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- 3. 3-3
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Copyright Notice
© Copyright. All worldwide rights reserved. No part of this publication may be reproduced,
stored in a retrieval system or transmitted in any form by any other means whatsoever: i.e.
photocopy, electronic, mechanical recording or otherwise without the prior written permission of
ST Aerospace Ltd.
Knowledge Levels — Category A, B1, B2, B3 and C Aircraft
Maintenance Licence
Basic knowledge for categories A, B1, B2 and B3 are indicated by the allocation of knowledge levels indicators (1,
2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2
basic knowledge levels.
The knowledge level indicators are defined as follows:
LEVEL 1
A familiarisation with the principal elements of the subject.
Objectives: The applicant should be familiar with the basic elements of the subject.
The applicant should be able to give a simple description of the whole subject, using common words and
examples.
The applicant should be able to use typical terms.
LEVEL 2
A general knowledge of the theoretical and practical aspects of the subject.
An ability to apply that knowledge.
Objectives: The applicant should be able to understand the theoretical fundamentals of the subject.
The applicant should be able to give a general description of the subject using, as appropriate, typical
examples.
The applicant should be able to use mathematical formulae in conjunction with physical laws describing the
subject.
The applicant should be able to read and understand sketches, drawings and schematics describing the
subject.
The applicant should be able to apply his knowledge in a practical manner using detailed procedures.
LEVEL 3
A detailed knowledge of the theoretical and practical aspects of the subject.
A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive
manner.
Objectives: The applicant should know the theory of the subject and interrelationships with other subjects.
The applicant should be able to give a detailed description of the subject using theoretical fundamentals
and specific examples.
The applicant should understand and be able to use mathematical formulae related to the subject.
The applicant should be able to read, understand and prepare sketches, simple drawings and schematics
describing the subject.
The applicant should be able to apply his knowledge in a practical manner using manufacturer's
instructions.
The applicant should be able to interpret results from various sources and measurements and apply
corrective action where appropriate.
- 5. 3-5
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Table of Contents
Module 1.3 Geometry ________________________________________________________9
Trigonometry _____________________________________________________________9
Trigonometrical Relationships _______________________________________________9
The Sine Curve _________________________________________________________14
The Cosine Curve _______________________________________________________14
The Tan Curve__________________________________________________________15
Other Trigonometric Functions _____________________________________________16
To Find the Length of an Unknown Side ______________________________________17
Coordinates and Graphs___________________________________________________25
The x and y Axis ________________________________________________________25
Graphical Representations of an Equation ____________________________________31
The Straight Line ________________________________________________________43
Derivation of the Equation y = mx + c ________________________________________49
Cartesian and Polar Coordinates____________________________________________55
Cartesian Coordinates ____________________________________________________55
Polar Coordinates _______________________________________________________55
Converting _____________________________________________________________56
- 7. 3-7
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Module 1.3 Enabling Objectives and Certification Statement
Certification Statement
These Study Notes comply with the syllabus of EASA Regulation (EC) No.2042/2003 Annex III
(Part-66) Appendix I, as amended by Regulation (EC) No.1149/2011, and the associated
Knowledge Levels as specified below:
Objective
Part-66
Reference
Licence Category
B1 B2
Geometry 1.3
(a) 1 1
Simple geometrical constructions
(b) 2 2
Graphical representations; nature and uses of
graphs, graphs of equations/functions
(c) 2 2
Simple trigonometry; trigonometrical
relationships, use of tables and rectangular and
polar coordinates
- 9. 3-9
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Module 1.3 Geometry
Trigonometry
Trigonometrical Relationships
1. By using Pythagoras, you are now able to partially solve right-angled triangles, i.e. you
can find the third side of a right-angled triangle when given its other 2 sides. This
chapter is concerned with establishing the basic trigonometrical concepts which will later
enable you to completely solve right-angled triangles, i.e. to find all their 6 elements
(angles and sides).
2. Similar triangles are triangles which are the same shape, one is simply an enlargement
of the other. Two important properties of similar triangles are:
a) their corresponding angles are equal.
b) their corresponding sides are proportional.
Consider the triangles:
(1) (2)
3. The above triangles are similar since they are equiangular and the ratios of their
corresponding sides are constant, i.e.
a)
BC
AB
EF
DE
3
5
6
10
3
5
b)
AC
AB
DF
DE
4
5
8
10
4
5
c)
BC
AC
EF
DF
3
4
6
8
3
4
5
10
4
3
8
6
A C
B
D F
E
- 10. 3-10
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
4. Now consider the following similar triangles:
In both cases side 'c' is the hypotenuse.
Taking angle A as the reference: Taking angle B as the reference:
a) Side 'a' is the side opposite a) Side 'b' is the side opposite
b) Side 'b' is the adjacent side b) Side 'a' is the adjacent side
Since the triangles are similar, the ratios of corresponding sides are constant, i.e., the
ratios
a
c
b
c
and
a
b
, are the same for all similar right-angled triangles.
5. In a right-angled triangle the ratio:
a)
side opposite the angle
hypotenuse
is called the SINE of the reference angle
sin A =
opposite
hypotenuse
a
c
b)
side adjacent to the angle
hypotenuse
is called the COSINE of the reference angle
cos A =
adjacent
hypotenuse
b
c
c)
side opposite the angle
side adjacent that angle
is called the TANGENT of the angle.
tan A =
opposite
adjacent
a
b
c
b
a
A C
B
c
a
b
A C
B
- 11. 3-11
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
The above are the fundamental trigonometrical ratios for right-angled triangles and must
be remembered. A convenient method to help you to remember them is ‘SOHCAHTOA’
or ‘SOHCAHTOA’ where S=sin, C=cos and T=tan
Example:
For the triangle shown find:
a) sine of angle B b) cosine of angle B c) tangent of angle B
a) sin B =
opp
hyp
3
5
0 6
.
b) cos B = 0.8
5
4
hyp
adj
c) tan B =
opp
adj
3
4
0 75
.
6. We will now investigate how the values of sin, cos and tan vary with the magnitude of the
angle.
a) When angle A is very small:
(1) sin A =
opp
hyp
a
c
and is very small.
When angle A is zero, sin A = 0
(2) cos A =
adj
hyp
b
c
and b c.
B
A C
5
4
3
B
A
C
a
b
c
~
- 12. 3-12
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
When angle A is zero, cos A = 1
(3) tan A =
opp
adj
a
b
and is very small.
When angle A is zero, tan A = 0
b) When angle A is large:
(1) sin A =
c
a
hyp
opp
and a c.
(2) cos A =
c
b
hyp
adj
and is small.
When angle A = 90o
, cos A = 0.
(3) tan A =
b
a
adj
opp
and is very large.
When angle A = 90o
, tan A = ∞.
We can summarise the above:
ANGLE 0o
90o
sin 0 1
cos 1 0
tan 0 ∞
Note: The maximum value of sin and cos is 1, but the maximum value of tan is infinity
(∞).
B
A C
c a
b
- 13. 3-13
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
8. We have seen that trigonometrical ratios vary as the angle varies and have calculated
values for 0o
and 90o
. We will now calculate the values for 30o
and 60o
. Consider the
equilateral triangle ABC of sides 2 units.
In triangle ABD, A = 60o
, B = 30o
and D = 90o
side d = 2 (given)
side b = 1 (half of AC)
side a = 2 1
2 2
(Pythagoras)
a = 3
Thus, in right-angled triangle ABD:
a) sin 60o
=
opp
hyp
3
2
17321
2
0 8660
.
.
b) cos 60o
=
adj
hyp
1
2
0 5000
.
c) tan 60o
=
opp
adj
3
1
17321
.
d) sin 30o
=
opp
hyp
1
2
0 5000
.
e) cos 30o
=
adj
hyp
3
2
17321
2
0 8660
.
.
f) tan 30o
=
opp
adj
1
3
0 5774
.
60
0
30
0
B
D
A
3
2
1
Line BD bisects ABC and is
perpendicular to AC
60
0
30
0
B
C
D
A
2
2
2
- 14. 3-14
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
9. We can now collect all our information and show graphically how the basic
trigonometrical ratios change as the angle increases from zero to 900
.
The functions all give graphs which are important. You should know how to sketch them
all and know how to use them.
The Sine Curve
x° 0 30 60 90 120 150 180 210 240 270 300 330 360
sin x° 0.00 0.50 0.86 1.00 0.86 0.50 0.00 -0.50 -0.86 -1.00 -0.86 -0.50 0.00
This is the curve drawn when you put all
the figures on the graph from the table
above. As you can see, this curve is in a
wave form. This wave can continue past
360° and go into the negatives.
The Cosine Curve
x° 0 30 60 90 120 150 180 210 240 270 300 330 360
cos x° 1.00 0.86 0.50 0.00 -0.50 -0.87 -1.00 -0.87 -0.50 0.00 0.50 0.87 1.00
If you look at this curve you can see it is
actually the same as the sine curve
except it is a different section (i.e. this
peaks at 0° where the sine curve peaks at
90°).
- 15. 3-15
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
The Tan Curve
x° 0 30 60 90 120 150 180 210 240 270 300 330 360
tan x° 0.00 0.58 1.73 ∞ -1.73 -0.58 0.00 0.58 1.73 ∞ -1.73 -0.58 0.00
The tan curve is very different from the
others. It is a non-continuous which
breaks as the value at the breaking point
(when x=90o
or x=270o
) is infinity. Again
this curve can be continued with the
section from x=90o
to x=270o
repeated.
From the curves we can see there is always more than one possible value for any
number you are working out the inverse of ( sin-1
0.5 = 30° or 150° ). The problem is that
your calculator only gives you one of the values ( the one below 90°). You must
remember the curves to find the position of the second angle.
10. You can, of course, use a graph to find the sin and cos of angles between 0 and 90o
. For
tan, this is only practical (because of length of axis) up to about 45o
. You should note
from the curves of y = sin and y = cos that there is a definite relationship between
sin and cos, e.g.:
a) sin 30o
= cos 60o
= 0.5000
b) sin 45o
= cos 45o
= 0.7071
c) sin 60o
= cos 30o
= 0.8660
d) sin 80o
= cos 10o
and so on.
- 16. 3-16
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Other Trigonometric Functions
Although less often used, other trigonometrical terms can be derived from the basic terms sin
and cos. These terms are called cot (cotangent), sec (secant) and cosec (cosecant). They are
determined as follows:
tan =
cos
sin
cot =
sin
cos
Reciprocal relations:
sin =
cosec
1
cosec =
sin
1
cos =
sec
1
sec =
cos
1
tan =
cot
1
cot =
tan
1
Square relations (also known as the Fundamental Identities):
sin² + cos² = 1
sec² - tan² = 1
cosec² - cot² = 1
- 17. 3-17
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
To Find the Length of an Unknown Side
1. So far we have evaluated the sine, cosine and tangent of angles, given the 3 sides of a
right-angled triangle. In the following text it is shown how to solve completely a right-
angled triangle, given any side and 2 angles.
2.
From the triangle shown:
a) sin =
opp
hyp
b) cos =
adj
hyp
sin =
a
c
cos =
b
c
a = c sin b = c cos
c) tan =
opp
adj
d) By Pythagoras:
tan =
a
b
c2
= a2
+ b2
a = b tan
B
c
a
A C
- 18. 3-18
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
3. The following examples involve the use of trigonometry, or combinations of trigonometry
and Pythagoras, to solve right-angled triangles.
a) In the right-angled triangle ABC, find angle A and side c
(1) To find angle A.
Note: In terms of angle a, we are given the side opposite and the side
adjacent.
Since
opp
adj
tan, this is the ratio we use.
tan A =
opp
adj
tan A =
12
5
tan A = 2.4
A = 67o
23' (after using a calculator or tables)
(2) To find side c
Note: If we use trig. to find side ‘c’, it necessitates our using angle A which we have just
found. If we have made an error in calculating angle A, this would also result in an error
in side ‘c’. By using Pythagoras, we use only given information and thus the possibility of
'carrying' an error is eliminated.
B
12
A C
5
- 19. 3-19
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
c2
= a2
+ b2
c = a b
2 2
c = 12 5
2 2
c = 144 25
c = 169
c = 13
- 21. 3-21
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
1. For the triangle shown, find the sine, cosine and tangent of A and C.
A
10
8
C
B
6
- 23. 3-23
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1. sin A = 0.6 sin C = 0.8
cos A = 0.8 cos C = 0.6
tan A = 0.75 tan C = 1.33
- 25. 3-25
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Coordinates and Graphs
The x and y Axis
An equation involving two variables can be represented by a graph drawn on ‘Coordinates
Axes’. Coordinate axes (illustrated below) consist of a horizontal line (usually referred to as the
x axis) and a vertical line (usually referred to as the y axis). The point of intersection of these
two lines is called the origin (usually denoted by the letter ‘0’).
Along the x and y axes we can mark off units of measurement (not necessarily the same on
both axes). The origin takes the value zero on both axes. The x axis takes positive values to the
right of the origin and negative values to the left of the origin. The y axis takes positive values
above the origin and negative values below the origin.
Any point on this diagram can be defined by its coordinates (consisting of two numbers). The
first, the x coordinate, is defined as the horizontal distance of the point from the y axis; the
second, the y coordinate, is defined as the vertical distance of the point from the x axis.
In general, a point is defined by its coordinates which are written in the form (a, b).
Example:
The point (3, 2) may be plotted on the coordinate axes as follows:-
- 26. 3-26
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Example:
Consider the following diagram
The points A, B, C, D, E and F above are defined by their coordinates as follows:
A (1, 4) D (-4, 1)
B (3, 2) E (-5, -3)
C (2, 1) F (3, -2)
- 27. 3-27
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
1. Plot the following points on coordinate axes.
i) (2, 3) ii) (1,4) iii) (5, 0)
iv) (0, 2) v) (3,-1) vi) (-2, 4)
vii) (-1, -3) viii) (0,-4) ix) (-5, 0)
x) (-4, 1) xi) (-3, -1) xii) (3, -3)
- 31. 3-31
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Graphical Representations of an Equation
An equation involving two variables can be represented, on coordinate axes, by means of a
graph.
For a given range of values of x, the corresponding y values can be calculated from the
equation being considered. The points obtained can then be plotted and joined together to form
the graph.
Before plotting the points on a graph, the axes must be drawn in a way that takes into account
the range of the x-values and the range of the y-values. If graph paper is used (which is
desirable) you should use a scale that involves a sensible number of units per square i.e. you
should use steps of, for example, 1, 2, 5 or 10 etc. units per square depending on the question.
You should avoid using steps along the axes of, for example, 7 or 9 units per square as this can
complicate the graph unnecessarily.
- 32. 3-32
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Example
Draw the graph of y = 2x + 1 for 0 ≤ x ≤ 5.
By taking x values of 0, 1, 2, ……..5, we can calculate the corresponding y values, as shown
below, by first evaluating the component parts of the equation.
x: 0 1 2 3 4 5
2x 0 2 4 6 8 10
+1 1 1 1 1 1 1
y: 1 3 5 7 9 11
We then plot the points obtained, each point being defined by its x coordinate and its
corresponding y coordinate. The points are then joined together to form the graph.
In this example the points to be plotted are (0, 1), (1, 3), (2, 5), (3, 7), (4, 9), (5, 11).
- 33. 3-33
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Example:
Draw a graph of y = x2
- 8x + 12 for 0 ≤ x ≤ 6
We again take x values covering the given range, and calculate the corresponding y values
from the given equation.
x: 0 1 2 3 4 5 6
x2
0 1 4 9 16 25 36
-8x 0 -8 -16 -24 -32 -40 -48
+12 +12 12 +12 +12 +12 +12 +12
y: 12 5 0 -3 -4 -3 0
We now plot the points obtained and join them together to form the graph. In this example the
points to be plotted are (0, 12), (1, 5), (2, 0), (3, -3), (4, -4), (5, -3), (6, 0).
N. B. For a more detailed graph we could, of course, include more points. e.g. by taking x
values of
0, ½, 1, 1½, 2, 2½, ……….5½, 6
and calculating the corresponding y values, we could plot nearly twice as many points as we did
in the above example
y = x
2
– 8x + 12
- 34. 3-34
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Example:
Draw a graph of y = x2
+ 1 for -3 ≤ x ≤ + 3
Again, taking x values covering the given range, we first calculate the corresponding y values
from the given equation.
x: -3 -2 -1 0 1 2 3
x2
9 4 1 0 1 4 9
+1 +1 +1 +1 +1 +1 +1 +1
y: 10 5 2 1 2 5 10
We now plot the points obtained and join them together to form the graph.
- 35. 3-35
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
1. Draw graphs of the following functions for 0 ≤ x ≤ 5
i) y = 2x + 5
ii) y = 5x + 1
iii) y = 3x - 5
iv) y = x2
- 6x + 5
v) y = x2
- 7x + 12
vi) y = 3x2
- 21x + 30
2. Draw graphs of the following functions for -3 ≤ x ≤ 3
i) y = 2x2
+ 7
ii) y = 3x2
- 12
iii) y = x3
- 7
iv) y = 4x3
- 16x2
- 16x + 64
v)
5
x
1
y
- 43. 3-43
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
The Straight Line
A straight line is defined as the shortest distance between two points.
The equation of a straight line is given by
y = mx + c
where m represents the slope of the line and c is the point where the line crosses the y-axis (i.e.
they intercept). The point where the line crosses the x-axis is called the x intercept.
Example:
In this example, m = 2 and c = 0
Note that whenever c = 0, the line will pass through the origin.
Example:
In this example, m = -3 and c = 6
As c = 6, we know that this line cuts the y axis at y = 6 (this can be verified by substituting x = 0
into the equation of the line, as x = 0 along the y axis).
Similarly, as y = 0 along the x axis, we can substitute y = 0 into the equation of the line to find
where the line intersects with the x axis (i.e. the x intercept).
We have, when y = 0
6 - 3x = 0
3x = 6
x = 2
- 44. 3-44
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Hence the line cuts the x axis at x = 2
We can now say that the y intercept = 6 and the x intercept = 2
Example:
In this example, m = 4 and c = -2
We know, immediately, that they intercept is -2 (i.e. the value of c)
To find the x intercept, we substitute y = 0 into the equation of the line.
i.e. 0 = -2 + 4x
4x = + 2
x= ½
Hence the x intercept is x = ½
Special Cases
A straight line parallel to the x-axis takes the form y = constant.
Similarly, a straight line parallel to the y-axis takes the form x = constant.
These cases are illustrated below:
Straight line parallel to the x axis Straight line parallel to they axis
y = - 2 + 4x
- 45. 3-45
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
For each of the following equations identify the gradient and the y-intercept.
i) y = 4x + 5
ii) y = 9x
iii) y = 8
iv) y =
7
3x
5
4
v)
13
5x
6
y
vi) 3y = 9x + 6
vii) 8y = x - 8
- 47. 3-47
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
m =gradient, c = y-intercept
i) m = 4, c = 5
ii) m = 9, c = 0
iii) m = 0, c = 8
iv) m = - 7
3
, c = 5
4
v) m = - 13
5
, c = 13
6
vi) m = 3, c = 2
vii) m = 8
1
, c= -1
- 49. 3-49
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Derivation of the Equation y = mx + c
Given the coordinates of two points, (x1, y1) and (x2, y2) say, we can calculate the equation of
the straight line that passes through these points.
Two methods of calculating this equation are illustrated below:
Example:
The question is:
Find the equation of the straight line that passes through the points (1, 4) and (3, 10).
Method 1
The general equation of a straight line is given by y = mx + c
and it is necessary to find numerical values for m and c.
If the straight line in question passes through the two given points, then each of these points
must satisfy the equation of this straight line. That is, we can substitute the coordinates of each
point as follows:
y = mx + c
substituting (1, 4) we have
4 = m + c (1)
likewise, substituting (3, 10) we have
10 = 3m + c (2)
Now (1) and (2) give us two equations in two unknowns, m and c, (i.e. simultaneous equations)
which we can solve.
- 50. 3-50
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
We have
4 = m + c (1)
10 = 3m + c (2)
subtracting (1) from (2) to eliminate c we obtain 6 = 2m
m = 3
substituting this value of m back into (1) we obtain
4 = m + c
4 = 3 + c
c = 4 - 3
c = 1
If we now substitute these numerical values of m and c into the equation y = mx + c, we obtain
the equation of the straight line passing through the points (1, 4) and (3, 10).
That is
y = 3x + 1
Method 2
In general, we can consider any two points (x1 y1) and (x2, y2). The straight line passing through
these points can be written as
y – y1 = m(x – x1)
where m =
1
2
1
2
x
x
y
y
(m is the gradient of the line)
Applying this to the points (1, 4) and (3, 10) we have x1 = 1; y1 = 4; x2 = 3; y2 = 10;
and we hence obtain:
m =
1
3
4
10
=
2
6
= 3
and our line becomes
y - 4 = 3(x - 1)
y - 4 = 3x - 3
y = 3x + 1
N.B. In this example, the point (1, 4) corresponded to (x1, y1) and the point (3, 10) corresponded
to (x2, y2).
If we had worked through this example with the point (3, 10) corresponding to (x1, y1), and the
point (1, 4) corresponding to (x2, y2), the answer would have been exactly the same.
- 51. 3-51
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
1. Calculate the equation of the straight line that passes through the following points.
i. (1, 3) and (3, 7)
ii. (0, 2) and (5, 22)
iii. (1, -5) and (-1, -9)
2. Calculate the equation of the straight line that passes through the following points:
i. (1, 7) and (3, 11)
ii. (0, 0) and (1, 6)
iii. (3, -2) and (2, 1)
iv. (0, 0) and (-2, 8)
v. (6, 1) and (4, -1)
vi. (0, -3) and (-2, 1)
vii. (2, 6) and (7, 6)
viii. (-5, -47) and (-2, -26)
ix. (3, 1) and (3, -2)
x. (1, 1¼) and (2, 2¾)
- 53. 3-53
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1.
i. y = 2x + 1
ii. y = 4x + 2
iii. y = 2x - 7
2.
i. y = 2x + 5
ii. y = 6x
iii. y = -7x + 2
iv. y = - 4x
v. y = x - 5
vi. y = -2x – 3
vii. y = 6
viii. y = 7x – 12
ix. x = 3 (y = mx + c does not work with lines of infinite gradient)
x. y =
2
3x -
4
1
- 55. 3-55
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Cartesian and Polar Coordinates
To pinpoint where you are on a map or graph there are two main systems:
Cartesian Coordinates
Using Cartesian Coordinates you mark a point by how far along and how far up it is (x and y
coordinates):
Polar Coordinates
Using Polar Coordinates you mark a point by how far away, and what angle it is (r and θ
coordinates):
- 56. 3-56
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Converting
To convert from one to the other, you need to solve the triangle:
To Convert from Cartesian to Polar
If you have a point in Cartesian Coordinates (x, y) and need it in Polar Coordinates (r, θ), you
need to solve a triangle where you know two sides.
Example: What is (12, 5) in Polar Coordinates?
Use Pythagoras Theorem to find the long side (the hypotenuse):
r2
= 122
+ 52
r = √ (122
+ 52
)
r = √ (144 + 25) = √ (169) = 13
Use the Tangent Function to find the angle:
tan θ =
12
5
θ = tan-1
12
5
= 22.6°
So, to convert from Cartesian Coordinates (x, y) to Polar Coordinates (r, θ):
r = √ (x2
+ y2
)
θ = tan-1
x
y
- 57. 3-57
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
To Convert from Polar to Cartesian
If you have a point in Polar Coordinates (r, θ), and need it in Cartesian Coordinates (x, y) you
need to solve a triangle where you know the long side and the angle:
Example: What is (13, 23°) in Cartesian Coordinates?
Use the Cosine Function for x:
cos (23°) =
13
x
Rearranging and solving:
x = 13 × cos (23°) = 13 × 0.921 = 11.98
Use the Sine Function for y:
sin (23°) =
13
y
Rearranging and solving:
y = 13 × sin (23°) = 13 × 0.391 = 5.08
So, to convert from Polar Coordinates (r, θ) to Cartesian Coordinates (x, y) :
x = r × cos( θ )
y = r × sin( θ )
- 59. 3-59
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Worksheet
1. Convert the following cartesian coordinates into polar coordinates:
(a) (3, 4)
(b) (10, 10)
(c) (10, 0)
2. Convert the following polar coordinates into cartesian coordinates
(a) 13 cm, 67.4°
(b) 50 m, 60°
(c) √8 ft, π/2 radians
- 61. 3-61
Module 1.3 Geometry
For Training Purposes Only
ST Aerospace Ltd
© Copyright 2013
Answers
1.
(a) 5, 53°
(b) 14.14, 45°
(c) 10, 0°
2.
(a) x = 5 cm, y = 12 cm
(b) x = 25 m, y = 43.3 m
(c) x = 2 m. y = 2 m
