finish this homework in 30 min and get 30 bucks

EXTREEMLY URGENT, NEEDS TO BE COMPLETED IN 30 MINUTES FROM THE MINUTE THIS IS POSTED!!I HAVE ATTACHED TWO FILES, ONE HAS THE QUESTIONS THE OTHER HAS THE TEXT FOR REFRENCE.

SEND: Work for Submission

Show all your workings clearly.

1. Two ships are observed from point O.

At a particular time their positions A and B are

as shown on the right.

The distance between the ships at this time is

A. 3.0 km

B. 3.2 km

C. 4.5 km

D. 9.7 km

E. 10.4 km

[VCAA 2005 Further Maths Exam 1]

2. The bearing of an aeroplane, X, from a control tower, T, is 055°. Another aeroplane, Y, is due east of control tower T. The bearing of aeroplane X from aeroplane Y is 302°.

The size of the angle TXY is

[VCAA 2005 Further Maths Exam 1]

A. 32°
B. 35°
C. 55°
D. 58°
E. 113°

A
B
C
4 km
6 km
N
3. A hiker walks 4 km from A on a bearing of 30º to a point B,
then 6 km on a bearing of 330º to a point C. The distance AC
in km is

A
B
C

D 6 sin 60º

4. Ship A and Ship B can both be seen from the lighthouse. Ship A is 5 km from the lighthouse,
on a bearing of 028o. Ship B is 5 km from Ship A, on a bearing of 130o.

(a) Two angles, x and y, are shown in the diagram.
(i) Determine the size of the angle x in degrees.
(ii) Determine the size of the angle y in degrees.
(b) Determine the bearing of the lighthouse from
Ship A.
(c) Determine the bearing of the lighthouse from
Ship B.

5. Starting from the camp at C, Tim takes a bearing of a mountain at M and notes it to be 25°.
He then walks 5 km to the hut at H and takes a second bearingof the same mountain and it is 345°.
(a) Work out the angles in the triangle CHM. Prove that it is a right angled triangle.
(b) From the mountain at M:
(i) what is the bearing of the camp? (ii) what is the bearing of the hut?
(c) How far is it (i) from the camp to the mountain (ii) from the hut to the mountain?
(d) Tim walks back to camp from the hut. What bearing does he follow?

(a) What is the angle of depression from D to B, correct to the nearest minute?
(b) How far did the boat travel from C to B, correct to the nearest metre?

7. Genie Construction is building a new shopping plaza on a plot of land that is a trapezium with the two parallel sides pointing north. The following is a diagram, which is not drawn to scale, of the plot:

160 m

230 m

Part 1

(a) Find the area and the perimeter of the site ABCD.

(b) Prove that the bearing of D from A is (to the nearest degree) 80

Part 2

A car park is to be made by running a straight line on a bearing of 25 until it meets the edge of the plot at E.

25
E

Find the area and the perimeter of the car park ABE.

22
6448cos120

++
o
52

75
0

25
0

5 km
345
0

C
M
H

160 m
230 m
A
D
C
B
120 m
N
o
o
o

A
D
C
B
120 m
N
25


o
30
sin
4

22
6448cos120

+-
o

PAGE

tice th

t 1.2º is

the same as 1º 12

‘

>Question 1: Estimating magnitude (size) of angles

o
Estimate the size of angle AB

in each of the diagrams shown.

‘

3 km

3.3 km

Question 2: Drawing angles

Draw an accurate diagram to represent an angle with

magnitude of:

75o

30o

30
0

120

A
B

N
O

210

300

C
Question 3: Calculations of sizes of angles an

lengths

Start

Finish

Bearing

8 km

8 (

opposite

)
6
(

adjacent

)



2338

6 km

8 km
d

Bearing



538



56
0

90
0

 35

56
0

4 km

35
0

3 km
Start
Finish

4 km
3 km
a
b
34
0

55
0

Start
Finish

5 km

8 km

Start
Finish

110

180

Camp

Springs

Hut

5 km

7 km

50
0

140

50
0

180
0

–140

= 40

50
0

140
0





50
0

=90

i. Which trig ratio would you use to find the distance DE?

ii.

What is the size of the following angles?

1. (DGE

2. (HGE

3. (GEF

iii. (GFE = 290.

What is the size of:

1. (FGH

2. (FGE

Using the above diagram, state whether each statement is TRUE or FALSE

i. GF2 = GE2 + EF2

ii.
sin139sin
distance GFdistance EF
EGF
=
°Ð

iii.
222
2cos41
EFGEGFGEGF
=+

–

´´´°

iv. area of triangle EFG = ½ ( EF (

400

Answers to Skills Check / Extra Resources

1 a.

400

115

o
o
o
90
40
50
=
+
=
+
b
a

o
50
=
a
2 a.

o
o
40
140
180
=
–
=
b

5 km
7 km
d

140
0

40
0

50
0

7 5



40
0

alternate



360

 40

0
=

320



50
0

3 a.

i. tan ratio

The triangle has a right angle at D and so

tan41
400
DE
=°
. Alternatively (since (DGE = 490)
tan49
DE
400
=°

ii.

1. (DGE = 490
2. (HGE = 410
3. (GEF = 1390
iii. 1. (FGH = 290
2. (FGE = 120

i. GF2 = GE2 + EF2 FALSE

The triangle is not right-angled and so Pythagoras’ rule cannot be used.

ii.
sin139sin
distance GFdistance EF
EGF
=
°Ð
TRUE

This is the correct application of the sine rule.

iii.
222
2cos41
EFGEGFGEGF
=+-´´´°
FALSE
The angle required is (FGE = 120, so using the cosine rule will give

222
2cos12
EFGEGFGEGF
=+-´´´°

iv. area of triangle EFG = ½ ( EF ( 400 TRUE

EF is the base of the triangle and 400 is the height perpendicular to EF,
so the rule Area = ½ ( base ( height can be used.

More information:

http://www.mathsisfun.com/angles.html

http://www.mathsisfun.com/algebra/trig-sine-law.html

http://www.mathsisfun.com/algebra/trig-cosine-law.html

http://www.mathsisfun.com/algebra/trig-solving-triangles.html

http://www.mathsisfun.com/algebra/trig-area-triangle-without-right-angle.html

More information:

https://www.mathsonline.com.au/

There are many lessons on this website covering the geometry/trigonometry module.

For some examples (once you have logged on), go to:

11 & 12 General > Measurement > Further Trigonometry > The Sine Rule: Finding a Side

11 & 12 General > Measurement > Further Trigonometry > The Cosine Rule: Finding a Side

Mt. Hederick

115
0

2 km

Hut 1

SEQ week \r 3 \h \* MERGEFORMAT Week 11

Applications of Geometry

& Trigonometry (I)

Angles of elevation & depression	Page 2	Triangulation	Page 14
Bearings & Directions	Page 5	Work for submission	Page 17
Examples of bearings	Page 9	Answers to exercises	Page 23

Reference:
ESSENTIAL Further Mathematics

Chapter 14

Key knowledge

· Angles of elevation and depression.

· Basic geometric and trigonometric concepts associated with orienteering and navigation.

· Concepts associated with triangulation.

Key skills

· Use angles of elevation and depression to solve different problems.

· Basic geometric and trigonometric concepts associated with problem solving.

· Use the triangulation concepts to solve problems in practical situations.

Tasks this week relate to outcomes 1, 2 and 3.

LESSON 1

ANGLES OF ELEVATION AND DEPRESSION

30
0

25
0

ANGLES OF ELEVATION:

Looking up from the horizontal gives

an angle of elevation. Angle of elevation

is measured from the horizontal upwards.

330

115
0

– 25

= 90

360
0

–330

=30

ANGLES OF DEPRESSION:

Looking down from the horizontal gives

an angle of depression. Angle of depression

is measured from the horizontal downwards.

Angles of elevation and depression are in a vertical plane

We can see from the following diagram that the angle of depression given from one location can give us the angle of elevation from the other position using the alternate angle law.

55
0

adjacent

opposite

55
0

180
0

–55

–90

= 45

a
30
0

270

Watch a powerpoint!

Insert your DECV course CD into your computer and then click:

Bearing of first hut from

second hut = 270

+ a

a
270
0

Hut1

Hut2

Further Mathematics Unit 4 > Further Resources > Angles of Elevation and Depression

Example 1

A pilot flying a plane at altitude 400 m, sees a small boat at an angle of depression of 1.2º

(a)
Draw a diagram

(b)
Find the horizontal distance of the boat to the plane.

Answers

A
B
C
80
0

160

A
B
C
60
0

110
0

(a)

(b)
We have a right- angled triangle. We are looking for side d (the adjacent side)

A
50
0

B
C
80
0

320
0

Use the tan ratio.

tan 1.2º =

adjacent
opposite
=
d
400

d =
0
2
.
1
tan
400

=
020947
.
0
400
= 19 096 m to the nearest metre
Example 2
A boat sights a lighthouse light 75 metres above sea level at an angle of elevation of 7.1º. How far is the boat from the lighthouse?
Answers

120
0

A
C
B
B
N
N

We have a right-angled triangle. We are looking for side d (the adjacent side)

6 km
9 km
60
0

150
0

P
Q
R
Use the tan ratio: tan 7.1º =
adjacent
opposite
=
d
75

d =
0
1
.
7
tan
75

=
124556
.
0
75
= 602 metres to the nearest metre.

Study Essential, Chapter 14.1 and the Examples 1, 2 and 3

Practice Exercise 1

(Answers at the end of this week)

1. Use your calculator to change:
(a)
20.6º into degrees and minutes (b)
42º 16′ into degrees, to one decimal place

(c)
14.8º into degrees and minutes
(d)
65º 43′ into degrees, to one decimal place.
2. From the top of a vertical cliff 130 m high, the angle of depression of a buoy at sea is 18º.
What is the distance of the buoy from the base of the cliff?

75
0

A
50
0

320
0

B
C
5 km

3. A 120 m tall tower is on top of a hill. From the base of the hill, the angle of elevation of the base of the tower is 9º, and 12º for the top of the tower.
(a)
Find (TAB, (ABC, (TBA, and (ATC
(b)
Find distance AB, to the nearest metre.
(c)
Find distance BC, the height of the hill, to the nearest metre.
4. A 2 m tall man standing at the edge of a 50 metre tall cliff sights two buoys with angles of depression 18º and 20º.
(a)
Draw a diagram.
(b)
Work out the distance between the buoys, correct to one decimal place.
5. A surveyor notes a church spire at an angle of elevation of 18º. She walks 150 metres to a fence and notes a second angle of elevation of 33º.
(a)
Draw a diagram.
(b)
Find the height of the spire to the nearest half metre.
(c)
Find the distance from the fence to the church, to the nearest half metre.
6. Try questions 1, 3, 5, 6, 9,10,11 and 18 of Exercise 14A, Essential textbook.
(Answers for these questions are at the back of your textbook. Contact your teacher if you need some help with solving these questions)

LESSON 2

BEARINGS AND DIRECTIONS

True Bearing
Trigonometry is used to determine directions and to calculate distances. The direction of one object from another in a horizontal plane can be determined by using bearings. The bearings are usually quoted in terms of an angle measured clockwise from North (and sometimes a capital T) and are known as True bearings.
In diagram 1, A is at 5 km from C on a bearing of 56º. This is a bearing of 056ºT. This bearing represents a direction of 56º (always measured in a clockwise direction) from North.

In diagram 2, B is at a distance of 3.3 km from C on a bearing of

310

º T. This is a bearing of 310ºT. This bearing represents a direction of 310º (again measured in a clockwise direction) from North.

35
0

S
8 km
125
0

330
0

T
V

38
0

b = 8
C A
B
60
0

S
A
N
145
0

60
0

A
N
145
0

60
0

B
180
0
145
0
=35
0

180
0
60
0
35
0
=85
0

A
B
S
20
35
0

85
0

60
0

A
145
0

35
0

360
0
35
0
=325
0

35
0

60
0

85
0

S
35
0

160 m
230 m
A
D
C
B
120 m
N

A
D
C
B
120 m
N
25


A
18
0

18
0

130 m
B
C

9
0

3
0

A
B
T
C
180
0
– 90
0
– 9
0
= 81
0

180
0
– 81
0
= 99
0

180
0
– 99
0
– 3
0
= 78
0

8 km
Start
Finish
5 km


N
d

Start
32
0

Bearing


Compass Bearing

Sometimes bearings are also quoted in terms of an angle measured East or West, or North or South.
In diagram 1, 056ºT could be expressed as N 56º E or E 34º N.
In diagram 2, the bearing 310ºT, can be expressed as W 40º N or N 50º W.
A bearing of N56º E means that you are initially pointing in a Northern direction; you rotate 56º towards the East and then head off in that direction.
A bearing of E34ºN means that you are initially pointing in a Eastern direction; you rotate 34º towards the North and then head off in that direction.

6 km
Start
Finish
10 km



30º 58′


Example 1

8
4
90º – 40º = 50º
90º – 75º = 15º

40
0

8 km
75
0

4 km
Start
Finish

Example 2

Watch two video lessons on How to Find the Bearing!
Insert your DECV course CD into your computer and then click:

8
4
50
0

15
0

x
y
(i) Further Mathematics Unit 4 > Further Resources > Bearings and Angles (I)

(ii) Further Mathematics Unit 4 > Further Resources > Bearings and Angles

e
8
50
0

15
0

d
(II)

Courtesy: www.mathsonline.com.au

Example 2

125
0

30
0

180º – 125º
= 55º
90º – 30º = 60º
Simon travels 6 km due North and then 8 km due East.
(a)
What is the bearing of the finish from the start?
(b)
To travel directly back to the start
(i)
What bearing will Simon need to follow?
(ii)
How far will he have to travel?

2 km
9 km
125
0

30
0

55
0

60
0

2 km
9 km
a
a
b  a
b
Start
Finish
Answers

Draw a diagram. The bearing of the finish point from the starting point is the angle ( shown.
We have the opposite and adjacent sides, so we use the tan ratio

320
0

55
0

5 km
10 km
tan
q
=
adjacent
opposite
=
6
8
= 1.3333

q
= tan–1 1.3333 = 53º 7′
8
4
¢
¢
= 53º 8’ (to nearest minutes).
The bearing of the finish point from the starting point is:
053º T (to nearest degree).
(b)

Finish
(i)
Draw a diagram to show the bearing we are looking for. To travel directly back to the start, Simon will follow the bearing shown.

The bearing is (180º + a)
a + b = 90º and we can find angle b from the triangle

b = 180º – 90º – 53º8′
= 36º 52′

a = 90º – 36º 52′

= 89º 60′ – 56º 52′

= 53º8′
The bearing is 180º + 53º8’= 233º 8′
Travelling back, Simon has to follow the bearing of 233º T (nearest degree)
(ii)
d is the distance travelled back to the
start. Use Pythagoras’ theorem:

d2 = 62 + 82 = 36 + 64 = 100

d =
100
= 10 km

Simon has to travel 10 km back to his starting position.

Example 3
Tanya hikes 4 km on a bearing of 56º, then 3 km on a bearing of 35º
(a)
Draw a diagram showing her journey.
(b)
How far north is she from her starting point?
Answers
(a)
Draw a diagram

55
0

90º – 55º = 35º
270
0

320º – 270º
= 50º

50
0

35
0

10 km
5 km
f
e

110
0

B
C
110
0

180
0

60
0

B
C
110
0

180
0
110
0
= 70
0

60
0

A
B
C
80
0

160
0

80
0

180
0
160
0
= 20
0

(b)
We have two right angled triangles
The distance north of the starting point is (a + b)
sin 34º =
4
a

Þ
a = 4 × sin 34º = 2.2368
sin 55º =
3
b

Þ
b = 3 × sin 55º = 2.4575

a + b = 2.2368 + 2.4575 = 4.6943

»
4.7 km (one decimal place).

Tanya is 4.7 km north from the starting point.

Practice Exercise 2

(Answers at the end of this week)

B
160
0

180
0

160
0

1. Costa travelled 5 km due south, then 8 km due west.
(a)
To travel directly back to the start

(i)
what bearing will Costa need to follow?
(ii)
How far will he have to travel on this bearing?

(b)
Tomas decides to take the shortest route from

A
50
0

B
C
320
0

80
0
50
0
= 30
0

50
0

360
0
320
0
= 40
0

40
0

start to finish. What bearing should he follow?
2. Tamara travelled 6 km due west, then 10 km due north.
(a)
To travel directly back to the start, what
bearing will she need to follow?
(b)
Shawn takes the shortest route from the start to the

finish. What bearing did he follow?
3. Samantha travelled 8 km on a bearing of 40º then 4 km on a bearing of 75º.
(a)
Draw a diagram of the journey.
(b)
How far north from the start was she?
(c)
How far east from the start was she?
4. Trent travelled 2 km on a bearing of 125º then 9 km on a bearing of 30º.
(a)
Draw a diagram of the journey.

(b)
How far north of the start is he?
5. Chris travels 5 km on a bearing of 320º, then 10 km on a bearing of 55º.
(a)
Draw a diagram of the journey

(b)
How far east of the start is he?

LESSON 3

MORE EXAMPLES OF BEARING

We’ll need to know some angle rules. There are two angle rules that we use in bearings

30
0

50
0
+ 40
0
= 90
0

180
0
90
0
30
0

= 60
0

A
B
C

A
50
0

B
180
0

50
0

B
C
40
0

180
0
 40
0
= 140
0

30
0

90
0

60
0

C
40
0

A
B

Let’s see some examples of how we use these rules.

Example 1
The hiking club hikes 7 km on a bearing

A
C
B
N
N
45
0

120
0
45
0
= 75
0

45
0

180
0
45
0
75
0

= 60
0

of 50° to the mineral springs, then 5 km
on a bearing of 140° to the hut.
(a)
Show that the angle at the springs is 90°
(b)
How far will the hike back to the camp be?
(c)
They are going to hike from the hut directly back to camp. What bearing should they follow?

Answers

A
C
45
0

180
0

(a) Draw a diagram and work out the angles.
Angle at the spring
b
a
+

A
B
60
0

180
0

B
C

60
0

150
0

P
Q
R
60
0

180
0
150
0

= 30
0

8 km
S
T
V
65
0

o
30
(b)
d is the distance back to the hut.

d is the hypotenuse.

Use Pythagoras’ theorem

d2 = 72 + 5 2 = 49 + 25 = 74

d =

8.6

74
=
(one decimal place)

The hike back from the hut to the camp is about 8.6km.

o
82
(c)

o
68

The bearing back to camp

= 320

° minus angle
a

We can use trig ratio to find angle a since we have proven that the triangle is a right-angled triangle. We have all three sides, so we can use either sin, cos or tan. I’ll use tan because
this uses the sides given in the question.

tan
a
=
1.4

5
7

adjacent
opposite

=
=

a
= tan (1 1.4 (
a
= 54° 28(
The bearing back to camp = 320° ( 54° 28(
= 319° 60( ( 540 28(

= 2650 32(

Example 2

Tod took a bearing from the first hut, of Mt.Hederick and noted it was 250. He then walked 2 km on a bearing of 115° to the next hut. He took a second bearing of Mt Henderick, and noted that it was 330°.
(a)
(i)
Show that the triangle is a right-angled triangle.

(ii)
Find the angle at Mt Hederick

(b)
(i)
How far is the first hut from Mt. Hedrick?

(ii)
How far is the second hut from Mt. Hederick?
(c)
What is the bearing of the first hut from the second hut?
Answers

(a)
Work out the angles

(i)

(ii)

(b)

(i)
Distance
d
is from the first hut to Mt. Hedrick
tan 55° =
d
2

hypotenuse
opposite
=

(
2
d

tan
1
=
o
d is now on the top line

o
55

tan
1
( 2 = d
d =
place.

decimal

one

to
km

1.4

55
tan

tan
2
=
¸
=
o

(ii) e2 = d2 + 22 = (1.4)2 + 4 = 1.96 + 4 = 5.96
e =
96
.
5
= 2.4 km to one decimal place.
(c)

Still confused with Bearings? Watch these video lessons
Insert your DECV course CD into your computer and then click:

75
0

25
0

5 km
345
0

C
M
H
(i) Further Mathematics Unit 4 > Further Resources > Compass Question (I)

(ii) Further Mathematics Unit 4 > Further Resources > Compass Question (II)

Courtesy: www.mathsonline.com.au

Practice Exercise 3

(Answers at the end of this week)

1. In the diagrams below work out
(i)
the angle at B
(ii)
the bearing of B from C

(a)

(b)

2. In the following diagram work out:
(a)
The angle at B
(b) The bearing of C from B
(c) The bearing of A from B
(d) The bearing of A from C

3. The bearing of B from A is 120°
C is Northeast of A
C is due north of B. What is the bearing of
(a)
A from C?
(b)
A from B?
(c)
B from C?

4. Tom and Tina hike 10 km from their camp on bearing of 040°, then 8 km on a bearing of 130°
(a) Work out the angles in the triangle. Show it is a right-angled triangle.
(b) From the hut, what is the bearing of the camp?
(c) How far is the camp from the hut?
5. The walking group walks 6 km from P on a bearing of 060°
to Q, then 9 km on a bearing of 150° to R.
(a) Show that the angle at Q is a right angle.
(b) They walk directly from R to P:
(i) How far is this, correct to one decimal place?
(ii) What bearing do they follow?
6. Carey starts at point A, and notes the bearing of the hill at B to be 50°. He hikes 5 km on a bearing of 075° to the hut at C. He takes a second bearing of the hill to be 320°.
(a) Find the angles in the triangle, and show that it is a right angled triangle.
(b) From the hill at B:
(i) What is the bearing of A?
(ii) What is the bearing of C?
(c) What is the bearing of A from C?
(d)
(i) How far is B from C
(ii) How far is A from B?
7. Chris starts at point S and takes a bearing of 125
o
to point V The bearing of the tower T from S is 035
o
. He moves 8 km to V and takes a second bearing of the tower T from V. It is 330°.
(a) Find the angles in the triangle and show that it is a right angled triangle.
(b) From the tower at T:
(i) what is the bearing of S?
(ii) what is the bearing of V?
(c) What is the bearing of S from V?
(d)
(i) How far is T from S?
(ii) How far is V from T?

LESSON 4

TRIANGULATION

Have you wondered how the surveyors can measure the distance between two mountains in
the countryside or how we can determine how far is that sail ship in the sea. These two cases
are examples of distances which cannot be measured directly because of physical restrictions. It is not possible to walk straight from one mountain to the next line nor to walk over the water to measure the distance to the ship.

Triangulation is a mathematical technique used to measure such distances. This method involves the use of the theory of solving triangles.

When we can use triangulation

Triangulation can be used in the following situations:
· A single point can be observed but can’t be reached or measured due to physical restrictions.
· Two points can be observed and neither can be reached or measured directly due to physical restrictions.
· Surveying using triangulation involves using a base line of known length and measuring the angles of points of interest (which may be inaccessible) from both ends of the base line

How we can use the triangulation method

In both situations we need to have:
· Two references points from which we can record the bearing of our inaccessible point or points.
· The distance between our two chosen references points.
· The bearing of one of these references points from the other must also be measured.

Example 1

1. Points A and B mark the position of two lighthouses. The lighthouse at A is 20 km north of the light house at B. The light keeper at A sees a ship on a bearing of 1450. The light keeper at B sees the same ship on a bearing of

600

(a) How far is the ship from each light house?
(b) The ship’s captain takes bearings of each light house.
(i) What is the bearing of the lighthouse at A from the ship?
(ii) What is the bearing of the lighthouse at B from the ship?

Answers

(a)

C
sin
c

A
sin
a
=

85
sin
20

sin
a
o
o
=

a =
km

11.5

35
sin

85
sin
20
=
´
o
o

The ship is 11.5 km from lighthouse B.

C
sin
c

sin
b
=

o
o
85
sin
20

60
sin
b
=

b =
km

17.4

60
sin

sin
20
=
´
o
o

The ship is 17.4 km from lighthouse A.

(b) (i)
From S, the bearing of
the lighthouse A is:
360° ( 35° = 325°

(b) (ii)
From S, the bearing of
lighthouse Bis:
360° ( 35° ( 85° = 240°

Study Essential Chapter 14.1 ( Triangulation) and Example 6

Practice Exercise 4

(Answers for question one at the end of this week and answers for question 2 are at the back of your textbook.

1. A surveyor has measured the angles to
An inaccessible point, from both ends of
a base line AB Angle CAB equals
o
68
and
angle ABC equals
o
82
. If AB =112 metres,
Find:
(a) The length of AC
(b) The length of BC.
(c) The area of triangle ABC.
2. Try the following questions from Exercise 14A,
of your textbook:
(i) Question 7, page 375.
(ii) Question 12, page 376.
(iii) Question 13, page 376.
(iv) Question 15, page 376.
(v) Question 17, page 376.

By now, your supervisor should have received the materials for the second School-Assessed Coursework (SAC 2) for Unit 4. Please check with your supervisor. If he/she has not yet received them, he/she must contact your teacher as soon as possible.

SEND: Work for Submission

Show all your workings clearly.
1.
Two ships are observed from point O.

At a particular time their positions A and B are
as shown on the right.
The distance between the ships at this time is

A.
3.0 km
B.
3.2 km

C.
4.5 km

D.
9.7 km

E.
10.4 km
2.
The bearing of an aeroplane, X, from a control tower, T, is 055°. Another aeroplane, Y, is
due east of control tower T. The bearing of aeroplane X from aeroplane Y is 302°.

The size of the angle TXY is

A.
32°

B.
35°
C.
55°

D.
58°

E.
113°
3.
A hiker walks 4 km from A on a bearing of 30º to a point B,
then 6 km on a bearing of 330º to a point C. The distance AC
in km is

o
30
sin
4

22
6448cos120

+-
o

22
6448cos120

++
o

D
6 sin 60º
E

4. Ship A and Ship B can both be seen from the lighthouse. Ship A is 5 km from the lighthouse,
on a bearing of 028o. Ship B is 5 km from Ship A, on a bearing of 130o.
(a) Two angles, x and y, are shown in the diagram.
(i) Determine the size of the angle x in degrees.
(ii) Determine the size of the angle y in degrees.
(b) Determine the bearing of the lighthouse from
Ship A.
(c) Determine the bearing of the lighthouse from
Ship B.
5. Starting from the camp at C, Tim takes a bearing of a mountain at M and notes it to be 25°.
He then walks 5 km to the hut at H and takes a second bearingof the same mountain and it is 345°.
(a) Work out the angles in the triangle CHM. Prove that it is a right angled triangle.
(b) From the mountain at M:
(i) what is the bearing of the camp? (ii) what is the bearing of the hut?
(c) How far is it (i) from the camp to the mountain (ii) from the hut to the mountain?
(d) Tim walks back to camp from the hut. What bearing does he follow?

6.
The base of a lighthouse D, is at the top of a cliff 168 metres above sea level. The angle of depression from D to a boat at C is 28o. The boat heads towards the base of the cliff, A, and stops at B. The distance AB is 128 metres.
(a)
What is the angle of depression from D to B, correct to the nearest minute?
(b)
How far did the boat travel from C to B, correct to the nearest metre?
7.
Genie Construction is building a new shopping plaza on a plot of land that is a trapezium
with the two parallel sides pointing north. The following is a diagram, which is not drawn to
scale, of the plot:

Part 1

(a) Find the area and the perimeter of the site ABCD.
(b) Prove that the bearing of D from A is (to the nearest degree) 80
o

Part 2

A car park is to be made by running a straight line on a bearing of 25
o
until it meets the edge of the plot at E.

Find the area and the perimeter of the car park ABE.

SEND: Work for Submission – Exam Practice

In Further Mathematics there are two end-of-year examinations. Examination 1 is a set of multiple-choice questions covering the core and the three modules. Geometry and Measurement is one of the modules.
In Exam 1, there are a total of 40 questions to be completed in 90 minutes with nine of these questions covering the Geometry and Measurement module. Each question should take, on average, 2 minutes. One mark is given for each correct answer.
In order to obtain practice working under such conditions, we suggest you complete the five multiple-choice questions below.
Tear out this sheet and include it with the rest of your submission for this week.

Restrict your time to 10 minutes only.

It is not necessary to show your working as credit is given for correct answers only.

Circle the letter beside the correct answer.

1
A man walks 4 km due east followed by 6 km due south. The bearing he must take to return to the start is closest to:

A
034o
B
056o
C
304o
D
326o
E
346o
2
A boat sails at a bearing of 265o from A to B.
What bearing would be taken from B to return to A?

A
005o
B
085o
C
090o
D
355o
E
275o
3
From a point on a cliff 200 m above sea level, the angle of depression to a boat is 40o.
The distance from the foot of the cliff to the boat to the nearest metre is:

A
238 m

B
168 m

C
153 m

D
261 m

E
311 m

4
A boat sails from a harbour on a bearing of 045o for 100 km. It then takes a bearing of 190o for 50 km. How far from the harbour is it, correct to the nearest km?

A
51 km

B
82 km

C
66 km

D
74 km

E
3437 km

5
A hiker walks 3.2 km on a bearing of 120( and then takes a bearing of 055( and walks 6 km. His bearing from the start is:

A
013(
B
077(
C
235(
D
257(
E
330(

Answers to Exercises

PRACTICE EXERCISE 1

1.
(a)
20.6º = 20º 36′
(b) 42º 16′ = 42.3º
(c) 14.8º = 14º 48′ (d) 65º 43′ = 65.7º
2. The distance of the buoy from shore = AB

tan 18º =
AB
130

0.3249 =
AB
130

AB =
3249
.
0
130

= 400.123
»
400 m.
3.
(a)

(b)

A
Sin
120

=
o
78
Sin
AB

AB =
o
3

sin
120
× sin 78º = 2242.77
»
2243 m

(c)

sin 9º =
2243
BC

BC = sin 9º × 2243 = 350.88
»
351 m

Height of the hill is 351m.
4.

(a)
(b) Find BD before finding AB, the distance between the buoys
Sin 20º =
BD
52

Þ
BD =
o
20
Sin
52

= 152.0378

o
2
Sin
AB

=
o
18
Sin
BD

AB =
o
18
Sin
0378
.
152

× sin 2º = 17.17
»
17.2 metres
Distance between the buyys is 17.2km

(a)

(b) Find BD before finding DC, the height of the spire.

o
18
Sin
BD

=
o
15
Sin
150

BD =
o
15
Sin
150

× sin 18º = 179.0924
Referring to triangle BDC, sin 33º =
BD
DC

sin 33º =
0924
.
179
DC

DC = 179.0924 × sin 33º
= 97.54
»
97.5 metres
The height of the spire is 97.5 m

Þ
BC = 179.0924 × cos 33º

= 150.199

»
150 metres to the nearest half metre

PRACTICE EXERCISE 2

1.
(a)

(i) The bearing required is angle b we find
the angle
q
in the triangle, then
( = 90º –
q

tan
q
=
8
5
= 0.625

q
= tan–1 0.625 = 32º 0′
( = 90º – 32º = 58º
To travel back to the start, Costa needs to follow the bearing
of 058º T
(ii) Let d = distance travelled directly back to the start
d2 = 82 + 52 = 89
Þ
d =
89
= 9.433
»
9.4 km
(b) Angle ( = 180º – 90º – 32º = 58º
Bearing = 180º + 58º = 238º
The bearing that Tomas has to follow from start to finish
is 238ºT
2
(a)
To travel straight back to the start, follow a bearing shown by angle (.
( +
q
= 180º
Þ
( = 180º –
q

Find angle
q

tan
q
=
10
6
= 0.6

q
= tan–1 0.6 = 30º 58′
Bearing (= 180º – 30º 58′
= 179º 60′ – 30º 58′
= 149º 2′

The required bearing is 149º 2′

(b)
To travel directly from start to finish, Shawn has to follow the
to follow the bearing shown
Bearing = 270º + angle (

(= 180º – 90º – 30º 58′
(= 59º 2′
Bearing = 270º + 59º 2′ = 329º 2′
3.
(a)

(b)
Distance travelled north = x + y
sin 50º =
8
x

Þ
x = 8 × sin 50º = 6.12836
sin 15º =
4
y

EMBED Equation.3
Þ
y = 4 × sin 15º = 1.03527
x + y = 7.1636
»
7.2 km.

Samantha is 7.2 km north from the starting point.
(c)
Distance travelled east = e + d
cos15º =
4
e

Þ
e = 4 × cos15º
= 3.8637
cos50º =
8
d

Þ
d = 8 × cos50º = 5.1423
e + d = 9.006
»
9 km.

Samantha is 9 km east from the starting point.
4.
(a)

(b)

Distance travelled north from the start is (b – a)
sin 60º =
9
b

Þ
b = 9 × sin 60º
= 7.794229
cos55º =
2
a

Þ
a = 2 × cos55º
= 1.147152
b – a = 6.647
»
6.6km

Trent is 6.6 km north from the starting point
5
(a)

(b)
The distance travelled east from the start equals (f – e)
cos35º =
10
f

EMBED Equation.3
Þ
f = 10 × cos35º = 8.19152
cos50º =
5
e

Þ
e = 5 × cos 50º= 3.213938
f – e = 4.977582
»
5.0 km ( Chris is 5 km east from the
starting point)
PRACTICE EXERCISE 3

(a)
(i)
(ii)
Angle B = 60° + 70° = 130° From C, bearing of B is
180° + 110° = 290°
(b) (i)

(ii)

Angle B = 80° + 20° = 100°
From C, the bearing of B is :
180° + 160° = 340°
2.
(a)
Angle at B = 50° + 40° = 90°

(b) (c)

From B, the bearing of C = 140°
From B, the bearing of A is:
180° + 50° = 230°

(d) From C, the bearing of A
360° ( 40° ( 60° = 260°

3.
North east is 45°
(a) From C, the bearing of A is 180° + 45° = 225° (b) From B, the bearing of A is: 360° ( 60° = 300°)

4.
(a)
Angle at M = 40° + 50° = 90°
(b) From Hut H, the bearing of camp C is:
360° ( 50° ( ( = 310° ( (,
Find angle (.
tan ( =
8
10
= 1.25
( = tan ( 1 1.25 = 51° 20(
Bearing = 310° ( 51° 20(
= 258° 40(
(c) HC2 = 102 + 82 = 164
HC =
164
= 12.8 km

5.
(a)
Angle at Q = 60° + 30° = 90°
(b)

(ii)

PR2 = 92 + 62 = 117 ( Pythagoras’ theorem)
Bearing is 360° ( 30° ( ( = 330° ( (
PR =
117
= 10.8 km (one decimal place)
Find (

tan ( =
9
6
= 0.666

( = tan (1 0.666 . . . = 33° 41(

Bearing = 329° 60( ( 33° 41( = 296° 19(

They have to follow a bearing of 296° 19(
6.
(a)

(b)

(i)

(ii)
From B the bearing of A is 180°+ 50°= 230°
From B the bearing of C is 140°

(c)

From C the bearing of A is 255°
(d)

(i) cos 65° =
5
BC
BC = 5 ( cos 65 = 2.1km
(ii) sin 65° =
5
AB
AB = 5 ( sin 65° = 4.5 km.
7.
(a)

(b)
(i)
(ii)
From T, the bearing of S From T, the bearing of V
is 180° + 35° = 215° 180° ( 30° = 150°
(c)

From V, the bearing of S is 360° ( 25° ( 30° = 305°

(d)
tan 65° =
ST
8

ST =
o
65
tan
8
= 3.7 km
sin 65° =
TV
8

TV =
°
65

sin
8
= 8.8 km
PRACTICE EXERCISE 4

1.
(a) Angle ACB =
o
o
o
o
30
82
68
180
=
–
–

To find AC, we use the sine rule:

C
AB
B
AC
sin
sin
=

o
o
30
sin
112
82
sin
=
AC

m
AC
AC
222
82
.
221
30
sin
82
sin
112
82
sin
112
30
sin
»
=
´
=
´
=
´
o
o
o
o

(b) To find BC we can use the cosine rule:

m
BC
BC
BC
AC
AB
AC
AB
BC
208
56
.
43199
563
.
43199
3746
.
0
222
112
2
222
112
68
cos
2
2
2
2
2
2
2
2
»
=
=
´
´
´
–
+
=
´
´
´
–
+
=
o

=
927
.
0
222
112
2
1
´
´
´

= 11526.7
2
m

2. Answers for these questions are at the back of your textbook.
Checklist
This week you should have submitted this work to me.
Please tick the items you have sent, and keep this as your record:
· Work for submission.
· Work for submission – Exam Practice.
END OF WEEK 11

315 Clarendon Street, Thornbury 3071

Telephone (03) 8480 0000

FAX (03) 9416 8371 (Despatch)

Toll free (1800) 133 511

SCHOOL NO.

STUDENT NUMBER ___________________

SCHOOL NAME _______________________

STUDENT NAME ______________________

Fix your student barcode
label over this space.

84111
[ DOCVARIABLE “SubjectCode” \* MERGEFORMAT 84111]

SUBJECT

DOCVARIABLE “Subject” \* MERGEFORMAT Further Mathematics Unit 4

DOCVARIABLE “SubjectCode” \* MERGEFORMAT
[ DOCVARIABLE “SubjectCode” \* MERGEFORMAT ZX]

YEAR/LEVEL

12 DOCVARIABLE “Year” \* MERGEFORMAT

WEEK

TEACHER

________________________

PLEASE ATTACH WORK TO BE SENT.

NOTE: Please write your number on each page of your work which is attached to this page.

SEND

Please check that you have attached:
· Work for submission.
· Work for submission – Exam Practice.
If you have not included any of these items, please explain why not.
_____________________________________________________________________________
_____________________________________________________________________________
_____________________________________________________________________________

Use the space on the back of this sheet if you have any questions you would like to ask, or problems with your work that you would like to share with your teacher.

YOUR QUESTIONS AND COMMENTS

DISTANCE EDUCATION CENTRE TEACHER’S COMMENTS

DISTANCE EDUCATION CENTRE TEACHER

SKILLS CHECK for Week 11
Refresh your memory by attempting the questions below. You will need these skills when working through this week’s lessons. Answers are given on the next page.

GH is parallel to DF

400 m

410

300

750

1300

N.B. If you have further queries, please contact your teacher.

Boat B

1.20

Plane P

400 m

400 m
Opposite side

adjacent side (d)

�

75 m
Opposite side

d
adjacent side

7.10

75 m

7.10

Notice that 7.1� EMBED Equation.3 ��is the same as � EMBED Equation.3 ��.

9 0

12 0

120 m

North

�

Diagram 2

North

56 0

5 km

East

Clockwise direction

Diagram 1

When dealing with problems that involve bearings, always draw a set of axes at each reference point

�

The bearing of A from O is 030º T The bearing of C from O is 210º T
The bearing of B from O is 120º T The bearing of D from O is 300º T

�

180º = 179º 60′
179º 60′ – 143º 8′ = 36º 52′
Subtract degrees and minutes separately

�

550

�

300

600

Rule 1:
Alternate angles are equal

1800

1200

600

1800

1200

�

Rule 2:
Angles on a straight line add up to 1800

�

� EMBED Equation.3 ��

� EMBED Equation.3 ��(alternate angles)
� EMBED Equation.3 ��(supplementary angles)

�

330o

The angle at Mt. Hederick is 25° + 30° = 55°

�

The angle at the first
hut is 90o, hence the triangle is a right-angled triangle.

�

The bearing of the first hut from the second hut is (270° + a)
Angle a = 90° ( 35° ( 30°
= 90° ( 65° = 25°
The bearing is
270° + 25° = 295°

�

350

Hut 1

Hut 2

�

400

1300

8 km

10 km

�

In this diagram point B is the inaccessible point, but
we know the following:
The distance between A and C.
The bearing of A from C and C from A.
The bearings of B from A and C

With the information above, we can calculate the
Distance AB &AC or even calculate the area of
Triangle ABC

�

In this diagram, we know the following:
The distance between A and B.
The bearing of A from B or of B from A.
The bearings of C and D from A.
The bearings of C and D from B

Points D & C are inaccessible points, but with the
information above, we can calculate distances AD,
AC, BD and BC

�

20 km

�

� EMBED MS_ClipArt_Gallery ��

[VCAA 2005 Further Maths Exam 1]

4 km

6 km

�

160 m

230 m

25�EMBED Equation.3��

�

120 m

3 0

99 0

78 0

9 0

2243 m

18 0

20 0

1800 – 200 = 1600

1800 – 180 – 1600 = 20

50 + 2 = 52 m

20 0

52 m

18 0

2 0

160 0

152.0378 m

1800 – 180 – 1470 = 150

18 0

1800 – 330 = 1470

150 m

Brick fence

18 0

33 0

Top of
Spire

150 m

Church

18 0

147 0

15 0

b =150

33 0

179.0929

�

Start

Finish

�
�

Start

�

600

�

500

�

400

1300

400

1300

500

10 km

8 km

�

6 km

9 km

6 km

9 km

900

1500

300

900

250

1800 (900 (250= 650

500

750(500= 250

3200

500

3600 (3200 = 400

400

500

400

1800 (400 = 1400

400

500

1800

400

650

3600( 400 ( 650 = 2550

5 km

250

650

350

1250( 350 = 900

3300

350

3600( 3300 = 300

300

8 km

1800( 900 (650= 250

650

900

350

1800

1800( 300 = 1500

900

350

300

250

300

250

�

� EMBED Equation.3 ��

112m

_953102613.unknown

_955801827.unknown

_1205050541.unknown

_1207137146.unknown

_1352105575.unknown

_1352107187.unknown

_1352107245.unknown

_1352108282.unknown

_1352105586.unknown

_1207383186.unknown

_1298460140.unknown

_1330340713.unknown

_1298460157.unknown

_1207647060.unknown

_1207739142.unknown

_1207384438.unknown

_1207379443.unknown

_1207381186.unknown

_1207379540.unknown

_1207379220.unknown

_1205691745.unknown

_1205916909.unknown

_1207127137.unknown

_1207127475.unknown

_1205916999.unknown

_1205918235.unknown

_1205918389.unknown

_1205918144.unknown

_1205916931.unknown

_1205691919.unknown

_1205916397.unknown

_1205916657.unknown

_1205916127.unknown

_1205916067.unknown

_1205691826.unknown

_1205050691.unknown

_1205050788.unknown

_1205050608.unknown

_955802029.unknown

_980102952.unknown

_1018700082.unknown

_1204887750.unknown

_1204887862.unknown

_1204841936.unknown

_1204841972.unknown

_980102953.unknown

_956056519.unknown

_956056556.unknown

_980102951.unknown

_955807649.unknown

_956050891.unknown

_955807580.unknown

_955801876.unknown

_955801908.unknown

_955801864.unknown

_953109510.unknown

_953109819.unknown

_953110133.unknown

_953110216.unknown

_953110376.unknown

_953110496.unknown

_953110599.unknown

_953110287.unknown

_953110166.unknown

_953109975.unknown

_953110050.unknown

_953109886.unknown

_953109702.unknown

_953109763.unknown

_953109632.unknown

_953103321.unknown

_953109360.unknown

_953109392.unknown

_953109314.unknown

_953103105.unknown

_953103158.unknown

_953102718.unknown

_952928933.unknown

_952929745.unknown

_952930027.unknown

_952930307.unknown

_952932062.unknown

_953030278.unknown

_953036892.unknown

_953037549.unknown

_953030895.unknown

_952934020.unknown

_952934135.unknown

_952931693.unknown

_952931989.unknown

_952930419.unknown

_952930239.unknown

_952930302.unknown

_952930111.unknown

_952929858.unknown

_952929935.unknown

_952929797.unknown

_952929038.unknown

_952929575.unknown

_952929691.unknown

_952929493.unknown

_952928995.unknown

_952929011.unknown

_952928953.unknown

_952927578.unknown

_952928518.unknown

_952928720.unknown

_952928892.unknown

_952928695.unknown

_952928467.unknown

_952928486.unknown

_952928427.unknown

_952928387.unknown

_952856815.unknown

_952857079.unknown

_952857177.unknown

_952924890.unknown

_952856992.unknown

_952852662.unknown

_952852834.unknown

_952847309.unknown

_952852505.unknown

_948192794.unknown

Turn in your highest-quality paper
Get a qualified writer to help you with

“ finish this homework in 30 min and get 30 bucks ”

Get high-quality paper

Guarantee! All work is written by expert writers!

Still stressed from student homework?

Get quality assistance from academic writers!

Order now