和PQ;∠OPB的顶点是点P,边是PO和PB。
答案:∠OAB的顶点是点A,边是AO和AB;∠APQ的顶点是点P,边是PA和PQ;∠OPB的顶点是点P,边是PO和PB。
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