1

Contents

1. General Information2. Finding Sides 3. Finding Angles4. Bearings5. Sine Rule6. Area Formula7. Cosine Rule - Side8. Cosine Rule – Angle9. Radial Surveys

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General Mathematic(HSC)

3

3.1 Calculating Trig Ratios (1/11)

You can use your calculator to find trigonometric ratios.

When finding the angle we need to show working

Example:Example:

Sin Sin θθ = 0.649 = 0.649

θθ = Sin = Sin-1-1(0.649)(0.649)

= 40.4662= 40.4662oo

Use Use

DMS or DMS or o o ’ ’’’ ’’= 40= 40oo 27’ 58’’ 27’ 58’’

Degrees

Minutes

Seconds

4

3.1 The Trigonometric Ratios (2/11)

α

OppositeOppositeAdjacentAdjacent HypotenuseHypotenuse

5

3.1 The Trigonometric Ratios (3/11)

OppositeOppositeAdjacentAdjacent HypotenuseHypotenuse

β

6

3.1 The Trigonometric Ratios (4/11)

θθ

oo hh

aa

SSin θ =

CCos θ =

TTan θ =

oh

ah

oa

SS00HHCCAAHHTT00AA

omeomeldldagsagsananlwayslwaysideideheirheirldldgege

7

3.1 The Trigonometric Ratios (5/11)

θθ

oo hh

aa

θθ

oo hh

aa

8

3.1 The Trigonometric Ratios (6/11)

4545oo

13cm13cm

xx

h

Cos 45o = x13a

Because we have aa and hhwe must use CosCos.

x13x1313x13x

x = 13 x Cos 45o

≈ 9.192 388

≈ 9.2cm

9

3.1 The Trigonometric Ratios (7/11)

3535oo

10m10mxx

h

Sin 35o = x10

o

Because we have oo and hhwe must use SinSin.

x10x1010x10x

x = 10 x Sin 35o

≈ 5.735 764

≈ 5.7m

10

3.1 Finding an Unknown Side (8/11)

4040oo

12m12m

pp

aTan 40o = p

12

oBecause we have oo and aa

we must use TanTan.

x12x1212x12x

p = 12 x Tan 40o

≈ 10.069 195

≈ 10.1m

11

3.1 Finding an Unknown Side (9/11)

5050oo

3m3m ddh

Sin 50o = 3d

o

Because we have oo and hhwe must use SinSin.

x dx dd xd x

d = 3 ÷ Sin 50o ≈ 3.916 221

≈ 3.9 m

d x Sin 50o = 3÷ Sin 50÷ Sin 50oo ÷ Sin 50÷ Sin 50oo

12

3.1 Finding an Unknown Side (10/11)

6060oo

7m7m

mm aCos 60o = 7

m

h

Because we have aa and hhwe must use CosCos.

x mx mm xm x

m = 7 ÷ Cos 60o ≈ 14.0 m

m x Cos 60o = 7÷ Cos 60÷ Cos 60oo ÷ Cos 60÷ Cos 60oo

13

3.1 Finding an Unknown Side (11/11)

6060oo

7m7m

ww aTan 60o =10

m

o

Because we have oo and aawe must use TanTan.

x wx ww xw x

w = 10 ÷ Tan 60o ≈ 5.773 502≈ 5.8 m

w x Tan 60o = 10÷ Tan 60÷ Tan 60oo ÷ Tan 60÷ Tan 60oo

14

3.1 Finding Angles (1/3)

θθoo

7m7m3m3mh

o

Because we have oo and hhwe must use SinSin.

Sin θo = 37

θo = Sin-1( )37

= 25.376 933 525Use Use

DMS or DMS or o o ’ ’’’ ’’Degrees

Minutes Seconds

= 25o 22’ 37’’

15

3.1 Finding Angles – (2/3)

θθoo

5m5m

2m2m

h

a

Because we have aa and hhwe must use CosCos.

Cos θo = 25

θo = Cos-1( )25

= 66.421 821 522Use Use

DMS or DMS or o o ’ ’’’ ’’Degrees

Minutes Seconds

= 66o 25’ 31’’

16

3.1 Finding Angles (3/3)

θθoo

6m6m

3m3m

a

o

Because we have oo and aawe must use TanTan.

Tan θo =36

θo = Tan-1( )36

= 26.565 051 177Use Use

DMS or DMS or o o ’ ’’’ ’’ = 26o 33’ 54’’

17

3.2 Bearing (1/4)

•Bearings are used to describe direction.

•Compass bearings have four main directions and four middledirections

NE

SESW

NW

E

N

S

W

•True Bearings are more specific, using a 3-digit angle clockwise from North.

Web|Flash Web|FlashPractice 1 Practice 2

Web|FlashPractice 3

18

3.2 Bearing Ex 1 (2/4)

Ship AA is 17km west of a lighthouse L.

AA..

..BB

..LL

Ship BB is due south of the lighthouse L.

Ship BB is SESE of ship A.

45o

NN

17 km

Calculate distance d from Ship AA to ship B.

ddCos 45o =17d

d = 17 ÷ Cos 45o

≈ 24 km

19

3.2 Bearing Ex 2 (3/4)

A Ship sails for 20 km on a bearing of 130sails for 20 km on a bearing of 130oo..

130130oo

How far south is the ship from its starting point??

20 km20 kmdd 5050oo

Cos 50o = d20

d = 20 x Cos 50o

≈ 12.9 km

20

3.2 Bearing Ex 3 (4/4)

Mary walks 3.4 km east, then 1.3 km south.

3.4 km3.4 km

1.3 km1.3 km

θθoo

Find Mary’s bearing from her starting position.

Tan θo =1.33.4

θo = Tan-1 ( )1.3 3.4

9090oo

≈ 21o

Bearing = 90o + 21o

= 111o

21

3.3 The Sine Rule – Finding a Side (1/1)

aSin ASin A

AA

aa

BB

bb

CC

cc

= bSin BSin B

= cSin CSin C

Web|Flash

To use the Sine Rule to find a side …we need two angles and …the sides opposite them.

22

3.4 The Sine Rule – Finding an Angle (1/2)

Sin Aaa

AA

aa

BB

bb

CC

cc

= Sin Bbb

= Sin Ccc

To use the Sine Rule to find an angle …we need two angles and …the sides opposite them.

23

3.4 The Sine Rule – Finding an Angle (2/2)

Sin Aaa

AA

5 cm5 cm4040oo

4 cm4 cm= Sin B

bbSin A

55= Sin 40

44

Sin A = 5 x Sin 4044

A = 5 x Sin 4044

Sin-1 ( )

X 5 X 5

≈ 54o

24

3.5 Find the Area using Sine (1/)

aa

bb

CC

Web|Flash

Area = ab x Sin C12

Area = ab x Sin C

25

3.5 Find the Area using Sine (2/)

12

= x 4 x 412

x Sin 60o

≈ 6.9 cm2

60o

26

3.5 Find the Area using Sine (3/)

12

= x 4 12

= 7.2 cm2

x 3.6

Area = bh

27

3.5 Find the Area using Sine (3/)

Why the Difference?Area = ab x Sin C1

20.5 x 3.95 x 3.95 x Sin 59.5o

a = 4,b = 4,c = 60o

Lower Limit = ≈ 6.70.5 x 4.05 x 4.05 x Sin 60.5oUpper Limit = ≈ 7.1

Area = bh12

0.5 x 3.95 x 3.55

b = 4,h = 3.6

Lower Limit = = 7.00.5 x 4.05 x 3.65Upper Limit = = 7.4

The error in both calculations overlap so both answers could be correct within the error

limits.

Error!!!Error!!!

28

3.6 The Cosine Rule – Find a Side (1/2)

AA

aa

BB

bb

CC

cc

Web|Flash

c2 = a2 + b2 – 2ab Cos C

b2 = a2 + c2 – 2ac Cos B

a2 = b2 + c2 – 2bc Cos A

29

3.6 The Cosine Rule – Find a Side (2/2)

77

66

3030oo

cc

Web|Flash

c2 = a2 + b2 – 2ab Cos C

c2 = 62+ 72– 2 x 6 x 7x Cos 30

c = 62 + 72 – 2 x 6 x 7 x Cos 30

= 3.500552254≈ 3.5

30

3.7 The Cosine Rule – Find an Angle (1/2)

AA

aa

BB

bb

CC

cc

Cos C =a2 + b2 – c2

2ab

Cos B =a2 + c2 – b2

2ac

Cos A =b2 + c2 – a2

2bc

Cos-1( )2x7x5

2x7x5

31

3.7 The Cosine Rule – Find an Angle (2/2)

AA

77

BB

55

CCoo

66Cos C =a2 + b2 – c2

2ab

Cos C = + 5272 – 62

Cos C = + 5272 – 62Cos-1

C = 57.12165044o

C ≈ 57o

32

3.8 The Radial Survey (1/3)

A survey where angles and distances are measured from a point

030o

140o

245o

310o

N

45m

65m35m

50m

110o

105o65o

360o-310o+30o

65o

33

3.8 The Radial Survey – Perimeter (2/3)

030o

140o

245o

310o

45m

65m35m

50m

110o

105o65o

65o

d1

d2

d3

d4

d2 = a2 + b2 – 2ab Cos C

d12 = 502 + 452 – 2x50x45 Cos 65

d1 = 502 + 452 – 2x50x45 Cos 65

d1 ≈ 51.217358

d1 ≈ 51 m

Repeat for dd22, dd33 and dd44.

dd22 ≈ 91 m dd33 ≈ 81 m

dd44 ≈ 47 m

Perimeter ≈ dd11++ dd2 2 ++ dd3 3 ++ dd44

≈ 51 + 91 +81 +47

≈ 270 m

34

3.8 The Radial Survey – Area (3/3)030o

140o

245o

310o

45m

65m35m

50m

110o

105o65o

65o

A1

A2

A3

A4

Area = ab x Sin C12

A1 = x 50 x 45 x Sin 6512

≈ 1019.596 260

≈ 1020

Repeat for AA22, AA33 and AA44.

AA22 ≈ 1374 m2 AA33 ≈ 1099 m2

AA44 ≈ 793 m2

Area ≈ AA11++ AA2 2 ++ AA3 3 ++ AA44

≈ 1020 + 1374 +1099 +793

≈ 4286 m2