Download Trigonometry and Pythagoras Theorem

Trigonometry & Pythagoras Theorem
Mathematics Skills Guide
This is one of a series of guides designed to help you increase your confidence in handling
Mathematics. This guide 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 guides 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 maths Skills Adviser from the Skills Team on the email shown
below.
 Access more maths Skills Guides and resources at the website below.
 Look at one of the many textbooks in the library.
Web: www.hull.ac.uk/skills
Email: [email protected]
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:
a 2  b2  c 2
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  h 2 .
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: 58  c
This can now be left as
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
http://libguides.hull.ac.uk/skills
.
Exercise 1
1 For each of the following triangles find the length of the hypotenuse:
a)
b)
c)
c
c
c
6
12
8
4
7
9
2 For each of the following triangles find the length of the unknown side:
a)
b)
c)
12
20
13
4
a
b
c
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
Angle
Opposite
Angle
Adjacent
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.
sin 
opposite
includes all of the necessary information as the side opposite the angle
hypotenuse
is known and the hypotenuse is the side that is to be determined.
Substituting the values in gives: sin 30  5 .
c
Rearrange: c  sin 30  5  c 
5
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
Hence the length of side c is 10.
Another example:
12
45
b
0.5
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
75
10
f
8
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
12
so is less accurate than 5 .
12
Another example:
use

16
Here the known sides are the adjacent and
cosine.
cos   6  0.375 .
the hypotenuse, so we
16
6
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
12  12  2 .
Using the formulae for sin, cos and tan on either angle, we can now find that:
cos 45  1
sin 45  1
2
2
tan 45  1  1
1
You may wish to confirm these answers yourself.
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 b2  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:
tan 30  1
sin 30  1
cos 30  3
2
3
2
and
cos 60  1
sin 60  3
tan 60  3  3
2
2
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
2
are exact
answers, 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 A cos A
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.)
We would appreciate your comments on this worksheet, especially if you’ve
found any errors, so that we can improve it for future use. Please contact
the Maths Skills Adviser by email at [email protected]
The information in this leaflet can be made available in an alternative format on request
using the email address above.