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Pythagoras’ theorem
and trigonometry (1)
CHAPTER
19.1 Pythagoras’ theorem
Pythagoras was a famous mathematician in Ancient Greece.
The theorem which is named after him is an important result about
right-angled triangles.
A
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.
C
B
The right-angled triangle in the diagram on
the 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.
25 cm2
16 cm2
4
5
3
9 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:
In a right-angled triangle, the area of the square on the hypotenuse
is equal to the sum of areas of the squares on the other two sides.
Area of square R area of square P area of square Q
Pythagoras’ theorem can be used to find
the length of the third side of a right-angled triangle
when the lengths of the other two sides are known.
For this, the theorem is usually stated in terms of the
lengths of the sides of the triangle.
R
Q
P
That is
c
b
c2 a2 b2
a
D
Pythagoras’ theorem can also be written
DE 2 EF 2 DF 2
(DE 2 means that the length of the side DE is squared.)
F
E
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Pythagoras‘ theorem and trigonometry (1)
CHAPTER 19
19.2 Finding lengths
Pythagoras’ theorem can be used to work out the length of the hypotenuse of a right-angled triangle
when the lengths of the two shorter sides are given.
Example 1
Work out the length of the hypotenuse in this triangle.
c cm
15 cm
Solution 1
c2 a2 b2
8 cm
State Pythagoras’ theorem.
c2 82 152
Substitute the given lengths.
c2 64 225
Work out 82 and 152 and add the results.
c2 289
c 289
17
Find 289
Length of hypotenuse 17 cm
The answer is sensible because the hypotenuse
is longer than the other two sides.
It is important to be able to apply Pythagoras’ theorem when the triangle is in a different position.
Example 2
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.
X
8.6 cm
13.9 cm
Y
Z
Solution 2
The hypotenuse is the side opposite the right angle.
Angle X is the right angle so the hypotenuse is YZ.
YZ2 XY 2 XZ 2
State Pythagoras’ theorem.
YZ2 8.62 13.92
Substitute the given lengths.
YZ2 73.96 193.21
Work out 8.62 and 13.92 and add the results.
YZ2 267.17
YZ 267.17
16.34…
YZ 16.3 cm (to 3 s.f.)
296
Find 267.17
Write down at least 4 figures.
Give the final answer correct to 3 significant figures.
19.2 Finding lengths
CHAPTER 19
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.
Example 3
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.
A
5.8 cm
C
Solution 3
17.4 cm
Angle A is the right angle so the hypotenuse is BC.
BC2 AC2 AB2
State Pythagoras’ theorem.
17.42 5.82 AB2
B
Substitute the given lengths.
302.76 33.64 AB2
Work out 17.42 and 5.82
302.76 33.64 AB2
Subtract 33.64 from both sides.
269.12 AB2
AB 269.12
16.40…
Find 269.12
Write down at least 4 figures.
AB 16.4 cm (to 3 s.f.)
Give the final answer correct to 3 significant figures.
Exercise 19A
1 Work out the length of the sides marked with letters in these triangles.
a
b
c
12 cm
a
6 cm
8 cm
c
b
40 cm
5 cm
9 cm
2 Work out the length of the sides marked with letters in these triangles.
a
b
c
a
25 cm
37 cm
12 cm
20 cm
c
25 cm
b
24 cm
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
7.9 cm
a
4.8 cm
b
c
d
6.2 cm
10.6 cm
9.1 cm
7.4 cm
d
8.3 cm
4.8 cm
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Pythagoras‘ theorem and trigonometry (1)
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
10.7 cm
4.8 cm
11.3 cm
b
a
8.1 cm
c
c
d
1.8 cm
12.4 cm
8.3 cm
2.1 cm
5 a In triangle ABC
angle 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 ABC
angle 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.
d
B
3.4 cm
A
C
12.1 cm
B
16.3 cm
5.9 cm
C
A
c In triangle PQR
angle 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.
5.9 cm
R
P
13.1 cm
Q
d In triangle PQR
angle 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.
R
P
9.6 cm
11.2 cm
Q
e In triangle XYZ
angle 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.
X
12.6 cm
16.5 cm
Y
Z
f In triangle DEF
angle 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 their
lengths.
ii Work out the length of DE.
Give your answer correct to 3 significant figures.
298
19.3 Applying Pythagoras’ theorem
CHAPTER 19
19.3 Applying Pythagoras’ theorem
Pythagoras’ theorem can be used to solve problems.
Example 4
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
7.2 km
5.7 km
Finish
Draw a sketch
of the boat’s
journey
Remember that the points
of the compass are
N
W
E
d km
S
Start
d2 5.72 7.22
d2 32.49 51.84
d2 84.33
d 84.33
9.183…
Distance 9.18 km (to 3 s.f.)
The sketch is of a right-angled triangle so that
Pythagoras’ theorem can be used.
The distance between the starting point and the
finishing point is the length of the hypotenuse of the
triangle, marked d km in the sketch.
Isosceles triangles can be split into two right-angled triangles and Pythagoras’ theorem can then
be used.
Example 5
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.
A
8 cm
B
8 cm
M
C
6 cm
Solution 5
Pythagoras’ theorem cannot be used in triangle ABC as this triangle is not right-angled.
A
a
As M is the midpoint of the base of the isosceles triangle,
the line AM is the line of symmetry of triangle ABC. So
AM is perpendicular to the base and angle AMB 90°.
8 cm
B
3 cm
h cm
M
By Pythagoras
AB2 AM2 BM2
82 h2 32
64 h2 9
64 9 h2
55 h2
h 55
7.416…
h 7.42
Height of triangle 7.42 cm (to 3 s.f.)
BM 3 cm as M is the midpoint of BC.
Draw a sketch of triangle ABM.
Triangle ABM is right-angled with hypotenuse AB. The
height, AM, of the triangle is marked h cm on the sketch.
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Pythagoras‘ theorem and trigonometry (1)
CHAPTER 19
b
Area 12 6 7.416…
Area 22.24…
Area 22.2 cm2 (to 3 s.f.)
area of a triangle 21 base height
For triangle ABC base 6 cm and
height 7.416… cm.
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.
5m
3.6 m
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.
6 cm
9 cm
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.
A
10 km
C
4 Work out the area of the triangle.
Give your answer correct to 3 significant figures.
B
21 km
14.5 cm
10.6 cm
5 Work out the perimeter of the triangle.
Give your answer correct to 3 significant figures.
11.7 cm
8.3 cm
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 of
ii CQ.
i NQ
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.
300
A
2.5 m
P
B
N 2m C
5m
Q
19.4 Coordinates, line segments and Pythagoras’ theorem
CHAPTER 19
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.
39 cm
12 cm
x
y
9 cm
a
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.
5.4 cm
4.7 cm
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?
3.9 cm
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 cm
Triangle 3
10 cm, 17 cm and 18 cm
Triangle 4
20 cm, 21 cm and 29 cm
Triangle 5
8 cm, 17 cm and 20 cm
Triangle 6
33 cm, 56 cm and 65 cm
Find by calculation which of these triangles are right-angled triangles.
19.4 Coordinates, line segments and Pythagoras’ theorem
y
6
B
5
4
M
3
2
1
O
A
1
2
3
4
5
6
7
8
9 x
The diagram shows the straight line joining the points A(2, 1) and B(8, 5). A line joining two points is
called a line segment. So in the diagram, AB is the line segment joining the points A and B.
Midpoint of a line segment
The midpoint M of the line segment AB has coordinates (5, 3).
28
Notice that the x-coordinate of A is 2, the x-coordinate of B is 8 and 5, the x-coordinate of M.
2
15
Similarly the y-coordinate of A is 1, the y-coordinate of B is 5 and 3, the y-coordinate of M.
2
In general the x-coordinate of the midpoint of a line segment is the mean of the x-coordinates of its
endpoints and the y-coordinate of its midpoint is the mean of the y-coordinates of its endpoints.
That is
ap bq
the midpoint of the line joining (a, b) and (p, q) is the point , 2
2
301
Pythagoras‘ theorem and trigonometry (1)
CHAPTER 19
Example 6
Find the midpoint of the line joining
a (3, 5) and (13, 7)
b (5, 8) and (9, 13).
Solution 6
a 3 13 16
16
8
2
5 7 12
12
6
2
Midpoint is (8, 6)
b 5 9 4
4
2
2
8 13 5
5
2.5
2
Midpoint is (2, 2.5)
The x-coordinates are 3 and 13
The y-coordinates are 5 and 7
The x-coordinates are 5 and 9
The y-coordinates are 8 and 13
Length of a line segment
The 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 used
to find the length of AB.
AB2 82 42
AB2 64 16 80
AB 80
8.94 (to 3 s.f.)
y
6
B
5
4
3
4
2
1
A
O
C
8
1
2
3
4
5
6
7
8
9 10 x
Example 7
Find the length of the line joining
a A(3, 2) and B(15, 7)
b P(9, 4) and Q(7, 5)
Solution 7
a y
B
7
5
2
O
A
12
3
AB2 122 52
AB2 144 25 169
AB 169
13
302
C
Draw a sketch showing A and B and
complete the right-angled triangle ABC.
AC 15 3 12
BC 7 2 5
15 x
Use Pythagoras’ theorem to find the length of AB.
19.4 Coordinates, line segments and Pythagoras’ theorem
b
P (9, 4)
Draw a sketch showing P and Q and
complete the right-angled triangle PQR.
9
R
QR 7 9 7 9 16
16
Q (7, 5)
PQ2 162 92
PQ2 256 81 337
PQ 337
18.4 (to 3 s.f.)
CHAPTER 19
PR 4 5 4 5 9
Use Pythagoras’ theorem to find the length of PQ.
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 of
ii BC
iii AC.
i AB
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 has
coordinates (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 the
i 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.
ii C(1, 10)
iii D(7, 16)
iv E(4, 15).
i B(16, 7)
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 of
ii BC
iii CD
iv DA.
i AB
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).
(p a)2
(q b)2
Show that the length of the line segment AB is 303
Pythagoras‘ theorem and trigonometry (1)
CHAPTER 19
19.5 Trigonometry – introduction
Trigonometry 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 know
accurate distances and accurate angle sizes.
It is also used in many other areas of surveying, engineering and architecture.
Trigonometric ratios
Here are two right-angled triangles.
The triangle with hypotenuse 2 is an enlargement with scale factor 2 of
the triangle with hypotenuse 1, that is, its sides are twice as long.
p 2s and q 2c
1
s
70°
So if s and c are known for the right-angled triangle
c
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 any
standard scientific calculator.
The length s is called the sine of 70° written sin 70°. Not all calculators are the same
but the following key sequence to find sin 70° applies to many calculators.
p
2
70°
q
Make sure that the angle mode of your calculator is degrees, usually shown by D on the
calculator screen.
Press
Key in
Press
sin
70
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 cos
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
1
So for any right-angled triangle
sin 70°
70°
cos 70°
1
p
2
sin x°
x°
cos x°
p
r
70°
q
x°
or
304
q
p r sin xº and q r cos xº
p
q
sin xº and cos xº r
r
19.6 Finding lengths using trigonometry
CHAPTER 19
The hypotenuse (hyp) of a right-angled triangle is the side opposite the right angle and is the
longest side of the triangle. The sides of the triangle are named according to their position relative to
the angle given or the angle to be found. If this angle is xº then the side opposite this angle is called
the opposite side (opp). The side next to this angle is called the adjacent side (adj).
hyp
opp
x°
adj
The results above become
opp hyp sin xº and adj hyp cos xº
or
opp
adj
sin xº and cos xº hyp
hyp
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
opp
tan x° or opp adj tan x°
adj
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
Example 8
Work out the length of each of the marked sides. Give each answer correct to 3 significant figures.
a
b
c
b
34°
13 cm
14.2 cm
24°
c
50°
a
Solution 8
a
13 cm
hyp
7 cm
13 cm is the hypotenuse
a is adjacent to the 50º angle.
adj and hyp are involved so use cos
adj
cos or adj hyp cos
hyp
50°
a adj
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.
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Pythagoras‘ theorem and trigonometry (1)
CHAPTER 19
b
b opp
24°
14.2 cm is the hypotenuse
b is opposite to the 24º angle.
opp
sin or opp hyp sin
hyp
14.2 cm
hyp
b 14.2 sin 24º 14.2sin 24º
b 14.2 0.4067…
b 5.7756…
b 5.78 cm
Give your answer correct to 3 significant figures.
c
Opposite and adjacent are involved so use
opp
tan or opp adj tan
adj
34°
When using tan to find a length it is easier to find
the opposite side. Relative to the angle of 34º,
7 cm is the opposite side and c is the adjacent side.
c
7 cm
The third angle in the triangle is
(180 90 34)º 56º
Relative to the angle of 56º, c is the opposite
side and 7 cm is the adjacent side.
c
opp
56°
7 cm
adj
c 7 tan 56º 7 tan 56º
c 7 1.4825…
c 10.3779…
c 10.4 cm
Give your answer correct to 3 significant figures.
Example 9
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.
C
Solution 9
C
A
In triangle ABC, BC is the hypotenuse and
8.4 cm is adjacent to the 37° angle so use
adj
cos or adj hyp cos
hyp
37°
8.4 cm
B
8.4 BC cos 37°
8.4
BC 10.5179…
cos 37°
BC 10.5 cm
A
37°
8.4 cm
Substitute the known values in adj hyp cos
Make BC the subject.
Give your answer correct to 3 significant figures.
As BC is the hypotenuse its length must be
greater than 8.4 cm so this is a sensible answer.
306
B
19.6 Finding lengths using trigonometry
CHAPTER 19
Exercise 19D
1 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°
f cos 18.9°
g tan 4°
h sin 14.7°
i cos 75.3°
e tan 45°
2 Work out the lengths of the sides marked with letters. Give each answer correct to
3 significant figures.
a
b
c
16 cm
a
20°
c
26 cm
37°
15.4 cm
49°
b
d
e
56°
f
e
9.4 cm
7.8 cm
f 28°
73°
10.4 cm
d
g
h
i
h
g
55°
24.9 cm
63°
14.9 cm
i
75°
27.3 cm
3 Triangle PQR is right-angled at Q. In each part calculate the length of QR.
P
Give each answer correct to 3 significant figures.
a PQ 7.3 cm, angle QPR 68º
b PR 17.2 m, angle QRP 39º
Q
c 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
ii AC
i BC
Give each answer correct to 3 significant figures.
13 cm
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
50°
A
to the nearest cm2.
R
B
28°
C
D
5 Calculate the length of BC in these triangles. Give each answer correct to 3 significant figures.
a C
b C
25°
A
A
64°
10.6 cm
12.5 cm
B
B
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Pythagoras‘ theorem and trigonometry (1)
CHAPTER 19
19.7 Finding angles using trigonometry
Example 10
Work out the size of each of the angles marked with letters.
Give each answer correct to 1 decimal place.
a
b
16.1 cm
15.9 cm
c
c
7.5 cm
b
9.7 cm
11.7 cm
6.2 cm
a
Solution 10
a
15.9 cm
hyp
15.9 cm is the hypotenuse
11.7 cm is opposite angle a.
11.7 cm
opp
opp
sin hyp
a
11.7
sin a 0.7358…
15.9
a 47.379…°
a 47.4°
Use your calculator to find sin1 0.7358… which is 47.379…°
Give your answer correct to 1 decimal place.
b
7.5 cm
adj
b
16.1 cm
hyp
7.5
cos b 0.4658…
16.1
b 62.235…°
b 62.2°
16.1 cm is the hypotenuse
7.5 cm is adjacent to angle b.
adj
cos hyp
Use your calculator to find cos1 0.4658… which is 62.235…°
Give your answer correct to 1 decimal place.
c
c
6.2 cm is opposite angle c
9.7 cm is adjacent to angle c.
9.7 cm
adj
opp
tan adj
6.2 cm
opp
6.2
tan c 0.6391…
9.7
c 32.585…°
c 32.6°
308
Use your calculator to find tan1 0.6391… which is 32.585…°
Give your answer correct to 1 decimal place.
19.8 Trigonometry problems
CHAPTER 19
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.
b sin xº 0.43
c cos xº 0.6
a cos xº 0.5
f tan xº 2.03
g sin xº 0.047
e sin xº 0.8516
d tan xº 0.96
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
13 cm
18 cm
5 cm
11 cm
c
b
a
14 cm
9 cm
d
e
17 cm
f
13.8 cm
18.3 cm
15.8 cm
e
f
20 cm
d
12.8 cm
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.
A
B
4 In triangle ACD the point B lies on AD so that CB and
C
AD are perpendicular.
9.8 cm
a Using triangle ABC calculate the size of angle ACB.
7.4 cm
Give your answer correct to 1 decimal place.
A
b Using triangle BCD calculate the size of angle BCD.
11.6 cm
B
Give your answer correct to 1 decimal place.
c Hence calculate the size of angle ACD. Give your answer to the nearest degree.
C
D
19.8 Trigonometry problems
Trigonometry can be used to solve problems. Sometimes Pythagoras’ theorem is needed as well.
Some questions involve bearings (see Section 2.8).
Example 11
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.
309
Pythagoras‘ theorem and trigonometry (1)
CHAPTER 19
Solution 11
a
N
e km
D
n km
A
67°
Draw a diagram showing the positions
of Ayton (A) and Beeville (B).
B
From B draw a line west to meet at D
the ‘north’ line from A.
40 km
In the right-angled triangle ABD, the length of AD is
the distance that B is north of A (n km). The length of
DB is the distance that B is east of A (e km).
In triangle ABD the 40 km is the hypotenuse and e km
is opposite the 67° angle
opp
sin or opp hyp sin
hyp
e 40 sin 67° 36.82…
Distance east 36.8 km
Give your answer correct to 3 significant figures.
n km is adjacent to the 67° angle
n 40 cos 67° 15.629…
adj
cos or adj hyp cos
hyp
Distance north 15.6 km
b
Give your answer correct to 3 significant figures.
N
D
n
A
e
67°
B
60 km
Mark the point C (for Ceeham) on the
diagram 60 km east of B.
C
Ceeham is 15.6 km north of Ayton.
40 km
Ceeham is 60 36.8 96.8 km east of Ayton.
N
D
Draw triangle ADC.
96.8 km
15.6 km
C
The distance between Aytown and
Ceeham is the length of AC.
A
AC2 AD2 DC2 15.62 96.82
Find the length of AC using Pythagoras’ theorem.
AC2 9613.6
AC 98.04…
Distance between Aytown and
Ceeham is 98 km
Give your answer to the nearest km.
To find the bearing of C from A
calculate the size of angle DAC.
96.8
tan DAC 6.205…
15.6
opp
tan adj
DAC 80.8…°
DAC 81°
Bearing of Ceeham from Aytown is 081°
310
Give your answer to the nearest degree.
A bearing must have 3 figures.
19.8 Trigonometry problems
CHAPTER 19
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°.
3.5 cm
14.7 cm
y° x cm
12.4 cm
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.
5m
4.8 m
3 The diagram shows the plans for the sails of a boat.
Work out the length of the side marked
a a
b b
c c
a 15°
1.5 m
6m
b
6m
65°
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.
T
40°
A
20 m
B
8m C
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
b Q is east of P.
a Q is north 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.
O
8.4 cm
A
70°
B
9.7 cm
C
311
Pythagoras‘ theorem and trigonometry (1)
CHAPTER 19
7 A, B and C are three buoys marking the course
of a yacht race.
a Calculate how far B is
ii east of A.
i north of A
b Calculate how far C is
ii east of B.
i north of B
c Hence calculate how far C is
ii east of A.
i north of A
d Calculate the distance and bearing of C from A.
N
C
N
72°
25 km
B
40° 17 km
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 a
bearing 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 cm
10 cm
70°
10 The diagram shows an isosceles trapezium.
5.8 cm
a Work out the distance, h cm, between the two parallel
6.7 cm
h cm
sides of the trapezium.
35°
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
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
c
a
c2 a2 b2
b
trigonometric ratios for right-angled triangles
hyp
x°
adj
opp hyp sin xº
adj hyp cos xº
opp adj tan xº
312
opp
opp
sin xº hyp
adj
cos xº hyp
opp
tan xº adj
6.7 cm
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 the
end points of the segment
find the coordinates of the midpoint of a line segment, given the coordinates of the end
points of the segment
use trigonometry in right-angled triangles to find the length of an unknown side and to find
the size of an unknown angle
apply Pythagoras’ theorem and trigonometry to right-angled triangle problems, including
bearings.
Chapter 19 review questions
1 ABC is a right-angled triangle.
AB 8 cm, BC 11 cm
Calculate the length of AC.
Give your answer correct to
3 significant figures.
A
Diagram NOT
accurately drawn
8 cm
B
C
11 cm
(1388 March 2003)
2 Angle MLN 90°
LM 3.7 m
MN 6.3 m
Work out the length of LN.
Give your answer correct to
3 significant figures.
L
Diagram NOT
accurately drawn
3.7 m
M
N
6.3 m
(1388 March 2004)
3 Work out the length in centimetres of AM.
Give your answer correct to
2 decimal places.
A
7 cm
B
Diagram NOT
accurately drawn
7 cm
C
M
8 cm
(1388 March 2003)
4 Ballymena is due west of Larne.
N
Ballymena
Woodburn is 15 km due south of Larne.
Ballymena is 32 km from Woodburn.
a Calculate the distance of Larne from
Ballymena.
Give your answer in kilometres, correct
to 1 decimal place.
b Calculate the bearing of Ballymena from Woodburn.
Larne
Diagram NOT
accurately drawn
15 km
32 km
Woodburn
(1385 June 1998)
313
Pythagoras‘ theorem and trigonometry (1)
CHAPTER 19
5 Angle ABC 90°
Angle ACB 24°
AC 6.2 cm
Calculate the length of BC.
Give your answer correct to 3 significant figures.
A
Diagram NOT
accurately drawn
6.2 cm
24°
B
C
(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.
Diagram NOT
accurately drawn
O
9 cm
15 cm
(1385 November 2001)
7 The diagram shows triangle ABC.
BC 8.5 cm
Angle ABC 90°
Angle ACB 38°
Work out the length of AB.
Give your answer correct to 3 significant figures.
A
Diagram NOT
accurately drawn
38°
B
C
8.5 cm
(1387 June 2003)
8 ABD and DBC are two right-angled triangles.
AB 9 m
Angle ABD 35°
Angle DBC 50°
Calculate the length of DC.
Give your answer correct to 3 significant figures.
A
Diagram NOT
accurately drawn
9m
35°
D
B
50°
C
(1385 November 2002)
9
314
The diagram shows the positions of three telephone
masts 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.
N
A
Diagram NOT
accurately drawn
8 km
N
B
x°
5 km
C
(1385 June 1999)
Chapter 19 review questions
CHAPTER 19
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 m
Work out the length of BC.
Give your answer correct to
3 significant figures.
D
Diagram NOT
accurately drawn
3.2 m
1.9 m
C
B
(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°
k
Show that sin C where k is an integer.
20
Write down the value of k.
13
2.1 m
A
A
D
Diagram NOT
accurately drawn
B
C
(1388 January 2003)
Comptown
N
Diagram NOT
accurately drawn
7.4 km
Ambletown
Bowtown
9.6 km
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.
14
(1388 January 2003)
B
6 cm
Diagram NOT
C accurately drawn
8 cm
A
40°
D
ABCD is a quadrilateral.
angle BCD 90°,
angle BAD 40°
Angle BDA 90°,
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)
315