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Mathematics
Worksheet
Trigonometry &
Pythagoras’ Theorem
This is one of a series of worksheets designed to help you increase your confidence in
handling Mathematics. This worksheet contains both theory and exercises which cover:1.
2.
3.
4.
5.
Pythagoras’ Theorem
Introduction to trigonometry
Using trigonometry to find an unknown side
Using trigonometry to find an unknown angle
Trigonometric diagrams and identities
There are often different ways of doing things in Mathematics and the methods suggested in
the worksheets may not be the ones you were taught. If you are successful and happy with
the methods you use it may not be necessary for you to change them. If you have problems
or need help in any part of the work then there are a number of ways you can get help.
For students at the University of Hull
 Ask your lecturers



You can contact a Mathematics Tutor from the Study Advice Service on the ground
floor of the Brynmor Jones Library if you are in Hull or in the Keith Donaldson Library if
you are in Scarborough; you can also contact us by email
Come to a Drop-In session organised for your department
Look at one of the many textbooks in the library.
For others
 Ask your lecturers
 Access your Study Advice or Maths Help Service
 Use any other facilities that may be available.
If you do find anything you may think is incorrect (in the text or answers) or want further help
please contact us by email.
Tel:
01482 466199
Web: www.hull.ac.uk/studyadvice
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1. Pythagoras’ Theorem
Pythagoras’ Theorem is used to find the lengths of unknown sides in triangles. It can be used in
the following conditions:
1. The triangle is a right-angled triangle, i.e. contains an angle of 90 degrees.
2. Two of the three sides are already known.
Two main uses of Pythagoras’ Theorem are converting a vector to magnitude and direction form
and finding the resultant force given a force in the horizontal and vertical directions.
The rule is:
a2  b2  c2
In words:
The square of the hypotenuse is equal to the sum of the
squares of the other two sides.
Note that this is sometimes expressed as a 2  b2  h2 .
Here a and b are the two shorter sides of the triangle - the ones which are
attached to the right-angle. c or h is the hypotenuse, the longest side;
the side that lies opposite the right-angle.
c
b
a
Examples
1 Finding the hypotenuse
If a  3, b  7 find c .
c
7
3
Substitute the known values into Pythagoras’ Theorem: 32  72  c2
Evaluate the Left Hand Side (LHS) 9  49  c2  58  c2 .
Take the positive square root of both sides:
This can now be left as
58  c
58  c or a calculator may be used to find
58  7.616 to 3 d.p.
2 Finding a shorter side
If a  5, c  13 find b .
13
b
5
Substitute the known values into Pythagoras’ Theorem: 52  b2  132
This becomes 25  b2  169 . Subtract 25 from both sides to get b 2 on its own:
b2  169  25  144 . Take the positive square root of both sides: b  144  12 .
So b  12 .
Pythagoras’ Theorem is often used to find the length of vectors. The theorem can be extended to 3
dimensions by squaring all 3 components and adding, then square rooting. For more information
on this, please refer to Vectors 1, available from www.hull.ac.uk/studyadvice.
Exercise 1
1 For each of the following triangles find the length of the hypotenuse:
a)
b)
c)
c
c
c
12
6
8
4
7
9
2 For each of the following triangles find the length of the unknown side:
a)
b)
c)
12
20
4
13
b
c
a
b
6
7
2. Introduction to trigonometry
Basic trigonometry uses the rules sine, cosine and tangent. These functions are actually infinite
series and would prove very difficult and time-consuming to calculate to a reasonable degree of
accuracy by hand. Fortunately scientific calculators are able to deal with these functions.
The most common use of sine, cosine and tangent is with right-angled triangles. They are used to
find unknown sides and angles.
These functions are reliant on either knowing an angle and a side or the lengths of two sides.
The formulae for sine, cosine and tangent are:
sin  
opposite
hypotenuse
cos  
adjacent
hypotenuse
tan  
opposite
adjacent
Where  is used to denote the angle of interest, and sin , for example, is the value of the sine
function acting on  .
These rules are often remembered by writing down
SOH
CAH
TOA
You may find it useful to include a ‘/’ between the second and third letters in each row to remind
you of the division.
Hypotenuse, adjacent and opposite refer to the lengths of the sides of the triangle. Note that whilst
the position of the hypotenuse is fixed (it is always the side opposite the right angle), the positions
of the opposite and adjacent sides are dependant on the location of the angle that is being used.
Hypotenuse
Hypotenuse
Opposite
Angle
Adjacent
Angle
Adjacent
Opposite
The position of the angle of interest determines the labels on the sides.
3. Using trigonometry to find an unknown side
Trigonometry can be used to find an unknown side of a triangle when you know only one angle and
the length of one side.
Given a right-angled triangle such as:
c
5
30
How can the length of side c be found?
An angle and a side are known. Look for a trigonometric formula which includes both the known
side and the unknown side.
opposite
includes all of the necessary information as the side opposite the angle is
hypotenuse
known and the hypotenuse is the side that is to be determined.
sin 
Substituting the values in gives: sin 30  5 .
c
5
Rearrange: c  sin 30  5  c 
sin 30
(for help with rearranging equations see Algebra 3)
All that remains is to substitute in the value of sin 30 (found via your calculator) and work out the
value of the fraction…
c  5  5  10 .
sin 30 0.5
Hence the length of side c is 10.
Another example:
12
45
b
Here we use cosine, as the side we need is adjacent to the known angle, and the hypotenuse is
known.
So cos 45  b , rearranging, 12  cos 45  b  12  1  12  8.49
12
2
2
Notes:
Remember to check that your calculator is in degrees if using degrees or in radians if using
radians. This can normally be altered via the mode button.
Depending on the make and age of the calculator being used it may be necessary to type in either
sin 30 or 30 sin to get the value of sin 30 .
Always work with the numbers as they are shown on the calculator screen until the final result is
produced. Then this figure can be rounded. Rounding figures part-way through the calculation will
result in a less accurate answer.
Exercise 2
1 Find the lengths of the missing sides in the following triangles:
a)
b)
c)
35
11
60
b
e
c
d
a
8
75
10
f
4. Using trigonometry to find an unknown angle
Trigonometry can be used to find an unknown angle of a right-angled triangle when you know only
the length of two sides.
Given a right-angled triangle such as:
12
5

How do we find the size of angle  ?
We can use the same formulae as we have been using for finding unknown sides.
In the above triangle the sides that are known are, in relation to  , the opposite side and the
hypotenuse.
So, use a formula which uses both the opposite side and the hypotenuse.
sin 
opposite
can be used here.
hypotenuse
Substituting in the values of the known sides gives: sin  5  0.42 .
12
To get from sin  0.42 to a value for  you need the inverse sine operation.
This is on most calculators as sin1 and is usually accessed using a 2nd function or shift key, then
the sin key.
sin1( 5 )  24.62 to 2 d.p. So   24.62
12
Note that I used 5 in the calculation rather than 0.42. This is because 0.42 is rounded and so is
12
less accurate than 5 .
12
Another example:

16
6
Here the known sides are the adjacent and the hypotenuse, so we use
cosine.
cos   6  0.375 .
16
So,   cos1(0.375)  67.98 to 2 d.p.
Notes:
Remember to check that your calculator is in degrees if you are using degrees or in radians if you
are using radians. This can normally be altered via the mode button.
Depending on your calculator you may need to type in either 2nd/shift sin 30 or 30 2nd/shift sin to
get the value of sin1 30 .
Exercise 3
Find the size of the marked angles
a)
b)
c)
4
3
9
11
6
8
5. Trigonometric diagrams and identities
Diagrams
There are 2 diagrams that can be memorised in order to recall certain values of sin, cos and tan
quickly.
The first is a right-angled isosceles triangle with two sides of length 1.
Because this is an Isosceles triangle, it has 2 angles the same.
The size of these angles is 180  90  45
1
2
1
Hence this triangle will provide us with the values for cos 45, sin 45, and tan 45. Using
Pythagoras’ Theorem, we find that the length of the hypotenuse is equal to
Using the formulae for sin, cos and tan on either angle, we can now find that:
cos 45  1
2
sin 45  1
You may wish to confirm these answers yourself.
2
tan 45  1  1
1
12  12  2 .
The second diagram is used to find sin, cos and tan for angles of 30 and 60
It is half of an equilateral triangle with sides of length 2.
2
30
2
2
60
2
1
This gives us angles of 30 and 60 as the angles of an equilateral triangle are all 60 . Here we
know the length of the hypotenuse, but one of the shorter sides is unknown. Using Pythagoras’
Theorem we get b 2  22  12  4  1  3 , hence the length of the missing side is
3.
Using the formulae for sin, cos and tan, we can now find that:
cos 30  3
2
tan 30  1
3
sin 30  1
2
and
cos 60  1
2
sin 60  3
2
tan 60  3  3
1
Again, you may wish to check these.
Notes:
If you have 2 angles that add up to 90 , then the sine of the first will equal the cosine of the
second, for example see sin 60 and cos 30 .
It is best to leave in the ‘surds’ or square-root signs. Results such as
3
are exact answers,
2
rounding them off will only make them less accurate.
Identities
There are a number of identities in trigonometry. Identities are facts that will always be true, no
matter whether numbers (or in this case angles) are changed. These can be used to simplify long
algebraic arguments.
The one you are most likely to encounter is:
sin2   cos2   1
This can be seen using a diagram:
If you are given a right-angled triangle with hypotenuse length 1, then the
adjacent side can be written as hyp  cos by rearranging the formula for
cosine and the opposite side as hyp  sin by rearranging the sine rule.
Since the hypotenuse in this case is 1, these simplify to cos and sin .

Using Pythagoras’ Theorem, the squares of these sides must sum to give
the square of the hypotenuse, hence
1
sin2   cos2   1
Other identities which may be of use are:
sin( A  B )  sin A cos B  cos A sin B
cos( A  B )  cos A cos B sin A sin B (Note the
sign)
cos 2 A  cos2 A  sin2 A
sin 2 A  2 sin Acos A
For further identities, please refer to ‘a useful guide to facts and formulae’ produced by
Loughborough University, copies of which are available (free) from the Study Advice Desk.
Answers
Exercise 1
1. a) 10
b) 13.89 to 2 d.p.
c) 9.85 to 2 d.p.
2. a) 11.31 to 2 d.p. b) 19.08 to 2 d.p. c) 10.95 to 2 d.p.
Exercise 2
1. a  9.53 to 2 d.p., b  5.5
c  9.77 to 2 d.p., d  5.60 to 2 d.p.
e  10.35 to 2 d.p., f  2.68 to 2.d.p.
Exercise 3
a)   69.44 b)   48.19 c)   54.90 (all to 2 d.p.)
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