232-42072155.AUS M Trig Polar Coords Solutions AUS

7/27/2019 232-42072155.AUS M Trig Polar Coords Solutions AUS

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Trigonometry

&

PolarCoordina

tes

www.mathlecs.com.au


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Trigonometry & Polar Coordinates

100% Trigonometry & Polar Coordinates Soluons

Mathletics 100% 3P Learning

1SERIES TOPIC NUMBER

M 3

Knowing MoreSolutions

5. FindthecoordinatesofthepointsA,B,CandDcorrectto3decimalplaces.

a

c

b

d

A:

C:

B:

D:

(to d.p.)

, ,

. , .

cos sinA x y A

A

45 45

0 707 0 707 3

` c c=

=

^ ^

^

h h

h

(to d.p.)

, ,

. , .

cos sinC x y C

C

250 250

0 342 0 940 3

` c c=

= - -

^ ^

^

h h

h

(to d.p.)

, ,

. , .

cos sinB x y B

B

170 170

0 985 0 174 3

` c c=

= -

^ ^

^

h h

h

(to d.p.)

, ,

. , .

cos sinD x y D

D

315 315

0 707 0 707 3

` c c=

= -

^ ^

^

h h

h

cos

sin

x

y

45

45

45

c

c

c

i=

=

=

cos

sin

x

y

250

250

250

c

c

c

i=

=

=

cos

sin

x

y

170

170

170

c

c

c

i=

=

=

cos

sin

x

y

315

315

315

c

c

c

i=

=

=


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M4


6. Completethetableforthefollowingpointsroundedto2d.p.whereappropriate:

Point r iPolarcoordinates

(r,i)x= rcosi y = rsini (x, y) Coordinates

A 2 30c A(2, 30c) 2 30

1.73

cos 3c =

=

sin2 30 1c = . ,A 1 73 1^ h

B 4 85c B(4, 85c) .cos4 85 0 35c = .sin4 85 3 98c = . , .B 0 35 3 98^ h

C 2 110c C(2, 110c) .cos2 110 0 68c = - .sin2 110 1 88c = . , .C 0 68 1 88-^ h

D 4 230c D(4, 230c) .cos4 230 2 57c = - .sin4 230 3 06c = - . . .D 2 57 3 06- -^ h

E 3 300c E(3, 300c)

.

.

cos cos3 300 3 60

3 0 50

1 50

c c=

=

=

^ h

3 300 3 60

.

sin sin

2

3 3

2 60

c c= -

= -

= -

. , .E 1 50 2 60-^ h

F 5 340c F(5, 340c) .cos3 340 2 82c = .sin3 340 1 03c = - . , .F 2 82 1 03-^ h

1

2

3

4

5

-1

-2

-3

-4

-5

1 2 3 4 5- 5 - 4 - 3 - 2 -1


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M6


,ri^ h cosx r i= siny r i= ,x y^ h

,2 45c

^ h cosx 2 45 1c

= = siny 2 45 1c

= = ,1 1^ h

,3 60c^ h .cosx 3 602

31 5c= = =

2

3.siny 3 60 3 2 60c= = =c m . , .1 50 2 60^ h

,5 130c^ h .cosx 5 130 3 21c= = - .siny 5 130 3 83c= = . , .3 21 3 83-^ h

,4 200c^ h 4 200 3.76cosx c= = - .siny 4 200 1 37c= = - . , .3 76 1 37- -^ h

,6 315c^ h .cos cosx 6 315 6 452

64 24c c= = = = .siny 6 315 4 24c= = - . , .4 24 4 24-^ h

,7 270c^ h cosx 7 270 7 0 0c= = =^ h siny 7 270 7c= = - ,0 7-^ h

9. Completethetablebelowtondtherectangularcoordinates(to2d.p.):


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M 7

Using Our KnowledgeSolutions

i360c

iArclengthcoveredby i iinRadians

30c

360

30

12

1

c

c

= r r121

2 6# r

r

= r r6 6'r r

=

60c360

60

6

1

c

c= 2 r r

6

1

3# r

r= r r

3 3'

r r=

90c360

90

4

1

c

c= 2 r r

4

1

2# r

r= r r

2 2'

r r=

012 c360

120

3

1

c

c= 2 r r

3

1

3

2# r

r= r r

3

2

3

2'

r r=

2 04 c360

240

3

2

c

c= 2 r r

3

2

3

4# r

r= r r

3

4

3

4'

r r=

027 c360270

43

c

c= 2 r r

43

23# r r= r r

23

23'r r=

030 c360

300

6

5

c

c= 2 r r

6

5

3

5# r

r= r r

3

5

3

5'

r r=

036 c360

3601

c

c= 2 r r

1

12# r r= 2 2r r'r r=

10.Usethecircleandtheaboveformulatocompletethetablebelow(ignorethelastcolumnfornow):

11.Completethelastcolumnofthetableonthepreviouspage.Usethetabletodrawalltheanglesinradians

ontheaxesbelow:

radians306

cr

=

radians063

cr

=

radians902

cr

=

radians0123

2c

r=

radians2 043

4c

r= radians030

3

5c

r=

radians360 2c r=

radians0272

3c

r=


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M8


12. Usethisformulatoconvertthefollowinganglesfromdegreestoradians(leaveanswersasfracons):

a

a

a

c

c

c

e

e

b

b

d

d

f

f

b

b

rad

60 60180

3

#c cc

r

r

=

=

3

2

3

2 180120#

cc

r r

r= =

2702

3

2

3 180#

cc

r r

r= =

rad

40 40180

9

2

#c cc

r

r

=

=

rad

225 225180

4

5

#c cc

r

r

=

=

rad

90 90180

2

#c cc

r

r

=

=

4 4

18045#

cc

r r

r= =

3

4

3

4 180240#

cc

r r

r= =

1506

5

6

5 180#

cc

r r

r= =

3154

7

4

7 180#

cc

r r

r= =

rad

100 100180

9

5

#c cc

r

r

=

= rad

315 315180

4

7

#c cc

r

r

=

=

13. Todevelopaformulatoconvertradianstodegrees,makeicthesubjectoftheformulaabove:

(degrees) =(radians) #180cr

14. Usethisformulatoconvertthefollowinganglesfromradianstodegrees(1decimalplace):

15.

Find the arc length L.

cmL r 44

#ir

r= = =

cmL r 6 3

4

8#ir

r= = =

cm.L r 4 86

54#i

rr= = =

cm2 2L r r r#i r r= = =

Find the perimeter of the following sector.

6

FindL and Pin the following diagram. Find the arc length of the enre circle below.

(Hint: Convert the angle to radians rst)

This is the formula for circumference of a circle.


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M 9


Idenfy the major and minor segments on the diagram

16. Usethediagramontheletoanswerthefollowingquesons:

a

b

c

d

Find the area of the sector: cm6r2

1

2

16

3

2 2 2

#ir

r= =^ h

3r

6 cm

Minor segment

Major segment

Find the area of the minor segment

Area sector cm cm ( d.p.)6 6 18.85 2r2

1

2

1

3

2 2 2 2i

rr= = = =^ h

Area triangle cm or cm ( d.p.)60 18 9 15.59 2sin sinr2

1

2

16

2

33

2 2 2 2ci= = = =^ ch m

Find the area of the triangle

Area minor segment = Area sector - area triangle cm cm ( d.p.)6 9 3.26 23 22r= - =

17. Findthearclengthofasectorsubtendinganangleof6

5randaradiusof12cm.

6

5rcm10

L r

126

5

i

r

r

=

=

=

` j


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M10


18. Asectorwitharadiusof4cmhasanareaof cm4 2r .Whatistheangleinsidethesector?

Use the formula for the area of a sector, and solve for i.

rad

4

4

4 8

8 4

2

14

2

16

8

4

2

2` r i

r i

r i

i r

ir

ir

=

=

=

=

=

=

^ h

19. Findtheoverlappingareaofthe2circlesbelowifx3

r= and y

3

2r= (to2d.p.):

x y

12 cm

3 cm

The overlapping area is the sum of2 minor segments. Use the

formula for the area of minor segment for each of these.

Area

cm

12 3sin sin2

1

3 3 2

1

3

2

3

2

723 2

3

2

9

3

2

2

3

24 36 3 34

93

27 34

144 9

274

1533

2 2

2

`r r r r

r r

r r

r

r

= - + -

= - + -

= - + -

= -+

= -

` `

c c

`

`

j j

m m

j

j

20. Whatistheperimeteroftheoverlappingareainthepreviousqueson?

The perimeter of the overlapping area is the sum of 2 arc lengths. Use the arc length formula for each side.

Perimeter

cm6

123

33

2

3

12 6

3

18

`r r

r r

r

r

= +

=+

=

=

` `j j

First substute A 4r= and r=4.


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M 11


21. FindtheareaoftheregionbelowifMisthecentreofthecircleandABisthediameter:

23c

23c

16cm

D

B

A

M

C

By symmetry the two triangles are exactly the same,

since AMB+ is a straight angle so

AMD AMC23 23 180c c c+ ++ = + =

AMD AMC

180 23

157

`

c c

c

+ +=

= -

=

Total area

cm ( d.p.).

sin sin2

116 157

2

116 157

2

116

180

46

202 79 2

2 2 2

2

` c cr

= + +

=

^ ^ ^ `h h h j

Convert the angle of46c to radians: rad46180

46c

r=


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M12

Thinking MoreSolutions

i sini cosi tani

0 0 1 0

6r

2

1

2

3

3

1

3r

2

3

2

13

4r

2

1

2

11

2r

1 0 Undened

4

5r

2

1-

2

1- 1

2

3r-1 0 Undened

22. Usetheexactraosabovetocompletethetable:

23. Completetheradianvaluesforthequadrantsontheaxes:

24. Completethefollowingtables:

rad2

90cr ^ h

rad2

3270c

r ^ h

rad0 0c^ hrad 180cr ^ h

a bRuleindegrees Ruleinradians

sin cos90c i i- =^ h sin cos2

ri i- =` j

cos sin90c i i- =^ h cos sin2

ri i- =` j

tan cot90c i i- =^ h tan cot2

ri i- =` j

Ruleindegrees Ruleinradians

sin sin180c i i- =^ h sin sinr i i- =^ h

cos cos180c i i- = -^ h cos cosr i i- = -^ h

tan tan180c i i- = -^ h tan tanr i i- = -^ h


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M 13


c dRuleindegrees Ruleinradians

sin sin180c i i+ = -^ h sin sinr i i+ = -^ h

cos cos180c i i+ = -^ h cos cosr i i+ = -^ h

tan tan180c i i+ =^ h tan tanr i i+ =^ h

Ruleindegrees Ruleinradians

sin sin360c i i- = -^ h sin sin2r i i- = -^ h

cos cos360c i i- =^ h cos cos2r i i- =^ h

tan tan360c i i- = -^ h tan tan2r i i- = -^ h

24.

25. Usewhatyouknowaboutquadrantstocompletethefollowingtable:

Expressionofi i(radians) Quadrantofi SimplicaonRule

sin6

r

6r

1st

sin6

r

cos2

r

2r

1st

cos2

r

sin4

3r

4

3r2nd

sin sin sin4

3

4

3

4

rr

r r= - =` j

tan

12

11r

12

11r2nd

tan tan tan

12

11

12 12

rr

r r= - = -

` j

cos4

5r

4

5r3rd

cos cos cos4

5

4 4

rr

r r= + = -` j

tan3

2r

3

2r2nd

tan tan tan3

2

3 3

rr

r r= - = -` j

cos10

9r

10

9r2nd

cos cos cos10

9

10 10

rr

r r= - =-` j

sin 10

19r

10

19r4

th

sin sin sin10

192 10 10

rr

r r= - = -` j

tan4

7r

4

7r4

thtan tan tan

4

72

4 4

rr

r r= - = -` j

sin10

r

10

r1

stsin

10

r

cos7

8r

7

8r3rd

cos cos cos7

8

7 7

rr

r r= + = -` j

tan8

15r

8

15r4

thtan tan tan

8

152

8 8

rr

r r= - = -` j


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M14


iasafracon iasadecimal Quadrant

2

r1.5707... 1

st

6r

0.5235... 1st

3

4r4.1887... 3

rd

4

3r2.3561... 2

nd

4

7r5.4977.. 4

th

Point r iPolarCoordinates

,r i^ hcosx r i= siny r i=

,x y^ hCoordinates

A 56r

,56r` j cos5

6 2

5 3r= 5 sin

6 2

5r= ,

25 3

25c m

B 42r ,

24 r` j cos4

20

r= 4sin

24

r= ,0 4^ h

C 36

7r ,673 r` j 3 cos

6

7

2

3 3r= - 3 sin

6

7

2

3r=- ,

23 3

23

- -c m

D 54

7r,47

5r` j cos5

4

7

2

5r= 5sin

4

7

2

5r=- ,

2

5

2

5-c m

E 2 r ,2 r^ h cos2 2r = - 2 sin 0r = ,2 0-^ h

26. Canyoucompletethetablebelow?

4

5

-

-

-3

-4

-5

-

--

-

3

2

1

1 2 3 4 55 4 2 1

1

-3

27. Completethetableforthefollowingpoints:


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M 15


,x y^ h Quandrant r x y2 2= + tanAx

y1=

- ` j i ,r i^ h

,3 4-^ h 4th 5r 3 42 2= + - =^ h . ...tan3

40 927

1 -= -

-

` j.A2 5 36r+ = , .5 5 36^ h

3 , 6^ h 1st 3r 6 32 2

= + =^ ^h h3

0.9 ...tan tan6

2 551 1

= =- -c m .A 0 96= , .3 0 96^ h

,6 8-^ h 2nd r 6 8 102 2= - + - =^ ^h h 0.927...tan6

81

-=-

- ` j .A 2 21r+ = , .10 2 21^ h

,12 9- -^ h 3rd r 12 9 152 2= - + - =^ ^h h 0.927...tan9

121

-

-=

- ` j .A 4 07r+ = , .15 4 07^ h

5, 0-^ h - 5r 5 02 2= - + =^ ^h h tan5

00

1

-=

- ` j r ,5 r^ h

,0 5^ h - 5r 0 52 2= + =^ ^h h undefinedtan0

51=

- ` j2r ,5

2r` j

,1 3- -^ h 3rd r 1 3 222

= - + - =^ ^h h tan1

3

3

1 r

-

-=

- c m A3

4r

r+ = ,2

34r` j

2 , 7-^ h 4th r 2 7 32 2

= + - =^ ^h h . ...tan2

71 0799

1 -= -

- c m .A 2 5 20r+ = , .3 5 20^ h

,r i^ h cosx r i= siny r i= ,x y^ h

,43r` j cos4

34

2

12

r= =` j sin4

34

2

32 3

r= =c m ,2 2 3^ h

,24r` j 1cos2

4

r= sin2

41

r= ,1 1^ h

,345r` j cos3

4

5

2

3r=- sin3

4

5

2

3r=- ,

2

3

2

3- -c m

,28

15r` j . ...cos2 815 1 8477r = . ...sin2 815 0 7653r = - (to d.p.). , . 21 85 0 77-^ h

,12r` j cos

20

r= sin

21

r= ,0 1^ h

,3 r^ h cos3 3r = - sin3 0r = ,3 0-^ h

,523r` j cos5

2

30

r= 5 5sin

2

3r= - 0, 5-^ h

28. Usetherulesforeachquadranttocompletethefollowingtable(to2d.p.):

29. Completethetablebelowtondtherectangularcoordinates(Setcalculatorstoradianmode):


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M16

Notes


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232-42072155.AUS M Trig Polar Coords Solutions AUS

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Transcript of 232-42072155.AUS M Trig Polar Coords Solutions AUS