Angles 2 11 - Wikispaces 11...11.1 Triangles This shows that the angles in this triangle add up to 180° but ... Solution 2 Angle sum of ... The angles at a point add up to 360° and

189

11.1 Triangles

This shows that the angles in this triangle add up to 180° but it is not a proof.That comes later in this chapter.

The angles on a straight line add up to 180° and so the angles in this triangle add up to 180°.

Work out the size of angle x.

Solution 172° � 57° � 129°

180° � 129° � 51°

x � 51°

Sometimes both the fact that the angle sum of a triangle is 180° and the angle facts fromChapter 3 are needed.

Work out the size of a angle x b angle y.Give reasons for your answers.

Solution 2

Sum of angles on a straight line is 180°.Angle sum of triangle is 180°.

b 180° � 53° � 127°y � 127°

a 67° � 60° � 127°180° � 127° � 53°

x � 53°

Example 2

State the size of angle x.

Take the result away from 180°, as the angle sum of a triangle is 180°.

Add 72° and 57°

Example 1

The angle sum of a triangle is 180°.

11C H A P T E R

Angles 2

c a

b

c a

b

cab

Draw a triangle on paperand label its angles a, band c.

Tear off its corners. Fit angles a, b and ctogether. They make astraight line.

x

72° 57°

x y

60°

67°

190

CHAPTER 11 Angles 2

Exercise 11A

In this exercise, the triangles are not accurately drawn.

In Questions 1–12, find the size of each of the angles marked with letters and show yourworking.

1 2 3 4

5 6 7 8

9 10 11 12

In Questions 13–15, find the size of each of the angles marked with letters and show yourworking.Give reasons for your answers.

13 14 15

11.2 Equilateral triangles and isosceles triangles

z

24°48°

y

42° 58°w x137°

84°

t u

v

114°41°

q

s r

81°

123°m n p

59°

119°k

l

48° 57°

j30°

60°

h i36°

34°

g f 76°

43°e

64°28°

d

27°32°

c

57°

b119°

39°a

37° 66°

60°

60° 60°

An equilateral trianglehas three equal sides andthree equal angles.As the angle sum of atriangle is 180°, the sizeof each angle is 180 � 3 � 60°.

An isosceles triangle hastwo equal sides and theangles opposite the equalsides are equal.

A triangle whose sidesare all different lengthsis called a scalenetriangle.

Work out the size of a angle x b angle y.Give reasons for your answers.

Solution 3a x � 41°

b 41° � 41° � 82°

180° � 82° � 98°

y � 98°

Work out the size of angle x.Give reasons for your answer.

Solution 4

180° � 146° � 34°

34° � 2° � 17°

x � 17°

Exercise 11B

In this exercise, the triangles are not accurately drawn.


1 2 3 4

5 6 7 8

9 10 11 12v

128°

t

u

62°

s r 80°

q

p n 106°

m l k

73°

i

j

58°h

fg

29° e116° d

58°

c69° a

b

Isosceles triangle with equal angles opposite equal sides.

Angle sum of triangle is 180°.

Example 4

Angle sum of triangle is 180°.

Isosceles triangle with equal angles opposite equal sides.

Example 3

191

11.2 Equilateral triangles and isosceles triangles CHAPTER 11

41° x

y

146°

x

192

CHAPTER 11 Angles 2

In Questions 13–15, find the size of each of the angles marked with letters and show your working.Give reasons for your answers.

13 14 15

11.3 Quadrilaterals

The angles at a point add up to 360° and so this shows that the angles in this quadrilateraladd up to 360°.

To prove this result, draw a diagonal of the quadrilateral.The diagonal splits the quadrilateral into two triangles.The angle sum of each triangle is 180°.So the angle sum of the quadrilateral is 2 � 180° � 360°.

Work out the size of angle x.

Solution 576° � 118° � 98° � 292°

360° � 292° � 68°

x � 68° State the size of angle x.

Take the result away from 360, as the angle sum of a quadrilateral is 360°.

Add 76, 118 and 98

Example 5

The angle sum of a quadrilateral is 360°.

y

68°x

42°w104°

d

a b

c

d

a b

c

d

ab

c

A quadrilateral is ashape with four straightsides and four angles.To find the angle sum ofa quadrilateral, draw aquadrilateral on paperand label its angles a, b, c and d.

Tear off its corners. Fit angles a, b, c and dtogether at a point.

118°

76°

98°

x

193

11.3 Quadrilaterals CHAPTER 11

a Write down the size of angle x.

b Work out the size of angle y. Give reasons for your answer.

Solution 6a x � 75°

b 121° � 72° � 75° � 268°

360° � 268° � 92°

y � 92°

The diagram shows a kite.

a Write down the size of angle x.

b Work out the size of angle y. Give reasons for your answer.

Solution 7

Exercise 11C

In this exercise, the quadrilaterals are not accurately drawn.


1 2 3 4

5 6 7 8

9 10 11 124

113° 129°

s

u

t

75°

126°84°

p q

r83°

94° 41°m

n

116°

82°58°

k l

147°

143°

j

i

66°

112°

48°g

h

58°

f

115°

96°

e

118°

121°74°

d67°

109° 124°

c113°

64° 71°

b83° 98°

76° a

Angle sum of a quadrilateral is 360°.

b 83° � 109° � 109° � 301°

360° � 301° � 59°

y � 59°

a x � 109°

Example 7

Angle sum of a quadrilateral is 360°

Where two straight lines cross, the opposite angles are equal.

Example 6

72°

121°

75°

x

y

109°

x

83° y

A kite has a line of symmetry. Angle x is areflection of the 109° angleand so the twoangles are equal.

109°

x

83° y

13 The diagram shows a kite. 14 The diagram shows a kite.

a Write down the size of angle v. Work out the value of x.

b Work out the size of angle w.

15 The diagram shows an isosceles trapezium.

a Write down the value of a.

b Work out the value of b.

In Questions 16–20, find the sizes of the angles marked with letters and show your working.Give reasons for your answers.

16 17 18

19 20

11.4 PolygonsA polygon is a shape with three or more straight sides.

Some polygons have special names.

A 3-sided polygon is called a triangle.A 4-sided polygon is called a quadrilateral.A 5-sided polygon is called a pentagon.A 6-sided polygon is called a hexagon.An 8-sided polygon is called an octagon.A 10-sided polygon is called a decagon.

To find the sum of the angles of a polygon, split it into triangles.

For example, for this hexagon, draw as many diagonals as possible from one corner.

This splits the hexagon into four triangles.The angle sum of a triangle is 180° and so the sum of the angles of a hexagon is 4 � 180° � 720°.

Sometimes, these angles are called interior angles to emphasise that they are inside the polygon.

56° 38°

n

143°

m

l

61°

134°

113° i j

k

69°

74°

68°

h

g

116°

80°

55°

e

f

126°

37°w

v

119°47°

x

x

194

CHAPTER 11 Angles 2

62°

b b

a

195

11.4 Polygons CHAPTER 11

Using this method, the sum of the interior angles of any polygon can be found.

The number of triangles into which the polygon can be split up is always two less than thenumber of sides.

Find the sum of the angles of a 12-sided polygon (dodecagon).

Solution 812 � 2 � 10

10 � 180 � 1800

The sum of the angles � 1800°

A polygon with all its sides the same length and all its angles the same size is called aregular polygon.

So a square is a regular polygon, because all its sides are the same length and all its angles are 90°, but a rhombus is not a regular polygon.Although its sides are all the same length, its angles are not all the same size.

Here are three more regular polygons.

a regular pentagon a regular hexagon a regular octagon

Regular octagons tessellatewith squares.

Bees’ honeycomb is madeup of regular hexagons.

The Pentagon in Washington DCis the headquarters of the

US Department of Defence.

State the sum of the angles in degrees.

Multiply the number of triangles by 180.

Subtract 2 from the number of sides to find the number of triangles.

Example 8

Number of sides Number of triangles Sum of the interior angles

4 2 360°

5 3 540°

6 4 720°

7 5 900°

8 6 1080°

9 7 1260°

10 8 1440°

196

CHAPTER 11 Angles 2

Find the size of each interior angle of a regular decagon.

Solution 910 � 2 � 8

8 � 180 � 1440

1440 � 10 � 144

Each interior angle is 144°

The diagram shows a regular 9-sided polygon (nonagon) with centre O.

a Work out the size of i angle x ii angle y.

b Use your answer to part a ii to work out the size of each interior angle of the polygon.

Solution 10a i x � 360° � 9

x � 40°

(40° is the angle at the centre of a regular 9-sided polygon.)

ii 180° � 40° � 140

140° � 2° � 70°

y � 70°

b 2 � 70° � 140°

Each interior angle is 140°.

Exercise 11D

In this exercise, the polygons are not accurately drawn.

1 Find the sum of the angles of a 15-sided polygon.

2 Find the sum of the angles of a 20-sided polygon.

State the size of each interior angle.

State the size of angle y.

State the size of angle x.

Example 10

State the size of each interior angle.

All 10 interior angles are the same size. So divide 1440 by 10

Subtract 2 from the number of sides to find the number of triangles.

Example 9

Each corner of the polygon could be joined to the centreO to make 9 equal angles at O. The total of all 9 anglesis 360°, as altogether they make a complete turn.

The angle sum of a triangle is 180° and sothe sum of the two base angles is 140°.

The triangle is isosceles and sothe two base angles are equal.

Because the polygon is regular, it has nine linesof symmetry and each interior angle is twicethe size of each base angle of the triangle.

Multiply the number of triangles by 180to find the sum of all 10 interior angles.

x

O

y

197

11.4 Polygons CHAPTER 11

3 A polygon can be split into 17 triangles by drawing diagonals from one corner.How many sides has the polygon?


4 5 6

7 8 9

10 The diagram shows a pentagon.All its sides are the same length.

a Work out the value of g.

b Is the pentagon a regular polygon?Explain your answer.

11 Work out the size of each interior angle of

a a regular pentagon b a regular hexagon c a regular octagon.

12 Work out the size of each interior angle of a regular 15-sided polygon.

13 Work out the size of each interior angle of a regular 20-sided polygon.

14 Work out the size of the angle at the centre of a regular pentagon.

15 Work out the size of the angle at the centreof a regular 12-sided polygon.

16 The angle at the centre of a regular polygon is 60°.How many sides has the polygon?

17 The angle at the centre of a regular polygon is 20°.

a How many sides has the polygon?

b Work out the size of each interior angle of the polygon.

f

129°114°

134°

122°

137°

140°

162°

e 153°

124°128°136°

118°

d104°

121°

126°132°

118°

c

82°

147° 123°

88°

131°

b

97°

104° 121°

a

117°

109°

94°

81°

g g

60°

Australia’s 50 centcoin is a regular

12-sided polygon(dodecagon)

198

CHAPTER 11 Angles 2

18 a Work out the angle at the centre of a regular octagon.

b Draw a circle with a radius of 5 cm and, using your answer to part a , draw aregular octagon inside the circle.

19 a Work out the angle at the centre of a regular 10-sided polygon.

b Draw a circle with a radius of 5 cm and, using your answer to part a , draw aregular 10-sided polygon inside the circle.

20 The diagram shows a pentagon. 21 The diagram shows a hexagon.Work out the size of Work out the size of

a angle h b angle i. a angle j b angle k.

22 Craig says, ‘The sum of the interior angles of this polygon is 1000°’.Explain why he must be wrong.

23 The diagram shows a quadrilateral.

a Work out the size of each of the angles marked with letters.

b Work out l � m � n � p

24 The diagram shows a pentagon.

a Work out the size of each of the angles marked with letters.

b Work out q � r � s � t � u

25 The diagram shows a hexagon.

a Work out the size of each of the angles markedwith letters.

b Work out u � v � w � x � y � z

11.5 Exterior anglesA polygon’s interior angles are the angles inside the polygon.

Extend a side to make an exterior angle, which is outside the polygon.

At each vertex (corner), the interior angle and the exteriorangle are on a straight line and so their sum is 180°.

interior angle � exterior angle � 180°

The sum of the exterior angles of any polygon is 360°.

68°

97°

142°124°

87°

j k83°

127°

116°

93°

h i

106°

124°

94°

58°

n

pl

m

117°

145°

124°73°

81°

s

t

uq

r

163°u

85°z

123°129°

152°

68°

w

v

x

y

interiorangle exterior angle

199

11.5 Exterior angles CHAPTER 11

To show this, imagine someone standing at P on thisquadrilateral, facing in the direction of the arrow.

They turn through angle a, so that they are facing inthe direction PQ, and then walk to Q.

At Q, they turn through angle b, so that they arefacing in the direction QR, and then walk to R.

At R, they turn through angle c, so that they are facing in the direction RS, and then walk to S.

At S, they turn through angle d.

They are now facing in the direction of the arrow again and so they have turned through 360°.

The total angle they have turned through is also the sum of the exterior angles of thequadrilateral.

So a � b � c � d � 360°

The same argument can be used with any polygon, not just a quadrilateral.

The sizes of four of the exterior angles of a pentagon are 67°, 114°, 58° and 73°.Work out the size of the other exterior angle.

Solution 11

67° � 114° � 58° � 73° � 312°

360° � 312° � 48°

Exterior angle � 48°

For a regular 18-sided polygon, work outa the size of each exterior angle, b the size of each interior angle.

Solution 12a 360° � 18 � 20°

b 180° � 20° � 160°

The size of each interior angle of a regular polygon is 150°. Work outa the size of each exterior angle, b the number of sides the polygon has.

Solution 13a 180° � 150° � 30°

b 360° � 30 � 12

Example 13

Example 12

State the size of the exterior angle.

Subtract the result from 360.

Add the four given exterior angles.

Example 11

Q

R

SP

bc

da

Because the polygon is regular, all 18 exterior anglesare equal. Their sum is 360° and so divide 360° by 18.

At a corner, the sum of the interior angle and theexterior angle is 180°. So subtract 20° from 180°.

At a corner, the sum of the interior angle and theexterior angle is 180°. So subtract 150° from 180°.

Because the polygon is regular, all the exterior anglesare 30°. Their sum is 360° and so divide 360 by 30.

200

CHAPTER 11 Angles 2

Exercise 11E

1 At a vertex (corner) of a polygon, the size of the interior angle is 134°.Work out the size of the exterior angle.

2 At a vertex of a polygon, the size of the exterior angle is 67°.Work out the size of the interior angle.

3 The sizes of three of the exterior angles of a quadrilateral are 72°, 119° and 107°.Work out the size of the other exterior angle.

4 The sizes of five of the exterior angles of a hexagon are 43°, 109°, 58°, 74° and 49°.Work out the size of the other exterior angle.

5 Work out the size of each exterior angle of a regular octagon.

6 Work out the size of each exterior angle of a regular 9-sided polygon.

7 For a regular 24-sided polygon, work out

a the size of each exterior angle, b the size of each interior angle.

8 For a regular 40-sided polygon, work out

a the size of each exterior angle, b the size of each interior angle.

9 The size of each interior angle of a regular polygon is 168°. Work out

a the size of each exterior angle, b the number of sides the polygon has.

10 The size of each interior angle of a regular polygon is 170°.Work out the number of sides the polygon has.

11.6 Corresponding angles and alternate anglesParallel lines are always the same distance apart. They never meet. (Section 3.5) In diagrams, arrows are used to show that lines are parallel.

Other pairs of corresponding angles have been shaded in the diagrams below.

The F shape formed bycorresponding angles can behelpful in recognising them.

In the diagram, a straight line crosses two parallellines. The shaded angles are called correspondingangles and are equal to each other.

201

11.6 Corresponding angles and alternate angles CHAPTER 11

In the diagram, a straight line crosses two parallel lines.

The shaded angles are called alternate anglesand are equal to each other.

Write down the letter of the angle which isa corresponding to the shaded angle,b alternate to the shaded angle.

Solution 14a Angle q is the corresponding angle to

the shaded angle.b Angle s is the alternate angle to the shaded angle.

a Find the size of angle x.b Give a reason for your answer.

Solution 15a x � 78° b Alternate angles.

a Find the size of angle p.b Give a reason for your answer.c Find the size of angle q.d Give a reason for your answer.

Solution 16a 180° � 67° � 113° b The sum of the angles on a

p � 113° straight line is 180°.c q � 113° d Corresponding angles.

Example 16

Example 15

Notice that they form a Z shape.

Notice that they form an F shape.

Example 14

Another pair of alternate angles has beenshaded in this diagram.

The Z shape formed by alternate anglescan be helpful in recognising them.

s p

qr

x78°

p

q

67°

202

CHAPTER 11 Angles 2

Exercise 11F

In this exercise, the diagrams are not accurately drawn.

1 Write down the letter of the angle which isa corresponding to the shaded angle,b alternate to the shaded angle.

2 Write down the letter of the angle which isa corresponding to the shaded angle,b alternate to the shaded angle.

In Questions 3–5, find the sizes of the angles marked with letters and state whether the pairs of angles are corresponding or alternate.

3 4 5

In Questions 6–20, find the sizes of the angles marked with letters.Give reasons for your answers.

6 7 8

9 10 11

12 13 14

15 16 17 53°

62° i

j

h

69°52°

eg

f154°

75°

a b

c

d

82°

61°

x

z

y

76°

47° uw

v

42°

59°t

s

r

52°np

q

67°

79°

l m

76°

j

k

i h126°

fg

82°

132°d

e

c

96°b

118°

75°a

uv

wx

y wv

x

203

11.7 Proofs CHAPTER 11

18 19 20

11.7 ProofsIn mathematics, a proof is a reasoned argument to show that a statement is always true.The proofs which follow make use of corresponding and alternate angles.

Proof 1

An exterior angle of a triangle is equal to the sum of the interior angles at the other two vertices

The diagram shows a triangle PQR.

Extend the side PQ to S.

At Q draw a line QT parallel to PR.

Then angle x � angle a (corresponding angles)

and angle y � angle b (alternate angles)

Adding, x � y � a � bx � y is the exterior angle of the triangle and a � b is the sum of the interior angles at the other two vertices and so the statement is true.

Proof 2

The angle sum of a triangle is 180°

This proof starts in the same way as Proof 1.

The diagram shows a triangle PQR.

Extend the side PQ to S.

At Q draw a line QT parallel to PR.

Then angle x � angle a (corresponding angles)

and angle y � angle b (alternate angles)

Adding, x � y � a � bAs x, y and c are angles on a straight line, their angle sum is 180°, that is

x � y � c � 180°

So a � b � c � 180° which proves that the statement is true.

Proof 3

The opposite angles of a parallelogram are equal

Draw a diagonal of the parallelogram.

angle a � angle c (alternate angles)

angle b � angle d (alternate angles)

Adding, a � b � c � d which proves that the statement is true.

78° 37°

m

64°

l

59°

k

a

b

R

PQ

yx

S

T

a c

b

R

PQ

yx

S

T

a

cd

b

a Find the size of angle w. b Give a reason for your answer.

Solution 17a 63° � 44° � 107° b Exterior angle of a triangle.

w � 107° (As the full reason is long, it may be shortened to this.)

a Find the size of angle x. b Give a reason for your answer.

c Find the size of angle y. d Give reasons for your answer.

Solution 18a x � 71° b Opposite angles of a parallelogram are equal.

c 2 � 71° � 142° d Angle sum of a quadrilateral is 360°.360° � 142° � 218° Opposite angles of a parallelogram are equal.

218° � 2 � 109°y � 109°

Exercise 11G

In this exercise, the diagrams are not accurately drawn.Find the size of each of the angles marked with letters.Give reasons for your answers.

1 2 3 4

5 6 7 8

9 10

11.8 BearingsBearings are used to describe directions.Bearings are measured clockwise from North.When the angle is less than 100°, one or two zeros are written in front of the angle, sothat the bearing still has three figures.

r 47°

p

q36°42°

n

39°127°

km

l

i j

117°

h g

118°

e

f64°

d

137°

c

78° 104°b 71°

38°

a47°

62°

Example 18

Example 17

204

CHAPTER 11 Angles 2

w 44°

63°

x

y71°

Folkestone and Dover are shown on the map.

The bearing of a ship from Folkestone is 117°.

The bearing of the ship from Dover is 209°.

Draw an accurate diagram to show the positionof the ship.

Mark the position with a cross X. Label it S.

Solution 21Draw a line on a bearing of 117° from Folkestone.

Draw a line on a bearing of 209° from Dover by

measuring an angle of 151° (360° � 209°)

anticlockwise from North.

(Alternatively, measure an angle of 29° clockwise

from South.)

Put a X where the lines cross.

Label the position S.

Example 21

Measure the bearing of Q from P.

Solution 20To find the angle clockwisefrom North with a semi-circular protractor measure the shadedanticlockwise angle (38°) and subtract it from 360°.

360° � 38° � 342°

The bearing of Q fromP is 342°.

Measure the bearing of B from A.

Solution 19From North, measurethe angle clockwise.

The angle is 52°.

So the bearing is 052°.

Example 20Example 19

205

11.8 Bearings CHAPTER 11

B

A

N

B

A

N

Q

P

N

Q

P

N

P

Q

N

N

Folkstone

Dover

N

Folkstone

Dover

S

N

209°

N

117°

206

CHAPTER 11 Angles 2

Sometimes, answers to questions have to be worked out, not found using a protractor.

The bearing of B from A is 061°.

Work out the bearing of A from B.

Solution 22The bearing of A from B is the reflex angle at B.

y � 61° (alternate angles)

Bearing � 180° � 61°� 241°

Exercise 11H

In Questions 1–4, measure the bearing of Q from P.

1 2

3 4

N

P

Q

N

P

Q

Example 22

61°

N

N

Diagram NOTaccurately drawn

A

B

61°

N

N


A

yB

N

P

Q

N

PQ

207

11.8 Bearings CHAPTER 11

5 Draw diagrams similar to those in Questions 1–4 to show the bearings

a 026° b 217° c 109° d 334°.

6 The diagram shows two points, A and B.The bearing of a point L from A is 048°.The bearing of L from B is 292°.On the diagram on the resource sheet find the position of L by making an accurate drawing.

7 The diagram shows two points, P and Q.The bearing of a point M from P is 114°.The bearing of M from Q is 213°.On the diagram on the resource sheet find the position of M by making an accurate drawing.

8 Cromer and Great Yarmouth are shown on the map.The bearing of a ship from Cromer is 052°.The bearing of the ship from Great Yarmouth is 348°.On the diagram on the resource sheet find the position of the ship by making an accurate drawing.Mark the position of the ship with a cross X.Label it S.

11 The bearing of B from A is 074°.The bearing of C from B is 180°.AB � AC.Work out the bearing of C from A.

12 The diagram shows the positions ofYork, Scarborough and Hull.The bearing of Scarborough fromYork is 052°.The bearing of Hull from York is 118°.The distance between York and Scarborough is the same as the distance between York and Hull.Work out the bearing of Hull from Scarborough.

N

T

S146°

N Diagram NOTaccurately drawn

N

Q

P

38°

N


10 The bearing of T from S is 146°.Work out the bearing of S from T.

9 The bearing of Q from P is 038°.Work out the bearing of P from Q.

P Q

A B

N

Cromer

N

GreatYarmouth

74°

N

C

B

A

NDiagram NOTaccurately drawn

52°

N

York

Hull

Scarborough


208

CHAPTER 11 Angles 2

Chapter summary

You should know and be able to use these facts


An equilateral triangle has three equal angles and three equal sides.

An isosceles triangle has two equal sides and the angles opposite the equal sides are equal.

A triangle whose sides are all different lengths is called a scalene triangle.

A quadrilateral is a shape with four straight sides and four angles.

The angle sum of a quadrilateral is 360°.

A polygon is a shape with three or more straight sides.

A 5-sided polygon is called a pentagon.

A 6-sided polygon is called a hexagon.

An 8-sided polygon is called an octagon.

A 10-sided polygon is called a decagon.

The angle sum of a polygon can be found by subtracting 2 from the number of sides and multiplying the result by 180°.

A polygon with all its sides the same length and all its angles the same size is called a regular polygon.

At a vertex, interior angle � exterior angle � 180°.

The sum of the exterior angles of any polygon is 360°.To find the size of each exterior angle of a regular polygon, divide 360° by the number of sides.

Where a straight line crosses two parallel lines,the corresponding angles are equal.

Where a straight line crosses two parallel lines,the alternate angles are equal.

Bearings are measured clockwise � from North.

You should also know these proofs

An exterior angle of a triangle is equal to the sum of the interior angles at the other two vertices.


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60° 60°

60°

interiorangle exterior angle

209

Chapter 11 review questions CHAPTER 11

Chapter 11 review questionsIn Questions 1–12, find the size of each of the angles marked with a letter.The diagrams are not accurately drawn.

1 2 3 4

5 6 7 8

9 10 11 12

13 In triangle ABC, AB � AC and angle C � 50°.

a Write down the special name of triangle ABC.

b Work out the value of y.(1385 June 1999)

14 Calculate the value of x.

(4400 Nov 2004)

15 Work out the value of a.

(1388 Mar 2002)

16 Work out the size of each exterior angle of a regular 10-sided polygon.

17 a Work out the sum of the interior angles of a 9-sided polygon.

The size of each exterior angle of a regular polygon is 20°.

b Work out how many sides the polygon has.

102°78°

43°

63°

a


150°

25°

45°

x


53°

t113°r

s

64°

q p

57°

m

n

141°

92°

117° 128°

114°

l112°

94°

j

k109°

64° 84°

i

118° g

h

38°ef

56°

d

58° 94° b

c71°

64° a

50°

yA

B C


18 The diagram shows a regular hexagon.

a Work out the value of x.

b Work out the value of y.

(1385 June 2001)

19 ABCDE is a regular pentagon.AEF and CDF are straight lines.Work out the size of angle DFE.Give reasons for your answer.

(1388 Mar 2004)

20 a i Write down the size of the angle marked x.

ii Give a reason for your answer.

b i Write down the size of the angle marked y.

ii Give a reason for your answer.(1384 Nov 1996)

21 AC � BCAB is parallel to DCAngle ABC � 52°

a Work out the value of i p ii qThe angles marked p° and r° are equal.

b What geometrical name is given to this type of equal angles? (1384 Nov 1997)

22 The diagram represents the positions of Wigan and Manchester.

a Measure and write down the bearing of Manchester from Wigan.

b Find the bearing of Wigan from Manchester.

(1385 June 1998)

23 Measure the bearing of A from B.

(1388 March 2004)

210

CHAPTER 11 Angles 2

y

x


A

B

CD

F

E Diagram NOTaccurately drawn

A

B x y

D E

C


55°

75°

p°

A

Bq°r°

D

C52°

N

Wigan

Manchester

N

N

B

A

24 Work out the bearing of

i B from P,

ii P from A.

(1387 Nov 2004)

25 The diagram shows the position of each of three buildings in a town.

The bearing of the Hospital from the Art Gallery is 072°.The Cinema is due East of the Hospital.The distance from the Hospital to the Art Gallery is equal to the distance from theHospital to the Cinema.Work out the bearing of the Cinema from the Art Gallery.

(1387 Nov 2003)

N

CinemaHospital

72°

N

Artgallery


211

Chapter 11 review questions CHAPTER 11

N

63°138°

P

A

B


Angles 2 11 - Wikispaces 11...11.1 Triangles This shows that the angles in this triangle add up to 180° but ... Solution 2 Angle sum of ... The angles at a point add up to 360° and

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