Angles, Straight Lines and Symmetry - Oxford Open Learning
Mathematics GCSE
Lesson Fifteen
Aims
Context
Module Five: Basic Geometry
Angles, Straight Lines and Symmetry
The aim of this lesson is to enable you to: •
recognise and apply the basic ideas of Geometry, particularly with reference to angles and straight lines
•
recognise and describe different types of angles
•
define and recognise different types of geometrical shape and symmetry
•
apply your situations.
knowledge
of
these
to
practical
Geometry is a very ancient branch of mathematics and is closely related to practical usage in many fields of life. The use of symmetry is a powerful tool in solving many problems in geometry. It can reduce the amount of work we need to do in significant ways. This lesson is therefore very useful to you for your general understanding of mathematics and for your success in the examination.
Oxford Open Learning
1
Lesson Fifteen
Angles, Straight Lines and Symmetry
Introduction The word ‘geometry’ comes from the Greek and means ‘measuring the Earth’. It was developed in ancient times, particularly by Euclid, and many of his ideas are still valid today. Geometry is essentially a practical subject, and the GCSE examinations place great emphasis on your ability to draw and recognise geometrical figures. You will need a protractor, ruler and, later, a pair of compasses. As you work through the geometry modules, be prepared to draw your own diagrams. This is by far the best way of understanding the basic ideas. You should be familiar with many of these ideas and this chapter will reinforce your basic knowledge.
Types of Angle An acute angle is less than 90°:
Acute angle
A right-angle is exactly 90°:
Right angle
An obtuse angle is greater than 90° but less than 180°:
Obtuse
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Mathematics GCSE
Module Five: Basic Geometry
A reflex angle is greater than 180° but less than 360°:
Reflex angle
These four definitions can be summarised using the notation of inequalities. The angle is denoted by x degrees. Angle Acute Right angle Obtuse Reflex
Condition 0 < x < 90 x = 90 90 < x < 180 180 < x < 360
Angles on a Straight Line Y
X
There are two right-angles, or 180 degrees, on a straight line. So the angles X° and Y° add up to 180°: X + Y = 180°. Vertically Opposite Angles
Q R
P S
Angles P and R are equal. Angles Q and S are equal.
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Lesson Fifteen
Angles, Straight Lines and Symmetry
Corresponding Angles P and Q are a pair of corresponding angles, and are equal to each other. P
Q
Alternate Angles R and S are a pair of alternate angles, and are equal to each other.
R
S
Parallel Lines B
A D
C
E F H
G
The parallel lines are both marked with a single arrow. A third line crosses both the parallel lines to make eight angles. However, these eight angles fall into two sets. The angles A, C, E and G are all equal to one another. The remaining angles, B, D, F and H are also equal. Furthermore, if any angle in one set is added to any angle in the other set, then the sum is 180°.
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Mathematics GCSE
Module Five: Basic Geometry
Remember that the third straight line crosses both parallel lines. When this third line crosses each line, it makes four angles, so there are four pairs of corresponding angles: A and E, B and F, C and G, D and H. Thus A = E, B = F, C = G and D = H, which we knew already. C and E are a pair of alternate angles, sometimes called ‘Z angles’. D and F are another pair of alternate angles. Example 1 Find the missing angles denoted by letters in the following diagram. Give reasons for your answers. M 72°
N P
R
S
Q
T
True, there are lots of new words to learn and important relationships between angles. However, in this diagram, the situation is straightforward. The eight angles fall into two sets. Angles S, Q, N are all equal to the angle marked as 72° (these are all the acute angles). What about the second set? Consider the angle M. This angle and the 72° together make a straight line which is 180°. The angle M is therefore 108°. (The actual arithmetic is just 180 – 72 = 108). Now that we know angle M, we know all the other angles in the second set. Angles M, P, R and T are all 108° (these are all the obtuse angles). Why are the angles in each set the same? We have already shown that angle M is 108°. This depended upon the fact that the angles on a straight line add up to 180°. Angle P is opposite to angle M, so must be 108°. Alternatively, angle P and the 72° make a straight line, and must add up to 180°. So angle P must be 108°.
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Lesson Fifteen
Angles, Straight Lines and Symmetry
Angle N is opposite the 72°, so must also be 72°. We only need one of the facts about parallel lines to find out the first of the angles Q, R, S, T. The angle marked 72° and Q are a pair of corresponding angles, and are equal. So Q is 72°. Q and R make a straight line, so that they add up to 180°. R must therefore be 108°. T is opposite to R and is therefore equal to R and must be 108°. Q and S are opposite and are equal, so that S must be 72°. We could have proceeded differently at various stages. For instance, to obtain the first angle out of Q, R, S and T, we could have used the fact that N and Q are equal because they are a pair of alternate angles. Rather than feel overwhelmed by the variety of facts and details, notice instead that: • •
there are usually different strategies available, but you only need one many of the important facts are inter-related.
Example 2 Angle BAC is 50°. Find the remaining three angles of the parallelogram ABCD. A
B
C
D
The situation is simple: all acute angles in the diagram are 50°, and all obtuse angles in the diagram are 130°. So angle ABD (top right angle of the parallelogram) is 130°, angle BDC (bottom right angle of the parallelogram) is 50° and angle ACD (bottom left angle of the parallelogram) is 130°. The next Example concentrates on each of the four angle relationships in turn. You need to be able to recognise each of these four relationships, and also to name them. This Example has the same style as GCSE examination work.
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Mathematics GCSE
Module Five: Basic Geometry
Example 3 Find the missing angles denoted by letters in the following diagrams. In each case give a reason for your answer. a
70° b
120°
53°
c
(a) a = 60° (b) b = 70° (c) c = 53° (d) d = 67°
67°
d
. because the angles on a straight line sum to 180°. because opposite angles are equal. because corresponding angles are equal. because alternate angles are equal.
The final Example is an old style of examination question. Example 4 Find the angle x in the following diagram. B
A 46° x
C 38° E
D
A special trick is required. Draw a line through C which is parallel to both AB and DE. The required angle x is now split into the two angles y and z. We now notice that y and angle ABC are equal because they are alternate angles. Also, angle z and angle CDE are equal because they are alternate angles. So y = 46° and z = 38°. We require angle x. But x = y + z = 46 + 38 = 84. So angle x is 84°.
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Lesson Fifteen
Angles, Straight Lines and Symmetry
B
A 46°
y
C z 38° E
Activity 1
D
Use your protractor to measure the following angles and write down the name of the angle, choosing from obtuse, acute, right or reflex.
(a)
(b)
(c)
(d)
8
(e)
Mathematics GCSE
Activity 2
Module Five: Basic Geometry
Complete the following checklist of angle properties. You need to know all of these properties and their names, so that you can recognise when to use them in calculations. (a) Angles at a point a
b
d
c
a + b + c + d = __ (b) Adjacent angles on a straight line a
b
a + b = __ (c) Opposite angles b a
c d
a = __
d = __
(d) Parallel lines: corresponding angles b a
a = __ (e)
Parallel lines: alternate angles a b
a = __ (f)
Parallel lines: interior angles a
b
a + b = __
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Lesson Fifteen
Activity 3
Angles, Straight Lines and Symmetry
Find the missing angles denoted by letters in the following diagrams.
1.
132°
a
2.
36°
41°
b
3.
43°
d c
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Mathematics GCSE
Module Five: Basic Geometry
4.
e
g
28°
f
5.
43° h
32°
6.
i
140°
130°
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Lesson Fifteen
Angles, Straight Lines and Symmetry
Bearings NNW
N
NNE
NW
NE
WNW
ENE
W
E ESE
WSW SW
SE SSW
S
SSE
This is a practical application of angles. Bearings are one method of showing direction of travel. It is probably the most accurate method, though many of you may be familiar with the idea of compass points. Both methods are still used, so we will begin by looking at 16 points of the compass.
The Bearing of P from Q This means the direction you need to travel if you want to travel from Q to P (in a straight line). The instructions for obtaining the “bearing of P from Q” are as follows: 1. Draw a North line through the SECOND letter (Q in this case) 2. Measure CLOCKWISE from this North line until you are facing the FIRST letter (P in this case). 3. The answer is a three figure bearing, so an angle of 30° would be written 030°.
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Mathematics GCSE
Module Five: Basic Geometry
Example 1 P is North East of Q. Find the bearing of: (a) P from Q (b) Q from P. N P
45° Q
(a)
Since P is North East of Q, the angle NQP is 45°. The “bearing of P from Q” is the angle the North line (QN) must rotate (clockwise) until it lies along QP. In this case, the angle is obviously 45°. However, a ‘three figure bearing’ is required: the answer if 045°.
(b)
The bearing of Q from P is different! It is best to start with a new diagram. There is a new North line, since the SECOND letter is now P. The “bearing of Q from P” is shown as the angle x: this is the angle that the new North line must rotate until it lies along PQ (when it faces in the direction of the FIRST letter). Extend the (new) North line ‘south’ of the point P. Split the angle x into two angles. One of these ‘angles’ is the straight line NPR: this is 180°. The other angle is angle QPR. The new North line PN is clearly parallel to the old North line (through Q). We recognise that angle QPR and the 45° are a pair of alternate angles, and are therefore equal. The angle x is therefore 45 + 180 = 225. So the bearing of Q from P is 225°. N
P x
45°
Q
R
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Angles, Straight Lines and Symmetry
There are two important points to notice from this example: 1. ‘the bearing of P from Q’ is different from ‘the bearing of Q from P’ 2. you must be able to recognise alternate angles. Example 2 B is South East of A. Find the bearing of: (a) B from A (a)
(b) A from B.
Draw a North line through the SECOND letter, which is A. Extend it ‘south’ of A to C: this is the dashed portion. Since B is South East of A, the angle CAB is 45°. Rotate the North line AN clockwise until it lies along AB (and therefore faces the FIRST letter, B). The bearing of B from A is the angle marked x. But the angle x and 45° make a straight line, and therefore add up to 180°. The angle x must be 180 – 45 = 135. The required bearing is 135°. N
x A
45° C (b)
B
Draw a new diagram. Draw a (new) North line through the SECOND letter, which is B. Measure clockwise until the North line BN lies along BA , when it faces towards the FIRST letter A. The required angle is the reflex angle marked y. Again, notice that the old and new North lines are parallel. There is again a pair of alternate angles: the acute angle NBA must be equal to 45°. The simple way to find y is to subtract 45 from 360 to give 315. The required bearing is 315°.
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Mathematics GCSE
Module Five: Basic Geometry
N
A
45° B
C
y
There are three overall rules: • • •
always mark in the North line at the point at which you are taking the bearing; always take the bearing in a clockwise direction; always give bearings as 3-figure numbers e.g. an angle of 15° is given as a bearing of 015°.
Example 3
Lighthouse
P
If you check with your protractor, you will find that the angle shown is 45°. So the bearing of the lighthouse from P is 045°. N
X
This time the bearing of the tree from X is 337°. Note that the bearing is the reflex angle because we must take it anticlockwise.
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Lesson Fifteen
Activity 4
Angles, Straight Lines and Symmetry
1.
Use your protractor to find the following bearings. Use the centre of the object marked with a dot when you take your measurements. The answers are given at the end of the lesson.
(a) Space station
•
Satellite from space station
Satellite
• (b)
Trawler
• Lightship from trawler
• (c) Airport Airport from aircraft
Aircraft
•
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Lightship
Mathematics GCSE
Activity 5
Module Five: Basic Geometry
1.
Without measuring, find the bearing of: (a) B from A (b) A from B.
N
B 78°
A
2.
Without measuring, find the bearing of: (a) D from C (b) C from D
N
C
D
163°
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Lesson Fifteen
Angles, Straight Lines and Symmetry
3.
Without measuring, find the bearing of: (a) F from E (b) E from F
N
E
71°
4
F
Without measuring, find the bearing of: (a) H from G (b) G from H.
N
G
H 139°
18
5.
The bearing of P from Q is 30°. Find the bearing of Q from P.
6.
The bearing of R from S is 141°. Find the bearing of S from R.
7.
The bearing of T from U is 195°. Find the bearing of U from T.
8.
The bearing of V from W is 347°. Find the bearing of W from V.
Mathematics GCSE
Module Five: Basic Geometry
Symmetry Most of us have an idea about symmetry. I think you will agree that Figure A below is reasonably symmetrical, but Figure B is certainly not.
F igu r e (a )
F igu r e (b )
However, in mathematics, symmetry has a more precise meaning than just a general impression.
Line Symmetry A shape has ‘line symmetry’ if it has an ‘axis of symmetry’ or mirror line. If the shape is folded along the axis of symmetry, then the two halves of the shape exactly coincide. The arrow below has line symmetry: the axis of symmetry is shown as a dashed line.
Some shapes have many lines of symmetry. A regular octagon, with eight sides, has eight lines of symmetry, which are shown as dashed lines.
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Angles, Straight Lines and Symmetry
Some shapes have no lines of symmetry. A good example is a parallelogram. There is no possible mirror line for this shape. The shape will not fold onto itself wherever it is folded. If you are not sure, try it!
Example The diagram shows only half a shape. However, the shape has line symmetry about the dotted line. Draw the whole shape.
There are various physical ways of reflecting a shape. It is possible to fold along the dotted line then use a pencil to mark heavily on the ‘wrong’ side of the paper. You should then see
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Mathematics GCSE
Module Five: Basic Geometry
the outline of the second half of the shape on the squared paper. You can go over this properly to complete the shape. It is also possible to use tracing paper. Trace the half shape shown. Then turn the tracing paper over and place it so the mirror lines on the original drawing and the tracing paper coincide. Now use a pencil on the wrong side of the tracing paper: if you mark this heavily enough, you will see the outline of the second half of the shape on the squared paper. For difficult shapes it can be useful to have another method. This method can also be useful in other areas of the Course. Take one vertex (corner) at a time. Reflect each vertex in the mirror line. To achieve this, imagine the vertex moving: • •
into the mirror line at right-angles then continuing out of the other side of the mirror line, still at right-angles, so that it is the same distance from the mirror line as before, but the opposite side.
All that needs to be done to complete the shape is to join up the new positions of the vertices. Whichever method you use, the complete shape should look like the following.
Plane Symmetry 3D shapes cannot be reflected in a straight line. However, they can be reflected in a plane. It is therefore possible for 3D shapes to have a ‘plane of symmetry’. The following shape has
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Lesson Fifteen
Angles, Straight Lines and Symmetry
two planes of symmetry, both of which are vertical. The planes of symmetry are shown shaded in the separate diagrams which follow.
The next 3D shape is a ‘triangular prism’. The top face, ABC, is a ‘scalene’ triangle. The triangle has no lines of symmetry. However, the triangular prism has a horizontal plane of symmetry which is shown shaded.
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Mathematics GCSE
Module Five: Basic Geometry
B
A
C
Rotational Symmetry The following diagram shows a regular pentagon (five sides). It is possible to rotate the pentagon about the centre O in such a way that the new shape coincides with the old shape. In fact, there are five ways of doing this, corresponding to five possible angles of rotation. One fifth of a full turn is 360 ÷ 5 = 72 °. Original vertices:
Anticlockwise rotation:
1 5 2 5 3 5 4 5
A
B
C
D
E
New positions of vertices: B C D E A
of a turn: 72° of at turn: 144° of a turn: 216° of a turn: 288°
Full turn: 360°
C
D
E
A
B
D
E
A
B
C
E
A
B
C
D
A
B
C
D
E
A
B
E
O O
C
D
We say that the regular pentagon has rotational symmetry of order five, because there are five different angles of rotation
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Lesson Fifteen
Angles, Straight Lines and Symmetry
that will make the pentagon coincide with itself (although the letters may be different). Another way of thinking about the ‘order of rotational symmetry’ is as follows. If the smallest rotation that makes the shape coincide with itself is one fifth of a full turn, then the order of rotational symmetry is five. If the smallest rotation were a third of a full turn, then the order of rotational symmetry would be three. If the smallest rotation were a tenth of a full turn, the order of rotational symmetry would be ten, and so on.
Activity 6
1. Draw copies of the following shapes and mark in any lines of symmetry.
2.
The square bathroom tile shown below has a vertical and a horizontal line of symmetry as shown: (a) Complete the shading for the rest of the tile. (b) When complete, the tile has another type of symmetry. Describe this.
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Mathematics GCSE
Activity 7
Module Five: Basic Geometry
Now tackle the following extended exercise.
1.
The following diagrams each show half of a shape which has line symmetry about the dotted line. Copy each diagram onto squared paper and complete each shape.
(a)
(b)
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Lesson Fifteen
Angles, Straight Lines and Symmetry
(c)
(d)
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Mathematics GCSE
Module Five: Basic Geometry
2.
Copy each of the following shapes and mark the line of symmetry.
(a)
(b)
(c)
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Lesson Fifteen
Angles, Straight Lines and Symmetry
(d)
3.
Sketch the following shapes and mark in all lines of symmetry. (a)
(b)
(c)
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Mathematics GCSE
Module Five: Basic Geometry
(d)
(e)
4.
Examine the following list of letters carefully, and then put each into one of the following three categories: no lines of symmetry, one line of symmetry, two lines of symmetry
ABCDEFGHIJKLM NOPQRSTUVWXYZ 5. Write down the order of rotational symmetry of each of the following shapes. (a)
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Angles, Straight Lines and Symmetry
(b)
(c)
(d)
6.
30
How many planes of symmetry do the following 3D objects have? (a)
a square-based (Egyptian) pyramid
(b)
a cube
(c)
a Toblerone packet.
Mathematics GCSE
Module Five: Basic Geometry
Suggested Answers to Activities Activity One (a)
30°
acute
(b)
18°
acute
(c)
112°
obtuse
(d)
322°
reflex
(e)
90°
right
(a)
a + b + c + d = 360°
(b)
a + b = 180°
(c)
a = c, b = d
(d)
a=b
(e)
a=b
(f)
a + b = 180°
Activity Two
Activity Three 1
a = 48°
3
c = 43°, d = 137°
5
h = 75°
2 4
b = 103°
e = 28°, f = 152°, g = 28° 6
i = 90°
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Angles, Straight Lines and Symmetry
Activity Four 337.5°
0°
22.5°
315°
45°
292.5°
67.5°
270°
90°
247.5°
112.5° 225°
135° 202.5°
1.
(a)
225°
(b)
135°
(c)
335°
180°
Activity Five 1 2 3 4 5 6 7 8
(a) 078° (a) 017° (a) 109° (a) 139° 210° 321° 015° 167°
(b) (b) (b) (b)
Activity Six 1.
32
258° 197° 289° 319°
157.5°
Mathematics GCSE
2.
Module Five: Basic Geometry
(a)
(b)
Rotational symmetry of order 2.
Activity Seven 1.
(a)
(b)
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Lesson Fifteen
(c)
(d)
2
34
(a)
Angles, Straight Lines and Symmetry
Mathematics GCSE
Module Five: Basic Geometry
(b)
(c)
(d)
35
Lesson Fifteen
3 (a)
(b)
36
Angles, Straight Lines and Symmetry
Mathematics GCSE
Module Five: Basic Geometry
(c)
(d)
(e)
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Lesson Fifteen
Angles, Straight Lines and Symmetry
4 no lines of symmetry one line of symmetry two lines of symmetry
FGJKLNPQRSZ ABCDEMTUVWY HIOX
NB
The answer might vary according to the typestyle used, e.g. for letters B and C.
5
(a) 4
(b) 16
(c) 2
6
(a) 4
(b) 9
(c) 3
38
(d) 5
Lesson Fifteen
Aims
Context
Module Five: Basic Geometry
Angles, Straight Lines and Symmetry
The aim of this lesson is to enable you to: •
recognise and apply the basic ideas of Geometry, particularly with reference to angles and straight lines
•
recognise and describe different types of angles
•
define and recognise different types of geometrical shape and symmetry
•
apply your situations.
knowledge
of
these
to
practical
Geometry is a very ancient branch of mathematics and is closely related to practical usage in many fields of life. The use of symmetry is a powerful tool in solving many problems in geometry. It can reduce the amount of work we need to do in significant ways. This lesson is therefore very useful to you for your general understanding of mathematics and for your success in the examination.
Oxford Open Learning
1
Lesson Fifteen
Angles, Straight Lines and Symmetry
Introduction The word ‘geometry’ comes from the Greek and means ‘measuring the Earth’. It was developed in ancient times, particularly by Euclid, and many of his ideas are still valid today. Geometry is essentially a practical subject, and the GCSE examinations place great emphasis on your ability to draw and recognise geometrical figures. You will need a protractor, ruler and, later, a pair of compasses. As you work through the geometry modules, be prepared to draw your own diagrams. This is by far the best way of understanding the basic ideas. You should be familiar with many of these ideas and this chapter will reinforce your basic knowledge.
Types of Angle An acute angle is less than 90°:
Acute angle
A right-angle is exactly 90°:
Right angle
An obtuse angle is greater than 90° but less than 180°:
Obtuse
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Mathematics GCSE
Module Five: Basic Geometry
A reflex angle is greater than 180° but less than 360°:
Reflex angle
These four definitions can be summarised using the notation of inequalities. The angle is denoted by x degrees. Angle Acute Right angle Obtuse Reflex
Condition 0 < x < 90 x = 90 90 < x < 180 180 < x < 360
Angles on a Straight Line Y
X
There are two right-angles, or 180 degrees, on a straight line. So the angles X° and Y° add up to 180°: X + Y = 180°. Vertically Opposite Angles
Q R
P S
Angles P and R are equal. Angles Q and S are equal.
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Lesson Fifteen
Angles, Straight Lines and Symmetry
Corresponding Angles P and Q are a pair of corresponding angles, and are equal to each other. P
Q
Alternate Angles R and S are a pair of alternate angles, and are equal to each other.
R
S
Parallel Lines B
A D
C
E F H
G
The parallel lines are both marked with a single arrow. A third line crosses both the parallel lines to make eight angles. However, these eight angles fall into two sets. The angles A, C, E and G are all equal to one another. The remaining angles, B, D, F and H are also equal. Furthermore, if any angle in one set is added to any angle in the other set, then the sum is 180°.
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Mathematics GCSE
Module Five: Basic Geometry
Remember that the third straight line crosses both parallel lines. When this third line crosses each line, it makes four angles, so there are four pairs of corresponding angles: A and E, B and F, C and G, D and H. Thus A = E, B = F, C = G and D = H, which we knew already. C and E are a pair of alternate angles, sometimes called ‘Z angles’. D and F are another pair of alternate angles. Example 1 Find the missing angles denoted by letters in the following diagram. Give reasons for your answers. M 72°
N P
R
S
Q
T
True, there are lots of new words to learn and important relationships between angles. However, in this diagram, the situation is straightforward. The eight angles fall into two sets. Angles S, Q, N are all equal to the angle marked as 72° (these are all the acute angles). What about the second set? Consider the angle M. This angle and the 72° together make a straight line which is 180°. The angle M is therefore 108°. (The actual arithmetic is just 180 – 72 = 108). Now that we know angle M, we know all the other angles in the second set. Angles M, P, R and T are all 108° (these are all the obtuse angles). Why are the angles in each set the same? We have already shown that angle M is 108°. This depended upon the fact that the angles on a straight line add up to 180°. Angle P is opposite to angle M, so must be 108°. Alternatively, angle P and the 72° make a straight line, and must add up to 180°. So angle P must be 108°.
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Lesson Fifteen
Angles, Straight Lines and Symmetry
Angle N is opposite the 72°, so must also be 72°. We only need one of the facts about parallel lines to find out the first of the angles Q, R, S, T. The angle marked 72° and Q are a pair of corresponding angles, and are equal. So Q is 72°. Q and R make a straight line, so that they add up to 180°. R must therefore be 108°. T is opposite to R and is therefore equal to R and must be 108°. Q and S are opposite and are equal, so that S must be 72°. We could have proceeded differently at various stages. For instance, to obtain the first angle out of Q, R, S and T, we could have used the fact that N and Q are equal because they are a pair of alternate angles. Rather than feel overwhelmed by the variety of facts and details, notice instead that: • •
there are usually different strategies available, but you only need one many of the important facts are inter-related.
Example 2 Angle BAC is 50°. Find the remaining three angles of the parallelogram ABCD. A
B
C
D
The situation is simple: all acute angles in the diagram are 50°, and all obtuse angles in the diagram are 130°. So angle ABD (top right angle of the parallelogram) is 130°, angle BDC (bottom right angle of the parallelogram) is 50° and angle ACD (bottom left angle of the parallelogram) is 130°. The next Example concentrates on each of the four angle relationships in turn. You need to be able to recognise each of these four relationships, and also to name them. This Example has the same style as GCSE examination work.
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Mathematics GCSE
Module Five: Basic Geometry
Example 3 Find the missing angles denoted by letters in the following diagrams. In each case give a reason for your answer. a
70° b
120°
53°
c
(a) a = 60° (b) b = 70° (c) c = 53° (d) d = 67°
67°
d
. because the angles on a straight line sum to 180°. because opposite angles are equal. because corresponding angles are equal. because alternate angles are equal.
The final Example is an old style of examination question. Example 4 Find the angle x in the following diagram. B
A 46° x
C 38° E
D
A special trick is required. Draw a line through C which is parallel to both AB and DE. The required angle x is now split into the two angles y and z. We now notice that y and angle ABC are equal because they are alternate angles. Also, angle z and angle CDE are equal because they are alternate angles. So y = 46° and z = 38°. We require angle x. But x = y + z = 46 + 38 = 84. So angle x is 84°.
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Lesson Fifteen
Angles, Straight Lines and Symmetry
B
A 46°
y
C z 38° E
Activity 1
D
Use your protractor to measure the following angles and write down the name of the angle, choosing from obtuse, acute, right or reflex.
(a)
(b)
(c)
(d)
8
(e)
Mathematics GCSE
Activity 2
Module Five: Basic Geometry
Complete the following checklist of angle properties. You need to know all of these properties and their names, so that you can recognise when to use them in calculations. (a) Angles at a point a
b
d
c
a + b + c + d = __ (b) Adjacent angles on a straight line a
b
a + b = __ (c) Opposite angles b a
c d
a = __
d = __
(d) Parallel lines: corresponding angles b a
a = __ (e)
Parallel lines: alternate angles a b
a = __ (f)
Parallel lines: interior angles a
b
a + b = __
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Lesson Fifteen
Activity 3
Angles, Straight Lines and Symmetry
Find the missing angles denoted by letters in the following diagrams.
1.
132°
a
2.
36°
41°
b
3.
43°
d c
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Mathematics GCSE
Module Five: Basic Geometry
4.
e
g
28°
f
5.
43° h
32°
6.
i
140°
130°
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Angles, Straight Lines and Symmetry
Bearings NNW
N
NNE
NW
NE
WNW
ENE
W
E ESE
WSW SW
SE SSW
S
SSE
This is a practical application of angles. Bearings are one method of showing direction of travel. It is probably the most accurate method, though many of you may be familiar with the idea of compass points. Both methods are still used, so we will begin by looking at 16 points of the compass.
The Bearing of P from Q This means the direction you need to travel if you want to travel from Q to P (in a straight line). The instructions for obtaining the “bearing of P from Q” are as follows: 1. Draw a North line through the SECOND letter (Q in this case) 2. Measure CLOCKWISE from this North line until you are facing the FIRST letter (P in this case). 3. The answer is a three figure bearing, so an angle of 30° would be written 030°.
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Mathematics GCSE
Module Five: Basic Geometry
Example 1 P is North East of Q. Find the bearing of: (a) P from Q (b) Q from P. N P
45° Q
(a)
Since P is North East of Q, the angle NQP is 45°. The “bearing of P from Q” is the angle the North line (QN) must rotate (clockwise) until it lies along QP. In this case, the angle is obviously 45°. However, a ‘three figure bearing’ is required: the answer if 045°.
(b)
The bearing of Q from P is different! It is best to start with a new diagram. There is a new North line, since the SECOND letter is now P. The “bearing of Q from P” is shown as the angle x: this is the angle that the new North line must rotate until it lies along PQ (when it faces in the direction of the FIRST letter). Extend the (new) North line ‘south’ of the point P. Split the angle x into two angles. One of these ‘angles’ is the straight line NPR: this is 180°. The other angle is angle QPR. The new North line PN is clearly parallel to the old North line (through Q). We recognise that angle QPR and the 45° are a pair of alternate angles, and are therefore equal. The angle x is therefore 45 + 180 = 225. So the bearing of Q from P is 225°. N
P x
45°
Q
R
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Angles, Straight Lines and Symmetry
There are two important points to notice from this example: 1. ‘the bearing of P from Q’ is different from ‘the bearing of Q from P’ 2. you must be able to recognise alternate angles. Example 2 B is South East of A. Find the bearing of: (a) B from A (a)
(b) A from B.
Draw a North line through the SECOND letter, which is A. Extend it ‘south’ of A to C: this is the dashed portion. Since B is South East of A, the angle CAB is 45°. Rotate the North line AN clockwise until it lies along AB (and therefore faces the FIRST letter, B). The bearing of B from A is the angle marked x. But the angle x and 45° make a straight line, and therefore add up to 180°. The angle x must be 180 – 45 = 135. The required bearing is 135°. N
x A
45° C (b)
B
Draw a new diagram. Draw a (new) North line through the SECOND letter, which is B. Measure clockwise until the North line BN lies along BA , when it faces towards the FIRST letter A. The required angle is the reflex angle marked y. Again, notice that the old and new North lines are parallel. There is again a pair of alternate angles: the acute angle NBA must be equal to 45°. The simple way to find y is to subtract 45 from 360 to give 315. The required bearing is 315°.
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Mathematics GCSE
Module Five: Basic Geometry
N
A
45° B
C
y
There are three overall rules: • • •
always mark in the North line at the point at which you are taking the bearing; always take the bearing in a clockwise direction; always give bearings as 3-figure numbers e.g. an angle of 15° is given as a bearing of 015°.
Example 3
Lighthouse
P
If you check with your protractor, you will find that the angle shown is 45°. So the bearing of the lighthouse from P is 045°. N
X
This time the bearing of the tree from X is 337°. Note that the bearing is the reflex angle because we must take it anticlockwise.
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Lesson Fifteen
Activity 4
Angles, Straight Lines and Symmetry
1.
Use your protractor to find the following bearings. Use the centre of the object marked with a dot when you take your measurements. The answers are given at the end of the lesson.
(a) Space station
•
Satellite from space station
Satellite
• (b)
Trawler
• Lightship from trawler
• (c) Airport Airport from aircraft
Aircraft
•
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Lightship
Mathematics GCSE
Activity 5
Module Five: Basic Geometry
1.
Without measuring, find the bearing of: (a) B from A (b) A from B.
N
B 78°
A
2.
Without measuring, find the bearing of: (a) D from C (b) C from D
N
C
D
163°
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Lesson Fifteen
Angles, Straight Lines and Symmetry
3.
Without measuring, find the bearing of: (a) F from E (b) E from F
N
E
71°
4
F
Without measuring, find the bearing of: (a) H from G (b) G from H.
N
G
H 139°
18
5.
The bearing of P from Q is 30°. Find the bearing of Q from P.
6.
The bearing of R from S is 141°. Find the bearing of S from R.
7.
The bearing of T from U is 195°. Find the bearing of U from T.
8.
The bearing of V from W is 347°. Find the bearing of W from V.
Mathematics GCSE
Module Five: Basic Geometry
Symmetry Most of us have an idea about symmetry. I think you will agree that Figure A below is reasonably symmetrical, but Figure B is certainly not.
F igu r e (a )
F igu r e (b )
However, in mathematics, symmetry has a more precise meaning than just a general impression.
Line Symmetry A shape has ‘line symmetry’ if it has an ‘axis of symmetry’ or mirror line. If the shape is folded along the axis of symmetry, then the two halves of the shape exactly coincide. The arrow below has line symmetry: the axis of symmetry is shown as a dashed line.
Some shapes have many lines of symmetry. A regular octagon, with eight sides, has eight lines of symmetry, which are shown as dashed lines.
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Lesson Fifteen
Angles, Straight Lines and Symmetry
Some shapes have no lines of symmetry. A good example is a parallelogram. There is no possible mirror line for this shape. The shape will not fold onto itself wherever it is folded. If you are not sure, try it!
Example The diagram shows only half a shape. However, the shape has line symmetry about the dotted line. Draw the whole shape.
There are various physical ways of reflecting a shape. It is possible to fold along the dotted line then use a pencil to mark heavily on the ‘wrong’ side of the paper. You should then see
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Mathematics GCSE
Module Five: Basic Geometry
the outline of the second half of the shape on the squared paper. You can go over this properly to complete the shape. It is also possible to use tracing paper. Trace the half shape shown. Then turn the tracing paper over and place it so the mirror lines on the original drawing and the tracing paper coincide. Now use a pencil on the wrong side of the tracing paper: if you mark this heavily enough, you will see the outline of the second half of the shape on the squared paper. For difficult shapes it can be useful to have another method. This method can also be useful in other areas of the Course. Take one vertex (corner) at a time. Reflect each vertex in the mirror line. To achieve this, imagine the vertex moving: • •
into the mirror line at right-angles then continuing out of the other side of the mirror line, still at right-angles, so that it is the same distance from the mirror line as before, but the opposite side.
All that needs to be done to complete the shape is to join up the new positions of the vertices. Whichever method you use, the complete shape should look like the following.
Plane Symmetry 3D shapes cannot be reflected in a straight line. However, they can be reflected in a plane. It is therefore possible for 3D shapes to have a ‘plane of symmetry’. The following shape has
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Lesson Fifteen
Angles, Straight Lines and Symmetry
two planes of symmetry, both of which are vertical. The planes of symmetry are shown shaded in the separate diagrams which follow.
The next 3D shape is a ‘triangular prism’. The top face, ABC, is a ‘scalene’ triangle. The triangle has no lines of symmetry. However, the triangular prism has a horizontal plane of symmetry which is shown shaded.
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Mathematics GCSE
Module Five: Basic Geometry
B
A
C
Rotational Symmetry The following diagram shows a regular pentagon (five sides). It is possible to rotate the pentagon about the centre O in such a way that the new shape coincides with the old shape. In fact, there are five ways of doing this, corresponding to five possible angles of rotation. One fifth of a full turn is 360 ÷ 5 = 72 °. Original vertices:
Anticlockwise rotation:
1 5 2 5 3 5 4 5
A
B
C
D
E
New positions of vertices: B C D E A
of a turn: 72° of at turn: 144° of a turn: 216° of a turn: 288°
Full turn: 360°
C
D
E
A
B
D
E
A
B
C
E
A
B
C
D
A
B
C
D
E
A
B
E
O O
C
D
We say that the regular pentagon has rotational symmetry of order five, because there are five different angles of rotation
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Lesson Fifteen
Angles, Straight Lines and Symmetry
that will make the pentagon coincide with itself (although the letters may be different). Another way of thinking about the ‘order of rotational symmetry’ is as follows. If the smallest rotation that makes the shape coincide with itself is one fifth of a full turn, then the order of rotational symmetry is five. If the smallest rotation were a third of a full turn, then the order of rotational symmetry would be three. If the smallest rotation were a tenth of a full turn, the order of rotational symmetry would be ten, and so on.
Activity 6
1. Draw copies of the following shapes and mark in any lines of symmetry.
2.
The square bathroom tile shown below has a vertical and a horizontal line of symmetry as shown: (a) Complete the shading for the rest of the tile. (b) When complete, the tile has another type of symmetry. Describe this.
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Mathematics GCSE
Activity 7
Module Five: Basic Geometry
Now tackle the following extended exercise.
1.
The following diagrams each show half of a shape which has line symmetry about the dotted line. Copy each diagram onto squared paper and complete each shape.
(a)
(b)
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Lesson Fifteen
Angles, Straight Lines and Symmetry
(c)
(d)
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Mathematics GCSE
Module Five: Basic Geometry
2.
Copy each of the following shapes and mark the line of symmetry.
(a)
(b)
(c)
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Lesson Fifteen
Angles, Straight Lines and Symmetry
(d)
3.
Sketch the following shapes and mark in all lines of symmetry. (a)
(b)
(c)
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Mathematics GCSE
Module Five: Basic Geometry
(d)
(e)
4.
Examine the following list of letters carefully, and then put each into one of the following three categories: no lines of symmetry, one line of symmetry, two lines of symmetry
ABCDEFGHIJKLM NOPQRSTUVWXYZ 5. Write down the order of rotational symmetry of each of the following shapes. (a)
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Lesson Fifteen
Angles, Straight Lines and Symmetry
(b)
(c)
(d)
6.
30
How many planes of symmetry do the following 3D objects have? (a)
a square-based (Egyptian) pyramid
(b)
a cube
(c)
a Toblerone packet.
Mathematics GCSE
Module Five: Basic Geometry
Suggested Answers to Activities Activity One (a)
30°
acute
(b)
18°
acute
(c)
112°
obtuse
(d)
322°
reflex
(e)
90°
right
(a)
a + b + c + d = 360°
(b)
a + b = 180°
(c)
a = c, b = d
(d)
a=b
(e)
a=b
(f)
a + b = 180°
Activity Two
Activity Three 1
a = 48°
3
c = 43°, d = 137°
5
h = 75°
2 4
b = 103°
e = 28°, f = 152°, g = 28° 6
i = 90°
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Lesson Fifteen
Angles, Straight Lines and Symmetry
Activity Four 337.5°
0°
22.5°
315°
45°
292.5°
67.5°
270°
90°
247.5°
112.5° 225°
135° 202.5°
1.
(a)
225°
(b)
135°
(c)
335°
180°
Activity Five 1 2 3 4 5 6 7 8
(a) 078° (a) 017° (a) 109° (a) 139° 210° 321° 015° 167°
(b) (b) (b) (b)
Activity Six 1.
32
258° 197° 289° 319°
157.5°
Mathematics GCSE
2.
Module Five: Basic Geometry
(a)
(b)
Rotational symmetry of order 2.
Activity Seven 1.
(a)
(b)
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Lesson Fifteen
(c)
(d)
2
34
(a)
Angles, Straight Lines and Symmetry
Mathematics GCSE
Module Five: Basic Geometry
(b)
(c)
(d)
35
Lesson Fifteen
3 (a)
(b)
36
Angles, Straight Lines and Symmetry
Mathematics GCSE
Module Five: Basic Geometry
(c)
(d)
(e)
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Lesson Fifteen
Angles, Straight Lines and Symmetry
4 no lines of symmetry one line of symmetry two lines of symmetry
FGJKLNPQRSZ ABCDEMTUVWY HIOX
NB
The answer might vary according to the typestyle used, e.g. for letters B and C.
5
(a) 4
(b) 16
(c) 2
6
(a) 4
(b) 9
(c) 3
38
(d) 5
