6.3.2　角的比较与运算

.3.2角的比较与运算第1课时角的比较与和差运算课后·知能演练一、基础巩固1.计算72°22'+50°40'30″的结果是()A.123°62'30″ B.123°2'30″C.122°2'30″ D.123°12'30″2.计算52°10'20″-37°49'40″的结果为()A.14°20'40″ B.14°40'20″C.13°20'40″ D.15°20'40″3.小正方形网格如图所示,点A,B,C,D,O均为格点,那么∠AOB________∠COD.(填“>”“<”或“=”)

4.按图填空.(1)∠AOC-∠AOB=________;

(2)∠BOC+∠COD=________;

(3)∠AOE-∠BOE=________;

(4)∠BOE-________=∠BOD.

5.计算:(1)153°29'42″+26°40'32″;(2)132°25'-55°43'20″;(3)35°12'-34°48'+60°30'.二、能力提升6.如图所示,在正方形网格中,∠ABC________∠DEF(填“>”“<”或“=”).

7.计算7.56°-7°5'6″的值是________.

三、思维拓展8.已知∠A=24.1°+6°,∠B=56°-26°30',∠C=18°12'+11.8°,试通过计算,比较∠A,∠B和∠C的大小.

第2课时角的等分课后·知能演练一、基础巩固1.计算15°23'×4的结果是()A.61°92' B.60.92°C.60°32' D.61°32'2.如图,在直线AD上任取一点O,过点O作射线OB,OE平分∠DOB,OC平分∠AOB,∠BOC=26°时,∠BOE的度数为()A.61° B.62° C.63° D.64°3.已知∠AOB=90°36',射线OC在∠AOB内部.(1)若OC是∠AOB的平分线,则∠COA的大小为________;

(2)若∠COA=2∠COB,则∠COA的大小为________.

4.计算:(1)18°23'×6;(2)34°25'×4+35°42';(3)21°36'20″÷5.二、能力提升5.如图,已知P是直角三角尺ABC斜边AB上的一个动点,CD,CE分别是∠ACP和∠BCP的平分线,试探究:当点P在斜边AB上移动时,∠DCE的大小是否会发生变化?请说明理由.三、思维拓展6.已知∠ABC=50°,射线BD在射线BA的上方,BE平分∠CBD.(1)如图1,当射线BE在射线BA的下方,且∠ABD=30°时,求∠ABE的度数.请补全下面的解答过程;解:因为∠ABC=50°,∠ABD=30°,所以∠CBD=∠________+∠ABD=________°.

因为BE平分∠CBD,所以∠DBE=________∠CBD=________°,

所以∠ABE=∠DBE-∠________=________°.

(2)如图2,当射线BE在射线BA的上方,且∠ABE=15°时,求∠ABD的度数.【课后·知能演练】1.B解析:72°22'+50°40'30″=122°62'30″=123°2'30″.2.A解析:52°10'20″-37°49'40″=51°69'80″-37°49'40″=14°20'40″.3.>解析:如图,取点E,连接OE,根据网格可知,∠AOB=∠EOD.因为∠EOD>∠COD,所以∠AOB>∠COD.4.(1)∠BOC(2)∠BOD(3)∠AOB(4)∠DOE5.解:(1)153°29'42″+26°40'32″=179°69'74″=180°10'14″.(2)132°25'-55°43'20″=131°84'60″-55°43'20″=76°41'40″.(3)35°12'-34°48'+60°30'=34°72'-34°48'+60°30'=24'+60°30'=60°54'.6.>解析:因为∠ABC=45°,∠DEF<45°,所以∠ABC>∠DEF.7.28'30″解析:因为7.56°=7°+0.56°=7°+0.56×60'=7°+33.6'=7°+33'+0.6'=7°+33'+0.6×60″=7°+33'+36″=7°33'36″,所以7.56°-7°5'6″=7°33'36″-7°5'6″=28'30″.8.解:因为∠A=24.1°+6°=30.1°=30°+0.1°=30°+0.1×60'=30°+6'=30°6',∠B=56°-26°30'=55°60'-26°30'=29°30',∠C=18°12'+11.8°=18°12'+11°+0.8°=18°12'+11°+0.8×60'=18°12'+11°+48'=29°60'=30°,所以∠A>∠C>∠B.6.3.2角的比较与运算【课后·知能演练】1.D解析:15°23'×4=60°92'=61°32'.2.D解析:因为OC平分∠AOB,∠BOC=26°,所以∠AOB=2∠BOC=52°.所以∠BOD=180°-52°=128°.因为OE平分∠DOB,所以∠BOE=12∠DOB=12×128°=3.(1)45°18'(2)60°24'解析:(1)因为OC是∠AOB的平分线,∠AOB=90°36',所以∠COA=45°18'.(2)若∠COA=2∠COB,则∠COB=12∠COA因为∠COA+∠COB=90°36',所以32∠COA=90°36',所以∠COA=60°24'4.解:(1)18°23'×6=18°×6+23'×6=108°+138'=110°18'.(2)34°25'×4+35°42'=137°40'+35°42'=173°22'.(3)21°36'20″÷5=4°+96'20″÷5=4°+19'+80″÷5=4°+19'+16″=4°19'16″.5.解:∠DCE的大小不会发生变化,理由如下:因为CD,CE分别是∠ACP和∠BCP的平分线,所以∠DCP=12∠ACP,∠PCE=12∠所以∠DCE=∠DCP+∠PCE=12∠ACP