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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.
Question 1: Estimating magnitude (size) of angles
Estimate the size of angle ABC in each of the diagrams shown.
a.
b.
A
C
C
c.
A
A
A
C
B
B
A
C
B
A
A
B
C
BA
B
A
Question 2: Drawing angles
A
A
Draw an accurate diagram to represent an angle with a magnitude of:
a.
30o
b. 75o
c.
130o
Question 3: Calculations of sizes of angles and lengths
G
H
GH is parallel to DF
400 m
410
D
a.
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
F
E
b.
Using the above diagram, state whether
each statement is TRUE or FALSE
i. GF2 = GE2 + EF2
ii.
distance GF
sin139

distance EF
sin EGF
iii. EF2  GE2  GF2  2  GE  GF  cos 41
iv. area of triangle EFG = ½  EF  400
Answers to Skills Check / Extra Resources
1 a.
600
400
b.
2 a.
b.
1150
c.
c.
A
A
300
750
B
A
1300
B
A
3 a.
i. tan ratio
C
B at D andAso
The triangle has a right angle
A
400
 tan 41 .
Alternatively (since DGE = 490)
DE
ii. 1. DGE = 490
2. HGE = 410
3. GEF = 1390
C
DE A
400
 tan 49
iii. 1. FGH = 290
2. FGE = 120
b.
i. GF2 = GE2 + EF2 FALSE
The triangle is not right-angled and so Pythagoras’ rule cannot be used.
ii.
distance GF

distance EF
TRUE
sin139
sin EGF
This is the correct application of the sine rule.
iii. EF2  GE2  GF2  2  GE  GF  cos 41 FALSE
The angle required is FGE = 120, so using the cosine rule will give
EF2  GE2  GF2  2  GE  GF  cos12
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
N.B. If you have further queries, please contact your teacher.
A
Week 11
Applications of Geometry
& Trigonometry (I)
Angles of elevation & depression
Bearings & Directions
Examples of bearings
Page 2
Page 5
Page 9
Triangulation
Work for submission
Answers to exercises
Page 14
Page 17
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.
Week 11 Page 2
LESSON 1
ANGLES OF ELEVATION AND DEPRESSION
ANGLES OF ELEVATION:
Looking up from the horizontal gives
an angle of elevation. Angle of elevation
is measured from the horizontal upwards.
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.
Watch a powerpoint!
Insert your DECV course CD into your computer and then click:
Further Mathematics Unit 4 > Further Resources > Angles of Elevation and Depression
Week 11 Page 3
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)
Plane P
1.20
400 m
400 m
Opposite side
1.20
adjacent side (d)
Boat B
(b)
We have a right- angled triangle. We are looking for side d (the adjacent side)
Notice that 1.2º is
the same as 1º 12'
Use the tan ratio.
tan 1.2º =
d =
opposite 400
=
adjacent
d
400 = 400 = 19 096 m to the nearest metre
tan 1.20 0.020947
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
75 m
75 m
Opposite side
7.10
7.10
B
d
adjacent side
B
We have a right-angled triangle. We are looking for side d (the adjacent side)
Use the tan ratio: tan 7.1º =
d=
opposite 75
=
adjacent
d
Notice that 7.1  is the same as 7  6' .
75
75 =
= 602 metres to the nearest metre.
tan 7.10 0.124556
Week 11 Page 4
Study Essential, Chapter 14.1 and the Examples 1, 2 and 3
Practice Exercise 1
(Answers at the end of this week)
1.
2.
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.
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?
120 m
T
3.
A 120 m tall tower is on top of a hill.
From the base of the hill, the angle of
B
elevation of the base of the tower is
9º, and 12º for the top of the tower.
(a)
Find TAB, ABC, TBA,
12 0
90
and ATC
C
A
4.
5.
(b)
Find distance AB, to the nearest metre.
(c)
Find distance BC, the height of the hill, to the nearest metre.
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.
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)
(b)
(c)
6.
Draw a diagram.
Find the height of the spire to the nearest half metre.
Find the distance from the fence to the church, to the nearest half metre.
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)
Week 11 Page 5
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 310º T. This is a bearing of
310ºT. This bearing represents a direction of 310º (again measured in a clockwise direction)
from North.
North
B
3.3 km
B
North
A
56 0
C
3100
5 km
3 km
East
C
Clockwise
direction
A
Diagram 1
Diagram 2
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.
When dealing with problems that involve bearings, always draw a set of
axes at each reference point
Week 11 Page 6
Example 1
N
D
A
N
300
2100
O
3000
1200
O
B
C
The bearing of A from O is 030º T
Example
2 of B from O is 120º T
The bearing
The bearing of C from O is 210º T
The bearing of D from O is 300º T
Watch two video lessons on How to Find the Bearing!
Insert your DECV course CD into your computer and then click:
(i) Further Mathematics Unit 4 > Further Resources > Bearings and Angles (I)
(ii) Further Mathematics Unit 4 > Further Resources > Bearings and Angles (II)
Courtesy: www.mathsonline.com.au
Example 2
Simon travels 6 km due North and then 8 km due East.
N
(a)
What is the bearing of the finish from
the start?
(b)
To travel directly back to the start
8 km
Finish
6 km
(i)
(ii)
What bearing will Simon need to follow?
How far will he have to travel?
Start
Bearing
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
tan  =
opposite 8
= = 1.3333
adjacent 6
 = tan–1 1.3333 = 53º 7' 48 = 53º 8’ (to nearest minutes).
The bearing of the finish point from the starting point is:
053º T (to nearest degree).
8 ( opposite)
6
(adjacent)

Week 11 Page 7
(b)
(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.
Bearing

The bearing is (180º + a)
a + b = 90º and we can find angle b from the
triangle
1800
b
a
538
b = 180º – 90º – 53º8'
= 36º 52'
180º = 179º 60'
a = 90º – 36º 52'
179º 60' – 143º 8' = 36º 52'
= 89º 60' – 56º 52'
Subtract degrees and minutes separately
= 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:
8 km
2338
d2 = 62 + 82 = 36 + 64 = 100
d = 100 = 10 km
Simon has to travel 10 km back to his starting position.
6 km
d
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
Finish
350
350
3 km
560
560
Start
4 km
900  350 = 550
900  560 = 340
5
Week 11 Page 8
(b)
We have two right angled triangles
The distance north of the starting point is (a + b)
a
 a = 4 × sin 34º = 2.2368
4
b
sin 55º =
 b = 3 × sin 55º = 2.4575
3
sin 34º =
a + b = 2.2368 + 2.4575 = 4.6943
 4.7 km (one decimal place).
3 km
b
550
4 km
340
Tanya is 4.7 km north from the starting point.
a
Practice Exercise 2
(Answers at the end of this week)
1.
Start
Costa travelled 5 km due south, then 8 km due west.
(a)
(i)
(ii)
To travel directly back to the start
what bearing will Costa need to follow?
How far will he have to travel on this bearing?
5 km
Finish
(b)
2.
3.
4.
5.
8 km
Tomas decides to take the shortest route from
start to finish. What bearing should he follow?
Finish
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?
10 km
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?
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?
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?
Start
Week 11 Page 9
MORE EXAMPLES OF BEARING
LESSON 3
We’ll need to know some angle rules. There are two angle rules that we use in bearings
N
300
300
60
N 0
0
60
0
0
300
30
1800
0
70
0
180
0
0
60
1
2
0
1200
110
Rule 1:
Alternate angles are equal
0
1
8
0
0
Rule 2:
Angles6 on a straight line
add up0 to 1800
0
0
Let’s see some examples of how we use these rules.
Example 1
The hiking club hikes 7 km on a bearing
of 50° to the mineral springs, then 5 km
on a bearing of 140° to the hut.
(a)
(b)
(c)
Springs
Show that the angle at the springs is 90°
How far will the hike back to the camp be?
They are going to hike from the hut directly back to
camp. What bearing should they follow?
500
7 km
1400
5 km
Hut
Camp
Answers
(a) Draw a diagram and work out the angles.
Angle at the spring   
140
500

0
1400
500
500
1800 –1400 = 400

  50  (alternate angles)
  180  140  40 (supplementary angles)
500 + 400
=900
    50  40  90
Week 11 Page 10
(b)
d is the distance back to the hut.
d is the hypotenuse.
7 km
Use Pythagoras’ theorem
5 km
d2 = 72 + 5 2 = 49 + 25 = 74
d = 74  8.6 km (one decimal place)
d
The hike back from the hut to the camp is about 8.6km.
(c)
1400
7
500 400

40  alternate
5
500 400

3600  400 = 3200
The bearing back to camp = 320° minus angle 
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.
opposite 7
tan  =
  1.4
adjacent 5
 = tan 1 1.4   = 54° 28
The bearing back to camp = 320°  54° 28
= 319° 60  540 28
= 2650 32
Mt. Hederick
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)
(ii)
Show that the triangle is a right-angled triangle.
Find the angle at Mt Hederick
(b)
(i)
(ii)
How far is the first hut from Mt. Hedrick?
How far is the second hut from Mt. Hederick?
(c)
What is the bearing of the first hut from the second hut?
250
Hut 1
Hut 1
1150
2 km
330o
Week 11 Page 11
Answers
(a)
Work out the angles
(i)
The angle at Mt. Hederick
is 25° + 30° = 55°
(ii)
The angle at the first
hut is 90o, hence the triangle is
a right-angled triangle.
250
300
250
1150 – 250 = 900
3600 –3300 =300
300
3300
(b)
(i)
Distance d is from the first hut to Mt. Hedrick
d
tan 55° =

opposite
2

hypotenuse d
1
d


2
tan 55
550
adjacent
d is now on the top line
1
2=d
tan 55 
2
 2  tan 55   1.4 km to one decimal place.
d=
tan 55
(ii)
e
2
opposite
e2 = d2 + 22 = (1.4)2 + 4 = 1.96 + 4 = 5.96
e = 5.96 = 2.4 km to one decimal place.
(c)
550
Hut1
1800 –550 –900 = 35
4500
Hut 1
300
a
Hut2
2700
Bearing of first hut from
second hut = 2700 + a
a
Hut 2
2700
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°
Week 11 Page 12
Still confused with Bearings? Watch these video lessons
Insert your DECV course CD into your computer and then click:
(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)
B
110
600
B
800
0
1600
A
C
A
2.
C
B
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
3200
500
C
A
800
N
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?
C
N
A
1200
B
B
M
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?
1300
8 km
400
C
10 km
H
Week 11 Page 13
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.
Q
1500
600
(a) Show that the angle at Q is a right angle.
P
6 km
9 km
(b) They walk directly from R to P:
(i) How far is this, correct to one decimal place?
(ii) What bearing do they follow?
R
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
C
(b) From the hill at B:
(i) What is the bearing of A?
(ii) What is the bearing of C?
0
50
3200
5 km
A
750
(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  to point V The bearing of the tower T
from S is 035  . 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?
T
350
S
1250
8 km
V
3300
Week 11 Page 14
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.
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,
Example
AC,
BD and1BC
B
A
380
600
C
b=8
Week 11 Page 15
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
N
1450
A
20 km
(a) How far is the ship from each light house?
S
(b) The ship’s captain takes bearings of each light house.
600
(i) What is the bearing of the lighthouse at A from the ship?
B
(ii) What is the bearing of the lighthouse at B from the ship?
Answers
N
A
1450
A
1800 1450 =350
350
20
b
1800 600 350 =850
850
S
600
B
B
a
c

sin A sin C
(a)
a
sin 35
a=


20
sin 85
b
c

sin B sin C
20
sin 85
b

sin 60

  sin 35  11.5 km
b=
(b) (i)

20
sin 85 
20
sin 85 
 sin 60   17.4 km
(b) (ii)
A
1450
From S, the bearing of
360°  35° = 325°

The ship is 17.4 km from lighthouse A.
The ship is 11.5 km from lighthouse B.
the lighthouse A is:
S
600
From S, the bearing of
350
350
350
lighthouse Bis:
350
S
360°  35°  85° = 240°
850
600
3600 350 =3250
B
S
Week 11 Page 16
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 68 and
angle ABC equals 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.
Week 11 Page 17
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.
2.
10.4 km
[VCAA 2005 Further Maths Exam 1]
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°
[VCAA 2005 Further Maths Exam 1]
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
A
N
C
4
sin 30
B
62  42  48 cos 120
C
62  42  48 cos 120
6 km
B
D 6 sin 60º
E
52
4 km
A
Week 11 Page 18
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?
M
250
750
C
5 km
H
3450
Week 11 Page 19
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:
D
N
A
160 m
120 m
B
230 m
m
C
2
3
0
Part 1
m
(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 
1
6
0
Week 11 Page 20
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.
D
E
A
N
25 
2
5
120 m
B
Find the area and the perimeter of the car park ABE.
C
Week 11 Page 21
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.
return to A?
A 005o
B 085o
C 090o
D 355o
E 275o
What bearing would be taken from B to
Week 11 Page 22
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
Week 11 Page 23
Answers to Exercises
PRACTICE EXERCISE 1
1.
(a) 20.6º = 20º 36'
(c) 14.8º = 14º 48'
(b) 42º 16' = 42.3º
(d) 65º 43' = 65.7º
2. The distance of the buoy from shore = AB
130
tan 18º =
AB
130
0.3249 =
AB
130
AB =
0.3249
= 400.123  400 m.
C
180
130 m
180
A
B
3.
(a)
T
0
0
0
180 – 99 – 3 = 78
0
1800 – 810 = 990
B
1800 – 900 – 90 = 810
3
0
T
78 0
120 m
90
C
(b)
B
A
120 = AB
Sin A
Sin 78
AB = 120  × sin 78º = 2242.77  2243 m
sin 3
99 0
30
A
B
(c)
BC
sin 9º = 2243
BC = sin 9º × 2243 = 350.88  351 m
Height of the hill is 351m.
2243 m
90
C
A
4.
(a)
1800 – 180 – 1600 = 20
D
D
50 + 2 = 52 m
A
52 m
20 0
20 0
18 0
B
1800 – 200 = 1600
C
B
C
Week 11 Page 24
D
(b) Find BD before finding AB, the distance between the buoys
52
52  BD =
Sin 20º = BD
= 152.0378
Sin 20 
AB = BD
Sin 2
Sin 18
AB = 152 .0378
× sin 2º = 17.17  17.2 metres
Sin 18
Distance between the buyys is 17.2km
20
152.0378 m
18 0
A
160 0
B
D
D Top of
Spire
5.
(a)
1800 – 180 – 1470 = 150
18 0
18 0
A
A
33 0
150 m
B
Brick fence
(b) Find BD before finding DC, the height of the spire.
BD = 150
Sin 15 
Sin 18
BD = 150  × sin 18º = 179.0924
Sin 15
Referring to triangle BDC, sin 33º = DC
BD
DC
sin 33º = 179 .0924
A
DC = 179.0924 × sin 33º
= 97.54  97.5 metres
The height of the spire is 97.5 m
1800 – 330 = 1470
B
150 m
C
Church
D
15 0
b
a
18 0
b =150
147 0
B
D
(c) cos 33º = 179BC
.0924  BC = 179.0924 × cos 33º
= 150.199
 150 metres to the nearest half metre
179.0929
B
33 0
C
PRACTICE EXERCISE 2
1.
(a)
(i) The bearing required is angle b we find
the angle  in the triangle, then
 = 90º – 
tan  = 85 = 0.625
 = 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
 d = 89 = 9.433  9.4 km
Start
d
N
5 km


Finish
8 km
Week 11 Page 25
(b)
Start
Angle  = 180º – 90º – 32º = 58º
Bearing

Bearing = 180º + 58º = 238º
The bearing that Tomas has to follow from start to finish
320
is 238ºT
2
(a)
To travel straight back to the start, follow a bearing shown by
angle .

Finish
 +  = 180º   = 180º – 
Find angle 

10 km
tan  = 6 = 0.6
10
 = tan–1 0.6 = 30º 58'
Start
Bearing  = 180º – 30º 58'
6 km
= 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
30º 58'
to follow the bearing shown
Bearing = 270º + angle 
= 180º – 90º – 30º 58'

= 59º 2'
Bearing = 270º + 59º 2' = 329º 2'
3.
(a)
750
4 km
Finish
4
8
400 8 km
90º – 75º = 15º
Start
90º – 40º = 50º
(b)
Distance travelled north = x + y
4
y
0
15
8
x
500
sin 50º = x
8
 x = 8 × sin 50º = 6.12836
y
sin 15º = 4
 y = 4 × sin 15º = 1.03527
x + y =47.1636  7.2 km.
Samantha is 7.2 km north from the starting point.
Week 11 Page 26
(c)
Distance travelled east = e + d
150
cos15º = e
4
e
 e = 4 × cos15º
= 3.8637
8
 d = 8 × cos50º = 5.1423
500
cos50º = d
8
d
e + d = 9.006  9 km.
Samantha is 9 km east from the starting point.
4.
(a)
1250
1250
30
9 km
0
300
90º – 30º = 60º
180º – 125º
= 55º
2 km
(b)
Distance travelled north from the start is (b – a)
Finish
9 km
Start
a
550
sin 60º = b
9
ba
b
2 km
600
a
 b = 9 × sin 60º
= 7.794229
cos55º = a
2
 a = 2 × cos55º
= 1.147152
b – a = 6.647  6.6km
Trent is 6.6 km north from the starting point
5
(a)
Finish
10 km
550
Finish
5 km
550
3200
90º – 55º = 35º
320º – 270º
= 50º
Start
2700
Start
(b)
10 km
f
cos35º = 10
 f = 10 × cos35º = 8.19152
0
f
cos50º = e
5
 e = 5 × cos 50º= 3.213938
35
50
0
The distance travelled east from the start equals (f – e)
f – e = 4.977582  5.0 km ( Chris is 5 km east from the
e
5 km
starting point)
Week 11 Page 27
PRACTICE EXERCISE 3
1.
(a)
(i)
(ii)
1800 1100 = 700
1100
B
1100
B
600
600
C
C
1100
1800
A
600
Angle B = 60° + 70° = 130°
(b)
From C, bearing of B is
180° + 110° = 290°
(i)
(ii)
80
B
B
0
160
1600
0
800
A
1800 1600 = 200
1600
1800
C
Angle B = 80° + 20° = 100°
2.
(a)
From C, the bearing of B is :
180° + 160° = 340°
Angle at B = 50° + 40° = 90°
B
B
400
500 + 400 = 900
360 320 = 40
0
500
500
0
0
C
3200
C
80 50 = 30
0
A
0
0
30
0
1800 900 300
= 600
A
(b)
B
(c)
1800
B
500
180  40 = 140
0
0
0
500
400
C
From B, the bearing of C = 140°
500
A
From B, the bearing of A is:
180° + 50° = 230°
Week 11 Page 28
B
(d) From C, the bearing of A
360°  40°  60° = 260°
900
400
600
3.
C
300
N
A
C
N
450
450
1800 450 750
= 600
A
1200 450 = 750
B
North east is 45°
(b) From B, the bearing of A is: 360°  60° = 300°)
(a) From C, the bearing of A is 180° + 45° = 225°
C
1800
A
450
600
B
A
4.
(a)
M
1800
C
(c) From C, the bearing of B is 180° (due South)
B
1300
H
400
H
400
C
Angle at M = 40° + 50° = 90°
(b) From Hut H, the bearing of camp C is:
360°  50°   = 310°  ,
Find angle .
10
tan  =
= 1.25
8
 = tan  1 1.25 = 51° 20
Bearing = 310°  51° 20
= 258° 40
1300
M
500
500
H
C
Week 11 Page 29
M
(c) HC2 = 102 + 82 = 164
HC =
164 = 12.8 km
8 km
10 km
H
C
5.
(a)
Q
600
1500
Q
1800 1500
= 300
600
6 km
P
9 km
P
R
Angle at Q = 60° + 30° = 90°
R
(b)
(ii)
Q
6 km
900
1500
Q
9 km
300
300
P
P
R
R
PR2 = 92 + 62 = 117 ( Pythagoras’ theorem)
Bearing is 360°  30°   = 330°  
PR =
Find 
117 = 10.8 km (one decimal place)
6
= 0.666
9
 = tan 1 0.666 . . . = 33° 41
Bearing = 329° 60  33° 41 = 296° 19
They have to follow a bearing of 296° 19
tan  =
6.
(a)
500
500
400
900
C
320
A
5000
75 500= 250
B
3600 3200 = 400
B
C
0
250
A
1800 900 250= 650
Week 11 Page 30
(b)
(i)
(ii)
B
180
1800 400 = 1400
0
B
500
500
400
400
C
A
From B the bearing of A is 180°+ 50°= 230°
(c)
From B the bearing of C is 140°
B
400
650
C
3600 400  650 = 2550
A
From C the bearing of A is 255°
(d)
B
650
C
250
5 km
A
(i) cos 65° =
BC
5
BC = 5  cos 65 = 2.1km
7.
(a)
(ii) sin 65° =
AB
5
AB = 5  sin 65° = 4.5 km.
T
T
350
300
650
350
S
3600 3300 =
300
1250 350 =
900
S
900
1800 900 650= 250
8 km
V
3300
V
Week 11 Page 31
(b)
(i)
T
1800 300 = 1500
T
(ii)
1800
350
350
S
From T, the bearing of S
is 180° + 35° = 215°
From T, the bearing of V
180°  30° = 150°
V
(c)
T
350
300
900
S
300
250
250
V
From V, the bearing of S is 360°  25°  30° = 305°
T
(d)
8
tan 65° =
ST
8
ST =
 = 3.7 km
tan 65
8
sin 65° =
TV
8
TV =
= 8.8 km
sin 65 
650
S
8 km
V
Week 11 Page 32
PRACTICE EXERCISE 4
C
1.
(a) Angle ACB = 180  68  82  30
To find AC, we use the sine rule:




30 
AC
AB

sin B sin C
AC
112


sin 82
sin 30 
AC  sin 30   112  sin 82 
AC 
A
68
82 
112m
112  sin 82 
 221.82  222m
sin 30 
(b) To find BC we can use the cosine rule:
BC 2  AB 2  AC 2  2  AB  AC  cos 68 
BC 2  112 2  222 2  2  112  222  0.3746
BC 2  43199.563
BC  43199.56  208m
(c) Area of triangle ABC =
=
1
2
1
2
AB  AC  sin 68 
 112  222  0.927
= 11526.7 m
2
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:

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
Work for submission - Exam Practice.
END OF WEEK 11
B
Week 11 Page 33
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Further Mathematics Unit 4
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12
WEEK
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Week 11 Page 34
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