Pythagoras’ theorem and trigonometry (1) 19 19.1 Pythagoras’ theorem Pythagoras was a famous mathematician in Ancient Greece. The theorem which is named after him is an important

295

19.1 Pythagoras’ theoremPythagoras was a famous mathematician in Ancient Greece.The theorem which is named after him is an important result about right-angled triangles.

Here is a right-angled triangle ABC.

The angle at C is the right angle.The side, AB, opposite the right angle is called the hypotenuse.It is the longest side in the triangle.

The right-angled triangle in the diagram onthe right has sides of length 3 cm, 4 cm and 5 cm.Squares have been drawn on each side of the triangle and each square has been divided up into squares of side 1 cm.

The area of the square on the side of length 3 cm is 9 cm2.The area of the square on the side of length 4 cm is 16 cm2.The area of the square on the side of length 5 cm (the hypotenuse) is 25 cm2.

Notice that 25 cm2 � 9 cm2 � 16 cm2

that is, 52 � 32 � 42

In other words 52 (the area of the square on the hypotenuse) is equal to the sum of 32 and 42

(the areas of the squares on the other two sides added together).

This is an example of Pythagoras’ theorem. It is only true for right-angled triangles.

Pythagoras’ theorem states:

Area of square R � area of square P � area of square Q

Pythagoras’ theorem can be used to findthe length of the third side of a right-angled trianglewhen the lengths of the other two sides are known.For this, the theorem is usually stated in terms of thelengths of the sides of the triangle.

That is

Pythagoras’ theorem can also be written

DE 2 � EF 2 � DF 2

(DE 2 means that the length of the side DE is squared.)

c2 � a2 � b2

a

b c

R

P

Q

In a right-angled triangle, the area of the square on the hypotenuseis equal to the sum of areas of the squares on the other two sides.

A

C B

19C H A P T E R

Pythagoras’ theorem and trigonometry (1)

D

F E

9 cm2

16 cm2

25 cm2

4

3

5

19.2 Finding lengthsPythagoras’ theorem can be used to work out the length of the hypotenuse of a right-angled trianglewhen the lengths of the two shorter sides are given.

Work out the length of the hypotenuse in this triangle.

Solution 1c2 � a2 � b2

c2 � 82 � 152

c2 � 64 � 225

c2 � 289

c � �289� � 17

Length of hypotenuse � 17 cm

It is important to be able to apply Pythagoras’ theorem when the triangle is in a different position.

In triangle XYZ, angle X � 90°, XY � 8.6 cm and XZ � 13.9 cm.Work out the length of YZ.Give your answer correct to 3 significant figures.

Solution 2

YZ2 � XY2 � XZ2

YZ2 � 8.62 � 13.92

YZ2 � 73.96 � 193.21

YZ2 � 267.17

YZ � �267.17�� 16.34…

YZ � 16.3 cm (to 3 s.f.) Give the final answer correct to 3 significant figures.

Find �267.17� Write down at least 4 figures.

Work out 8.62 and 13.92 and add the results.

Substitute the given lengths.

State Pythagoras’ theorem.

The hypotenuse is the side opposite the right angle.Angle X is the right angle so the hypotenuse is YZ.

13.9 cm

8.6 cm

X

Y

Z

Example 2

The answer is sensible because the hypotenuseis longer than the other two sides.

Find �289�

Work out 82 and 152 and add the results.



8 cm

15 cm c cm

Example 1

296

CHAPTER 19 Pythagoras‘ theorem and trigonometry (1)

Pythagoras’ theorem can also be used to work out the length of one of the shorter sides in a right-angled triangle when the lengths of the other two sides are known.

In triangle ABC, angle A � 90°, BC � 17.4 cm and AC � 5.8 cm.Work out the length of AB. Give your answer correct to 3 significant figures.

Solution 3

BC2 � AC2 � AB2

17.42 � 5.82 � AB2

302.76 � 33.64 � AB2

302.76 � 33.64 � AB2

269.12 � AB2

AB � �269.12�� 16.40…

AB � 16.4 cm (to 3 s.f.)

Exercise 19A

1 Work out the length of the sides marked with letters in these triangles.a b c

2 Work out the length of the sides marked with letters in these triangles.a b c

3 Work out the length of the sides marked with letters in these triangles.Give each answer correct to 3 significant figures.a b c d

4.8 cm

10.6 cm

d6.2 cm

8.3 cm

c4.8 cm

9.1 cm

b

7.9 cm

7.4 cm

a

20 cm

25 cmc

12 cm37 cm

b24 cm

25 cma

8 cm

6 cm

c

40 cm

9 cm

b

12 cm

5cm

a

Give the final answer correct to 3 significant figures.

Find �269.12� Write down at least 4 figures.

Subtract 33.64 from both sides.

Work out 17.42 and 5.82



Angle A is the right angle so the hypotenuse is BC.17.4 cm

5.8 cm

B

A C

Example 3

297

19.2 Finding lengths CHAPTER 19

�

4 Work out the length of the sides marked with letters in these triangles.Give each answer correct to 3 significant figures.

a b

c d

5 a In triangle ABCangle A � 90°, AB � 3.4 cm and AC � 12.1 cm.Work out the length of BC.Give your answer correct to 3 significant figures.

b In triangle ABCangle A � 90°, AB � 5.9 cm and BC � 16.3 cm.Work out the length of AC.Give your answer correct to 3 significant figures.

c In triangle PQRangle R � 90°, PR � 5.9 cm and QR � 13.1 cm.Work out the length of PQ.Give your answer correct to 3 significant figures.

d In triangle PQRangle R � 90°, PQ � 11.2 cm and QR � 9.6 cm.Work out the length of RP.Give your answer correct to 3 significant figures.

e In triangle XYZangle X � 90°, XY � 12.6 cm and XZ � 16.5 cm.Work out the length of YZ.Give your answer correct to 3 significant figures.

f In triangle DEFangle E � 90°, DF � 10.1 cm and EF � 7.8 cm.

i Draw a sketch of the right-angled triangle DEF and label sides DF and EF with theirlengths.

ii Work out the length of DE.Give your answer correct to 3 significant figures.

16.5 cm

12.6 cm

Z

Y

X

9.6 cm 11.2 cm

R

Q

P

13.1 cm

5.9 cmR

Q

P

16.3 cm5.9 cm

A

B

C

3.4 cm

12.1 cmA C

B

8.3 cm

2.1 cmd12.4 cm

1.8 cm

c

11.3 cm

8.1 cm

b

4.8 cm10.7 cm

a

298


19.3 Applying Pythagoras’ theoremPythagoras’ theorem can be used to solve problems.

A boat travels due north for 5.7 km. The boat then turns and travels due east for 7.2 km. Work out the distance between the boat’s finishing point and its starting point. Give your answer correct to 3 significant figures.

Solution 4

d2 � 5.72 � 7.22

d2 � 32.49 � 51.84d2 � 84.33d � �84.33�� 9.183…

Distance � 9.18 km (to 3 s.f.)

Isosceles triangles can be split into two right-angled triangles and Pythagoras’ theorem can then be used.

The diagram shows an isosceles triangle ABC. The midpoint of BC is the point M.In the triangle, AB � AC � 8 cm and BC � 6 cm.a Work out the height, AM, of the triangle.

Give your answer correct to 3 significant figures.b Work out the area of triangle ABC.

Give your answer correct to 3 significant figures.

Solution 5Pythagoras’ theorem cannot be used in triangle ABC as this triangle is not right-angled.a

By PythagorasAB2 � AM2 � BM2

82 � h2 � 32

64 � h2 � 964 � 9 � h2

55 � h2

h � �55� � 7.416…h � 7.42

Height of triangle � 7.42 cm (to 3 s.f.)

Triangle ABM is right-angled with hypotenuse AB. Theheight, AM, of the triangle is marked h cm on the sketch.

Draw a sketch of triangle ABM.

BM � 3 cm as M is the midpoint of BC.

As M is the midpoint of the base of the isosceles triangle,the line AM is the line of symmetry of triangle ABC. SoAM is perpendicular to the base and angle AMB � 90°.

3 cm

8 cmh cm

B M

A

8 cm8 cm

A

CB

6 cmM

Example 5

The sketch is of a right-angled triangle so thatPythagoras’ theorem can be used.

Remember that the points of the compass are

Draw a sketchof the boat’sjourney

5.7 km

Start

Finish7.2 km

d km

Example 4

299

19.3 Applying Pythagoras’ theorem CHAPTER 19

N

S

W E

The distance between the starting point and thefinishing point is the length of the hypotenuse of thetriangle, marked d km in the sketch.

b Area � �12� � 6 � 7.416…

Area � 22.24…

Area � 22.2 cm2 (to 3 s.f.)

Exercise 19B

1 The diagram shows a ladder leaning against a vertical wall.The foot of the ladder is on horizontal ground.The length of the ladder is 5 m.The foot of the ladder is 3.6 m from the wall.Work out how far up the wall the ladder reaches.Give your answer correct to 3 significant figures.

2 The diagram shows a rectangle of length 9 cm and width 6 cm.Work out the length of a diagonal of the rectangle.Give your answer correct to 3 significant figures.

3 Aiton (A), Beeville (B) and Ceaborough (C) are three towns as shown in this diagram.Beeville is 10 km due south of Aiton and 21 km due east of Ceaborough.Work out the distance between Aiton and Ceaborough.Give your answer correct to the nearest km.

4 Work out the area of the triangle.Give your answer correct to 3 significant figures.

5 Work out the perimeter of the triangle.Give your answer correct to 3 significant figures.

6 The diagram represents the end view of a tent,triangle ABC, two guy-ropes, AP and AQ, and a vertical tent pole, AN. The tent is on horizontal ground so that PBNCQ is a straight horizontal line.Triangles ABC and APQ are both isosceles triangles.

BN � NC � 2 m, AN � 2.5 m and AP � AQ � 5 m

a Work out the length of the side AC of the tent.Give your answer correct to 3 significant figures.

b Work out the length ofi NQ ii CQ.

Give your answers correct to 3 significant figures.

There is a tent peg at P and a tent peg at Q.

c Work out the distance between the two tent pegs at P and Q.Give your answer correct to 3 significant figures.

A

P B N C Q2 m

5 m2.5 m

11.7 cm

8.3 cm

10.6 cm14.5 cm

21 km

10 km

A

BC

6 cm

9 cm

3.6 m

5 m

For triangle ABC base � 6 cm andheight � 7.416… cm.

area of a triangle � �12� � base � height

300


�

7 The diagram shows two right-angled triangles.

a Work out the length of the side marked x.

b Hence work out the length of the side marked y.

8 The diagram shows two right-angled triangles.Work out the length of the side marked a.Give your answer correct to 3 significant figures.

9 The lengths of the sides of a triangle are 8 cm, 15 cm and 17 cm.

a Find the value of i 82 � 152 ii 172

b What do you notice about the two answers in a?

c What information does this give about the triangle?

10 Here are the lengths of sides of six triangles.Triangle 1 5 cm, 12 cm and 13 cm Triangle 2 9 cm, 40 cm and 41 cmTriangle 3 10 cm, 17 cm and 18 cm Triangle 4 20 cm, 21 cm and 29 cmTriangle 5 8 cm, 17 cm and 20 cm Triangle 6 33 cm, 56 cm and 65 cmFind by calculation which of these triangles are right-angled triangles.

19.4 Coordinates, line segments and Pythagoras’ theorem

The diagram shows the straight line joining the points A(2, 1) and B(8, 5). A line joining two points iscalled a line segment. So in the diagram, AB is the line segment joining the points A and B.

Midpoint of a line segmentThe midpoint M of the line segment AB has coordinates (5, 3).

Notice that the x-coordinate of A is 2, the x-coordinate of B is 8 and �2 �

2

8� � 5, the x-coordinate of M.

Similarly the y-coordinate of A is 1, the y-coordinate of B is 5 and �1 �

2

5� � 3, the y-coordinate of M.

In general the x-coordinate of the midpoint of a line segment is the mean of the x-coordinates of itsendpoints and the y-coordinate of its midpoint is the mean of the y-coordinates of its endpoints.

That is

the midpoint of the line joining (a, b) and (p, q) is the point ��a �

2

p�, �

b �

2

q��

1

1 2 3 4 5 6 7 8 9

2

3

4

5

6

y

O

A

M

B

x

a

5.4 cm

3.9 cm

4.7 cm

x

y

39 cm

12 cm

9 cm

301

19.4 Coordinates, line segments and Pythagoras’ theorem CHAPTER 19

Find the midpoint of the line joininga (3, 5) and (13, 7) b (�5, 8) and (9, �13).

Solution 6a 3 � 13 � 16

�1

2

6� � 8

5 � 7 � 12

�1

2

2� � 6

Midpoint is (8, 6)

b �5 � 9 � 4

�4

2� � 2

8 � �13 � �5

��

2

5� � �2.5

Midpoint is (2, �2.5)

Length of a line segmentThe diagram shows the points A(1, 1) and B(9, 5).The right-angled triangle ABC has been drawn so that AC � 8 and BC � 4

Pythagoras’ theorem can be usedto find the length of AB.

AB2 � 82 � 42

AB2 � 64 �16 � 80

AB � �80� � 8.94 (to 3 s.f.)

Find the length of the line joining

a A(3, 2) and B(15, 7) b P(�9, 4) and Q(7, �5)

Solution 7a

AB2 � 122 � 52

AB2 � 144 � 25 � 169

AB � �169� � 13

BC � 7 � 2 � 5

AC � 15 � 3 � 12

Draw a sketch showing A and B andcomplete the right-angled triangle ABC.

3 15

212

5

7

y

A

B

C

O x

Example 7

1

1 2 3 4 5 6 7 8

8

4

9 10

2

3

4

5

6

y

OA C

B

x

The y-coordinates are 8 and �13

The x-coordinates are �5 and 9

The y-coordinates are 5 and 7

The x-coordinates are 3 and 13

Example 6

302


Use Pythagoras’ theorem to find the length of AB.

b

PQ2 � 162 � 92

PQ2 � 256 � 81 � 337

PQ � �337� � 18.4 (to 3 s.f.)

Exercise 19C

1 Work out the coordinates of the midpoint of the line joining

a (3, 1) and (11, 7) b (2, 5) and (12, 29) c (�6, 9) and (8, 13)

d (�4, �6) and (6, 12) e (9, �15) and (�11, 6) f (0, �5) and (9, �11)

2 Work out the length of the line joining each of the pair of points in question 1

3 The point A has coordinates (5, 2), the point B has coordinates (8, 6) and the point C has coordinates (1, 5).

a Work out the length ofi AB ii BC iii AC.

b What does your answer to part a tell you about triangle ABC?

4 The point P has coordinates (5, 3), the point Q has coordinates (6, 6) and the point R hascoordinates (�6, 10).

a Work out the length of each side of triangle PQR.

b Use your answers to part a to show that triangle PQR is a right-angled triangle.

c Work out the area of triangle PQR.

5 The points A(2, 6) and B(18, 36) are the ends of a diameter of a circle.

a Find the coordinates of the centre of the circle.

b Work out thei diameter of the circle ii radius of the circle.

6 A circle has centre O(4, 2). The point A(9, 14) lies on the circle.

a Work out the radius of the circle.

b Determine by calculation which of the following points also lie on the circle.i B(16, 7) ii C(�1, �10) iii D(7, 16) iv E(4, 15).

7 The point A has coordinates (�3, �8) and the point B has coordinates (8, 9).

a Find the coordinates of the midpoint of the line segment AB.

b Work out the length of the line segment AB. Give your answer correct to 3 significant figures.

8 The points A (5, 1), B (29, 8), C (9, 23) and D (�15, 16) are the vertices of quadrilateral ABCD.

a Work out the length ofi AB ii BC iii CD iv DA.

b Explain what your answers to a tell you about the quadrilateral ABCD.

9 The point A has coordinates (a, b) and the point B has coordinates (p, q).

Show that the length of the line segment AB is �(p � a)�2 � (q �� b)2�

Use Pythagoras’ theorem to find the length of PQ.

PR � 4 � �5 � 4 � 5 � 9

QR � 7 � �9 � 7 � 9 � 16

Draw a sketch showing P and Q andcomplete the right-angled triangle PQR.

9

16

P (�9, 4)

Q (7, �5)R

303

19.4 Coordinates, line segments and Pythagoras’ theorem CHAPTER 19

�

19.5 Trigonometry – introductionTrigonometry means ‘triangle measure’.It is used to work out lengths and angles in triangles and in shapes that can be divided up into triangles.Trigonometry is important in bridge building and tunnel building where it is important to knowaccurate distances and accurate angle sizes.It is also used in many other areas of surveying, engineering and architecture.

Trigonometric ratiosHere are two right-angled triangles.The triangle with hypotenuse 2 is an enlargement with scale factor 2 ofthe triangle with hypotenuse 1, that is, its sides are twice as long.

p � 2s and q � 2c

So if s and c are known for the right-angled triangle with hypotenuse 1, p and q can be calculated.If the length of its hypotenuse is known, the lengths of the sides of any right-angled triangle which is an enlargement of these triangles can be calculated.

The values of s and c are known accurately and can be found on anystandard scientific calculator.The length s is called the sine of 70° written sin 70°. Not all calculators are the samebut the following key sequence to find sin 70° applies to many calculators.

Make sure that the angle mode of your calculator is degrees, usually shown by on the calculator screen.

The number 0.93969262 should appear on your calculator display.

So correct to 4 decimal places sin 70° � 0.9397

The length c is called the cosine of 70° and is written cos 70°. As above but using the button

correct to 4 decimal places cos 70° � 0.3420

Using the triangles opposite and writing s as sin 70º and c as cos 70º

p � 2sin 70º and q � 2cos 70º

So for any right-angled triangle

p � r sin xº and q � r cos xº

or sin xº � �p

r� and cos xº � �

q

r�

p

q

r

x°

sin x°

cos x°x°

1

70°

2

q

p70°

sin 70°

cos 70°

1

cos

�

Press

70

Key in

sin

Press

D

70°

2

q

p70°

1

c

s

304


The hypotenuse (hyp) of a right-angled triangle is the side opposite the right angle and is thelongest side of the triangle. The sides of the triangle are named according to their position relative tothe angle given or the angle to be found. If this angle is xº then the side opposite this angle is calledthe opposite side (opp). The side next to this angle is called the adjacent side (adj).

The results above become

opp � hyp � sin xº and adj � hyp � cos xº

or sin xº � �o

h

p

yp

p� and cos xº � �

h

a

y

d

p

j�

sin x° is used when opposite and hypotenuse are involved.cos x° is used when adjacent and hypotenuse are involved.

When opposite and adjacent are involved, a third result called tan x°, is needed where

tan x° � �o

a

p

d

p

j� or opp � adj � tan x°

tan x° means the tangent of x°.

SOHCAHTOA might help you to remember these results.

Sin Opp Hyp Cos Adj Hyp Tan Opp Adj

19.6 Finding lengths using trigonometry

Work out the length of each of the marked sides. Give each answer correct to 3 significant figures.

a b c

Solution 8a

a � 13 � cos 50º � 13cos 50°a � 13 � 0.6427…a � 8.356…a � 8.36 cm Give your answer correct to 3 significant figures.

cos � �h

a

y

d

p

j� or adj � hyp � cos

adj and hyp are involved so use cos

13 cm is the hypotenusea is adjacent to the 50º angle.

50°

a adj

13 cmhyp

34°

c

7 cm

24°

b

14.2 cm

50°a

13 cm

Example 8

x°

hypopp

adj

305

19.6 Finding lengths using trigonometry CHAPTER 19

b

b � 14.2 � sin 24º � 14.2sin 24ºb � 14.2 � 0.4067…b � 5.7756…b � 5.78 cm

c

c � 7 � tan 56º � 7 tan 56ºc � 7 � 1.4825…c � 10.3779…c � 10.4 cm

In triangle ABC, angle CAB � 90°, angle ABC � 37° and AB � 8.4 cm.Calculate the length of BC.Give your answer correct to 3 significant figures.

Solution 9

8.4 � BC � cos 37°

BC � �co

8

s

.

3

4

7°� � 10.5179…

BC � 10.5 cm

As BC is the hypotenuse its length must be greater than 8.4 cm so this is a sensible answer.


Make BC the subject.

37°

C

A B8.4 cm

37°

C

A B8.4 cm

Example 9


Relative to the angle of 56º, c is the oppositeside and 7 cm is the adjacent side.

The third angle in the triangle is(180 � 90 � 34)º � 56º

56°

copp

7 cmadj

When using tan to find a length it is easier to findthe opposite side. Relative to the angle of 34º, 7 cm is the opposite side and c is the adjacent side.

Opposite and adjacent are involved so use

tan � �o

a

p

d

p

j� or opp � adj � tan34°

c

7 cm


14.2 cm is the hypotenuseb is opposite to the 24º angle.

24°

b opp

14.2 cmhyp

306


Substitute the known values in adj � hyp � cos

sin � �o

h

p

yp

p� or opp � hyp � sin

In triangle ABC, BC is the hypotenuse and 8.4 cm is adjacent to the 37° angle so use

cos � �h

a

y

d

p


Exercise 19D1 Use a calculator to find the value of each of the following.

Give each answer correct to 4 decimal places, where necessary.a sin 20° b sin 72° c cos 60° d tan 86° e tan 45°f cos 18.9° g tan 4° h sin 14.7° i cos 75.3°

2 Work out the lengths of the sides marked with letters. Give each answer correct to 3 significant figures.a b c

d e f

g h i

3 Triangle PQR is right-angled at Q. In each part calculate the length of QR.Give each answer correct to 3 significant figures.a PQ � 7.3 cm, angle QPR � 68ºb PR � 17.2 m, angle QRP � 39ºc PR � 12.6 cm, angle QPR � 59º

4 In triangle ABD the point C lies on AD so that BC and AD are perpendicular.a Using triangle ABC, work out the length of

i BC ii ACGive each answer correct to 3 significant figures.

b Work out the length of CD correct to 3 significant figures.c Hence calculate the length of AD correct to 3 significant figures.d Calculate the area of triangle ABD. Give your answer correct

to the nearest cm2.

5 Calculate the length of BC in these triangles. Give each answer correct to 3 significant figures.a b

A

C

B12.5 cm

25°

A

C

B10.6 cm

64°

Q R

P

63°

i

24.9 cm

55°

h

14.9 cm75°

g

27.3 cm

28°f10.4 cm

73°

e 7.8 cm56°

d

9.4 cm

37°

c

15.4 cm49°b

26 cm20°

a16 cm

307

19.6 Finding lengths using trigonometry CHAPTER 19

A DC

B

13 cm

50°

28°

�

19.7 Finding angles using trigonometry

Work out the size of each of the angles marked with letters.Give each answer correct to 1 decimal place.

a b c

Solution 10a

sin a � �1

1

1

5

.

.

7

9� � 0.7358…

a � 47.379…°

a � 47.4°

b

cos b � �1

7

6

.5

.1� � 0.4658…

b � 62.235…°

b � 62.2°

c

tan c � �6

9

.

.

2

7� � 0.6391…

c � 32.585…°

c � 32.6° Give your answer correct to 1 decimal place.

Use your calculator to find tan�1 0.6391… which is 32.585…°

tan � �o

a

p

d

p

j�

6.2 cm is opposite angle c9.7 cm is adjacent to angle c.c

9.7 cmadj

6.2 cmopp

Give your answer correct to 1 decimal place.

Use your calculator to find cos�1 0.4658… which is 62.235…°

cos � �h

a

y

d

p

j�

16.1 cm is the hypotenuse7.5 cm is adjacent to angle b.

b

7.5 cmadj16.1 cm

hyp

Give your answer correct to 1 decimal place.

Use your calculator to find sin�1 0.7358… which is 47.379…°

sin � �o

h

p

yp

p�

15.9 cm is the hypotenuse11.7 cm is opposite angle a.

a

15.9 cmhyp 11.7 cm

opp

c

9.7 cm

6.2 cm

b

7.5 cm16.1 cm

a

15.9 cm11.7 cm

Example 10

308


Exercise 19E

1 Use a calculator to find the value of x in each of the following.Give each answer correct to 1 decimal place where necessary.

a cos xº � 0.5 b sin xº � 0.43 c cos xº � 0.6 d tan xº � 0.96

e sin xº � 0.8516 f tan xº � 2.03 g sin xº � 0.047 h tan xº � �27�

2 Work out the size of each of the marked angles. Give each answer correct to 1 decimal place.

a b c

d e f

3 Triangle ABC is right-angled at B.Give each answer correct to 0.1°.

a AB � 8.9 cm and BC � 12.1 cm.Calculate the size of angle ACB.

b BC � 15.5 cm, AC � 24.7 cm.Calculate the size of angle BAC.

c AB � 6.3 cm, AC � 11.8 cm.Calculate the size of angle ACB.

4 In triangle ACD the point B lies on AD so that CB and AD are perpendicular.

a Using triangle ABC calculate the size of angle ACB.Give your answer correct to 1 decimal place.

b Using triangle BCD calculate the size of angle BCD.Give your answer correct to 1 decimal place.

c Hence calculate the size of angle ACD. Give your answer to the nearest degree.

19.8 Trigonometry problemsTrigonometry can be used to solve problems. Sometimes Pythagoras’ theorem is needed as well.Some questions involve bearings (see Section 2.8).

Two towns, Aytown and Beeville, are 40 km apart. The bearing of Beeville from Aytown is 067°.

a Calculate how far north and how far east Beeville is from Aytown. Give your answers correct to 3 significant figures.

Ceeham is 60 km east of Beeville.

b Calculate the distance between Aytown and Ceeham. Give your answer to the nearest km.

c Calculate the bearing of Ceeham from Aytown. Give your answer to the nearest degree.

Example 11

B

C

A D

7.4 cm9.8 cm

11.6 cm

B C

A

f 20 cm

13.8 cm

e

15.8 cm

17 cm

d

18.3 cm

12.8 cm

c5 cm

14 cmb

13 cm

9 cma

18 cm11 cm

309

19.8 Trigonometry problems CHAPTER 19

�

Solution 11a

e � 40 sin 67° � 36.82…

Distance east � 36.8 km

n � 40 cos 67° � 15.629…

Distance north � 15.6 km

b

AC2 � AD2 � DC2 � 15.62 � 96.82

AC2 � 9613.6

AC � 98.04…

Distance between Aytown andCeeham is 98 km

tan DAC � �9

1

6

5

.

.

8

6� � 6.205…

DAC � 80.8…°

DAC � 81°

Bearing of Ceeham from Aytown is 081° A bearing must have 3 figures.

Give your answer to the nearest degree.

tan � �o

a

p

d

p

j�

To find the bearing of C from Acalculate the size of angle DAC.

Give your answer to the nearest km.

Find the length of AC using Pythagoras’ theorem.

The distance between Aytown andCeeham is the length of AC.

Draw triangle ADC.

C

A

D15.6 km

96.8 km

N

Ceeham is 60 � 36.8 � 96.8 km east of Ayton.

Ceeham is 15.6 km north of Ayton.

Mark the point C (for Ceeham) on thediagram 60 km east of B.B

C

A

D e

n40 km

60 km

N

67°



In the right-angled triangle ABD, the length of AD isthe distance that B is north of A (n km). The length ofDB is the distance that B is east of A (e km).

From B draw a line west to meet at Dthe ‘north’ line from A.

Draw a diagram showing the positionsof Ayton (A) and Beeville (B).

B

A

De km

n km40 km

N

67°

310


In triangle ABD the 40 km is the hypotenuse and e kmis opposite the 67° angle

sin � �o

h

p

yp

p� or opp � hyp � sin

n km is adjacent to the 67° angle

cos � �h

a

y

d

p


Exercise 19F

Where necessary give lengths correct to 3 significant figures and angles correct to 1 decimal place.

1 a Calculate the length of the line marked x cm.

b Work out the size of the angle marked y°.

2 A ladder is 5 m long. The ladder rests against a vertical wall,with the foot of the ladder resting on horizontal ground.The ladder reaches up the wall a distance of 4.8 m.

a Work out how far the foot of the ladder is from the bottom of the wall.

b Work out the angle that the ladder makes with the ground.

3 The diagram shows the plans for the sails of a boat.Work out the length of the side markeda ab bc c

4 The diagram shows a vertical building standing on horizontal ground. The points A, B and C are in a straight line on the ground. The point T is at the top of the building so that TC is vertical. The angle of elevation of T from A is 40° as shown in the diagram.

a Work out the height, TC, of the building.

b Work out the size of the angle of elevation of T from B.

5 The points P and Q are marked on a horizontal field. The distance from P to Q is 100 m.The bearing of Q from P is 062°. Work out how far

a Q is north of P b Q is east of P.

6 The diagram shows a circle centre O.The line ABC is the tangent to the circle at B.

a Work out the radius of the circle.

b Work out the size of angle OCB.

40°8 mA B C

T

20 m

15°

1.5 m

6 m6 m

65°

a

b

c

4.8 m

5 m

y°14.7 cm

12.4 cm

3.5 cm

x cm

8.4 cm 70°

9.7 cmA B C

O

311

19.8 Trigonometry problems CHAPTER 19

�

7 A, B and C are three buoys marking the course of a yacht race.a Calculate how far B is

i north of A ii east of A.

b Calculate how far C isi north of B ii east of B.

c Hence calculate how far C isi north of A ii east of A.

d Calculate the distance and bearing of C from A.

8 A rock, R, is 40 km from a harbour, H, on a bearing of 040°. A port, P, is 30 km from R on abearing of 130°.

a Draw a sketch showing the points H, R and P and work out the size of angle HRP.

b Work out the distance HP.

c Work out the bearing of P from H.

9 The diagram shows an isosceles triangle.Calculate the area of the triangle.Give your answer to the nearest cm2.

10 The diagram shows an isosceles trapezium.

a Work out the distance, h cm, between the two parallel sides of the trapezium.

b Work out the length of the longer parallel side of the trapezium.

c Calculate the area of the trapezium. Give your answer to the nearest cm2

Chapter summary

35°

6.7 cm 6.7 cm

5.8 cm

h cm

70°

10 cm10 cm

72°

40° 17 km

25 km

A

B

CN

N

312


You should now know:

that in a right-angled triangle the side opposite the right angle is called the hypotenuse.It is the longest side in the triangle

Pythagoras’ theorem for right-angled triangles

trigonometric ratios for right-angled triangles

sin xº � �o

h

p

yp

p�

cos xº � �h

a

y

d

p

j�

tan xº � �o

a

p

d

p

j�

opp � hyp � sin xºadj � hyp � cos xº

opp � adj � tan xº

x°adj

hypopp

c2 � a2 � b2ac

b

�

�

�

Chapter 19 review questions1 ABC is a right-angled triangle.

AB � 8 cm, BC � 11 cmCalculate the length of AC.Give your answer correct to 3 significant figures.

(1388 March 2003)

2 Angle MLN � 90°LM � 3.7 mMN � 6.3 mWork out the length of LN.Give your answer correct to 3 significant figures.

(1388 March 2004)

3 Work out the length in centimetres of AM.Give your answer correct to 2 decimal places.

(1388 March 2003)

4 Ballymena is due west of Larne.Woodburn is 15 km due south of Larne.Ballymena is 32 km from Woodburn.

a Calculate the distance of Larne fromBallymena.Give your answer in kilometres, correctto 1 decimal place.

b Calculate the bearing of Ballymena from Woodburn. (1385 June 1998)

Diagram NOTaccurately drawn

15 km32 km

Ballymena Larne

Woodburn

N

B CM

A

7 cm 7 cmDiagram NOTaccurately drawn

8 cm

M N

L

3.7 m

6.3 m


A

B C11 cm

8 cm


313

Chapter 19 review questions CHAPTER 19

You should now be able to:

use Pythagoras’ theorem in right-angled triangles● to find the length of the hypotenuse when the lengths of the other two sides are known● to find the length of one of the shorter sides of the triangle when the lengths of the other

two sides are known

use Pythagoras’ theorem to find the length of a line segment, given the coordinates of theend points of the segment

find the coordinates of the midpoint of a line segment, given the coordinates of the endpoints of the segment

use trigonometry in right-angled triangles to find the length of an unknown side and to findthe size of an unknown angle

apply Pythagoras’ theorem and trigonometry to right-angled triangle problems, includingbearings.

�

�

�

�

�

5 Angle ABC � 90°Angle ACB � 24°AC � 6.2 cmCalculate the length of BC.Give your answer correct to 3 significant figures.

(1388 March 2004)

6 The diagram shows a rectangle drawn inside a circle.The centre of the circle is at O.The rectangle is 15 cm long and 9 cm wide.Calculate the circumference of the circle.Give your answer correct to 3 significant figures.

(1385 November 2001)

7 The diagram shows triangle ABC.BC � 8.5 cmAngle ABC � 90°Angle ACB � 38°Work out the length of AB.Give your answer correct to 3 significant figures.

(1387 June 2003)

8 ABD and DBC are two right-angled triangles.AB � 9 mAngle ABD � 35°Angle DBC � 50°Calculate the length of DC.Give your answer correct to 3 significant figures.

(1385 November 2002)

9 The diagram shows the positions of three telephonemasts A, B and C.Mast C is 5 kilometres due east of Mast B.Mast A is due north of Mast B and 8 kilometres from Mast C.

a Calculate the distance of A from B.Give your answer in kilometres, correct to 3 significant figures.

b i Calculate the size of the angle marked x°.Give your angle correct to 1 decimal place.

ii Calculate the bearing of A from C.Give your answer correct to 1 decimal place.

iii Calculate the bearing of C from A.Give your bearing correct to 1 decimal place. (1385 June 1999)


A

CB x°

8 km

5 km

N

N


A

C

BD35°

9 m

50°


8.5cm

A

B C38°


O9 cm

15 cm

Diagram NOTaccurately drawn6.2 cm

A

B C24°

314


10 A and B are points on a centimetre grid.A is the point (3, 2)B is the point (7, 8)

a Calculate the distance AB.Give your answer correct to 3 significant figures.

b Find the coordinates of the midpoint of AB.

11 ABCD is a trapezium.AD is parallel to BC.Angle A � angle B � 90°AD � 2.1 m AB � 1.9 m CD � 3.2 mWork out the length of BC.Give your answer correct to 3 significant figures. (1388 January 2003)

12 ABC is a right-angled triangle.D is the point on AB such that AD � 3DB.AC � 2DB and angle A � 90°

Show that sin C � where k is an integer.

Write down the value of k. (1388 January 2003)

13

Ambletown, Bowtown and Comptown are three towns.Ambletown is 9.6 km due west of Bowtown.Bowtown is 7.4 km due south of Comptown.Calculate the bearing of Ambletown from Comptown.Give your answer correct to 1 decimal place. (1388 January 2003)

14

ABCD is a quadrilateral.

Angle BDA � 90°, angle BCD � 90°, angle BAD � 40°

BC � 6 cm, BD � 8 cm

a Calculate the length of DC. Give your answer correct to 3 significant figures.

b Calculate the size of angle DBC. Give your answer correct to 3 significant figures.

c Calculate the length of AB. Give your answer correct to 3 significant figures.(1385 November 2000)

Diagram NOTaccurately drawn8 cm

6 cm

AD

C

B

40°

Diagram NOTaccurately drawn7.4 km

9.6 km

Comptown

BowtownAmbletown

N

k��20�


A

B

C

D


A

C

D

B

2.1 m

1.9 m3.2 m

315

Chapter 19 review questions CHAPTER 19

Documents

Pythagoras’ theorem and trigonometry (1) 19 19.1 Pythagoras’ theorem Pythagoras was a famous mathematician in Ancient Greece. The theorem which is named after him is an important