finish this homework in 30 min and get 30 bucks
pul0001PAGE
Notice that 1.2º is
the same as 1º 12'
Question 1: Estimating magnitude (size) of angles
o
Estimate the size of angle ABC in each of the diagrams shown.
' 6 7 o a. |
|
b. |
A B C 3 km 310 0
3.3 km |
c. |
|
Question 2: Drawing angles
Draw an accurate diagram to represent an angle with a magnitude of:
a. |
30o |
b. |
75o |
c. |
130o |
30
0
N
120
0
A
B
O
N
O
210
0
300
0
D
C
Question 3: Calculations of sizes of angles and lengths
N
Start
Finish
Bearing
6 km
8 km
8 ( opposite)
6
(adjacent)
2338
6 km
8 km
d
Bearing
538
b
a
180
0
90
0
56
0
= 34
0
56
0
35
0
90
0
35
0
= 55
0
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
10 km
70
0
110
0
180
0
Camp
Springs
Hut
50
0
140
0
5 km
7 km
50
0
140
0
50
0
180
0
–140
0
= 40
0
50
0
140
0
50
0
+ 40
0
=90
0
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
|
b. |
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. |
600 |
b. |
400 |
c. |
1150 |
o o o 90 40 50 = + = + b a o 50 = a 2 a. |
o o 40 140 180 = - = b |
b. |
5 km 7 km d |
c. |
140 0
40 0
50 0
7 5
|
40 0
alternate 40
360 0 40 0 = 320 0
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
|
||||
b. |
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 TRUEThis is the correct application of the sine rule. iii. 222 2cos41 EFGEGFGEGF =+-´´´° FALSEThe 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
25
0
115
0
2 km
Hut 1
Hut 1
SEQ week \r 3 \h \* MERGEFORMAT Week 11
Applications of Geometry
& Trigonometry (I)
Page 2 |
Page 14 |
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Page 5 |
Page 17 |
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Page 9 |
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
30
0
25
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
0
115
0
– 25
0
= 90
0
360
0
–330
0
=30
0
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
d
adjacent
2
opposite
e
55 0
180 0 –55 0 –90 0 = 45 0
a 30 0
270 0 Watch a powerpoint! Insert your DECV course CD into your computer and then click:
Bearing of first hut from second hut = 270 0 + 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
0
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 metreExample 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 310º 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
B
A
N
145
0
60
0
B
180
0
145
0
=35
0
180
0
60
0
35
0
=85
0
S
A
B
S
20
35
0
85
0
60
0
b
A
145
0
35
0
35
0
360
0
35
0
=325
0
S
35
0
60
0
85
0
S
35
0
B
160 m
230 m
A
D
C
B
120 m
N
o
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
AnswersDraw 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
tanq
=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 kmSimon 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
A
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 = 2.4575a + b = 2.2368 + 2.4575 = 4.6943
»
4.7 km (one decimal place).Tanya is 4.7 km north from the starting point.
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|
|
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 =
km
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
55
tan
1
=
o
d is now on the top line
o
55
tan
1
( 2 = dd =
place.
decimal
one
to
km
1.4
55
tan
2
55
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 |
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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 035o
. 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
35
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
B
sin b =
o o 85 sin 20
60 sin b = b = km
17.4
60 sin
85
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 |
|
|
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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
andangle 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.
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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
A
o
30
sin
4
B
22
6448cos120
+-
o
C22
6448cos120
++
o
D 6 sin 60º
E
52
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.
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SEND: Work for Submission – Exam Practice |
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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(
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Answers to Exercises |
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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 mHeight 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 metresDistance between the buyys is 17.2km
5.
(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 metresThe height of the spire is 97.5 m
(c) cos 33º =
0924
.
179
BC
Þ
BC = 179.0924 × cos 33º = 150.199
»
150 metres to the nearest half metrePRACTICE 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
= 180ºÞ
( = 180º –q
Find angleq
tanq
=10
6
= 0.6q
= 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.3Þ
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 = 7.794229cos55º =
2
a
Þ
a = 2 × cos55º = 1.147152b – a = 6.647
»
6.6kmTrent 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.3Þ
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 thestarting point)
PRACTICE EXERCISE 3
1.
(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°)
(c) From C, the bearing of B is 180° (due South)
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 kmsin 65° =
TV
8
TV =
°
65
sin
8
= 8.8 kmPRACTICE 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
(c) Area of triangle ABC =
o
68
sin
2
1
´
´
AC
AB
=
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.
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END OF WEEK 11
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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.
A
BA
CA
B
A
C
BA
CA
A
A
B
C
A
BA
CA
A
B
C
G
H
GH is parallel to DF
400 m
410
F
E
D
BA
CA
A
A
CA
300
BA
A
750
BA
1300
CA
N.B. If you have further queries, please contact your teacher.
Boat B
1.20
1.20
Plane P
400 m
400 m
Opposite side
adjacent side (d)
�
75 m
Opposite side
d
adjacent side
7.10
75 m
B
7.10
B
Notice that 7.1� EMBED Equation.3 ���is the same as � EMBED Equation.3 ���.
9 0
12 0
A
B
T
C
120 m
North
�
Diagram 2
North
A
B
56 0
5 km
East
C
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
40
�
�
�
300
N
N
300
300
00
300
600
600
Rule 1:
Alternate angles are equal
1800
1200
600
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
C
1300
8 km
M
H
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]
[VCAA 2005 Further Maths Exam 1]
A
B
C
4 km
6 km
N
�
160 m
230 m
25�EMBED Equation.3���
�
E
Q
Q
�
�
120 m
3 0
A
B
T
99 0
78 0
A
B
C
9 0
2243 m
A
B
C
D
18 0
20 0
1800 – 200 = 1600
1800 – 180 – 1600 = 20
50 + 2 = 52 m
B
C
D
20 0
52 m
A
B
D
18 0
2 0
160 0
152.0378 m
A
D
1800 – 180 – 1470 = 150
B
18 0
1800 – 330 = 1470
150 m
A
B
C
D
Brick fence
18 0
33 0
Top of
Spire
150 m
Church
A
B
D
18 0
147 0
15 0
a
b
b =150
B
C
D
33 0
179.0929
�
�
�
�
�
�
4
�
�
�
�
�
�
Start
Finish
�
�
Start
�
�
600
�
�
�
�
�
500
�
�
�
�
�
�
�
400
C
1300
M
H
400
H
C
M
H
1300
500
500
10 km
C
M
8 km
H
�
6 km
9 km
P
Q
R
6 km
9 km
P
Q
R
900
Q
1500
300
300
P
R
A
B
C
900
250
1800 (900 (250= 650
500
750(500= 250
3200
B
C
500
3600 (3200 = 400
400
500
A
C
400
B
1800 (400 = 1400
400
A
500
B
500
1800
A
B
C
400
650
3600( 400 ( 650 = 2550
B
5 km
A
C
250
650
350
S
1250( 350 = 900
3300
T
V
350
3600( 3300 = 300
300
8 km
1800( 900 (650= 250
S
T
V
650
900
350
S
T
350
1800
T
V
1800( 300 = 1500
S
T
V
900
350
300
250
300
250
�
C
� EMBED Equation.3 ���
� EMBED Equation.3 ���
� EMBED Equation.3 ���
B
A
112m