Maths A - Chapter 5

5syllabussyllabusrrefefererenceence
Strand:
Applied geometry
Core topic:
Elements of applied geometry
In thisIn this chachapterpter
5A Pythagoras’ theorem
5B Shadow sticks
5C Calculating trigonometric
ratios
5D Finding an unknown side
5E Finding angles
5F Angles of elevation and
depression
Right-angled
triangles and
trigonometry
MQ Maths A Yr 11 - 05 Page 157 Wednesday, July 4, 2001 4:39 PM

158 M a t h s Q u e s t M a t h s A Ye a r 1 1 f o r Q u e e n s l a n d
Introduction
Thousands of years before Christ, ancient civi-
lisations were able to build enormous structures
— like Stonehenge in England and the pyramids
in Egypt. How did they do it? Human ingenuity
allows us to achieve many things that at ﬁrst
seem impossible. How do we determine the
height of a tall object without physically scaling
it? Sometimes, even tasks that appear to be very
simple (for example, planning and designing a
staircase) present us with unexpected problems.
How in fact do we go about the practical task of
planning and designing a staircase?
The theorem of Pythagoras and the geometry
of right-angled triangles (trigonometry) can be
employed to answer these questions. As you pro-
gress through this chapter, solutions to problems
such as these will become clear.
To put the mathematics of Pythagoras into
perspective, let us ﬁrst consider the times in
which he lived.

C h a p t e r 5 R i g h t - a n g l e d t r i a n g l e s a n d t r i g o n o m e t r y 159
History of mathematics
P Y T H AG O R A S O F S A M O S ( c i rc a 5 8 0 B C – 5 0 0 B C )
During his life:
• Taoism is
founded
• Kung-Fu-tse
(Confucius)
is born
• Buddhism is
founded.
Pythagoras was a famous Greek mathemat-
ician and mystic but is now best known for
his theorem about the sides of a triangle. He
was born on Samos Island. It is believed that
he was born about 580 BC and died about
500 BC, but because of the way dates were
recorded then, various dates are given for his
life. Not much was known about his personal
life but it was known that he had a wife, son
and daughter.
When Pythagoras was a young man, he
travelled to Egypt and Babylonia (Mesopo-
tamia) where he learned much of his math-
ematics and developed an interest in
investigating it further.
He founded a cult with the idea that ‘the
essence of all things is a number’. This group
believed that all nature could be expressed in
terms of numbers. They found, however, that
some numbers could not be expressed as
rational numbers, such as . They kept this
information to themselves and there is a story
that they killed one member who told some-
body else about this fact.
Pythagoras showed that musical notes had
a mathematical pattern. He stretched a string
tightly and found that it produced a certain
sound and then found that if he halved the
length of the string, it produced a sound that
was in harmony with the ﬁrst. He also found
that if it was not exactly half, or a multiple of
a half then it was a clashing sound. This
approach is still used in musical instruments
today.
He is credited with the discovery now
known as ‘Pythagoras’ theorem’. This states
that ‘For a right-angled triangle, the square of
the hypotenuse (long side) is equal to the sum
of the squares of the two short sides’ and can
be written as c2
= a2
+ b2
using the diagram
below. Other people knew of this idea long
before Pythagoras. There is a Babylonian
tablet known as ‘Plimpton 322’ (believed to
have been made about
1500 years before
Pythagoras was born),
which has a set of values
of the type known as
Pythagorean triples.
Some examples of Pythagorean triples are:
3, 4 and 5
8, 15 and 17
and a more difﬁcult example: 20, 99 and 101.
For his investigations and demonstrations
with right angles, Pythagoras used a string in
which he had tied knots.
Questions
1. Where was Pythagoras born?
2. Where did he travel to and learn most of
his mathematics?
3. What is the formula for his famous
theorem?
4. What did he start to investigate about
patterns in nature?
5. What is the name of the ancient tablet
that contains Pythagorean triples?
6. What is a Pythagorean triple?
2
a
b
c

1 Name the hypotenuse in each of the following triangles.
a b
c d
2 State Pythagoras’ theorem.
3 Calculate the unknown lengths in the following right-angled triangles.
a b c
4 Complete each of the following to form ratios that are equivalent to the ratio 3 : 4 : 5.
a 9 : 12 : ____ b 1.5 : ____ : 2.5 c ____ : 1 : 1.25 d 7.5 : 10 : ____
5 The diagram at right depicts a tree in a ﬁeld on
a sunny day. Copy the diagram.
a Draw the position of the shadow cast by the tree.
b Mark the angle of elevation of the sun.
6 Find the value of the unknown angles in the following triangles.
a b
7 Express the following angles in degrees, minutes and seconds.
a 35.5° b 27.23° c 68.125°
8 Express the following angles in degrees, correct to 2 decimal places.
a 45°27′ b 84°35′22″ c 64°28″.
9 Deﬁne the trigonometric ratios:
a sine b cosine c tangent.
C
A
B
ED
F
H
G
I
K
J L
6 m
8 m
a
6 m
4 m
b
20 cm
a
34°
b
27°

10 Copy the diagrams below and label the sides as Opposite, Adjacent or Hypotenuse
with respect to the marked angle.
a b c
11 Using your calculator, determine the following (correct to 4 decimal places).
a sin 45° b cos 30° c tan 22.5°
d sin 67°15′ e tan 38°12′22″
12 Copy the following diagrams and mark the angles of elevation or depression, from the
observer to the object, on each.
a b
c
13 Solve the following for x.
a =
b =
Ground
Observer
Ship
Sea
Ground
Observer
Bird
5.2
4.5
-------
x
1.8
-------
5.2
4.5
-------
1.8
x
-------

Right-angled triangles
The properties of right-angled triangles
enable us to calculate lengths and angles
which are sometimes not able to be meas-
ured. The first property we will investigate
is Pythagoras’ theorem.
Pythagoras’ theorem
Pythagoras’ theorem allows us to calculate the length of a side
of a right-angled triangle, if we know the lengths of the
other two sides. Consider LABC at right.
AB is the hypotenuse (the longest side). It is opposite the
right angle.
Note that the sides of a triangle can be named in either of two ways.
1. A side can be named by the two capital letters given to the vertices at each end. This
is what has been done in the figure above to name the hypotenuse AB.
2. We can also name a side by using the lower-case letter of the opposite vertex. In the
figure above, we could have named the hypotenuse ‘c’.
Consider the right-angled triangle ABC (below)
with sides 3 cm, 4 cm and 5 cm. Squares have
been constructed on each of the sides. The area
of each square has been calculated (A = S2
) and
indicated.
Note that the area of the square on the
hypotenuse is equal to the sum of the areas of
the squares on the other two sides.
25 cm2
= 16 cm2
+ 9 cm2
Alternatively: (5 cm)2
= (4 cm)2
+ (3 cm)2
Which means: hypotenuse2
= base2
+ height2
hypotenuse
A
C B
Name the hypotenuse in the triangle at right.
THINK WRITE
The hypotenuse is opposite the right
angle.
The vertices at each end or the lower-
case letter of the opposite vertex can be
used to name the side.
The hypotenuse is QR or p.
Q
P
R1
2
1WORKEDExample
A = 9 cm2
A = 25 cm2
C B
A
3 cm
4 cm
5 cm
A = 16 cm2Pythagoras
calculations
GCpr
ogram
Pythagoras

This result is known as Pythagoras’ theorem. Pythagoras’ theorem states:
In any right-angled triangle, the square of the hypotenuse is equal to the sum of
the squares of the two shorter sides. That is,
Hypotenuse2
= base2
+ height2
This is the formula used to find the length of the hypotenuse in a right-angled tri-
angle when we are given the lengths of the two shorter sides.
In this example, the answer is a whole number because we are able to find
exactly. In most examples this will not be possible. In such cases, we are asked to write
the answer correct to a given number of decimal places.
By rearranging Pythagoras’ theorem, we can write the formula to find the length of a
shorter side of a triangle.
Since hypotenuse2
= base2
+ height2
it follows that base2
= hypotenuse2
− height2
and height2
= hypotenuse2
− base2
The method of solving this type of question is the same as in the previous example,
except that here we use subtraction instead of addition. For this reason, it is important
to look at each question carefully to determine whether you are finding the length of
the hypotenuse or one of the shorter sides.
Find the length of the hypotenuse in the triangle at right.
THINK WRITE
Write the formula. Hypotenuse2
= base2
+ height2
Substitute the lengths of the shorter
sides.
c2
= 152
+ 82
Evaluate the expression for c2
. = 225 + 64
= 289
Find the value of c by taking the square
root.
c =
= 17 cm
8 cm
15 cm
c
1
2
3
4 289
2WORKEDExample
E
XCEL Spread
sheet
Pythagoras
289
Find the length of side PQ in triangle PQR, correct to
1 decimal place.
THINK WRITE
Write the formula. Base2
= hypotenuse2
− height2
Substitute the lengths of the known
sides.
r2
= 162
− 92
Evaluate the expression. = 256 − 81
= 175
Find the answer by finding the square
root.
r =
= 13.2 m
QP
R
16 m
9 m
r
1
2
3
4 175
3WORKEDExample
Math
cad
Pythagoras

Because Pythagoras’ theorem applies only in right-angled triangles, we can use the
theorem to test whether a triangle is right-angled, acute-angled or obtuse-angled. If we
assume, as a starting point, that the triangle is right-angled, we could regard the longest
side as the hypotenuse. We could then calculate values for hypotenuse2
, base2
and
height2
. Interpreting the relationships between these three values, we could conclude
the following.
If hypotenuse2
= base2
+ height2
the triangle is right-angled.
If hypotenuse2
> base2
+ height2
the triangle is obtuse-angled (since the hypotenuse is longer than it should be for that
particular base and height).
If hypotenuse2
< base2
+ height2
the triangle is acute-angled (since the hypotenuse is too short for that particular base
and height).
Pythagorean triads (or Pythagorean triples) are sets of 3 numbers which satisfy
Pythagoras’ theorem. The ﬁrst right-angled triangle we dealt with in this section had
side lengths of 3 cm, 4 cm and 5 cm. This satisﬁed Pythagoras’ theorem, so the
numbers 3, 4 and 5 form a Pythagorean triad or triple. In fact, any multiple of these
numbers, for example 1. 6, 8 and 10
1.5, 2 and 2.5
would also form a Pythagorean triad or triple. Some other triads are:
5, 12, 13
8, 15, 17
9, 40, 41
Determine whether the triangle shown is right-angled,
acute-angled or obtuse-angled.
THINK WRITE
Assume that the longest side is a
hypotenuse.
Calculate the hypotenuse2
and base2
+
height2
separately.
hypotenuse2
= 72
= 49
base2
+ height2
= 52
+ 42
= 25 + 16
= 41
Write down an equality or inequality
statement.
hypotenuse2
> base2
+ height2
Write a conclusion. Therefore the triangle is obtuse-angled.
4 cm 5 cm
7 cm
1
2
3
4
4WORKEDExample

Pythagoras’ theorem can be used to solve more practical problems. In these cases, it is
necessary to draw a diagram that will help you decide the appropriate method for
finding a solution. The diagram simply needs to represent the triangle; it does not need
to show details of the situation described.
Is the set of numbers 4, 6, 7 a Pythagorean triad?
THINK WRITE
Find the sum of the squares of the two
smaller numbers.
42
+ 62
= 16 + 36
= 52
72
= 49
72
≠ 42
+ 62
Find the square of the largest number.
Compare the two results. The numbers
form a Pythagorean triad if the results
are the same.
Write your answer. 4, 6, 7 is not a Pythagorean triad
1
2
3
4
5WORKEDExample
E
XCEL Spread
sheet
Triads
The fire brigade attends a blaze in a tall building. They need to rescue a person from the
6th floor of the building, which is 30 metres above ground level. Their ladder is 32 metres
long and must be at least 10 metres from the foot of the building. Can the ladder be used
to reach the people needing rescue?
THINK WRITE
Draw a diagram and show all given
information.
Write the formula after deciding if you
are finding the hypotenuse or a shorter
side.
Hypotenuse2
= base2
+ height2
Substitute the lengths of the known
sides.
c2
= 102
+ 302
Evaluate the expression. = 100 + 900
= 1000
Find the answer by taking the square
root.
c =
= 31.62 m
Give a written answer. The ladder will be long enough to make the
rescue, since it is 32 m long.
1
10 m
30 m
c
Burning building
Fire engine
10 m
30 m c
2
3
4
5 1000
6
6WORKEDExample

1 Name the hypotenuse in each of the following triangles.
a b c
2 Find the length of the hypotenuse in each of the following triangles.
a b c
3 In each of the following, find the length of the hypotenuse, correct to 2 decimal places.
a b c
remember
1. Make sure that you can identify the hypotenuse of a right-angled triangle. It is
the side opposite the right angle, and the longest side.
2. If you are finding the length of the hypotenuse use
hypotenuse2
= base2
+ height2
.
3. If you are finding the length of a shorter side use either
base2
= hypotenuse2
− height2
or
height2
= hypotenuse2
− base2
.
4. Pythagoras’ theorem can be used to test whether a triangle is right-angled.
(a) if hypotenuse2
= base2
+ height2
, the triangle is right-angled
(b) if hypotenuse2
> base2
+ height2
, the triangle is obtuse-angled
(c) if hypotenuse2
< base2
+ height2
, the triangle is acute-angled.
5. Pythagorean triads or Pythagorean triples are sets of three numbers which
satisfy Pythagoras’ theorem — for example 3, 4, 5.
6. Read the question carefully to make sure that you give the answer in the correct
form.
7. Begin problem questions with a diagram and finish with an answer in words.
remember
5A
WORKED
Example
1 P
Q R
X
Y
Z
B
A C
WORKED
Example
2
12 cm
5 cm
x
150 mm
80 mm
m
60 m
11 m
z
9 cm
6 cm
a
4.9 m
4.9 m
8.6 km 11.3 km

4 Find the length of each unknown shorter side in the right-angled triangles below,
correct to 1 decimal place.
a b c
5 In each of the following right-angled triangles, ﬁnd the length of the side marked with
a pronumeral, correct to 1 decimal place.
a b
c d
6 Use the converse of Pythagoras’ theorem to determine if the following triangles are
right-angled, acute-angled or obtuse-angled.
a b c
7
The hypotenuse in LXYZ at right is:
8
Which of the following triangles is deﬁnitely right-angled?
A XY B XZ
C YZ D impossible to tell
A B
C D
WORKED
Example
3
12 cm
6 cm
p 2.2 m 2.9 m
q
4.37 m
2.01 m
t
8 cm
4 cm
m
10.5 cm
24.5 cm z
33 mm
34 mm a
37.25 m
52.75 m
p
WORKED
Example
4
10 cm
6 cm
8 cm 41 cm 40 cm
9 cm
31 m
38 m
16 m
mmultiple choiceultiple choice
X
Y
Z
24.5 m
32.5 m
20.5 m
5 m 24.5 m
25 m
24.5 m 84 m
87.5 m
25.4 m
24.5 m
35.3 m

9 Are the following sets of numbers Pythagorean triads?
10 Complete the following Pythagorean triads.
11 For each of the sets which were Pythagorean triads in question 9 state which side the
right angle is opposite.
12
Which of the following is a Pythagorean triad?
13
Which of the following is not a Pythagorean triad?
14 A television antenna is 12 m high. To support it, wires are attached to the ground 5 m
from the foot of the antenna. Find the length of each wire.
15 Susie needs to clean the guttering on her roof. She places her ladder 1.2 m back from
the edge of the guttering that is 3 m above the ground. How long will Susie’s ladder
need to be (correct to 2 decimal places)?
16 A rectangular gate is 3.5 m long and 1.3 m wide. The gate
is to be strengthened by a diagonal brace as shown at right.
How long should the brace be (correct to 2 decimal places)?
17 A 2.5-m ladder leans against a brick wall. The foot of the ladder is 1.2 m from the foot
of the wall. How high up the wall will the ladder reach (correct to 1 decimal place)?
18 Use the measurements in the diagram at right to ﬁnd the height of
the ﬂagpole, correct to 1 decimal place.
19 An isosceles, right-angled triangle has a hypotenuse of 10 cm. Calculate the length of
the shorter sides. (Hint: Call both shorter sides x.)
a 9, 12, 15 b 4, 5, 6 c 30, 40, 50 d 3, 6, 9
e 0.6, 0.8, 1.0 f 7, 24, 25 g 6, 13, 14 h 14, 20, 30
i 11, 60, 61 j 10, 24, 26 k 12, 16, 20 l 2, 3, 4
a 9, __, 15 b __, 24, 25 c 1.5, 2.0, __ d 3, __, 5
e 11, 60, __ f 10, __, 26 g __, 40, 41 h 0.7, 2.4, __
A 7, 14, 21 B 1.2, 1.5, 3.6 C 3, 6, 9
D 12, 13, 25 E 15, 20, 25
A 5, 4, 3 B 6, 9, 11 C 13, 84, 85
D 0.9, 4.0, 4.1 E 5, 12, 13
WORKED
Example
5
WORKED
Example
6
3.5 m
1.3 m
7.9 m
2.4 m

Resources: Paper, pencil, protractor, compass, calculator.
As mentioned previously, Pythagoras’ theorem states:
‘In a right-angled triangle, the square on the hypotenuse is equal to the sum of the
squares on the other two sides.’
In this investigation we consider replacing the word square/s with other geometric
shapes.
1 What would happen if the word square/s were to be replaced by the word
semicircle/s? Draw diagrams and investigate to see whether the statement
would still be true in all cases. (Make sure that the triangles you draw are
indeed right-angled.)
2 What would happen if we were to replace square/s with circle/s or equilateral
triangle/s? Remember that it must work for all situations.
3 What are your conclusions?
4 Prepare a formal report on your investigation. Support your conclusions with
detailed diagrams and mathematical evidence.
Constructing a staircase
Resources: Pen, paper, calculator.
With an understanding of the theorem of Pythagoras, you should now be able to
investigate the staircase problem.
Building stairs prompts the following questions:
1. How much timber should be ordered?
2. Where should the stairs start if we are to get the correct slope?
inv
estigat
ioninv
estigat
ion
3 cm
4 cm
5 cm
inv
estigat
ioninv
estigat
ion

We start by considering the general structure
of a staircase.
Mrs Kelly has contacted a local builder to
construct an external set of stairs from the
ground to an existing deck 2550 mm above
the ground.
1 Use the information in the diagram above
to determine:
a the number of treads, 170 mm apart,
required to reach a height of 2550 mm
(remember that no treads are necessary
at the top or base of the staircase)
b the horizontal distance from the deck where the footings should be
cemented (knowing that each tread is 250 mm wide)
c the length of each stringer
d the length of timber that must be
ordered if the treads and stringers
are constructed from 250-mm by
50-mm hardwood, and the stairs
are 1 metre wide.
2 Draw a diagram for Mrs Kelly, showing the structure of her staircase (with the
correct number of treads). Mark on your diagram all known measurements and
indicate the total length of timber required for the job.
Constructing a paper box
Resources: Paper, scissors.
How is it possible to determine the starting dimensions for constructing a box of a
particular size? Pythagoras’ theorem has the answer.
1 Take a square piece of paper of a reasonable size.
a Valley fold along all lines of symmetry.
Remember to crease all folds sharply.
Open the square out ﬂat.
b Fold each of the corners in a valley fold in
towards the centre.
Treads
Stringers × 2
Footings
level Ground
170 mm
250 mm
250 mm
50 mm
inv
estigat
ioninv
estigat
ion

c Valley fold the top, bottom, left and right sides
of the resulting shape in towards the centre,
opening each fold before making the next.
These folds will form the base of the box.
d Open two of the corner folds opposite each
other as indicated.
e Fold up the edges opposite each other (those
which already have centre folds) to make
one pair of the box’s sides.
f Fold in both other ends as shown, completing
the other two sides of the box.
2 With the base of the box completed, a lid could then
be fashioned from another piece of square paper
about 1 cm larger.

3 Having completed our box, the question now arises: ‘How is it possible to
determine the dimensions of square paper required to construct a box of a
particular size?’
a Open your box and look at all the creases formed. It should appear as
below.
b The base of the box is outlined with respect to the size of the original square
paper. Note that the box’s base is 2 ‘units’ in length, while the whole
diagonal of the paper is 8 ‘units’ in length. This means that the box’s base
is one-quarter the length of the diagonal of the original paper.
c What is the height of the box compared with the length of the diagonal of
the original square?
4 We now have the knowledge to make a box with
a side length of, say, 5 cm.
a The diagonal of our starting square must be
4 × 5 cm = 20 cm
b Using Pythagoras’ theorem, we can now determine
the length of the side of the square (S).
Hypotenuse2
= base2
+ height2
202
= S2
+ S2
400 = 2S2
= S2
200 = S2
So S =
= 14.1 cm
c So, starting with a piece of paper 14.1 cm square, after folding, the resulting
box would be 2.5 cm high, with a side length of 5 cm.
d Verify these measurements for yourself by folding such a piece of paper.
e Try this exercise with other paper sizes.
This is yet another example of the importance of geometry in many facets of our
daily lives.
S
S
20
cm
400
2
---------
200

Similar triangles and shadow sticks
Sometimes it is not possible to measure heights of objects such as trees, tall buildings
and mountains. It is, however, often possible to measure the length of a shadow cast by
these objects. If we compare the length of these shadows with the length of a shadow
cast by a stick of known height under the same conditions, it is possible to calculate
heights which are difﬁcult to measure.
Consider the situation below.
We need to determine the height of the building. Take a stick whose height we know
(or can measure) and place it, vertically, in the sunshine, near the building. The rays
from the sun are parallel, so we have two similar right-angled triangles. (The building
and the stick are at right angles to the ground.) We can measure both shadows.
The larger right-angled triangle formed by the building and its shadow is some scale
factor of the smaller right-angled triangle formed by the stick and the stick’s shadow.
Scale factor =
We can then apply the same scale factor to the height of the stick to determine the
height of the building. So,
Height of building = height of stick × scale factor
Using shadows to determine the
height of a tall object
Resources: Tall object, shadow stick, measuring tape, sunshine!
We shall apply the theory just developed, to determine the height of a tall object in
your school grounds.
1 Choose a tall, accessible object in your school grounds (tree, building, etc.) as
the object whose height you wish to determine. Let its height be B metres.
Building
Stick
Stick’s
shadow
Building’s shadow
Building
Stick
Stick’s
shadow
Building’s shadow
length of building shadow
length of stick shadow
--------------------------------------------------------------
inv
estigat
ioninv
estigat
ion

2 Measure the height of your shadow stick as S metres.
3 Place your shadow stick vertically in the sunshine near your tall object.
Measure the length of its shadow on the ground. Let this be s metres.
4 Measure the length of the building’s shadow on the ground. Let this be b
metres.
5 Draw a diagram representing this situation. Mark all known measurements on
the diagram.
6 Calculate the scale factor F (tall-object shadow length divided by stick shadow
length; that is, F = ).
7 The height of your tall object can then be determined by multiplying the height
of the stick by the scale factor; that is, B = S × F
8 Write up your activity, explaining all steps involved.
b
s
---
At the same time as a tree cast a shadow of 14 m,
a 168-cm-tall girl cast a shadow of 140 cm.
Find the height of the tree. Give the answer to
1 decimal place.
THINK WRITE
Identify the two similar triangles and
draw them separately. (We assume that
both the girl and the tree are
perpendicular to the ground.)
Identify the side of the triangle whose
length is required.
Calculate the scale factor.
Note: Measurements must be in the
same units.
Scale factor =
=
= 10
Apply the scale factor to the girl’s
height.
Height of tree = girl’s height × scale factor
= 1.68 m × 10
= 16.8 m
Write the ﬁnal answer, specifying units. Height of tree is 16.8 m
Shadow
(140 cm)
Sun’s rays
Girl
(168 cm)
14 metres
1
x
168 cm
140 cm 14 m
2
3
length of tree shadow
length of girl’s shadow
-------------------------------------------------------
14 m
1.4 m
-------------
4
5
7WORKEDExample

Shadow sticks
1 The following diagrams represent the measurements taken from shadows of sticks and
objects. Use the figures to determine the heights of the objects.
a b
c d
2 At the same time as a building cast a shadow of 14.3 metres, a 2-metre stick cast a
shadow of 5.3 metres. What is the height of the building?
3 A boulder on the shoreline cast a shadow of 15.8 m on the beach at the same time as a
1.5-metre stick cast a shadow of 3.7 m. Determine the height of the boulder.
4 Find the height of the flagpole shown in the diagram at right.
5 The shadow of a tree is 4 metres and at the same time
the shadow of a 1-metre stick is 25 cm. Assuming both
the tree and stick are perpendicular to the horizontal ground,
what is the height of the tree?
remember
1. To determine the height of objects too difficult to measure, we can use the
‘shadow stick’ method.
2. Scale factor = .
3. Height of object = height of stick × scale factor.
length of shadow of object
length of shadow of stick
----------------------------------------------------------------
remember
5B
WORKED
Example
7
a
3.6 m
8.2 m
2.5 m
b
4.5 m
20 m
3.5 m
c
1.2 m
4.2 m
1.5 m
d
5.8 m
15 m
3 m
0.9 m
9 m
1m
Guy wire

Questions 6 and 7 refer to the following
information.
A young tennis player’s serve is shown in
the diagram.
Assume the ball travels in a straight line.
6
The height of the ball just as it is hit, x, is
closest to:
A 3.6 m B 2.7 m C 2.5 m
D 1.8 m E 1.6 m
7
The height of the player, y, as shown is
closest to:
A 190 cm B 180 cm C 170
cm
D 160 cm E 150 cm
Calculating trigonometric ratios
We have already looked at Pythagoras’ theorem, which enabled us to ﬁnd the length of
one side of a right-angled triangle given the lengths of the other two. However, to deal
with other relationships in right-angled triangles, we need to turn to trigonometry.
Trigonometry allows us to work with the angles also; that is, deal with relationships
between angles and sides of right-angled triangles. For example, trigonometry enables
us to ﬁnd the length of a side, given the length of another side and the magnitude of an
angle.
So that we are clear about which lines and angles we are
describing, we need to identify the given angle, and name
the shorter sides with reference to it. For this reason, we
label the sides opposite and adjacent — that is, the sides
opposite and adjacent to the given angle. The diagram shows this relationship between
the sides and the angle, θ.
The trigonometric ratios are constant for a particular angle, and this is the reason the
shadow-stick method worked.
x
y1.1m
5 m 10 m
0.9 m
Work
SHEET 5.1
hypotenuse
opposite
adjacent
θ

Trigonometry uses the ratio of side lengths to calculate the lengths of sides and the size
of angles. The ratio of the opposite side to the adjacent side is called the tangent ratio.
This ratio is ﬁxed for any particular angle.
The tangent ratio for any angle, θ, can be found using the result:
In the investigation above, we found that for a 30° angle the ratio was approximately
0.58. We can ﬁnd a more accurate value for the tangent ratio on a calculator by using
the button.
Calculators require a particular sequence of button presses in order to perform this
calculation. Investigate the sequence required for your particular calculator.
For all calculations in trigonometry you will need to make sure that your calculator
is in DEGREES MODE. For most calculators you can check this by looking for a DEG
in the display.
Looking at the tangent ratio
Resources: Ruler, spreadsheet.
The tangent ratio for a given angle is a ratio of
sides in similar right-angled triangles, such as
those in the diagram. ∠BAC is common to each
triangle and is equal to 30°. We are going to
look at the ratio of the opposite side length to
the adjacent side length in each triangle.
1 Complete each of the following measurements.
a BC = mm AB = mm
b DE = mm AD = mm
c FG = mm AF = mm
d HI = mm AH = mm
2 Set up the spreadsheet below:
3 Enter values in columns 2 and 3 from the measurements above.
4 Provide a formula in column 4 to calculate the ratio of the opposite side length
to the adjacent side length. This is the tangent ratio.
5 The ∠BAC is common to each triangle. Consider the calculated values in
column 4. What is your conclusion?
inv
estigat
ioninv
estigat
ion
A
B
C
E
G
I
DFH
Triangle
Opposite side
length
Adjacent side
length
ABC
ADE
AFG
AHI
Opposite
Adjacent
-----------------------
tan θ
opposite side
adjacent side
-------------------------------=
tan

When measuring angles:
1 degree = 60 minutes
1 minute = 60 seconds
You need to be able to enter angles using both degrees and minutes into your calcu-
lator. Most calculators use a (Degrees, Minutes, Seconds) button or a
button. Check with your teacher to see how to do this.
The tangent ratio is used to solve problems involving the opposite side and the adjacent
side of a right-angled triangle. The tangent ratio does not allow us to solve problems
that involve the hypotenuse.
The sine ratio (abbreviated to sin; pronounced sine) is the name given to the ratio of
the opposite side and the hypotenuse.
Looking at the sine ratio
The tangent ratio for a given angle is a ratio of
the opposite side and the adjacent side in a
right-angled triangle. The sine ratio is the ratio
of the opposite side and the hypotenuse.
1 Complete each of the following
measurements as before. (As we saw earlier,
∠BAC is common to all these similar triangles and so in this exercise, we look
at the ratio of the side opposite ∠BAC to the hypotenuse of each triangle.)
a BC = mm AC = mm
b DE = mm AE = mm
c FG = mm AG = mm
d HI = mm AI = mm
DMS ° ’ ”
Using your calculator, ﬁnd the following, correct to 3 decimal places.
a tan 60° b 15 tan 75° c d tan 49°32′
THINK WRITE/DISPLAY
a Press and enter 60, then press . a tan 60° = 1.732
b Enter 15, press and , enter 75, then
press .
b 15 tan 75° = 55.981
c Enter 8, press and , enter 69, then
press .
c = 3.071
d Press , enter 49, press , enter 32,
press , then press .
d tan 49°32′ = 1.172
Note: Some calculators require that the angle size be entered before the trigonometric
functions.
8
tan 69°
------------------
tan =
× tan
=
÷ tan
=
8
tan 69°
-----------------
tan DMS
DMS =
8WORKEDExample
inv
estigat
ioninv
estigat
ion
A
B
C
E
G
I
DFH

In any right-angled triangle with equal angles, the ratio of the length of the opposite
side to the length of the hypotenuse will remain the same, regardless of the size of the
triangle. The formula for the sine ratio is:
The value of the sine ratio for any angle is found using the sin function on the calcu-
lator.
sin 30° = 0.5
Check this on your calculator.
2 Set up the spreadsheet below:
3 Enter values in columns 2 and 3 from the measurements above.
4 Provide a formula in column 4 to calculate the ratio of the length of the
opposite side to the length of the hypotenuse. This is the sine ratio.
5 The ∠BAC is common to each triangle. You should notice that the values in
column 4 are the same (or very close, allowing for measurement error).
Triangle
Opposite side
length
Hypotenuse
length
ABC
ADE
AFG
AHI
Opposite
Hypotenuse
------------------------------
sin θ
opposite side
hypotenuse
-------------------------------=
Find, correct to 3 decimal places:
a sin 57° b 9 sin 45° c d 9.6 sin 26°12′.
THINK WRITE/DISPLAY
a Press and enter 57, then press . a sin 57° = 0.839
b Enter 9, press and , enter 45,
then press .
b 9 sin 45° = 6.364
c Enter 18, press and , enter 44,
then press .
c = 25.912
d Enter 9.6, press and , enter 26,
press , enter 12, press , then
press .
d 9.6 sin 26°12′ = 4.238
Note: Check the sequence of button presses required by your calculator.
18
sin 44°
-----------------
sin =
× sin
=
÷ sin
=
18
sin 44°
-----------------
× sin
DMS DMS
=
9WORKEDExample

A third trigonometric ratio is the cosine ratio. This ratio compares the length of the
adjacent side and the hypotenuse.
The cosine ratio is found using the formula:
To calculate the cosine ratio for a given angle on your calculator, use the cos func-
tion. On your calculator check the calculation:
cos 30° = 0.866
Similarly, if we are given the sine, cosine or tangent of an angle, we are able to calcu-
late the size of that angle using the calculator. We do this using the inverse functions.
On most calculators these are the second function of the sin, cos and tan functions and
are denoted sin−1
, cos−1
and tan−1
.
Looking at the cosine ratio
1 Complete each of the following measurements.
a AB = mm AC = mm
b AD = mm AE = mm
c AF = mm AG = mm
d AH = mm AI = mm
2 Set up a spreadsheet of similar format to the two previous spreadsheets to
calculate the ratio of the length of the adjacent side to that of the hypotenuse.
3 You should again ﬁnd this ratio constant (or nearly so). This is the cosine ratio.
inv
estigationinv
estigat
ion
A
B
C
E
G
I
DFH
cos θ
adjacent side
hypotenuse
-------------------------------=
Find, correct to 3 decimal places:
a cos 27° b 6 cos 55° c d .
THINK WRITE/DISPLAY
a Press and enter 27, then press . a cos 27° = 0.891
b Enter 6, press and , enter 55,
then press .
b 6 cos 55° = 3.441
c Enter 21.3, press and , enter 74,
then press .
c = 77.275
d Enter 4.5, press and , enter 82,
press , enter 46, press , then
press .
d = 35.740
Note: Check the sequence requirements for your calculator.
21.3
cos 74°
------------------
4.5
cos 82°46′
--------------------------
cos =
× cos
=
÷ cos
=
21.3
cos 74°
------------------
× cos
DMS DMS
=
4.5
cos 82°46′
--------------------------
10WORKEDExample

Problems sometimes supply angles in degrees, minutes and seconds, or require
answers to be written in the form of degrees, minutes and seconds. On most calcu-
lators, you will use the DMS (Degrees, Minutes, Seconds) function or the
function.
If you are using a graphics calculator, the ‘angle’ function provides this facility. Check
with your teacher for the requirements of your particular calculator.
Find θ, correct to the nearest degree, given that sin θ = 0.738.
THINK WRITE/DISPLAY
Press [sin–1
] and enter .738,
then press .
Round your answer to the nearest
degree.
θ = 48°
Note: Check the sequence requirements for your calculator.
1 2nd F
=
2
11WORKEDExample
° ’ ”
Given that tan θ = 1.647, calculate θ to the nearest minute.
THINK WRITE/DISPLAY
Press [tan–1
] and enter 1.647,
then press .
Convert your answer to degrees and
minutes by pressing .
θ = 58°44′
1 2nd F
=
2
DMS
12WORKEDExample
remember
1. The tangent ratio is the ratio of the opposite side and the adjacent side.
tan θ =
2. The sine ratio is the ratio of the opposite side and the hypotenuse.
sin θ =
3. The cosine ratio is the ratio of the adjacent side and the hypotenuse.
cos θ =
4. The value of the trigonometric ratios can be found using the sin, cos and tan
functions on your calculator.
5. The angle can be found when given the trigonometric ratio using the sin−1
,
cos−1
and tan−1
functions on your calculator.
opposite side
adjacent side
-------------------------------
opposite side
hypotenuse
-------------------------------
adjacent side
hypotenuse
-------------------------------
remember

Calculating trigonometric
ratios
1 Calculate the value of each of the following, correct to 3 decimal places.
4 Calculate the value of each of the following, correct to 3 decimal places, if necessary.
6 Find θ, correct to the nearest degree, given that sin θ = 0.167.
7 Find θ, correct to the nearest degree, given that:
8 Find θ, correct to the nearest minute, given that cos θ = 0.058.
9 Find θ, correct to the nearest minute, given that:
Finding an unknown side
We can use the trigonometric ratios to ﬁnd the length of one
side of a right-angled triangle if we know the length of another
side and an angle. Consider the triangle at right.
In this triangle we are asked to ﬁnd the length of the opposite
side and have been given the length of the adjacent side.
a tan 57° b 9 tan 63° c d tan 33°19′
a sin 37° b 9.3 sin 13° c d
a cos 45° b 0.25 cos 9° c d 5.9 cos 2°3′
a sin 30° b cos 15° c tan 45°
d 48 tan 85° e 128 cos 60° f 9.35 sin 8°
g h i
a sin 24°38′ b tan 57°21′ c cos 84°40′
d 9 cos 55°30′ e 4.9 sin 35°50′ f 2.39 tan 8°59′
g h i
a sin θ = 0.698 b cos θ = 0.173 c tan θ = 1.517.
a tan θ = 0.931 b cos θ = 0.854 c sin θ = 0.277.
5C
WORKED
Example
8
8.6
tan 12°
-----------------
WORKED
Example
9
14.5
sin 72°
-----------------
48
sin 67°40′
-------------------------
WORKED
Example
10
6
cos 24°
------------------
4.5
cos 32°
------------------
0.5
tan 20°
-----------------
15
sin 72°
-----------------
19
tan 67°45′
-------------------------
49.6
cos 47°25′
--------------------------
0.84
sin 75°5′
----------------------
WORKED
Example
11
WORKED
Example
12
hyp
opp
adj
14 cm
30°
x

We know from the formula that: tan θ = . In this example, tan 30° = . From
our calculator we know that tan 30° = 0.577. We can set up an equation that will allow
us to find the value of x.
tan θ =
tan 30° =
x = 14 tan 30°
= 8.083 cm
In the example above, we were told to use the tangent ratio. In practice, we need to be
able to look at a problem and then decide if the solution is found using the sin, cos or
tan ratio. To do this we need to examine the three formulas.
tan θ =
We use this formula when we are finding either the opposite or adjacent side and are
given the length of the other.
sin θ =
The sin ratio is used when finding the opposite side or the hypotenuse when given
the length of the other.
cos θ =
The cos ratio is for problems where we are finding the adjacent side or the hypot-
enuse and are given the length of the other.
To make the decision we need to label the sides of the triangle and make a decision
based on these labels.
opposite
adjacent
--------------------
x
14
------
opp
adj
---------
x
14
------
Use the tangent ratio to find the value of h in the triangle at right,
correct to 2 decimal places.
THINK WRITE
Label the sides of the triangle opp, adj
and hyp.
Write the tangent formula. tan θ =
Substitute for θ (55°) and the adjacent
side (17 cm).
tan 55° =
Make h the subject of the equation. h = 17 tan 55°
Calculate. = 24.28 cm
17 cm
55°
h
1
hyp
opp
adj
17 cm
55°
h
2
opp
adj
---------
3
h
17
------
4
5
13WORKEDExample
Cabri Geo
metry
Trig
ratios
opposite side
adjacent side
-------------------------------
opposite side
hypotenuse
-------------------------------
adjacent side
hypotenuse
-------------------------------

To remember each of the formulas more easily, we can use this acronym:
SOHCAHTOA
We pronounce this acronym as ‘Sock ca toe her’. The initials of the acronym
represent the three trigonometric formulas.
sin θ = cos θ = tan θ =
Care needs to be taken at the substitution stage. In the previous two examples, the
unknown side was the numerator in the fraction, hence we multiplied to find the
answer. If after substitution, the unknown side is in the denominator, the final step is
done by division.
Find the length of the side marked x, correct to 2 decimal places.
THINK WRITE
Label the sides of the triangle.
x is the opposite side and 24 m is the
hypotenuse, therefore use the sin formula.
sin θ =
Substitute for θ and the hypotenuse. sin 50° =
Make x the subject of the equation. x = 24 sin 50°
Calculate. = 18.39 m
24 m
50°
x
1
hyp
opp
adj
24 m
50°
x
2
opp
hyp
---------
3
x
24
------
4
5
14WORKEDExample
S
O
H 
 
 
 
opp
hyp
---------
C
A
H 
 
 
 
adj
hyp
---------
T
O
A 
 
 
 
opp
adj
---------
Find the length of the side marked z in the triangle at right.
THINK WRITE
Label the sides opp, adj and hyp.
Choose the cosine ratio because we are
finding the hypotenuse and have been
given the adjacent side.
Write the formula. cos θ =
12.5 m
23°15'
z
1
opp
hyp
adj
12.5 m
23°15'
z
2
3
adj
hyp
---------
15WORKEDExample

Trigonometry is used to solve many practical problems. In these cases, it is necessary
to draw a diagram to represent the problem and then use trigonometry to solve the
problem. With written problems that require you to draw the diagram, it is necessary to
give the answer in words.
THINK WRITE
Substitute for θ and the adjacent side. cos 23°15′ =
Make z the subject of the equation. z cos 23°15′ = 12.5
z =
Note: In calculating the length of the hypotenuse, the process will always involve division
by the trigonometric function.
4
12.5
z
----------
5
12.5
cos 23°15′
--------------------------
6
A flying fox is used in an army training camp. The flying fox is supported by a cable that
runs from the top of a cliff face to a point 100 m from the base of the cliff. The cable
makes a 15° angle with the horizontal. Find the length of the cable used to support the
flying fox.
THINK WRITE
Draw a diagram and show information.
Label the sides of the triangle opp, adj
and hyp.
Choose the cosine ratio because we are
finding the hypotenuse and have been
given the adjacent side.
Write the formula. cos θ =
Substitute for θ and the adjacent side. cos 15° =
Make f the subject of the equation. f cos 15° = 100
f =
Give a written answer. The cable is approximately 103.5 m long.
1
15°
100 m
opp
hyp
adj
f
2
3
4
adj
hyp
---------
5
100
f
---------
6
100
cos 15°
------------------
7
8
16WORKEDExample

Finding an unknown side
1 Label the sides of each of the following triangles, with respect to the angle marked
with the pronumeral.
a b c
2 Use the tangent ratio to find the length of the side marked x
(correct to 1 decimal place).
3 Use the sine ratio to find the length of the side marked a
(correct to 2 decimal places).
4 Use the cosine ratio to find the length of the side marked d
(correct to nearest whole number).
remember
1. Trigonometry can be used to find a side in a right-angled triangle when we are
given the length of one side and the size of an angle.
2. The trig formulas are:
sin θ = cos θ = tan θ =
3. Take care to choose the correct trigonometric ratio for each question.
4. Substitute carefully and note the change in the calculation, depending upon
whether the unknown side is in the numerator or denominator.
5. Before using your calculator, check that it is in degrees mode.
6. Be sure that you know how to enter degrees and minutes into your calculator.
7. Written problems will require you to draw a diagram and give a written
answer.
opp
hyp
---------
adj
hyp
---------
opp
adj
---------
remember
5D
θ
α
γ
WORKED
Example
13
51 mm
71°
x
13 m
23°
a
35 cm
31°
d

5 The following questions use the tan, sin or cos ratios in their solution. Find the size of
the side marked with the pronumeral, correct to 1 decimal place.
a b c
6 Find the length of the side marked with the pronumeral in each of the following
a b c
7 Find the length of the side marked with the pronumeral in each of the following
a b c
d e f
g h i
j k l
WORKED
Example
14
68°
13 cmx
49°
48 m
y
41°
12.5 km
z
WORKED
Example
15
21°
4.8 m
t
77°
87 mm
p
36°
8.2 m
q
23°
2.3 m
a
39°
0.85 kmb
76°
8.5 km
x
116 mm
9°
m
16.75 cm
11°
d
13°
64.75 m
x
83°
44.3 m
x
20°
15.75 km
g
2.34 m
84°9'
m
84.6 km
60°32'
q
21.4 m
75°19't
26.8 cm
29°32'
r

8
Look at the diagram at right and state which of the following is correct.
9
Study the triangle at right and state which of the following
is correct.
10
Which of the statements below is not correct?
A The value of tan θ can never be greater than 1.
B The value of sin θ can never be greater than 1.
C The value of cos θ can never be greater than 1.
D tan 45° = 1
11
Study the diagram at right and state which of the statements is correct.
12 A tree casts a 3.6 m shadow when the sun’s angle of elevation is 59°. Calculate the
height of the tree, correct to the nearest metre.
13 A 10-m ladder just reaches to the top of a wall when it is leaning at 65° to the ground.
How far from the foot of the wall is the ladder (correct to 1 decimal place)?
14 The diagram at right shows the paths of two ships, A and B,
after they have left port.
If ship B sends a distress signal, how far must ship A sail to
give assistance (to the nearest kilometre).
15 A rectangular sign 13.5 m wide has a diagonal brace that
makes a 24° angle with the horizontal.
a Draw a diagram of this situation.
b Calculate the height of the sign, correct to the nearest metre.
16 A wooden gate has a diagonal brace built in for support. The gate stands 1.4 m high
and the diagonal makes a 60° angle with the horizontal.
a Draw a diagram of the gate.
b Calculate the length that the diagonal brace needs to be.
17 The wire support for a ﬂagpole makes a 70° angle with the ground. If the support is
3.3 m from the base of the ﬂagpole, calculate the length of the wire support (correct
to 2 decimal places).
A x = 9.2 sin 69° B
C x = 9.2 cos 69° D
A tan φ = B tan φ = C sin φ = D cos φ =
A w = 22 cos 36° B
C w = 22 cos 54° D w = 22 sin 54°
69°
9.2
x
x
9.2
sin 69°
-----------------=
x
9.2
cos 69°
------------------=
17
15
8
φ
8
15
------
15
8
------
15
17
------
8
17
------
36°
22 mm
w
w
22
sin 36°
-----------------=
WORKED
Example
16
60°
Port
23 km
A
B

18 A ship drops anchor vertically with an
anchor line 60 m long. After one hour
the anchor line makes a 15° angle with
the vertical.
b Calculate the depth of water, correct
to the nearest metre.
c Calculate the distance that the ship
has drifted, correct to 1 decimal place.
Find the size of the side marked with the pronumeral in each of the following. Where
necessary, give your answer correct to 1 decimal place.
1 2
3 4
5 6
7 8
9 10
1
15 cm
8 cm
a
10 m 20 m
b
30 km
40 km
c 21°
23 md
40°
39.2 cm
e
37°
42.1 m
f
50°
14.9 mm
g
42°
119 mm
h
17°
93.2 m i 45°
17 m
j

Finding angles
So far, we have concerned ourselves with finding side lengths. We are also able to use
trigonometry to find the sizes of angles when we have been given side lengths. We need
to reverse our previous processes.
Consider the triangle at right.
We want to find the size of the angle marked θ.
Using the formula sin θ = we know that in this triangle
sin θ =
=
= 0.5
We then calculate sin−1
(0.5) to find that θ = 30°.
As with all trigonometry it is important that you have your calculator set to degrees
mode for this work.
In many cases we will need to calculate the size of an angle, correct to the nearest
minute. The same method for finding the solution is used; however, you will need to
use your calculator to convert to degrees and minutes.
10 cm
5 cm
θopp
hyp
---------
5
10
------
1
2
---
Find the size of angle θ, correct to the nearest degree, in the
triangle at right.
THINK WRITE
Label the sides of the triangle and
choose the tan ratio.
tan θ =
Substitute for the opposite and adjacent
sides in the triangle.
=
Make θ the subject of the equation. θ = tan−1
Calculate. = 33°
6.5 m
4.3 m
θ
1
4.3 m
6.5 m
adj
hyp
opp
θ
opp
adj
---------
2
4.3
6.5
-------
3
4.3
6.5
-------
 
 
4
17WORKEDExample

The same methods can be used to solve problems. As with ﬁnding sides, we set the
question up by drawing a diagram of the situation.
Find the size of the angle θ at right, correct to the nearest minute.
THINK WRITE
Label the sides of the triangle and choose the
sin ratio.
sin θ =
Substitute for the opposite side and the
hypotenuse in the triangle.
=
Make θ the subject of the equation. θ = sin−1
Calculate and convert your answer to degrees
and minutes.
= 40°23′
4.6 cm
7.1 cm
θ
1 opp
hypadj
4.6 cm
7.1 cm
θ
opp
hyp
---------
2
4.6
7.1
-------
3
4.6
7.1
-------
 
 
4
18WORKEDExample
A ladder is leant against a wall. The foot of the ladder is 4 m from the base of the wall and
the ladder reaches 10 m up the wall. Calculate the angle that the ladder makes with the
ground.
THINK WRITE
Draw a diagram and label the sides.
Choose the tangent ratio and write the
formula.
tan θ =
Substitute for the opposite and adjacent sides. =
Calculate. = 68°12′
Give a written answer. The ladder makes an angle of 68°12′ with
the ground.
1
4 m
10 m
adj
opp
hyp
θ
2
opp
adj
---------
3
10
4
------
4
10
4
------
 
 
5
6
19WORKEDExample

Finding angles
1 Use the tangent ratio to find the size of the angle marked with the pronumeral in each
of the following, correct to the nearest degree.
a b c
2 Use the sine ratio to find the size of the angle marked with the pronumeral in each of
the following, correct to the nearest minute.
a b c
3 Use the cosine ratio to find the size of the angle marked with the pronumeral in each
of the following, correct to the nearest minute.
a b c
remember
1. Make sure that the calculator is in degrees mode.
2. To find an angle given the trig ratio, press or and then the
appropriate ratio button.
3. Be sure to know how to get your calculator to display an answer in degrees and
minutes. When rounding off minutes, check if the number of seconds is greater
than 30.
4. When solving triangles remember the SOHCAHTOA rule to choose the correct
formula.
5. In written problems draw a diagram and give an answer in words.
2nd F SHIFT
remember
5E
WORKED
Example
17
7 m
12 m
θ 11 m
3 m
φ
25 mm
162 mm
γ
WORKED
Example
18
24 m
13 m
θ
4.6 m
6.5 m
θ
9.7 km
5.6 km
α
15 cm
9 cm
θ
2.6 m4.6 m
α
27.8 cm
19.5 cm
β

4 In the following triangles, you will need to use all three trig ratios. Find the size of the
angle marked θ, correct to the nearest degree.
a b c
d e f
5 In each of the following ﬁnd the size of the angle marked θ, correct to the nearest
minute.
a b c
d e f
6
Look at the triangle drawn at right.
Which of the statements below is correct?
7
Consider the triangle drawn at right.
θ is closest to:
8
The exact value of sin . The angle θ =
A ∠ABC = 30° B ∠ABC = 60°
C ∠CAB = 30° D ∠ABC = 45°
A 41°55′ B 41°56′ C 48°4′ D 48°5′
A 30° B 45° C 60° D 90°
7 cm
11 cm
θ
15 cm
8 cm
θ
14 cm
9 cm
θ
3.6 m
9.2 m
θ
196 mm
32 mm
θ 14.9 m26.8 m
θ
30 m
19.2 m
θ
10 cm
63 cm
θ
2.5 m
0.6 m
θ
3.5 m
18.5 m
θ
16.3 m
8.3 m
θ
6.3 m
18.9 m
θ
10 cm
5 cm
A
BC
θ
12.5 m
9.3 m
θ
θ
3
2
-------=

9 A 10-m ladder leans against a wall 6 m high. Find the angle that the ladder makes
with the horizontal, correct to the nearest degree.
10 A kite is flying on a 40-m string. The kite is flying 10 m away
from the vertical as shown in the figure at right.
Find the angle the string makes with the horizontal, correct to
the nearest minute.
11 A ship’s compass shows a course due east of
the port from which it sails. After sailing
10 nautical miles, it is found that the ship is
1.5 nautical miles off course as shown in the
figure below.
Find the error in the compass reading, correct
to the nearest minute.
12 The diagram below shows a footballer’s shot
at goal.
By dividing the isosceles triangle in half
in order to form a right-angled triangle,
calculate, to the nearest degree, the angle
within which the footballer must kick to
get the ball to go between the posts.
13 A golfer hits the ball 250 m, but 20 m off
centre. Calculate the angle at which the
ball deviated from a straight line, correct
to the nearest minute.
WORKED
Example
19
40 m
10 m
kite
10 nm
Port
1.5 nm
7 m
30 m

1 Find the length of the hypotenuse in the triangle at right.
2 Find x, correct to 1 decimal place.
3 Find y, correct to 1 decimal place.
4 Find z, correct to 3 decimal places.
5 Calculate sin 56°,
6 Calculate 9.2 tan 50°,
7 Calculate ,
8 Find θ, given that sin θ = 0.5.
9 Find θ to the nearest degree, given
that cos θ = 0.299.
10 Find θ to the nearest minute, given
that tan θ = 2.
2
12 cm
5 cm
12 cm
12 cm
x
30 cm
20 cm
y
3.7 m
7.4 m
z
132
cos 8°45′
-----------------------

Angles of elevation and depression
The angle of elevation is measured upwards from a horizontal and refers to the angle
at which we need to look up to see an object. Similarly, the angle of depression is the
angle at which we need to look down from the horizontal to see an object.
We are able to use the angles of elevation and depression to calculate the heights and
distances of objects that would otherwise be difﬁcult to measure.
In practical situations, the angle of elevation is measured using a clinometer. Therefore,
the angle of elevation is measured from a person’s height at eye level. For this reason,
the height at eye level must be added to the calculated answer.
Angle of elevation
Horizontal
Angle of depression
Horizontal
From a point 50 m from the foot of a building, the angle of elevation
to the top of the building is measured as 40°.
Calculate the height, h, of the building, correct to the nearest metre.
THINK WRITE
Label the sides of the triangle opp,
adj and hyp.
Choose the tangent ratio because we
are ﬁnding the opposite side and
have been given the adjacent side.
Write the formula. tan θ =
Substitute for θ and the adjacent
side.
tan 40° =
Calculate. = 42 m
Give a written answer. The height of the building is approximately 42 m.
50 m
40°
h
1
50 m
40°
hyp
opp
h
adj
2
3
opp
adj
---------
4 h
50
------
5
6
7
20WORKEDExample
Bryan measures the angle of elevation to the top of a tree
as 64°, from a point 10 m from the foot of the tree. If the
height of Bryan’s eyes is 1.6 m, calculate the height of the
tree, correct to 1 decimal place.
10 m 1.6 m
64°
21WORKEDExample

A similar method for finding the solution is used for problems that involve an angle of
depression.
THINK WRITE
Choose the tangent ratio because we
are finding the opposite side and
have been given the adjacent side.
Substitute for θ and the adjacent side. tan 64° =
Calculate h. = 20.5 m
Add the eye height. Height of tree = 20.5 + 1.6
Height of tree = 22.1
Give a written answer. The height of the tree is approximately 22.1 m.
1
10 m
adj
hyp
opp
h
64°
2
3
opp
adj
---------
4
h
10
------
5
6
7
8
When an aeroplane in flight is 2 km from a runway,
the angle of depression to the runway is 10°.
Calculate the altitude of the aeroplane, correct to the
nearest metre.
THINK WRITE
Label the sides of the triangle
opp, adj and hyp.
Choose the tan ratio, because we
are finding the opposite side given
the adjacent side.
Substitute for θ and the adjacent
side, converting 2 km to metres.
tan 10° =
Calculate. = 353 m
Give a written answer. The altitude of the aeroplane is approximately 353 m.
10°
2 km
h
1
10°
2 km
h
adj
hyp
opp
2
3
opp
adj
---------
4
h
2000
------------
5
6
7
22WORKEDExample

Angles of elevation and depression can also be calculated by using known measure-
ments. This is done by drawing a right-angled triangle to represent a situation.
Angles of elevation and
depression
1 From a point 100 m from the foot of a building, the angle
of elevation to the top of the building is 15°.
Calculate the height of the building, correct to
1 decimal place.
2 From a ship the angle of elevation to an aeroplane is 60°.
The aeroplane is 2300 m due north of the ship.
Calculate the altitude of the aeroplane, correct to the
nearest metre.
A 5.2-m building casts a 3.6-m shadow.
Calculate the angle of elevation of the sun, correct to the nearest degree.
THINK WRITE
Choose the tan ratio because we
are given the opposite and
adjacent sides.
Substitute for opposite and adjacent. tan θ =
Calculate. = 55°
Give a written answer. The angle of elevation of the sun is approximately 55°.
θ
3.6 m
5.2 m
1
θ
3.6 m
adj
hyp
opp
5.2 m
2
3
opp
adj
---------
4
5.2
3.6
-------
5
5.2
3.6
-------
6
7
23WORKEDExample
remember
1. The angle of elevation is the angle at which you look up to see an object.
2. The angle of depression is the angle at which you look down to see an object.
3. Problems can be solved by using angles of elevation and depression with the
aid of a diagram.
4. Written problems should be given an answer in words.
remember
5F
WORKED
Example
20
100 m
15°
60°
2300 m

3 From a point out to sea, a ship sights the top of a lighthouse
at an angle of elevation of 12°. It is known that the top of the
lighthouse is 40 m above sea level.
Calculate the distance of the ship from the lighthouse,
correct to the nearest 10 m.
4 From a point 50 m from the foot of a building, Rod sights
the top of a building at an angle of elevation of 37°.
Given that Rod’s eyes are at a height of 1.5 m, calculate
the height of the building, correct to 1 decimal place.
5 Richard is flying a kite and sights the kite at an angle of
elevation of 65°. The altitude of the kite is 40 m and
Richard’s eyes are at a height of 1.8 m.
Calculate the length of string the kite is flying on, correct to
1 decimal place.
6 Bettina is standing on top of a cliff, 70 m above
sea level. She looks directly out to sea and sights a
ship at an angle of depression of 35°.
Calculate the distance of the ship from shore.
7 From an aeroplane flying at an altitude of
4000 m, the runway is sighted at an angle of
depression of 15°.
Calculate the distance of the aeroplane from the
runway, correct to the nearest kilometre.
8 There is a fire on the fifth floor of a building. The closest
a fire truck can get to the building is 10 m. The angle of
elevation from this point to where people need to be rescued
is 69°.
If the fire truck has a 30-m ladder, can the ladder be used to
make the rescue?
9 From a navy vessel, a beacon which is 80 m
above sea level is sighted at an angle of elevation
of 5°. The vessel sailed towards the beacon and
thirty minutes later the beacon is at an angle of
elevation of 60°.
Use the diagram at right to complete the following.
a Calculate the distance that the vessel was from the beacon, when the angle of
elevation to the beacon was 5° (the distance AC).
b Calculate the distance that the vessel sailed in the 30 minutes between the two
readings.
10 A 12 m high building casts a shadow 15 m long.
Calculate the angle of elevation of the sun, to the nearest
degree.
40 m
12°
x
WORKED
Example
21
1.5 m 50 m
37°
40 m
1.8 m
65°
x
WORKED
Example
22
35°
70 m
15°
4000 m
69°
10 m
5° 60°
80 m
A
D
C
B
WORKED
Example
23
θ
12 m
15 m

11 An aeroplane which is at an altitude of 1500 m is
4000 m from a ship in a horizontal direction, as
shown at right.
Calculate the angle of depression from the
aeroplane to the ship.
12 The angle of elevation to the top
of a tower is 12° from a point
400 m from the foot of the tower.
b Calculate the height of the tower,
correct to 1 decimal place.
c Calculate the angle of elevation
to the top of the tower, from
a point 100 m from the foot of
the tower.
Alternative method to determine the
height of a tall object
Resources: Measuring tape, clinometer or protractor, calculator.
In a previous investigation, you determined the height of a tall object using a
shadow stick, observing the lengths of shadows. An alternative method could be
employed using a clinometer. This is a device which measures angles. Your school
may have one. If not, a protractor could be used as a substitute (although the
accuracy of the reading would not be as great).
1 Locate a tall building or tree whose height you wish to determine.
2 Take a position some distance from its base, and measure your horizontal
distance from the base.
3 Use the measuring tape to measure your height at eye level.
4 Take the clinometer (or protractor) and measure the angle of elevation from
your eye level to the top of the tall object.
5 Draw a diagram representing the situation, labelling known lengths and angles.
6 Use a trigonometric ratio to determine the height of your tall building (or tree).
(Remember to allow for your height at eye level in your calculations.)
7 Write up this activity, showing detailed diagrams and procedures used to obtain
the height of your tall object.
θ
1500 m
4000 m
Work
SHEET 5.2
inv
estigat
ioninv
estigat
ion

Revisiting staircases
Resources: Pen, paper (graph paper if possible), calculator.
The mathematics used in constructing a staircase is more complex than you may at
first suspect. Earlier in this chapter we looked at a simple staircase. Let us now
examine the mathematics in more detail. The Home Building Code lays down
construction standards for staircases. As a starting point, it is necessary to
understand some key terms.
Rise and going
Rise and going refer to the staircase treads, and their arrangement and construction.
Rise
The rise is the vertical measurement from the upper surface of one tread to the
upper surface of the next tread.
Rise of flight
The rise of flight is the vertical measurement taken from a floor or landing to the
top of the next floor or landing; that is, the height of the staircase.
Going
The term going is the horizontal measurement from the front edge of one tread to
the front edge of the next.
Going of flight
The going of flight is the horizontal measurement for one flight, taken from the face
of the first or bottom step to the face of the last or top step in that flight.
The figure below shows rise (R) and going (G) labelled in a staircase.
The home building code sets the following maximums and minimums:
It also states that the sum of two times the rise plus the going must be a
maximum of 630 and a minimum of 585. Using R to represent the rise and G to
represent going, this could be written as:
2R + G = 585 to 630 mm
The building code also states that rise and going must be constant for any one
staircase. (Imagine walking up a staircase that had stair heights of varying sizes!)
inv
estigat
ioninv
estigat
ion
Rise R
Rise of flight
Going of flight
R
Going G
G
R G
R G
R
Max. (mm) Min. (mm)
Rise 190 115
Going 395 250
MQ Maths A Yr 11 - 05 Page 201 Friday, July 6, 2001 8:50 AM

The following figure shows a staircase with 9 rises. The number of treads used is
8, one less than the number of rises, because the top floor acts as the final step.
This staircase has 8 equal goings.
Task 1
A staircase is to have a rise of flight of 2844 mm. The builder wants to use a rise of
approximately 160 mm.
1 Within the building code regulations, determine:
a the number of rises
b an appropriate rise (R) measurement
c a suitable going (G) measurement.
2 Draw a diagram showing the design of the staircase. Include all measurements.
Provide calculations to confirm that your staircase conforms to the home
building code.
Task 2
Often, there are restrictions on the space available for the location of a staircase,
and the going must be kept to a minimum.
Tony and Mary are renovating their home and wish to install a staircase. The rise
of flight is 2608 mm and the restricted going of flight is 3945 mm.
1 Find a suitable rise, if a rise of approximately 160 mm is desired.
2 State the number of treads.
3 Determine the going.
4 Calculate the number of rises.
5 Check that the dimensions are in accordance with the building code.
6 Draw a diagram of your completed staircase, indicating all measurements.
Going
Going of flight
Rise of flight
Existing
top floor
Rise

• When using Pythagoras’ theorem to calculate the length of the hypotenuse of a
right-angled triangle, the formula is:
hypotenuse2
= base2
+ height2
• To find one of the shorter sides of a right-angled triangle, the formula is:
base2
= hypotenuse2
− height2
or height2
= hypotenuse2
− base2
• Pythagoras’ theorem can be used to test whether a triangle is right-angled.
If hypotenuse2
= base2
+ height2
, the triangle is right-angled.
If hypotenuse2
> base2
+ height2
, the triangle is obtuse-angled.
If hypotenuse2
< base2
+ height2
, the triangle is acute-angled.
• Pythagorean triads or Pythagorean triples are sets of three numbers that satisfy
Pythagoras’ theorem.
Shadow sticks
• Shadow sticks can be used to determine the height of objects too difficult to measure.
• Height of the object = height of shadow stick × scale factor
Trigonometry formulas for right-angled triangles
• tan θ =
• sin θ =
• cos θ =
• SOHCAHTOA — this acronym will help you remember trig formulas.
Steps to find a side of a right-angled triangle
• Label the sides of the triangle opposite, adjacent and hypotenuse.
• Choose the correct ratio.
• Substitute given information.
• Make the unknown side the subject of the equation.
• Calculate.
Steps to find an angle in a right-angled triangle
• Label the sides of the triangle opposite, adjacent and hypotenuse.
• Choose the correct ratio.
• Substitute given information.
• Make the unknown angle the subject of the equation.
• Calculate by using an inverse trig function.
Angles of elevation and depression
• The angle of elevation is the angle we look up from the horizontal to see an object.
• The angle of depression is the angle we look down from the horizontal to see an
object.
• Problems are solved using angles of elevation and depression by the same methods
as for all right-angled triangles.
summary
opp
adj
---------
opp
hyp
---------
adj
hyp
---------

1 Find the length of the side marked with a pronumeral, in each case writing your answer
a b c
d e f
2 To travel between the towns of Bolong and Molong, you need to travel west along a road for
45 km, then north along another road for another 87 km. Calculate the straight-line distance
between the two towns.
3 A rope is 80 m long and runs from a cliff top to the ground, 45 m from the base of the cliff.
Calculate the height of the cliff, to the nearest metre.
4 Classify the following triangles as acute, obtuse or right-angled.
a b
c d
5 Which of the following are Pythagorean triads?
a 9, 39, 40 b 3, 4, 5
c 0.3, 0.4, 0.5 d 6, 7, 10
6 A ﬂagpole casts a shadow of 15 metres at the same time as a 1.5-metre stick casts a shadow
of 2 metres. How high is the ﬂagpole?
7 Calculate each of the following, correct to 4 decimal places.
8 Calculate θ, correct to the nearest degree, given that:
a sin 46° b tan 76°42′ c 4.9 cos 56°
d 8.9 sin 67°3′ e f
a cos θ = 0.5874 b tan θ = 1.23 c sin θ = 0.8.
CHAPTER
review
5A
9.2 m
9.2 m
m
32 cm
26 cm
n
p4.8 m
3.2 m
q
17.25 cm
7.25 cm
r
1.9 km
1.3 km t
0.6 m
2.4 m
5A
5A
5A
5 cm
13 cm
12 cm
3 cm
5 cm
4 cm
7 m
19 m
9 m
8 cm
16 cm
15 cm
5A
5B
5C
5.69
cos 75°
------------------
2.5
tan 9°55′
----------------------
5C

9 Calculate θ, correct to the nearest minute, given that:
10 Find the length of each side marked with a pronumeral, correct to 1 decimal place.
a b c
d e f
g h i
j k l
11 A rope that is used to support a flagpole makes an angle of 70° with the ground. If the rope
is tied down 3.1 m from the foot of the flagpole, find the height of the flagpole, correct to 1
decimal place.
a cos θ = 0.199 b tan θ = 0.5 c sin θ = 0.257. 5C
5D
q9°
6 cm
x78°
3.9 m m
22°
12.6 cm
n
22°
12.6 cm
q
7.8 cm
32°
t
6.8 m
65°
g
26°42'
2.9 m
h
77°18'
4.8 cm
z 83°30' 138 mm
j
4.32 m
29°51'
k
38.5 m
16°8'
m
63 km
85°12'
5D

12 A dirt track runs off a road at an angle of 34° to the road. If I travel for 4.5 km along the dirt
track, what is the shortest direct distance back to the road (correct to 1 decimal place)?
13 A fire is burning in a building and people need to be rescued. The fire brigade’s ladder must
reach a height of 60 m and must be angled at 70° to the horizontal. How long must the
ladder be to complete the rescue?
14 Find the size of the angle marked θ in each of the following, giving your answer correct to
the nearest degree.
a b c
15 Find the size of the angle marked θ in each of the following, giving your answer correct to
the nearest minute.
a b c
16 A kite on an 80-m string reaches a height of 50 m in a strong wind. Calculate the angle the
string makes with the horizontal.
17 There is 50 m of line on a fishing reel.
When all the line is out, the bait sits on
the bed of a lake and has drifted 20 m
from the boat. Calculate the angle that
the fishing line makes with the vertical.
18 The top of a building is sighted at an angle of elevation of 40°, when an
observer is 27 m back from the base.
Calculate the height of the building, correct to the nearest metre.
19 Hakam stands 50 m back from the foot of an 80-m
telephone tower. Hakam’s eyes are at a height of 1.57 m.
Calculate the angle of elevation that Hakam must look to see
the top of the tower.
5D
5D
5E
θ
19 m16 m
θ
4.6 m
2.3 m θ 116 cm
43 cm
5E
θ
10.8 m
4.6 m
θ
2.9 m
6.1 m
θ
11.9 cm
13.8 cm
5E
5E
5F
40°
27 m
h
testtest
CHAPTER
yyourselfourself
testyyourselfourself
5
5F
θ
1.57 m 50 m
80 m

Maths A - Chapter 5

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Maths A - Chapter 5