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新人教版高中数学必修第一册:课时跟踪检测(三十一) 弧度制

新人教版高中数学必修第一册:课时跟踪检测(三十一)  弧度制
新人教版高中数学必修第一册:课时跟踪检测(三十一)  弧度制

课时跟踪检测(三十一) 弧度制

A 级——学考合格性考试达标练

1.

4

对应的角度为( ) A .75° B .125° C .135°

D .155°

解析:选C 由于1 rad =? ??

??

180π°,

所以3π4=34π×? ????

180π°=135°,故选C.

2.与角-π

6终边相同的角是( )

A.5π6

B.π3

C.11π6

D.2π3

解析:选C 与角-π6终边相同的角为2k π-π6,k ∈Z ,当k =1时,此角等于11π

6.故选

C.

3.把角-570°化为2k π+α(0≤α<2π,k ∈Z )的形式为( ) A .-3π-1

B .-4π+150°

C .-3k π-30°

D .-4π+5

6

π

解析:选D 因为-570°与5

6π的终边相同,所以把角-570°化为2k π+α(0≤α<2π)的

形式为-4π+5

6

π.

4.将分针拨快10分钟,则分针转过的弧度数是( ) A.π3 B .-

π3 C.π6

D .-

π6

解析:选B 分针拨快10分钟,决定了分针转动的方向是顺时针,即转过的弧度数是负的.因为分针拨快60分钟时转过弧度数为-2π,所以拨快10分钟转过的弧度数为-π

3

.

5.终边在y 轴的非负半轴上的角的集合是( )

A .{α|α=k π,k ∈Z } B.??????α??α=k π+π

2,k ∈Z

C .{α|α=2k π,k ∈Z }

D.????

??α??α=2k π+π

2,k ∈Z

解析:选D A 选项表示的角的终边在x 轴上,B 选项表示的角的终边在y 轴上;C 选项表示的角的终边在x 轴的非负半轴上;D 选项表示的角的终边在y 轴的非负半轴上.故选D.

6.-105°化为弧度为________,11π

3化为角度为________.

解析:-105°=-105×π180=-7

12π,

113π=11

3×180°=660°. 答案:-7

12

π 660°

7.已知一个扇形的弧所对的圆心角为54°,半径r =20 cm ,则该扇形的周长为________ cm.

解析:因为1°=

π180 rad ,所以54°=π180×54=3π10,则扇形的弧长l =αr =3π

10

×20= 6π(cm),故扇形的周长为(40+6π)cm.

答案:40+6π

8.圆的半径变为原来的3倍,而所对弧长不变,则该弧所对圆心角是原来圆弧所对圆心角的________倍.

解析:设原来圆的半径为r ,弧长为l ,弧所对的圆心角为α(0<α<2π),则现在的圆的半径为3r ,弧长为l ,设弧所对的圆心角为β(0<β<2π),于是l =αr =β·3r ,∴β=1

3

α.

答案:1

3

9.把下列各角化成2k π+α(0≤α<2π,k ∈Z )的形式,并指出是第几象限角. (1)-1 500°;(2)23

6

π.

解:(1)∵-1 500°=-1 800°+300°=-10π+5π

3,

∴-1 500°与5π

3

终边相同,是第四象限角.

(2)∵

236π=2π+11

6

π, ∴236π与11

6π终边相同,是第四象限角.

10.如图,扇形AOB 的面积是4 cm 2,它的周长是10 cm ,求扇形的圆心角α的弧度数及弦AB 的长.

解:设弧AB 长为l cm ,扇形半径为r cm ,

则由题意,得?????l +2r =10,12

l ·r =4,解得?????r =1,l =8(不合题意,舍去)或?????r =4,l =2.∴α=24=12(rad).

∴弦AB 的长为2r sin α

2=2×4×sin 14=8sin 1

4

(cm).

B 级——面向全国卷高考高分练

1.一个扇形的弧长与面积的数值都是6,则这个扇形的圆心角是( ) A .1 B .2 C .3

D .4

解析:选C 设扇形的圆心角的弧度数为α,半径为r ,由题意知?????αr =6,

12αr 2

=6,

解得α=3,

故选C.

2.若α3=2k π+π3(k ∈Z ),则α

2

的终边在( )

A .第一象限

B .第四象限

C .x 轴上

D .y 轴上

解析:选D ∵α

3=2k π+π3(k ∈Z ),∴α=6k π+π(k ∈Z ),∴α2=3k π+π

2(k ∈Z ).当k 为

奇数时,α2的终边在y 轴的非正半轴上;当k 为偶数时,α

2的终边在y 轴的非负半轴上.综

上,α

2

的终边在y 轴上,故选D. 3.若角α与角x +π4有相同的终边,角β与角x -π

4

有相同的终边,那么α与β间的关系为( )

A .α+β=0

B .α-β=0

C .α+β=2k π(k ∈Z )

D .α-β=π

2

+2k π(k ∈Z )

解析:选D ∵α=x +π4+2k 1π(k 1∈Z ),β=x -π4+2k 2π(k 2∈Z ),∴α-β=π

2+2(k 1-

k 2)π(k 1∈Z ,k 2∈Z ).

∵k 1∈Z ,k 2∈Z ,∴k 1-k 2∈Z . ∴α-β=π

2

+2k π(k ∈Z ).

4.已知某机械采用齿轮传动,由主动轮M 带着从动轮N 转动(如图所示),设主动轮M 的直径为150 mm ,从动轮N 的直径为300 mm ,若主动轮M 顺时针旋转

π

2

,则从动轮N 逆时针旋转( ) A.π8 B.π4 C.π2

D .π

解析:选B 设从动轮N 逆时针旋转θ rad ,由题意,知主动轮M 与从动轮N 转动的弧长相等,所以1502×π2=300

2×θ,解得θ=π4

,选B.

5.若角α的终边与8

5π角的终边相同,则在[0,2π)上,终边与α4

角的终边相同的角是

____________.

解析:由题意,得α=8π5+2k π(k ∈Z ),∴α4=2π5+k π2(k ∈Z ).令k =0,1,2,3,得α

4=

2π5,9π10,7π5,19π

10

. 答案:

2π5,9π10,7π5,19π

10

6.在直径为10 cm 的轮上,有一长为6 cm 的弦,P 为该弦的中点,轮子以每秒5 rad 的角速度旋转,则经过5 s 后点P 转过的弧长是________cm.

解析:点P 在以轮的中心为圆心,半径为4 cm 的圆上,5 s 后点P 转过的圆心角为 25 rad ,由弧长公式知,该弧长为100 cm.

答案:100 7.已知α=1 690°.

(1)把α写成2k π+β(k ∈Z ,β∈[0,2π))的形式;

(2)求θ,使θ与α终边相同,且θ∈(-4π,4π). 解:(1)1 690°=4×360°+250°=4×2π+25

18π.

(2)∵θ与α终边相同,∴θ=2k π+25

18π(k ∈Z ).

又θ∈(-4π,4π),∴-4π<2k π+25

18π<4π(k ∈Z ).

解得-9736

36(k ∈Z ),∴k =-2,-1,0,1.

∴θ的值是-4718π,-1118π,2518π,6118

π.

8.用弧度表示顶点在原点,始边与x 轴的非负半轴重合,终边在图中阴影部分的角的集合.

解:以射线OB 为终边的225°角与-135°角的终边相同,-135°=-

135×π180=-3π4,而135°=3π4,阴影部分(包括边界)位于-3π4与3π

4之间且跨越x 轴的非负

半轴.所以,终边在阴影部分(包括边界)的角的集合为

?????α????

????-3π4+2k π≤α≤3π

4+2k π,k ∈Z . C 级——拓展探索性题目应用练

如图,一长为 3 dm ,宽为1 dm 的长方形木块在桌面上作无滑动翻滚,翻滚到第四次时被一小木块挡住,使木块底面与桌面所成角为π

6,试求点A 走过的路程及走过的弧所

在的扇形的总面积.(圆心角为正)

解:在扇形ABA 1中,圆心角恰为π2,弧长l 1=π2·AB =π

2

·

3+1=π,面积S 1=12·π

2

·AB 2

=12·π2·4=π.在扇形A 1CA 2中,圆心角也为π2,弧长l 2=π2·A 1C =π2·1=π2,面积S 2=12·π

2·A 1C 2

=12·π2·12=π4.在扇形A 2DA 3中,圆心角为π-π2-π6=π3,弧长l 3=π3·A 2D =π3·3=33π,面积S 3=12·π3·A 2D 2=12·π3·(3)2

=π2,∴点A 走过的路程长l =l 1+l 2+l 3=π+π2+3π3

(9+23)π6,点A 走过的弧所在的扇形的总面积S =S 1+S 2+S 3=π+π4+π2=7π

4

.

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