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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. Pythagoras’ Theorem


2. Introduction to trigonometry

3. Using trigonometry to find an unknown side


4. Using trigonometry to find an unknown angle


5. Trigonom
etric 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 Stu
dy Advice Servi
ce
i
n the Brynmor
Jones Library

if you are in Hull;

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

Email: studyadvice@hull.ac.uk



Mathematics

Worksheet



Trigonometry &
Pythagoras’ Theorem


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
-
angle
d 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:


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
.





Here

and

are the two shorter sides of the triangle

-

the ones which are



attached to the right
-
angle.


or

is t
he hypotenuse, the longest side;





the

side tha
t lies opposite the right
-
angle.




E
xamples


1 Finding the hypotenuse





I
f

find
.







Substitute the known values into

P
ythagoras’
T
heorem:


Evaluate the Left Hand Side

(LHS)
.

Take the
positive
square root of both sides:

This can now be left as


or
a

calculator

may be used

to find

to 3 d.p.



2 Finding a shorte
r side



If

find
.








Substitute the known values into Pythagoras’ Theorem:

This becomes
.

Subtract 25 from both sides to
get

on

its own:
.

Take the positive square root of both sides:
.

So
.


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
.

a

b

c

3

7

c

5

b

13

Exercise 1

1 For each of
the following triangles find the length of the hypotenuse:

a)



b)




c)





2 For each of the following triangles find the length of the unknown side:

a)



b)



c)





2. Introduction to trigonometry


Basic trigonometry uses the rules sine, cosin
e 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 commo
n 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:






Where

is used to denote the angle of interest, and
, for example, is the value of the sine
function acting o
n
.


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 refe
r 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 be
ing used.







The position of the angle of interest determines the labels on the sides.





c

6

8

7

12

c

c

4

9

13

b

7

20
c

b

6

12

4

a

Hypotenuse

Opposite

Adjacent

Angle

Hypotenuse

Opposite

Adjacent

Angle

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 an
d
the length of one side.



Given a right
-
angled triangle such as:



How can
the length of side c

be found
?


A
n angle and a side

are known
. Look for a trig
onometric

formula which includes both the known
side and the unknown side.



includes
all of the necessary information

as the side opposite the angle is
known and the hypotenuse is the side that
is to be determined
.


Substituting the values in gives:
.

Rearrange:


(for h
elp with rearranging equations see Algebra 3)


All that remains is to substitute in the value of
(found via your calculator)
and work out the
value of the fraction…

.

Hence the length of side c is 10.



Another example:












Here we use cosine, as the side we need is
a
djacent to

the known angle, and the
hypotenuse
is
known.



So
, rearranging
,



Notes:

Remember to c
heck that
your

calculator is in degrees if using degrees or in radians if using
radians. This can normally be altered via the mode button.



5

c

1
2

b


Depending on
the make and age of the
calculator

being used

it may be necessary to type in

either

30

or
30


to get the value of
.



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)







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:






How do we find the size of angle
?


We can use the same formulae as w
e have been using for finding unknown sides.


In the above triangle t
he 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.



can be used here.

Substituting in the values of the known sides gives:
.

To get from

to a value for

you need the inverse sine operation.

This is on

most calculators as
and is usually accessed using a 2
nd

function or shift key, then
the sin key.


to 2 d.p. So

Note that I used
in the calculatio
n rather than 0.42. This is because 0.42 is rounded and so is
less accurate than
.






12


5




11

a

b

d

c

8

10

e

f

Another example:



Here the known sides are the
a
djacent and

the hypotenuse, so we use

cosine.


.


So,

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 ty
pe in either
2
nd
/shift

sin

30

or
30

2
nd
/shift

sin

to
get the value of
.



Exercise 3


Find the size of the marked angles


a)





b)




c)










5. Trigonometric diagrams and identities


Diagrams

There are 2 diagra
ms 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 Iso
sceles triangle, it has 2 an
gles the same.




The size of these
angles is




Hence this triangle will provide us with the values for


and

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:






You may wish to confirm these answers yourself.






6

16

1

1

8

3

6

4

9

11

The second diagram is used to find sin, cos and tan for angles of
and

It is half of an equilateral triangle with sides of length 2.








This gives us angles of

and

as the angles of an equilateral triangle are all
. Here we
know the length of the hypotenuse, but one of the shorter sides is unknown. Using Pythagoras’
Theorem we get
, hence t
he length of the missing side is
.


Using the formulae for sin, cos and tan, we can now find that:




and





Again, you may wish to check these.


Notes:

If you have 2 angles that add up to
, then the sine of the first will equal the cosine of the
second, for example see
and
.

It is best to leave in the ‘surds’ or square
-
root signs. Results such as


are exact answers,
rounding them off will only make them less accurate.



Identities

There are a number of identitie
s 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:




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

by rearranging the formula for

cosine and the opposite side as

by rearranging the sine rule.

Since the hypotenuse in this case is 1, these simplify to

and
.

Using Pythagoras’ Theorem, the squares of these sides must sum to give
the square of th
e hypotenuse, hence



Other identities which may be of use are:




(Note the

sign)


2

2

2

1

2



1





For fu
rther 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.

to 2 d.p.,


to 2 d.p.,

to 2 d.p.


to 2
d.p.,

to 2.d.p.


Exercise 3

a)

b)

c)

(all to 2 d.p.)





















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