원주각

Ⅳ. 원의 성질

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워 크 북 08

AB” O ACB=90˘

ACB CAB=180˘-(90˘+28˘)=62˘

BAD=96˘-62˘=34˘

ADC= ABC=28˘ APD

APC=34˘+28˘=62˘

09

AD” AB”

O ADB=90˘

ADE

EAD=180˘-(90˘+68˘)=22˘

x=2 CAD

x=2_22˘=44˘ 44˘

10

BAC= BDC=36˘

μAB=μ BC ACB= BDC=36˘

ABC

ABC=180˘-(36˘+36˘)=108˘

11

AB” O ACB=90˘

ABC= ADC=40˘ ACB

BAC=180˘-(90˘+40˘)=50˘

ADC : BAC=μAC : μ BC 40˘ : 50˘=8 : μ BC 4 : 5=8 : μ BC

μBC=10(cm)

12

ADB : CBD=μAB : μ CD x : CBD=4 : 1 4 CBD= x

CBD=;4!; x

DBE x=;4!; x+30˘

;4#; x=30˘ x=40˘

13

ACB : CAB : ABC=μAB : μ BC :μ CA

=3 : 1 : 5 ABC=180˘_

ABC=100˘ 100˘

5 3+1+5

68˘

x

A B

C D E

O

01

AOB=2 APB=2_53˘=106˘

OAB OA”=OB”

x=;2!;_(180˘-106˘)=37˘

02

AOB=2 APB=2_45˘=90˘

μAB=2p_12_;3ª6º0;=6p(cm)

03

x=;2!;_(360˘-110˘)=;2!;_250˘=125˘

y=;2!;_110˘=55˘

x- y=125˘-55˘=70˘ 70˘

04

OB”

AOB=2 APB=2_16˘=32˘

BOC=70˘-32˘=38˘

x=;2!; BOC=;2!;_38˘=19˘

05

O’A” OB”

AOB=2 ACB=2_72˘

AOB=144˘

PAO= PBO=90˘ AOBP

x=360˘-(90˘+144˘+90˘)=36˘ 36˘

06

ADB= ACB=23˘ DPB

68˘= x+23˘ x=45˘ 45˘

07

PB”

APB=;2!; AOB=;2!;_40˘=20˘

BPC=90˘-20˘=70˘

x= BPC=70˘

40˘

x A

B

P

Q C O 72˘

x A

B P

C O

70˘

16˘ x

A C

P Q

B O

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21 22

39 40

01 02 03^20˘ 04

05 06^62˘ 07 08^44˘

09 10 11 12^30˘

13 14^42˘ 15^31˘

02

AC”=AD” ACD

ADC=;2!;_(180˘-70˘)=55˘

ABCD

x=180˘-55˘=125˘

03 ABCE

( x+40˘)+85˘=180˘ x=55˘

ABCD

y=180˘-105˘=75˘

y- x=75˘-55˘=20˘ 20˘

04

AD” O

ABD=90˘

μAB=μ BC

BDC= ADB=25˘

ABCD O

(90˘+ x)+(25˘+25˘)=180˘ x=40˘

05

AC”

BAC=;2!; BOC=;2!;_110˘=55˘

CAE=140˘-55˘=85˘

ACDE O

x=180˘- CAE

=180˘-85˘=95˘

06 ACD

ADC=180˘-(63˘+55˘)=62˘

ABCD

ABE= ADC=62˘ 62˘

07

ABQP O

PQB=180˘-64˘=116˘

PQCD O'

x= PQB=116˘

x B A

C D

E

O 140˘

110˘

14

BC” μAB

;6!;

BCA=180˘_;6!;=30˘

BCA : CBD=μAB : μ CD

30˘ : CBD=3 : 5 3 CBD=150˘ CBD=50˘

BCP

CPD=30˘+50˘=80˘ 80˘

15

ABD= ACD A B C D

BAC=90˘-40˘=50˘

BAC+ BDC A B C D

BDC=110˘-70˘=40˘

BAC= BDC A B C D

BCD DBC=180˘-(50˘+80˘)=50˘

DAC+ DBC A B C D

ABC BAC=180˘-(40˘+60˘+40˘)=40˘

BAC+ BDC A B C D

16

A B C D DAC= DBC=25˘

BAC=80˘-25˘=55˘

x= BAC=55˘ 55˘

B A

P

C D

01

x=;2!; AOC=;2!;_126˘=63˘

ABCD O

y=180˘-63˘=117˘

y- x=117˘-63˘=54˘

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23 24

41 43

014'3 cm 02⁷ 03^{15 cm} 04

05 06^{16p cm} 07⁷ 08^{8 cm}

09 10^{12 cm} 11;;¡2∞;; cm 12^{4 cm}

1318'1å4 cm¤ 14 158'6

16 01

P’A”=PB”=x cm P’A”¥PB”=PC”¥PD”

x¤ =4_12 x¤ =48 x=4'3 ( x>0)

PB”=4'3(cm) 4'3 cm

02

PC”=x PD”=13-x

P’A”¥PB”=PC”¥PD” (17-3)_3=x_(13-x)

x¤ -13x+42=0 (x-7)(x-6)=0 x=7 PC”>PD”

PC”=7 7

03

AB”=x cm AB” : PA”=1 : 3 PA”=3x(cm) P’A”¥PB”=PC”¥PD” 3x_(3x+x)=12_(12+13) 12x¤ =300 x¤ =25 x=5 ( x>0)

PA”=3_5=15(cm) 15 cm

04

O PA”¥PB”=PE”¥PF”

O' PC”¥PD”=PE”¥PF”

PA”¥PB”=PC”¥PD” 5_(5+3)=4_(4+x) 40=16+4x 4x=24 x=6

08

PBC PCQ=40˘+48˘=88˘

ABCD CDQ= ABC=48˘

DCQ x=180˘-(48˘+88˘)=44˘ 44˘

ABCD ADP= ABC=48˘

ABQ PAQ=48˘+ x

PAD

40˘+(48˘+ x)+48˘=180˘ x=44˘

09

A+ C+180˘ ABCD

ABD ADB=180˘-(75˘+60˘)=45˘

ACB+ ADB ABCD

A+ C=180˘ ABCD

ABC+ CDE ABCD

ABD BAD=180˘-(25˘+35˘)=120˘

BAD= DCE ABCD

10

BCA= BAT=54˘

BOA=2 BCA=2_54˘=108˘

11

ABCD DAB=180˘-105˘=75˘

ABD ADB=180˘-(75˘+36˘)=69˘

ABE= ADB=69˘

DBE= DCB=105˘

36˘+ ABE=105˘ ABE=69˘

12

ACB=180˘_ =30 x= ACB=30˘

CAB=180˘_ =60˘ y= CAB=60˘

y- x=60˘-30˘=30˘ 30˘

13

AT”

BTA=90˘ BTA

BAT=180˘-(29˘+90˘)=61˘

ATP= ABT=29˘

ATP

61˘=29˘+ x x=32˘

29˘

x A T B

P O

2 1+2+3

1 1+2+3

14

AB”

ABC=90˘

CAB= CBT=48˘ ABC

ACB=180˘-(48˘+90˘)=42˘

ADB= ACB=42˘ 42˘

15

BPF= BAP=84˘ CPF= CDP=65˘

x=180˘-(84˘+65˘)=31˘ 31˘

T O

48˘

A

B

C D

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05

AB” CD” PC”=PD”

PC”=PD”=x cm

PA”¥PB”=PC”¥PD” (21-5)_5=x¤

x¤ =80 x=4'5 ( x>0) CD”=2 PC”=2_4'5=8'5(cm)

06

O r cm

PA”=r+;2!;r=;2#;r(cm) PB”=;2!;r(cm) PA”¥PB”=PC”¥PD” ;2#;r_;2!;r=8_6

;4#;r¤ =48 r¤ =64 r=8 ( r>0)

O 2p_8=16p(cm) 16p cm

07

PO”=x P’A”¥PB”=PC”¥PD”

(x-3)(x+3)=5_(5+3) x¤ -9=40 x¤ =49 x=7 ( x>0)

PO”=7 7

08

BAC= BDC A B C D

AB”=x cm PA”¥PB”=PD”¥PC”

12_(12+x)=10_(10+14) 144+12x=240 12x=96 x=8

AB”=8(cm) 8 cm

09

PA”=AB”=x cm PT” ¤ =P’A”¥PB”

(6'2)¤ =x_(x+x)

x¤ =36 x=6 ( x>0) PA”=6(cm)

10

PT” PQ” PQ”=PT”=18(cm)

PA”=x cm PT” ¤ =PA”¥PB”

18¤ =x_(18+9) 27x=324 x=12

PA”=12(cm) 12 cm

11

PT” ¤ =PA”¥PB” PT” ¤ =5_(5+15)=100 PT”=10(cm) ( PT”>0)

PBT PTA

PBT= PTA P PBTª PTA AA

PT” : PA”=BT” : TA” 10 : 5=15 : TA”

10 TA”=75 AT”=;;¡2∞;;(cm) ;;¡2∞;; cm

12

AO” O

B PA”=x cm

PT” ¤ =PA”¥PB”

8¤ =x_(x+6+6) x¤ +12x-64=0 (x-4)(x+16)=0

x=4 ( x>0)

PA”=4(cm) 4 cm

13

PT” ¤ =PB”¥PA” PT” ¤ =4_(4+14)=72 PT”=6'2(cm) ( PT”>0)

PT” O ATP=90˘

ATP

AT”=øπ18¤ -(6'2)¤ ='2ß5å2=6'7(cm)

ATP=;2!;_6'2_6'7=18'1å4(cm¤ ) 18'1å4 cm¤

14

QA”¥QB”=QC”¥QT”

QA”_3=2_9 QA”=6 PA”=x PT” ¤ =PA”¥PB”

6¤ =x_(x+6+3) x¤ +9x-36=0 (x-3)(x+12)=0 x=3 ( x>0)

PA”=3

15

PT” ¤ =PA”¥PB” x¤ =6_(6+10)=96 x=4'6 ( x>0)

PT'” ¤ =PA”¥PB” PT'”=PT”

y=4'6

x+y=4'6+4'6=8'6 8'6

16

O PT” ¤ =P’A”¥PB”

O' PT” ¤ =PC”¥PD”

P’A”¥PB”=PC”¥PD” 5_(5+AB”)=6_(6+9) 25+5AB”=90 5AB”=65 AB”=13

A

T B

P

O 6 cm 8 cm

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워 크 북

1. 대푯값과 산포도

Ⅰ. 통계

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26 28

01¹⁵ 02¹⁵ 03⁵⁷ 04² 05⁶⁴ 06¹³ 07^20.6

08'3ß.2 09¹²⁰

01

5 5 11 2 9 a 8

a=8

6 3 6 10 8 13 5 3 5 6 8 10 13

b= =7

a+b=8+7=15

15

02

x 5 1 x 5

x yy

=x 75+x=6x

5x=75 x=15 yy

15

03 a=

a= =50 yy

44 46 47 48 50 53 56 56

b= =49 yy

56

c=56 yy

a-b+c=50-49+56=57 yy

57

04 8

=8

33+x=40 x=7

9+6+8+10+x 5 48+50

2 400

8

44+47+50+56+48+53+46+56 8

12+15+16+14+18+x 6

6+8 2

=

=

=

= =2

2

05 5

=5 x+y+20=30

x+y=10 yy yy

2

=2²

=4

=4

x²+y²-10(x+y)+60=24 yy yy

x²+y²-10_10+60=24

x²+y²=64 yy

64 06

5 a b c d e 7

=7 a+b+c+d+e=35 m=

m=

m= =10 yy

5 a b c d e 3

=3²=9

n=

n=

n='9=3 yy

m+n=10+3=13 yy

13 (a-7)¤ +(b-7)¤ +y+(e-7)¤

æ≠1111111≠1111111445

(a+3-10)¤ +(b+3-10)¤ +y+(e+3-10)¤

æ≠1111111≠111111≠14411111444445 (a-7)¤ +(b-7)¤ +y+(e-7)¤

5 35+15

5

a+b+c+d+e+15 5

(a+3)+(b+3)+(c+3)+(d+3)+(e+3) 5

a+b+c+d+e 5

4+1+4+1+x¤ -10x+25+y¤ -10y+25 6

(-2)¤ +1¤ +2¤ +(-1)¤ +(x-5)¤ +(y-5)¤

6

(3-5)¤ +(6-5)¤ +(7-5)¤ +(4-5)¤ +(x-5)¤ +(y-5)¤

6 3+6+7+4+x+y

6 10

5

1+4+0+4+1 5

1¤ +(-2)¤ +0¤ +2¤ +(-1)¤

5

(9-8)¤ +(6-8)¤ +(8-8)¤ +(10-8)¤ +(7-8)¤

5

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07

=

=

= =11

=

=

= =20 6

20.6

08

=

=

=;1%0);=5 yy

=

=

=;1#0@;=3 2 yy

='∂∂3 2 yy

'3ß.2

09

25 70 80

25-(2+6+4+3)=10 yy

=

=

= =75 yy

=

=

= =120 yy

120 3000

25

800+600+0+400+1200 25

(-20)¤ _2+(-10)¤ _6+0¤ _10+10¤ _4+20¤ _3 25

1875 25

110+390+750+340+285 25

55_2+65_6+75_10+85_4+95_3 25

12+0+4+16 10

(-2)¤ _3+0¤ _5+2¤ _1+4¤ _1 10

9+25+7+9 10

3_3+5_5+7_1+9_1 10

412 20

162+75+5+72+98 20

(-9)¤ _2+(-5)¤ _3+(-1)¤ _5+3¤ _8+7¤ _2 20

220 20

4+18+50+112+36 20

2_2+6_3+10_5+14_8+18_2

20

1. 피타고라스 정리

Ⅱ. 피타고라스 정리

서술형훈련하기

29 32

01^{17 cm} 02^{7 cm} 0318'1å0 cm¤ 04^{34 cm¤}

052'2å9 cm 06^{26 cm¤} 0710 2'7 0815<x<21 092'4å1 cm 104'3 cm 112'2 128'6 cm 01

ABD

BD”="√25²-15²=20(cm) CD”=28-20=8(cm)

ADC

AC”="√15²+8²=17(cm)

17 cm

02

AB”=x cm ABC

AC”="√x²+x²='2x(cm) ACD

AD”=øπ('2x)²+x²='3x(cm) ADE

AE”=øπ('3x)²+x²=2x(cm) AEF

AF”=øπ(2x)²+x²='5x(cm) yy '5x=7'5 x=7

AB”=7(cm) yy

7 cm

03

A D BC”

H H' H’H'”=AD”=6(cm)

BH”=C’H'”

BH=;2!;_(12-6)

BH”=3(cm) yy

ABH

AH”="√7²-3²=2'1å0(cm) yy ABCD=;2!;_(6+12)_2'1å0

ABCD=18'1å0(cm²) yy

18'1å0 cm¤

H H'

A

B C

D

12 cm 6 cm

6 cm

7 cm 7 cm

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워 크 북 08

12-9<x<12+9 3<x<21

x>12 12<x<21 yy yy

x²>12²+9² x²>225

x>15 x>0 yy yy

15<x<21 yy

15<x<21

09 ABH

BH”="√10²-6²=8(cm) yy BCH

BC”="√8²+10²=2'4å1(cm) yy 10²+(2'4å1)²>16² AB”¤ +BC”¤ >AC”¤ ABC

yy 2'4å1 cm

10

BD” : CD”=3 : 1 BD”=3x cm CD”=x cm AD”¤ =BD”_CD”

(2'3)²=3x_x 3x²=12 x¤ =4 x=2 ( x>0) CD”=2(cm) BD”=3 CD”=3_2=6(cm)

ABD

AB”=øπ6²+(2'3)¤ =4'3(cm)

4'3 cm

11

AB”²+CD”²=AD”²+BC”² AB”²+5²=4²+7² AB”²=40

AB”=2'1ß0(cm) ( AB”>0) yy ABO

x=øπ(2'1å0)²-(4'2)²='8=2'2 yy 2'2

12 BC”

28p+20p=48p(cm²) yy

r cm

;2!_p_r

2=48p r²=96 r=4'6 ( r>0)

BC”=2r=2_4'6=8'6(cm) yy

8'6 cm 04

AEH™ BFE™ CGF™ DHG EH”=FE”=GF”=HG”

HEF= EFG= FGH= GHE EFGH

AH”=8-5=3(cm) AEH

EH”="√5²+3²='3å4(cm) EFGH=('3å4)²=34(cm²)

34 cm¤

05

ABC™ CDE BC”=DE”=7(cm) CD”=AB”=3(cm)

BD”=7+3=10(cm)yy

A DE”

H

AH”=BD”=10(cm) EH”=7-3=4(cm) yy AHE

AE”="√10²+4²=2'2å9(cm) yy 2'2å9 cm

06 ABC

BC”="√6¤ +4¤ =2'1å3(cm) yy

A DE” BC” DE”

F G

ABD+ AEC

= FBD+ FEC

=;2!; BDGF+;2!; FGEC

=;2!; BDEC

=;2!;_(2'1å3)¤ =26(cm²) yy

26 cm¤

07

x x="√6²+8²=10

8 x="√8²-6²=2'7

6 8 x

x 10 2'7

10 2'7 A

B C

D E

F

G 6 cm 4 cm A H

B C D

E

3 cm

7 cm

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2. 피타고라스 정리의 활용

Ⅱ. 피타고라스 정리

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33 39

012'3å4 cm 028'2 cm 03;;¡5¢;; cm 0412'3 cm¤

05^{36 cm} 06^{84 cm¤} 074'6 cm 086'2 cm 0912'3 cm¤ 102'1å3 11⁵ 12¹⁰ 13²⁷⁰ 1425'2 cm¤ 15^{96 cm¤} 1627'3 cm‹

1736'7 cm‹ 1872'3p cm‹194'1å0 cm 204'2å9p cm 216'3 cm

01

5x cm 3x cm 5x_3x=60 15x¤ =60

x²=4 x=2 ( x>0)

10 cm 6 cm

"√10²+6²=2'3å4(cm)

2'3å4 cm

02

O r cm

pr²=16p r²=16

r=4 ( r>0) yy

ABCD

2r=2_4=8(cm) yy

ABCD 8'2 cm yy

8'2 cm

03

BD”="√8²+6²=10(cm) yy ABD AB”²=BE”_BD”

6²=BE”_10 BE”=;;¡5•;;(cm)

DF”=;;¡5•;;(cm) yy

EF”=10-{;;¡5•;;+;;¡5•;;}

EF”=;;¡5¢;;(cm) yy

;;¡5¢;; cm

04

ABC BD”=CD” AD” ABC AD”= _8=4'3(cm)

ADE= _(4'3)²=12'3(cm²)

12'3 cm¤

05

A BC” H

BH”=;2!;_10=5(cm) ABC 60 cm²

;2!;_10_AH”=60

AH”=12(cm) yy

ABH

AB”="√5²+12²=13(cm) yy ABC

13+10+13=36(cm) yy

36 cm 06

A BC”

H BH”=x cm CH”=14-x(cm)

ABH ACH

AH”²=15²-x²=13²-(14-x)² 225-x²=169-(196-28x+x²) 28x=252 x=9

BH”=9(cm) yy

ABH

AH”="√15²-9²=12(cm) yy ABC=;2!;_14_12=84(cm¤ ) yy

84 cm¤

07

ABC AB” : BC”=1 : '3 8 : BC”=1 : '3 BC”=8'3(cm)

DBC CD” : BC”=1 : '2

CD” : 8'3=1 : '2 CD”=4'6(cm)

4'6 cm 08

ABC AB” : AC”=2 : 1

12 : AC”=2 : 1 AC”=6(cm) yy

ADC AD” : AC”='2 : 1

AD” : 6='2 : 1 AD”=6'2(cm) yy 6'2 cm A

B H C

15 cm

13 cm

14 cm

'3 4 '3

2

H A

B C

10 cm

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워 크 북 09

A BC”

H ABH

AB” : AH”=2 : '3 4 : AH”=2 : '3

AH”=2'3(cm) yy

ABCD=6_2'3=12'3(cm²) yy

12'3 cm¤

10

y=x²-6x+7=(x-3)²-2 y=x²-6x+7 (3 -2) P(3 -2)

P(3 -2) A(-1 4)

PA”=øπ(-1-3)¤ π+{4-(-2)}¤ =2'1å3

2'1å3

11

AB”=øπ(-1-2)¤ +π(6-5)²='1å0 BC”=øπ{0-(-1)}²+(π9-6)²='1å0

CA”=øπ(2-0)²+(π5-9)²='2å0 yy AB”²+BC”²=CA”² ABC B=90˘

yy

ABC=;2!;_'1å0_'1å0=5 yy

5

12

B x B'

B'(5 -4) yy

AP”+BP”

=AP”+B'P”

æAB'”

=øπ{5-(-3)}π¤ +(-4-2)¤

=10

AP”+BP” 10 yy

10

13

a 2a 5a

øπa²+(2a)²+(5a)²=3'3å0

30a²=270 a²=9 a=3 ( a>0)

3 6 15

y

O x

A

B

B' 2

4

-4 P -3 5

H 60˘

A

B C

D 4 cm

6 cm

3_6_15=270

270

14 AE”=a cm

'3a=5'6 a=5'2

AE”=5'2(cm) yy

EG”='2a='2_5'2=10(cm) yy

AEG=;2!;_10_5'2=25'2(cm¤ ) yy 25'2 cm¤

15

A’M”=MÚG”””=GN”=N”AÚ AMGN

yy a cm

MÚN”='2a(cm) AG”='3a(cm) yy

AMGN=;2!;_'2a_'3a AMGN= a²=8'6

a¤ =16 a=4 ( a>0) yy

6_(4_4)=96(cm²) yy

96 cm¤

16

a cm

a=6 a=3'6

_(3'6)‹ =27'3(cm‹ )

27'3 cm‹

17

yy BD”="√6²+6²=6'2(cm)

DH”=;2!;BD”=;2!;_6'2=3'2(cm) OHD

OH”=øπ9²-(3'2)²=3'7(cm) yy

;3!;_6²_3'7=36'7(cm‹ ) yy

36'7 cm‹

A

B H

C D O

6 cm

6 cm 9 cm

'2 12 '6 3

'6 2

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18

AOB AO” : AB”='3 : 2

AO” : 12='3 : 2 AO”=6'3(cm) yy OB” : AB”=1 : 2

OB” : 12=1 : 2 OB”=6(cm) yy

;3!;_p_6²_6'3=72'3p(cm‹ ) yy 72'3p cm‹

19

BG”

BG”=øπ(4+4+4)²+4²=4'1å0(cm)

4'1å0 cm

20

A’B"”

yy BÆ’B'”

BÆ’B'”=BÆ’'B"”=2p_5=10p(cm) yy A’B"”=øπ(10p+10p)²+(8p)²

A’B"””=4'2å9p(cm) yy

4'2å9p cm

21

A’A'” yy

x˘

2p_6_ =2p_2

x=120 yy

OAA' O

A’A'” H

AOH= A'OH=60˘

OAH OA” : AH”=2 : '3 6 : AH”=2 : '3 AH”=3'3(cm)

A’A'”=2AH”=2_3'3=6'3(cm) yy

6'3 cm 60˘ 60˘

6 cm O

A A'

H

x 360

6 cm x˘

O

A A'

8p cm B

A

B' A'

B"

A"

F B A

H D C

E G

4 cm

1. 삼각비

Ⅲ. 삼각비

서술형훈련하기

40 44

01 023'5 03² 04;3$;

05;5$; 06 07

084('3+1) 09^60˘ 10^1.4819 11

12⁰ 13 ^{1.3722 71} 14^23˘

15^1.723

'3 2 '2

2 '6

3 '2å1

7

01

AB” : AC”=2 : '3 AB”=2k, AC”='3k BC”=øπ(2k)¤ +('3k)¤ ='7k

sin B= = =

02 cos A=

;3@;= AB”=6 yy

BC”="√9¤ -6¤ =3'5 yy

3'5

03 tan A=3

AC”="√1¤ +3¤ ='1å0 yy sin A= =

cos A= = yy

sin A-cos A= -sin A-cos A= = sin A+cos A= + sin A-cos A= =

= ÷

= _ =2 yy

2 5

'1å0 2'1å0

5

'1å0 5 2'1å0

5 sin A+cos A sin A-cos A

2'1å0 5 4'1å0

10

'1å0 10 3'1å0

10

'1å0 5 2'1å0

10

'1å0 10 3'1å0

10 '1å0

10 1 '1å0

3'1å0 10 3 '1å0

A B

C

1 3

AB”

9 AB”

AC”

'2å1 7 '2å1

7 '3k '7k AC”

BC”

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워 크 북 09

2'3x-2y+13=0 y='3x+;;¡¡2£;;

tan a='3 yy

a=60˘ yy

60˘

10

sin 40˘= = sin 40˘=AB”=0.6428

tan 40˘= = sin 40˘=CD”=0.8391

sin 40˘+tan 40˘=0.6428+0.8391

=1.4819

1.4819

11

'3x-3y+6=0 y= x+2

tan a= a=30˘ yy

cos a_sin 3a-cos 3a_tan a

=cos 30˘_sin 90˘-cos 90˘_tan 30˘

= _1-0_ = yy

12

45˘<A<90˘

0<cos A<sin A<1

sin A-cos A>0, cos A-sin A<0 yy øπ(sin A-cos A)¤ -øπ(cos A-sin A)¤

=sin A-cos A-{-(cos A-sin A)}

=sin A-cos A+(cos A-sin A)

=0 yy

0

13

sin 35˘=x x=0.5736 cos 37˘=y y=0.7986

x+y=0.5736+0.7986=1.3722 cos x˘=0.8192 x=35 tan y˘=0.7265 y=36 x+y=35+36=71

1.3722 71 '3

2 '3

2 '3

3 '3

2 '3

3

'3 3 CD”

1 CD”

OD”

AB”

1 AB”

OA”

04 ABC

AC”="√20¤ -16¤ =12 x=90˘- B= C tan x=tan C=

tan x=;1!2^;=;3$;

;3$;

05 ABD

BD”="√12¤ +9¤ =15 yy

x=90˘- D= B yy

sin x=sin B=

sin x=;1!5@;=;5$; yy

;5$;

06

BH”="√5¤ +5¤ +5¤ =5'3 yy

FH”="√5¤ +5¤ =5'2 yy

cos x= = = yy

07

sin 30˘=;2!;

4x-10˘=30˘, 4x=40˘

x=10˘

cos(3x+15˘)=cos 45˘=

08 cos 30˘=

= BD”=4'3 yy

sin 30˘=

;2!;= AD”=4

DAC= DCA=45˘

CD”=AD”=4 yy

BC”=BD”+CD”

BC=4'3+4=4('3+1) yy

4('3+1) AD”

8 AD”

AB”

BD”

8 '3

2

BD”

AB”

'2 2 '2

2

'6 3 '6

3 5'2 5'3 FH”

BH”

AD”

BD”

x

x A

H

B C

D

12 9

15

AB”

AC” ^x _x

A

B C

D

E 20

16 12

http://zuaki.tistory.com

14

cos A= =;1£0ª0;=0 39

cos 67˘=0.3907 A=67˘ yy

C=90˘-67˘=23˘ yy

23˘

15

B=90˘-38˘=52˘ yy

cos 52˘=

0 6157=;1”0; x=6 157 yy

sin 52˘=

0 7880= y=7 880 yy

y-x=7 880-6 157=1 723 yy

1.723 y

10 AC”

AB”

BC”

AB”

AB”

AC”

2. 삼각비의 활용

Ⅲ. 삼각비

서술형훈련하기

45 49

01^27.8 0210'3 cm‹ 033(3+'3)m 042'7

056'6 06^{8 m} 073(3-'3)

0825'3 0950(3+'3) m 10^{16 cm¤}

11^120˘ 1215'3 cm¤ 1315'3 cm¤ 14^{32 cm¤}

1572'3 cm¤

01

C=90˘-35˘=55˘

x=20 sin 55˘=20_0 82=16 4 y=20 cos 55˘=20_0 57=11 4 x+y=16 4+11 4=27 8

27.8

02

ABD A=90˘

AB”=BD” sin 30˘

AB”=4_;2!;=2(cm) yy

AD”=BD” cos 30˘

AD”=4_'3=2'3(cm) yy

2

;2!;_2_2'3_5=10'3(cm‹ ) yy

10'3 cm‹

03

CH”=9(m) CEH

EH”=CH” tan 45˘

EH=9_1=9(m) yy

CHD

DH”=CH” tan 30˘

DH=9_ =3'3(m) yy

B

9+3'3=3(3+'3)(m) yy

3(3+'3) m

04

A BC”

H ABH

AH”=8 sin 30˘

AH”=8_;2!;=4 BH”=8 cos 30˘

BH=8_ =4'3 CH”=6'3-4'3=2'3 AHC

AC”=øπ4¤ +(2'3)¤ =2'7

2'7

05

C AB” H

BCH CH”=12 sin 60˘

CH”=12_ =6'3 yy

HCB=90˘-60˘=30˘

ACH=75˘-30˘=45˘ yy

AHC

AC”= =6'3÷

AC”=6'3_ =6'6 yy

6'6 2

'2

'2 2 CH”

cos 45˘

'3 2

'3 2 '3

3

45˘

A C 30˘

B D

9 m H

E

30˘

A

H C B

6'3 8

75˘

60˘

A

B C

12 H

http://zuaki.tistory.com

워 크 북 06

B AC”

H BCH

BH”=BC” sin 45˘

BH=4'2_

BH=4(m) yy

ABH

AB”= =4÷;2!;

AB”=4_2=8(m) yy

8 m

07

AH”=h ABH

BH”=h tan 45˘

BH”=h_1=h AHC CH”=h tan 30˘

CH”=h_ = h BC”=BH”+CH”

h+ h=6 h=6

h= =3(3-'3)

AH”=3(3-'3)

3(3-'3)

08

AH”=h ABH

BH”=h tan 60˘

BH”=h_'3='3h ACH

CH”=h tan 30˘

CH”=h_ = h yy

BC”=BH”-CH”

'3h- h=10 h=10

h=5'3

AH”=5'3 yy

ABC=;2!;_10_5'3

ABC=25'3 yy

25'3 2'3

3 '3

3

'3 3 '3

3

30˘

60˘

30˘ 120˘

A

B 10 C H

h

18 3+'3

3+'3 3 '3

3

'3 3 '3

3

45˘

45˘

60˘

30˘

A

B H C

6 h

BH”

sin 30˘

'2 2

09

AH”=h m ABH

BH”=h tan 45˘

BH=h_1 BH=h(m)

ACH CH”=h tan 30˘

CH=h_

CH= h(m) yy

BC”=BH”-CH”

h- h=100 h=100

h=

h=50(3+'3)

50(3+'3) m yy

50(3+'3) m

10

AB”=BC”

A= C=75˘

B=180˘-(75˘+75˘) B=30˘

ABC=;2!;_8_8_sin B ABC=;2!;_8_8_sin 30˘

ABC=;2!;_8_8_;2!;

ABC=16(cm¤ )

16 cm¤

11

ABC 10'3 cm¤

ABC=;2!;_5_8_sin(180˘-C)=10'3 yy 20_sin(180˘-C)=10'3

sin(180˘-C)=

180˘-C=60˘

C=120˘ yy

120˘

'3 2 300 3-'3

3-'3 3 '3

3 '3

3 '3

3

45˘ 60˘

A

B100 mC H 45˘

30˘ h m 45˘

A 30˘ B

C

4'2 m H

http://zuaki.tistory.com

12 ABC

AC”=4 tan 60˘=4_'3

AC”=4'3(cm) yy

ABCD

= ABC+ ACD

=;2!;_4_4'3+;2!;_4'3_7_sin 30˘

=;2!;_4_4'3+;2!;_4'3_7_;2!;

=8'3+7'3

=15'3(cm¤ ) yy

15'3 cm¤

13

ABCD=6_10_sin 60˘

ABCD=6_10_

ABCD=30'3(cm¤ ) AED=;2!; ABCD AED=;2!;_30'3 AED=15'3(cm¤ )

15'3 cm¤

14 ABCD

AC”=BD”=8(cm) yy

ABCD=;2!;_8_8_sin 90˘

ABCD=;2!;_8_8_1

ABCD=32(cm¤ ) yy

32 cm¤

15

ABCD 48cm

48_;4!;=12(cm) yy

ABCD=2 ABD

ABCD=2_[;2!;_12_12_sin (180˘-120˘)]

ABCD=2_{;2!;_12_12_ }

ABCD=72'3(cm¤ ) yy

72'3 cm¤

'3 2 '3

2

1. 원과 직선

Ⅳ. 원의 성질

서술형훈련하기

50 55

01^{13 cm} 024'7 cm 03^{4p cm¤} 042'1å0 cm 05^65˘ 06^{30 cm} 07^{2 cm} 08^{25p cm¤}

09^{36p cm¤} 108'3 cm 11^{30 cm} 125'2 cm 13^{3 cm} 14^{24 cm} 15^(30-4p)cm¤

16^{24 cm} 17^{162 cm¤} 18^{6 cm} 01

A’M”=;2!; AB”

A’M”=;2!;_24 A’M”=12(cm)

O r cm

O’M”=r-8(cm) OMA r¤ =12¤ +(r-8)¤

r¤ =144+r¤ -16r+64 16r=208

r=13 O 13 cm

13 cm

02

CD”

O OA”=8(cm) OD”=8-2

OD”=6(cm) yy

AOD AD”="√8¤ -6¤

AD”=2'7(cm) yy

AB”=2 AD”

AB”=2_2'7

AB”=4'7(cm) yy

4'7 cm

A B

C

D 2 cm

O 6 cm 8 cm

O A

B M C

8 cm 24 cm r cm

http://zuaki.tistory.com

워 크 북 03

O AB”

M

A’M”=;2!;AB”

A’M”=;2!;_2'3

A’M”='3(cm) yy

O r cm

MÚO”=;2R;(cm) yy

AOM r¤ =('3)¤ +{;2R;}2 r¤ =3+

=3 r¤ =4

r=2 ( r>0) yy

O

p_2¤ =4p(cm¤ ) yy

4p cm¤

04

A’M”=;2!;AB”

A’M”=;2!;_6=3(cm) AOM

O’M”="√7¤ -3¤

O’M”=2'1å0(cm)

CD”=2CN”=2_3=6(cm) AB”=CD”

ON”=O’M”=2'1å0(cm)

2'1å0 cm

05 AMON

A=360˘-(90˘+130˘+90˘)

A=50˘ yy

O’M”=ON”

AB”=AC”

ABC yy

B=;2!;_(180˘-50˘)

B=65˘ yy

65˘

3r¤

4 r¤

4

A M B

O 2'3 cm

r cm

06

OD”=OE”=OF”

AB”=BC”=CA”

ABC yy

AB”=2AD”=2_5=10(cm) yy

ABC

10+10+10=30(cm) yy

30 cm

07

PA≥ O

OAP=90˘

O r cm

OA”=r(cm) OP”=2+r(cm)

OPA

(r+2)¤ =(2'3)¤ +r¤

r¤ +4r+4=12+r¤

4r=8 r=2

O 2 cm

2 cm

08

PA≥ PB≥ O PAO= PBO PAO=90˘

AOB

=360˘-(90˘+70˘+90˘)

=110˘ yy

360˘-110˘=250˘ yy

p_6¤ _;3@6%0);=25p(cm¤ ) yy

25p cm¤

09

O AB”

H AB

AH”=;2!; AB”

AH”=;2!;_12=6(cm) yy

A H B

O

12 cm R cm r cm

70˘

A

B

P O

6 cm A B P

O 2 cm

2'3 cm r cmr cm

http://zuaki.tistory.com

R cm r cm OAH

R¤ =6¤ +r¤ R¤ -r¤ =36 yy

pR¤ -pr¤ =p(R¤ -r¤ )

=36p(cm¤ ) yy

36p cm¤

10

OP” AOP

BOP OA”=OB”

AP”=BP”

OAP= OBP=90˘

AOP™ BOP SAS AOP= BOP AOP=;2!; AOB AOP=;2!;_120˘

AOP=60˘

AOP AP”=4 tan 60˘

AP=4_'3=4'3(cm) PA”=PB”

PA”+PB”=2 PA”

=2_4'3=8'3(cm)

8'3 cm

11 POC PC”="√17¤ -8¤

PC”=15(cm) yy

AR”=AP” BR”=BQ” CP”=CQ”

ABC =AB”+BC”+CA”

=(AR”+BR”)+BC”+CA”

=(AP”+BQ”)+BC”+CA”

=(AP”+CA”)+(BQ”+BC”)

=CP”+CQ”

=2CP”

=2_15

=30(cm) yy

30 cm A

B

P O

4 cm 120˘

12

DC”=DE”+CE”=D’A”+CB”

DC”=5+10=15(cm) yy

D BC”

H

HC”=10-5

HC”=5(cm) yy

DHC DH”="√15¤ -5¤

DH”=10'2(cm) yy

O

;2!; AB”=;2!; DH”

;2!; AB”=;2!;_10'2

;2!; AB”=5'2(cm) yy

5'2 cm

13

AF”=x cm AD”=AF”=x(cm)

BE”=BD”=8-x(cm) CE”=CF”=7-x(cm) BC”=BE”+CE”

9=(8-x)+(7-x) 9=15-2x

2x=6 x=3

AF”=3(cm)

3 cm

14 AD”=x cm AF”=AD”=x(cm)

BD”=BE”=4(cm) CF”=CE”=2(cm) AB”=x+4(cm)

AC”=x+2(cm) yy

ABC

(x+4)¤ =6¤ +(x+2)¤

x¤ +8x+16=36+x¤ +4x+4 4x=24

x=6 yy

ABC

(6+4)+6+(6+2)=24(cm) yy

24 cm

A B

H C

D E

O 5 cm

10 cm

http://zuaki.tistory.com

워 크 북 01

BE”

AEB= ADB=37˘

BEC=;2!; BOC BEC=;2!;_84˘=42˘

AEC= AEB+ BEC

=37˘+42˘=79˘

79˘

02 BPC

BCD=40˘+ x yy

ADC= ABC= x yy

QCD

(40˘+ x)+ x=76˘

40˘+2 x=76˘

2 x=36˘ x=18˘ yy

18˘

03

O BA' A’'C”

A'CB=90˘

BA'C= BAC

BA'C=60˘ yy

A’'B”= =5'3÷

A’'B=5'3_ =10(cm) yy

O

;2!;_10=5(cm) yy

5 cm 2

'3

'3 2 5'3

sin 60˘

60˘

60˘

A

A'

B C

O

5'3 cm 84˘

37˘

A

B

C

D E

O

2. 원주각

Ⅳ. 원의 성질

서술형훈련하기

56 61

01^79˘ 02^18˘ 03^{5 cm} 04^60˘

05^{15 cm} 06^21˘ 07^108˘ 08^40˘

09^43˘ 10^22˘ 11^80˘ 128'3 cm¤

138'2 cm 14^{20 cm¤} 15^{8p cm} 164'3 cm 1712'2 cm 18³¹

15 ABC

AC”="√13¤ -12¤ =5(cm)

O r cm

AD”=AF”=r(cm) BE”=BD”=12-r(cm)

CE”=CF”=5-r(cm) yy

BC”=BE”+CE”

13=(12-r)+(5-r) 13=17-2r

2r=4 r=2 yy

;2!;_12_5-p_2¤ =30-4p(cm¤ ) yy (30-4p)cm¤

16

AH”=AE”=3(cm) AD”=3+2=5(cm) AB”+CD”=AD”+BC”

ABCD =AB”+BC”+CD”+DÆ’A”

=(AB”+CD”)+(BC”+DA”)

=2(BC”+DA”)

=2_(7+5)

=24(cm)

24 cm 17

O 6 cm

CD”=2_6=12(cm) yy

AD”+BC”=AB”+CD”=15+12=27(cm) yy ABCD=;2!;_(AD”+BC”)_CD”

ABCD=;2!;_27_12

ABCD=162(cm¤ ) yy

162 cm¤

18 CDE

ED”="√10¤ -8¤ =6(cm) yy

AE”=x cm

BC”=AD”=x+6(cm) yy

ABCE AE”+BC”=AB”+EC”

x+(x+6)=8+10 2x+6=18

2x=12 x=6

AE”=6(cm) yy

6 cm A

B C

D

E F O 12 cm

13 cm r cm

http://zuaki.tistory.com

04

BCA= BDC=35˘

BCD

50˘+(35˘+ ACD)+35˘=180˘

ACD+120˘=180˘

ACD=60˘

60˘

05 APD

DAP=84˘-60˘=24˘ yy

6 : μAB=24˘ : 60˘ μAB=15(cm) yy 15 cm

06

AC” μAB μ CD

O ;5!; ;1¡2;

ACB=180˘_;5!;=36˘

DAC=180˘_;1¡2;=15˘ yy

ACP

P=36˘-15˘=21˘ yy

21˘

07

;2!;

AOC=2 B=2_72˘=144˘

B+ D=180˘

72˘+ D=180˘ D=108˘

AOCD

x+ y=360˘-(144˘+108˘) x+ y=108˘

108˘

08

DAB= DCE=110˘

DAC=110˘-60˘=50˘ yy

DBC= DAC=50˘ yy

90˘

ABC=90˘

ABD=90˘-50˘=40˘ yy

40˘

A

B C P

O D

09 ABCD C=180˘-124˘

C=56˘ yy

P= x PBC

PBQ= x+56˘ yy

AQB

25˘+( x+56˘)=124˘

x+81˘=124˘

x=43˘ P=43˘ yy

43˘

10

ACB= ABT=68˘

AOB=2 ACB

=2_68˘=136˘

OAB

OAB=;2!;_(180˘-136˘)=22˘

22˘

11

AC” μAB : μ BC=3 : 5 ACB: CAB=3 : 5 yy ACB=3 a CAB=5 a

ABC

5 a+52˘+3 a=180˘

8 a+52˘=180˘

8 a=128˘ a=16˘

CAB=5_16˘=80˘ yy

CBT= CAB=80˘ yy

80˘

12

BAC= BCP=30˘

90˘

ACB=90˘ yy

ACB AB”=8 cm AC”=8 cos 30˘

AT”=8_ =4'3(cm) BC”=8 sin 30˘

BT=8_;2!;=4(cm) yy

ACB=;2!;_4'3_4=8'3(cm¤ ) yy 8'3 cm¤

'3 2

30˘

30˘

A

C B

P O

4 cm 52˘

A

B D C

O

T

http://zuaki.tistory.com

워 크 북 17

AB”=2 AO”

AB=2_9 AB=18(cm) AC”¤ =AO”¥AB”

AC”¤ =9_18=162

AC”=9'2(cm) ( AC”>0) yy

C’O'”, DB”

AO'C ABD A

ACO'= ADB

AO'C ABD (AA ) yy AC” : AD”=A’O'” : AB”

A’O'”=9+;2(;=;;™2¶;;(cm) 9'2 : AD”=;;™2¶;; : 18

AD”=12'2(cm) yy

12'2 cm

18

PT”¤ =PA”¥PB”

x¤ =5_(5+15) x¤ =100

x=10 ( x>0) yy

PA”¥PB”=PC”¥PD”

5_(5+15)=4_(4+y) 100=16+4y

4y=84 y=21 yy

x+y=10+21=31 yy

31

A B

C D

O O' 9 cm

13

AP” : PB”=1 : 3 PA”=x cm PB”=3x(cm)

PA”¥PB”=PC”¥PD”

x_3x=8_3 3x¤ =24 x¤ =8 x=2'2 ( x>0)

PA”=2'2(cm)

PB”=3x=3_2'2=6'2(cm) AB”=2'2+6'2=8'2(cm)

8'2 cm

14 AO”=5 cm

AB”=2 AO”=2_5=10(cm)

AP”=10-2=8(cm) yy

O CP”

D CP”=x cm PD”=PC”=x(cm) PA”¥PB”=PC”¥PD”

8_2=x_x x¤ =16 x=4 ( x>0)

CP”=4(cm) yy

ABC=;2!;_10_4=20(cm¤ ) yy

20 cm¤

15

O r cm

PC”=8+r(cm) PD”=8-r(cm) yy

PB”¥PA”=PD”¥PC”

6_(2+6)=(8-r)(8+r) 48=64-r¤ r¤ =16

r=4 ( r>0) yy

O

2p_4=8p(cm) yy

8p cm

16

PTA= ABT= APT APT

AP”=AT”=4(cm) PT”¤ =PA”¥PB”

PT”¤ =4_(4+8)=48 PT”=4'3(cm) ( PT”>0)

4'3 cm

A P B

C

D 5 cm O

2 cm

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M e em mo o

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