5
syllabus
syllabus
rref
efer
erence
ence
Strand:
Applied geometry
Core topic:
Elements of applied geometry
In this
In this
cha
chapter
pter
5A Pythagoras’ theorem
5B Shadow sticks
5C Calculating trigonometric
ratios
5D Finding an unknown side
5E Finding angles
5F Angles of elevation and
depression
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.
History
of mathematics
P Y T H A G O R A S O F S A M O S ( c i r c 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
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
E D
F
H
G
I
K
J L
6 m
8 m a
6 m
4 m b
20 cm
a
34°
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
R 1
2
1
WORKED
E
xample
A = 9 cm2
A = 25 cm2
C B
A
3 cm 4 cm
5 cm
A = 16 cm2 Pythagoras
calculations
G Cpro
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
2
WORKED
E
xample
EXCEL Spreadsheet 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
Q P
R
16 m
9 m
r 1
2
3
4 175
3
WORKED
E
xample
MathcadBecause 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
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
5
WORKED
E
xample
EXCEL Spreadsheet 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
Pythagoras’ theorem
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
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
m
multiple choiceultiple choice
X
Y Z
m
multiple choiceultiple choice
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
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
m
multiple choiceultiple choice
m
multiple choiceultiple choice
WORKED
Example
6
3.5 m
1.3 m
7.9 m
Pythagoras’ theorem
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?
inve stigatio
n
in
ve
sti
ga
ti
o
n
3 cm
4 cm 5 cm
inve stigatio
n
in
ve
sti
ga
ti
o
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
inve stigatio
n
in
ve
sti
ga
ti
o
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.
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
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
---inve stigatio
n
in
ve
sti
ga
ti
o
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
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
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
y
1.1 m
5 m 10 m
0.9 m
m
multiple choiceultiple choice
m
multiple choiceultiple choice
Work
SHEET
5.1
hypotenuse opposite
adjacent
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?
inve stigatio
n
in
ve
sti
ga
ti
o
n
A B
C E G I
D F H
Triangle
Opposite side length
Adjacent side length
ABC ADE AFG AHI
Opposite Adjacent
---tan θ opposite side adjacent side
---=
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
Resources: Ruler, spreadsheet.
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 =
8
WORKED
E
xample
inve stigatio
n
in
ve
sti
ga
ti
o
n
A B
C E G I
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
=
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
Resources: Ruler, spreadsheet.
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.
inve stigatio
n
in
ve
sti
ga
ti
o
n
A B
C E G I
D F H
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′
---10
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
11
WORKED
E
xample
° ’ ”
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
12
WORKED
E
xample
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
Calculating trigonometric
ratios
1 Calculate the value of each of the following, correct to 3 decimal places.
2 Calculate the value of each of the following, correct to 3 decimal places.
3 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.
5 Calculate the value of each of the following, correct to 2 decimal places.
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°
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
13
WORKED
E
xample
Cabri Geome try Trig ratios
opposite side adjacent side
---opposite 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
14
WORKED
E
xample
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
---15
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=
Calculate. = 13.60 m
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=
Calculate. = 103.5 m
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
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
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 (correct to 1 decimal place).
a b c
7 Find the length of the side marked with the pronumeral in each of the following (correct to 1 decimal place).
a b c
d e f
g h i
j k l
WORKED
Example
14
68° 13 cm x
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 km b
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
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°
m
multiple choiceultiple choice
69° 9.2 x x 9.2
sin 69°
---=
x 9.2
cos 69°
---=
m
multiple choiceultiple choice
17 15 8 φ 8 15 --- 15 8 --- 15 17 --- 8 17 ---m
multiple choiceultiple choice
m
multiple choiceultiple choice
36° 22 mm w
w 22
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.
a Draw a diagram of this situation.
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 m d
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°
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
θ
opp 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
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
hyp adj
4.6 cm
7.1 cm
θ
opp hyp
---2 4.6
7.1
---3 4.6
7.1
---
4
18
WORKED
E
xample
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. = Make θ the subject of the equation. θ= tan−1
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
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 m 4.6 m
α
27.8 cm
19.5 cm
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 θ 32 mm 14.9 m
26.8 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 8.3 m 16.3 m
θ
6.3 m
18.9 m
θ
m
multiple choiceultiple choice
10 cm 5 cm
A
B
C θ
m
multiple choiceultiple choice
12.5 m 9.3 m
θ
m
multiple choiceultiple choice
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
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°,
correct to 4 decimal places.
6 Calculate 9.2 tan 50°, correct to 2 decimal places.
7 Calculate , correct to 2 decimal places.
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
---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° =
Make h the subject of the equation. h= 50 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
opph
adj 2
3 opp
adj
---4 h
50
---5 6 7
20
WORKED
E
xample
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°
A similar method for finding the solution is used for problems that involve an angle of depression.
THINK WRITE
Label the sides 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 64° =
Make h the subject of the equation. h= 10 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 opph
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.
Write the formula. tan θ=
Substitute for θ and the adjacent side, converting 2 km to metres.
tan 10° =
Make h the subject of the equation. h= 2000 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
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
Label the sides opp, adj and hyp. Choose the tan ratio because we are given the opposite and adjacent sides.
Write the formula. tan θ=
Substitute for opposite and adjacent. tan θ =
Make θ the subject of the equation. θ= tan−1
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
23
WORKED
E
xample
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°
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 a
