Mathematics
Grade 12
SELF STUDY GUIDE
Trigonometry and Euclidean Geometry
1
TABLE OF CONTENTS
PAGE
1. Introduction
3
2. How to use this self-study guide
4
3. Trigonometry
5
3.1 Grade 10 Revision work
8
3.2 Grade 11 Revision work
12
3.3 Reduction Formulae
13
3.4 Typical EXAM questions
58
2
98
Euclidean Geometry
2.1 Practice Exercises
121
2
1.
Introduction
The declaration of COVID-19 as a global pandemic by the World Health Organisation led
to the disruption of effective teaching and learning in many schools in South Africa. The
majority of learners in various grades spent less time in class due to the phased-in
approach and rotational/ alternate attendance system that was implemented by various
provinces. Consequently, the majority of schools were not able to complete all the relevant
content designed for specific grades in accordance with the Curriculum and Assessment
Policy Statements in most subjects.
As part of mitigating against the impact of COVID-19 on the current Grade 12, the
Department of Basic Education (DBE) worked in collaboration with subject specialists from
various Provincial Education Departments (PEDs) developed this Self-Study Guide. The
Study Guide covers those topics, skills and concepts that are located in Grade 12, that are
critical to lay the foundation for Grade 12. The main aim is to close the pre-existing content
gaps in order to strengthen the mastery of subject knowledge in Grade 12. More
importantly, the Study Guide will engender the attitudes in the learners to learning
independently while mastering the core cross-cutting concepts.
3
2.
How to use this Self Study Guide?
•
This Self Study Guide summaries only two topics, Trigonometry and Euclidean
Geometry. Hence the prescribed textbooks must be used to find more exercises.
•
It highlights key concepts which must be known by all learners.
•
Deeper insight into the relevance of each of the formulae and under which
circumstances it can be used is very essential.
•
Learners should know the variables in each formula and its role in the formula.
Learners should distinguish variable in different formulae.
•
More practice in each topic is very essential for you to understand mathematical
concepts.
•
The learners must read the question very carefully and make sure that they
understand what is asked and then answer the question.
•
make sure that Euclidean Geometry is covered in earlier grades. Basic work
should be covered thoroughly. An explanation of the theorem must be
accompanied by showing the relationship in a diagram.
•
After answering all questions in this Self Study Guide, try to answer the previous
question paper to gauge your understanding of the concepts your required to know.
4
1. TRIGONOMETRY
Introduction to trigonometry
Naming of sides in a right-angled triangle with respect to given angles.
Hy
p
θ
Adjacent side to θ
Opposite side to θ
ot
en
us
es
id
e
θ
A
B
C
1.
•
•
•
Hy
po
ten
u
se
sid
e
(si
de
op
pt
o
Opposite side to θ
Adjacent side to θ
In DABC
:
)
AC is side opposite to 900
known as hypotenuse.
AB is opposite side to Ĉ
BC is adjacent side to Ĉ .
P
Q
R
In DPQR
:
•
•
•
RQ is side opposite to 900
known as hypotenuse.
PQ is opposite side to R̂ .
PR is adjacent side to R̂ .
5
2.
•
AC is side opposite to 900
known as hypotenuse..
AB is adjacent side to Â
BC is side oppositet to  .
•
•
•
RQ is side opposite to 900
known as hypotenuse.
•
PQ is adjacent side to Q̂ .
•
PR is opposite side to Q̂
DEFINITIONS OF TRIGONOMETRIC RATIOS
Trigonometric ratios can be defined in right-angled triangles ONLY.
D
θ
E
•
•
•
F
opp. sideto q DE
=
hypotenuse DF
adj. sideto q EF
cos q =
=
hypotenuse DF
opp. side to q DE
tan q =
=
adj. side to q EF
sin q =
•
•
•
hypotenuse DF
=
opp. sideto q DE
hypotenuse DF
sec q =
=
adj. sideto q EF
adj. sideto q EF
cot q =
=
opp. sideto q DE
cos ecq =
RECIPROCAL IDENTITIES
NB: ONLY EXAMINED IN GRADE 10
•
sin q =
1
cos ecq
•
cos ecq =
1
sin q
•
cos q =
1
s ecq
•
sec q =
1
cos q
•
tan q =
1
cot q
•
cot q =
1
tan q
6
Revision grade 8, 9 and 10 work (use of Pythagoras Theorem)
Pythagoras theorem is only used in right-angled triangles: “The square on the
hypotenuse is equal to the sum of the squares in the remaining two sides of a
triangle”.
Example 1
In the diagram below, DABC, Bˆ = 900 , AB = 3 cm, BC = 4cm
A
3 cm
θ
B
C
4 cm
1.1 Calculate the length of AC.
Solution
AC2 =AB2 +BC2
Pyth.theorem
= ( 3cm ) + ( 4 cm )
2
AC=5cm
2
ü correct substitution in Pyth. Theo
ü answer
(2)
7
1.2 Determine the values of the following trigonometric ratios:
1.2.1
1.2.2
1.3 1.3.1
sin q
cos q
sin q =
AB 3
ü
=
AC 5
correct ratio
cos q =
BC 4
= ü
AC 5
correct ratio
Determine the size of  in terms of θ
(1)
(1)
(1)
Aˆ = 1800 - (900 + q ) [Ðs in a D ]
Aˆ =1800 - 900 - q
Aˆ = 90° - q
size of Â
ü
1.3.2
(1)
(
Hence, or otherwise, determine the value of cos 900 - q
(
)
cos 90 - q =
0
(
adj. sideto 900 - q
)
)
(1)
(1)
hypotenuse
=
AB
BC
3
= ü
5
8
3.1 Revision grade 10
CARTESIAN PLANE AND IDENTITIES
(x ; y)
y
r
θ
x
NB r is always positive, whilst x and y can be positive or negative
Defining trig ratios in terms of x, y and r:
•
•
•
sin q =
y
r
cos ecq =
cos q =
x
r
sec q =
r
x
tan q =
y
x
cot q =
x
y
r
y
9
3.2 Revision grade 11
DERIVING IDENTITIES (using x, y and r)
2
æxö æ yö
1. cos 2 q + sin 2 q = ç ÷ + ç ÷
èrø èrø
x2 + y 2
=
r2
r2
= 2
r
=1
y
sin q
2.
= r
cos q x
r
2
(
NB x 2 + y 2 = r 2 Pyth
)
y
x
= tan q
=
\
\ cos2 q + sin 2 q = 1
sin q
= tan q
cos q
CO-RATIOS/FUNCTIONS
(
) rx = cosq
•
sin 900 - q =
•
y
cos 900 - q = = sin q
r
(
)
900 - q
r
y
x
θ
BASIC IDENTITIES
•
cos 2 q + sin 2 q = 1
•
tan q =
sin q
cos q
10
Grade 12 Identities
COMPOUND ANGLE IDENTITIES
•
cos (a - b ) = cos a .cos b + sin a .sin b
•
cos (a + b ) = cos a .cos b - sin a .sin b
•
sin (a - b ) = sin a .cos b - cos b .sin a
•
sin (a + b ) = sin a .cos b + cos b .sin a
DOUBLE ANGLE IDENTITIES
•
•
ìcos 2 a - sin 2 a ü
ï
ï
cos 2a = í1 - 2sin 2 a
ý
ï2 cos 2 a - 1 ï
î
þ
sin 2a = 2sin a cos a
Proofs for the compound angle identities are examinable
11
SIGNS OF TRIGONOMETRIC RATIOS IN ALL THE FOUR QUADRANTS
y
y +
= =+
r +
-x cos q =
= =r
+
y +
tan q =
= =-x -
sin q =
ONLY
SINE +
nd
2
quadrant
ONLY
TANGENT +
3rd quadrant
-y = =r
+
-x cos q =
= =r
+
-y tan q =
= =+
-x -
sin q =
y +
= =+
r +
x +
cos q = = = +
r +
y +
tan q = = = +
x +
sin q =
-y sin q =
= =r
+
x +
cos q = = = +
r +
-y tan q =
= =x
+
ALL
RATIOS +
1st quadrant
x
ONLY
COSINE +
4th quadrant
Example 1
If sin q = -
1.
12
and 900 £ q £ 2700 , determine the values of the following:
13
sin q
cos q
When answering this question, you need to define your trig ratio. Like sin q = -
12 y
= . Then you
13 r
will know that y = -12 and r = 13 , r will never be negative, then the negative sign will be taken by
y. Sine is negative in 3rd and 4th quadrants. 900 £ q £ 2700 is an angle between 2nd and 3rd quadrants.
To know which quadrant from the two conditions, we must choose the quadrant that satisfies both
conditions. Hence the 3rd quadrant.
12
y
x=?
y = -12
θ
x
r = 13
r 2 = x2 + y 2
Pythagoras theorem
x = ± (13) - ( -12 )
2
2
3 3quad
is neg]
= =-5
-5 [x [inxin
rd quad
is neg ]ü
2.
cos 2 q
æ -5 ö
cos q = ç ÷
è 13 ø
2
2
rd
-12
sin q
12
= 13 = ü
5
cos q
5
13
3.
=
25
ü
169
1 - sin 2 q
2
æ -12 ö
1 - sin q = 1 - ç
÷ ü
è 13 ø
2
= 1=
144
169
25
ü
169
13
CAST DIAGRAM
CAST: ALL STUDENTS TAKE COFFEE
900
2nd quadrant
1st quadrant
A
S
1800
00
360
C
T
0
th
4 quadrant
3rd quadrant
900 + q
900 - q
1800 - q
-1800 - q
-3600 + q
+
±1800
1800 + q
3600 - q
-
00
±3600
-900 + q
-1800 + q
-q
-900 - q
2700
2700
NB: We can do reduction for angles rotating clockwise by adding 3600 up until the angle
is in the range of 00 to 3600 .
3rd quadrant
SPECIAL ANGLES
300 300
2
450
2
2
1
3
60
450
0
600
1
sin 00 = 0
cos 00 = 1
tan 00 = 0
1
1
sin 300 =
1
2
3
2
1
tan 300 =
3
cos 300 =
1
2
1
cos 450 =
2
0
tan 45 = 1
sin 450 =
3
2
1
cos 600 =
2
3
tan 600 =
1
sin 600 =
14
sin 900 = 1
sin1800 = 0
sin 2700 = -1
sin 3600 = 0
cos 900 = 0
cos1800 = -1
cos 2700 = 0
cos 3600 = 1
tan 900 = undefined
tan1800 = 0
tan 2700 = undefined
tan 3600 = 0
REDUCTION FORMULAE
Identify in which quadrant the angle(s) lie first, then you will be able to know the sign of each
trigonometric ratio(s) referring to CAST diagram, then change the trig ratio to its co-function if
you are reducing by 90.
-q (4th quadrant)
900 - q (1st quadrant)
(
)
•
sin ( -q ) = - sin q
(
)
•
cos ( -q ) = cos q
•
tan ( -q ) = - tan q
•
tan ( -q ) = tan 3600 - q = - tan q
•
sin 900 - q = cosq
•
cos 900 - q = sin q
•
tan 90 - q =
(
0
(
)
) cos 90 - q
(
)
=
=
sin 900 - q
0
cos q
sin q
1
tan q
900 + q (2nd quadrant)
OR
00 and 3600
q - 900 (4th quadrant)
)
•
(
)
•
sin 900 + q = cosq
•
cos 900 + q = - sin q
)
Adding 3600 until the range is between
(
•
(
(
)
(
)
sin q - 900 = - cosq
cos q - 900 = sin q
15
•
(
)
tan 900 + q = -
1
tan q
1800 - q (2nd quadrant)
•
)
•
(
)
•
(
)
sin 1800 - q = sin q
•
cos 1800 - q = - cosq
•
tan 1800 - q = - tan q
1800 + q (3rd quadrant)
•
sin 1800 + q = - sin q
•
0
•
•
tan (1800 + q ) = tan q
•
•
3600 - q (4th quadrant)
(
)
(
)
(
)
•
sin 3600 - q = - sin q
•
cos 3600 - q = cosq
•
tan 3600 - q = - tan q
3600 + q (1st quadrant)
•
(
)
sin 3600 + q = sin q
(
)
(
)
(
) tan1 q
sin -q - 900 = - cosq
cos -q - 900 = - sin q
tan -q - 900 =
q - 1800 (3rd quadrant)
(
)
• cos (180 + q ) = - cosq
•
1
tan q
)
-q - 900 (3rd quadrant)
(
•
(
tan q - 900 = -
(
)
(
)
sin q -1800 = - sin q
OR
(
)
sin q - 1800 = sin éë - 1800 - q ùû
= - sin q
(
)
cos q -1800 = - cosq
tan (q -1800 ) = tan q
-q - 1800 (2nd quadrant)
•
•
•
(
)
(
)
(
)
sin -q -1800 = sin q
cos -q -1800 = - cosq
tan -q -1800 = - tan q
q - 3600 (1st quadrant)
•
(
)
sin q - 3600 = sin q
16
(
)
(
)
•
cos 3600 + q = cosq
•
tan 3600 + q = tan q
•
•
(
)
(
)
cos q - 3600 = cosq
tan q - 3600 = tan q
Worked-out Example 1
Write the following as ratios of θ:
Solutions
1.1.
cos(1800 - q )
1.1 cos(1800 - q ) = - cos q ü
1.2.
tan(q - 3600 )
1.2 tan(q - 3600 ) = tan q ü
1.3.
sin ( -q )
1.3 sin ( -q ) = - sin q ü
Worked-out Example 2
Express the following as ratios of acute angles
Solutions
2.1
2.1
tan1300
(
tan1300 = tan 1800 - 500
)
= - tan 500 ü
1300 cannot just be written in any way but in terms of 900 or 1800 . It is in the 2nd
quadrant and is greater than 900 but less than 1800 .
\1300 = 1800 - 500 OR 1300 = 900 + 400 . In this case we choose expression by 1800
as our ratios remain the same in 1800 . We can write 1300 = 900 + 400 but bear in
mind that our ratios change to their co-ratios when reducing by 900 .
2.2
cos ( -284)
0
2.2
cos ( -284) = cos(76 - 3600 )
0
= cos 760 ü
17
OR
cos ( -284) = cos(-284 + 3600 )
0
= cos 760 ü
Explanation in 2.1 also applies in 2.2 according the quadrant where the
angle lies.
Worked-out Example 3
Simplify the following expressions:
3.1
(
)
tan ( 360 - q ) .cos q .sin ( 360 + q )
cos ( -q ) cos(900 + q ) tan q + 1800
0
0
Solutions
(
)
tan ( 360 - q ) .cos q .sin ( 360 + q )
ü cos q
( cos q ) . ( - sin q ) . ( tan q )
( - tan q ) .cos q . ( sin q )
ü tan q
cos ( -q ) cos(900 + q ) tan q + 1800
0
=
0
=1
ü - sin q
ü - tan q
ü sin q
ü1
3.2
(
2sin 400.cos -50 0
sin 80
)
0
Solutions
ü sin 400
18
=
=
(
2sin 400.cos 400 - 900
)
ü 2sin 400 cos 400
2sin 400 cos 400
(
2sin 400. sin 400
0
2sin 40 .cos 40
ü tan 400
)
0
= tan 400
3.3
1
sin 2 x
2
é 1
ù
tan 5400 + x . ê 2 - tan 2 x ú
ë cos x
û
(
)
1
.2sin x.cos x
2
=
é
sin 2 x ù
2
1
cos
x
.
ê
cos 2 x úú
( tan x ) . ê
2
cos x
ê
ú
ëê
ûú
=
=
sin x.cos x
sin x é1 - sin 2 x ù
.
cos x êë cos 2 x úû
ü 2sin x.cos x
ü tan x ü
ü
sin 2 x
cos 2 x
sin x
cos x
ü 1 - sin 2 x = cos 2 x
sin x cos x
sin x cos 2 x
.
cos x cos 2 x
= sin x.cos x.
= cos 2 x
= cos x
cos x
sin x
ü simplification
ü cos x
19
Worked-out Example 4
If cos 350 = p , determine the following in terms of p:
Solutions
4.1
4.1
sin 350
cos 350 =
y
p x
=
1 r
r =1
y=?
350
x
x= p
According to the
definition of cosine, p
represents x and
1 represents r
NB: Consider finding the 3rd angle
before you start doing your calculations.
y=
(1) - ( p )
2
2
Pythagoras
= 1 - p2
ü y = 1 - p2
y
\sin 35 =
r
0
=
1 - p2
1
= 1 - p2
\sin 350 =
1 - p2
1
1 - p2
ü sin 35 =
1
0
20
4.2
(
tan 2150 + sin -550
(
)
)
= tan 1800 + 350 + (- sin 550 )
(
) (
= tan 350 + - sin 55 0
)
æ 1 - p2 ö æ p ö
÷+ç- ÷
=ç
ç
p ÷ è 1ø
è
ø
=
1- p - p
p
2
ü tan 350 ü
sin 550
üü
substitution
2
HOW TO USE DOUBLE ANGLE IDENTITIES [always change double angles to single
angle to make your expression to be in a more simplified form]
1.
If you see sin 2x always substitute it by 2sin cos x . There is only 1 option for
sin 2x .
2.
cos 2x has 3 options
2.1
If you see sin x before or after cos 2x , replace cos 2x by 1 - 2sin 2 x
2.2
If you see cos x before or after cos 2x , replace cos 2x by 2 cos 2 x - 1
2.3
If you see sin x cos x before or after cos 2x , replace cos 2x by cos 2 x - sin 2 x
2.4
If you see ±1 before or after cos 2x .try to eliminate it by replacing by
1 - 2sin 2 x or 2 cos 2 x - 1.
(
)
é cos 2 x - 1 = 1 - 2sin 2 x - 1 ù
ê
ú
ê
ú
= -2sin 2 x
ê
ú
OR
ê
ú
ê1 - cos 2 x = 1 - 2 cos 2 x - 1 ú
ê
ú
2
ê
ú
= -2 cos x
ê
ú
ê
ú
OR
ê
ú
2
êcos 2 x + 1 = 2 cos x - 1 + 1 ú
ê
ú
= 2 cos 2 x
ë
û
(
)
21
For example:
By doing so you will be left with 1 term. Replace cos 2x by the identity which will be the additive
inverse to ±1
TRIGONOMETRIC IDENTITIES INCLUDING DOUBLE ANGLES
Worked-out Example1: Prove that
sin 2 x
= tan x
cos 2 x + 1
•
•
sin 2 x
cos 2 x + 1
2sin x cos x
=
2 cos 2 x - 1 + 1
2sin x cos x
=
2 cos 2 x
sin x
=
cos x
= tan x
LHS =
•
Numerator has 1 option only
Denominator cos 2 x + 1 , then eliminate
1 by replacing cos 2 x by 2cos2 x - 1
Simplify
= RHS
sin 2 x - cos x
cos x
=
sin x - cos 2 x sin x + 1
Worked-out Example 2: Prove that
sin 2 x - cos x
sin x - cos 2 x
2sin x cos x - cos x
=
sin x - 1 - 2sin 2 x
•
sin 2x has only 1 option
cos x ( 2sin x - 1)
•
Denominator has sin x , then
replace cos 2 x by 1 - 2sin 2 x
Denominator must be written in
standard form
Factorise numerator and
denominator
LHS =
(
=
=
=
)
sin x - 1 + 2sin 2 x
cos x ( 2sin x - 1)
2sin x + sin x - 1
cos x ( 2sin x - 1)
2
( 2sin x - 1)( sin x + 1)
•
•
cos x
sin x + 1
= RHS
=
22
Worked-out Example 3: Prove that cos 4 x = 8cos 4 x - cos 2 x + 1
LHS = cos 4 x
= 2 cos 2 2 x - 1
(
)
= 2 ( 4 cos x - 4 cos x + 1) - 1
2
= 2 2 cos 2 x - 1 - 1
4
2
= 8cos 4 x - 8cos 2 x + 2 - 1
= 8cos 4 x - 8cos 2 x + 1
Worked-out Example 1 Expressions
1.1
Determine, without using a calculator, the value of the following
trigonometric expression:
Solutions
sin 2 x. cos( - x) + cos 2 x. sin(360° - x)
sin(180° + x)
Reduction of
angles CAST
diagram
sin 2 x.cos(- x) + cos 2 x.sin(360° - x)
sin(180° + x)
=
=
=
sin 2 x.cox + cos 2 x. ( - sin x )
( - sin x )
sin 2 x.cos x - cos 2 x sin x
- sin x
sin ( 2 x - x )
- sin x
sin x
=
- sin x
= -1
Recognition of
compound
angle
expression
23
Worked-out
Example 2
Solutions
2.1
2
Prove that cos15 =
0
( 3 + 1) without
4
using a calculator
LHS = cos150
(
= cos 450 - 300
)
= cos 450 cos 300 + sin 450 sin 30 0
=
=
2 3
2 1
.
+
.
2 2
2 2
2
( 3 + 1)
4
Express 150 in terms of special angles as you are told to prove without using a calculator. 150 is an
acute angle that can be expressed in terms of 150 = 450 - 300 or 150 = 600 - 450 . From there, these
angles are forming compound angles. We cannot reduce 150 as it is acute angle already. Compound
(
)
angle identity for cos 450 - 300 needs to be applied.
Worked-out Example 3
3.1
If tan 200 = k , determine, without using a calculator, expressions in terms of k :
700
1
1- k 2
200
k
Solutions
Express 400 in terms 200 as
they are related. 400 = 2 ´ 200 ,
then apply double angle
identities
24
3.1
sin 400
sin 400 = 2sin 200.cos 200
= 2.
1- k 2
.k
1
= 2k 1 - k 2
Express 350 in terms 700 as
1
they are related. 350 = ´ 700 ,
2
then apply double angle
identities
3.2
cos 350
cos 350
cos 700 = 2 cos 2 350 - 1
cos 2a = 2 cos 2 a - 1
cos 700 + 1 = 2 cos 2 350
a = 350
cos 700 + 1
= cos 2 350
2
cos 350 =
cos 700 + 1
2
=
1- k 2 +1
2
=
2 - k2
2
25
TRIGONOMETRIC EQUATIONS
Solving of equations if 00 £ x £ 3600
Any trigonometric function is positive in two quadrants and negative in two quadrants; so
there will always be two solutions if 00 £ x £ 3600 . The sign of the ratio tells us in which
quadrant the angle is. Reference angle is an acute angle that is always positive irrespective
of the ratio.
If sin x = + ratio , 0 £ ratio £ 1
If cos x = + ratio 0 £ ratio £ 1
If tan x = + ratio , ratio ³ 0
+
+
+
+
+
+
1st : x = ref Ð
or
1st : x = ref Ð
or
1st : x = ref Ð
or
2nd : x = 1800 - ref Ð
4th : x = 3600 - ref Ð
3rd : x = 1800 + ref Ð
NB:
1st : x ¹ 900 - ref Ð
1st : x ¹ 900 - ref Ð
1st : x ¹ 900 - ref Ð
,
2nd : x ¹ 900 + ref Ð
900 + x is in the 2nd quad but it
is not advisable to use 90 as
our ratios are changing to their
co-ratios
NB:
,
4th : x ¹ 2700 + ref Ð
2700 + x is in the 4th quad
but it is not advisable to
use 270 as our ratios are
changing to their co-ratios
as well as 90
NB:
,
3rd : x ¹ 2700 - ref Ð
2700 - x is in the 3rd quad but it
is not advisable to use 270
as our ratios are changing to
their co-ratios as well as 90
If sin x = - ratio ,
If cos x = - ratio
If tan x = -ratio
-1 £ ratio £ 0
-1 £ ratio £ 0
ratio £ 0
26
-
-
-
-
-
3rd : x = 1800 + ref Ð
or
2nd : x = 1800 - ref Ð
2nd : x = 1800 - ref Ð
or
or
4th : x = 3600 - ref Ð
3rd : x = 1800 + ref Ð
4th : x = 3600 - ref Ð
NB:
3: x ¹ 2700 - ref Ð
4th : x ¹ 2700 + ref Ð
NB:
2nd : x ¹ 900 + ref Ð
3rd : x ¹ 2700 - ref Ð
NB:
2nd : x ¹ 900 + ref Ð
4th : x ¹ 2700 + ref Ð
2700 - ref Ð is in the 3nd quad but
900 + ref Ð is in the 2nd quad
2700 + ref Ð is in the 3rd quad but it
it is not advisable to use 2700 as
our ratios are changing to their coratios
but it is not advisable to use
as our ratios are changing to
their co-ratios when reducing by
900 as well as 2700
is not advisable to use 2700 as our
ratios are changing to their co-ratios
as well as 900
27
Solve for x:
1.1
cos x =
Solutions
2
3
cos x =
2
3
cos x =
æ2ö
ref Ð = cos -1 ç ÷
è3ø
= 48,190
1st : x = ref Ð = 48,190
4th : x = 3600 - ref Ð = 311,810
2
3
æ2ö
x = ± cos -1 ç ÷
è3ø
= ±48,190 + 360.k , k Î !
x = -311,810 or - 48,190 or 48,190 or 311,810
\x = 48,190 or 311,810
1.2
sin x = -
1
2
sin x = -
1
2
æ1ö
ref Ð = sin ç ÷
è2ø
= 300
-1
When determining the reference angle,
ignore the negative sign to the ratio as
you will get negative angle which will not
preserve a reference angle.
3rd : x = 1800 + 300
= 2100
4th : x = 3600 - 300
x = 3300
GENERAL SOLUTIONS
note that k element of intergers
In determining general solutions, we do the same way as solving equations but consider the period
of sine and cosine graphs as they repeat their shapes after a period of 360 and tan graph
repeating itself after a period of 180.
28
sin q = t , 0 £ t £ 1
cos q = t , 0 £ t £ 1
tan q = t , t Î !
1: q = ref Ð + 3600.k
or
1: q = ref Ð + 3600.k
or
1: q = ref Ð + 1800.k , k Î "
2 :q = 1800 - ref Ð + 360.k , k Î !
4 :q = 3600 - ref Ð + 360.k , k Î !
sin q = t , - 1 £ t £ 0
cos q = t , - 1 £ t £ 0
tan q = t , t Î !
3rd : q = 1800 + ref Ð + 3600.k
or
1: q = 1800 - ref Ð + 3600.k
1: q = 1800 - ref Ð + 1800.k , k Î "
4th :q = 3600 - ref Ð + 360.k , k Î !
2 :q = 1800 + ref Ð + 360.k , k Î !
or
TAKE NOTE OF THESE GENERAL TYPES OF EQUATIONS
Determine the general solutions for the following equations by considering the following worked-out examples:
1
2
1
sin q =
2
2sin q = 3cos q
3
cos q = cos(600 - a )
4
sin q = cos a
2
5.1 2 cos q = sin q - 2
2
5.2 cos x + sin 2 x - 1 = 0
0
0
5.3 cos q - 3 sin q = 3 and q Î éë-810 ; - 540 ùû
29
1.
sin q =
We are used in this type of equation
from grade 10. The only thing that is
new in grade 11 is general solution
which has been already explained in
page 14.
1
2
q = 300 + 3600.k
OR
q = 1800 - 300 + 360.k , k Î !
2.
2sin q = 3cos q
3
sin q = cos q
2
3
tan q =
2
q = 56,310 + 1800.k , k Î !
3.
cos q = cos(600 - a )
NB: when trigonometric functions
are not the same but angles the
same. Then, divide both sides by
cos q to get tan q . Do not divide by
cos q
sin q as you will get
.
sin q
\ q = ± ( 600 - a ) + 3600.k , k Î!
4.
sin q = cos a
(
sin q = sin 900 - a
)
ref Ðq = 900 - a
q = 900 - a + 3600.k ,
k Î!
OR
Since the functions are the same drop
down the angles.
Since angles are not the same, we cannot
divide by cosine on both sides to get a
tangent, which angle will tangent be taking
between the 2? Then introduction of
co-functions will be applicable to make the
the functions to be the same.
q = 1800 - (900 - a ) + 360.k
= 900 + a + 3600.k
30
5.
NOTE: If an equation does not look like 1-4 type
and contains more than 2 terms.
Use identities and factorise.
OR
5.1
2 cos q = sin 2 q - 2
-(1 - cos 2 q ) + 2 cos q + 2 = 0
-1 + cos 2 q + 2 cos q + 2 = 0
cos 2 q + 2 cos q + 1 = 0
• Terms more than 2, then 1-4 types not
applicable
• sin 2 q can be written in the terms of cos q
using square identities, e.g.,
sin 2 q = 1 - cos 2 q .
( cos q + 1) = 0
2
cos q = -1
q = ±1800 + 3600.k , k Î !
5.2
•
cos 2 x + sin 2 x - 1 = 0
cos 2 x + 2sin x cos x - 1 = 0
cos x + 2sin x cos x - (cos x + sin x) = 0
2
2
•
Terms more than 2, then 1-4 types not
applicable
Change of double angle to single angle as
•
Since we have sin x and cos x we need the
2
cos 2 x + 2sin x cos x - cos 2 x - sin 2 x = 0
identity of 1 in terms of sin x and cos x .
2sin x cos x - sin 2 x = 0
sin x(2 cos x - sin x) = 0
sin x = 0 or 2 cos x - sin x = 0
x = 00 + 3600.k , k Î !
OR
2 cos x - sin x = 0
sin 2 x = 2sin x cos x
1 = cos 2 x + sin 2 x
•
•
Simplification will lead to 2 terms, then
factorise
sin x = 2cos x functins not the same but
ratios the same then divide by cos x on both
sides to get tan x on the left hand side.
sin x = 2 cos x
tan x = 2
x = 63, 440 + 1800.k
31
5.3 Solve for θ if cos q - 3 sin q = 3 and
•
•
•
Terms more than 2, then 1-4 types not
applicable
Take all terms having square root to the same
side.
Then, square both sides of the equation and
also show squaring on your calculations
Write equation in its standard form
Factorise
•
Since the interval is q Î é-8100 ; - 5400 ù for
q Î éë-8100 ; - 5400 ùû
•
cos q - 3 sin q = 3
•
cos q = 3 + 3 sin q
cos q = 3 + 6sin q + 3sin q
2
2
1 - sin 2 q = 3sin 2 q + 6sin q + 3
4sin 2 q + 6sin q + 2 = 0
ë
û
values of θ, k-values must be -3 to - 2
2sin 2 q + 3sin q + 1 = 0
( 2sin q + 1)( sin q + 1) = 0
1
or sin q = -1
2
q = 2100 + 3600.k or q = 2700 + 3600.k , k Î !
or
sin q = -
q = 3300 + 3600.k
If k = - 2 or - 3
then
\ q = - 5100 or - 4500 or - 3900 or - 7500 or - 8100
32
TRIGONOMETRIC GRAPHS
Basic trigonometric graphs/functions of sine and cosine have same characteristics except
their shapes.
3 basic/mother trigonometric graphs/functions are shown below:
y = sin x, 00 £ x £ 3600
0
y = cos x, 00 £ x £ 3600
0
33
y = tan x, 00 £ x £ 3600
0
NB You need to be able to sketch, recognise and interpret graphs of the following:
•
y = a sin k ( x + p ) + q
•
y = a cos k ( x + p ) + q
•
y = a tan k ( x + p ) + q
Observe the effects of a, k, p and q on the basic graphs as shown below
Effects of a
a affects the amplitude of sine and cosine graphs. If a < 0 the basic graph flips along the xaxis,
34
Effects of k
k indicates the contraction or expansion of the graph.
k affects the period of the graph in the following way:
For sine and cosine graphs, the period becomes
The period of the tangent graph is
3600
k
180 0
.
k
35
Effects of p
p shifts the graphs horizontally (4 graphs shifting)
Effects of q
36
q shifts the graph vertically
BASIC PROPERTIES OF TRIGONOMETRIC GRAPHS
Worked-out Example 1
y = sin x
y = cos x
y = tan x
Period
Amplitude
Range
3600
1
[1 - (-1)] = 1
2
- 1 £ y £ 1 or
1
[1 - (-1)] = 1
2
- 1 £ y £ 1 or
tangent doesn’t
have a min/max yvalue.
yÎR
3600
180 0
y Î [ -1;1]
y Î [ -1;1]
\ amplitude not
available
Worked-out Example 2
Trigonometric graph
Period
Amplitude
Range
1
y = sin q
2
3600
1
2
-
y = -3 sin 2q
3600
= 1800
2
3
-3 £ y £ 3 or y Î [-3 ; 3]
1
y = cos q
2
3600
= 7200
1
2
1
-1 £ y £ 1 or y Î [-1 ; 1]
3
y = 4 cos q
4
360
= 4800
3
4
4
-4 £ y £ 4 or y Î [ -4 ; 4]
1
1
é 1 1ù
£ y £ or y Î ê - ; ú
2
2
ë 2 2û
37
Worked-out Example 1
1.
Use the sine graph given below to answer the following questions:
1.1
What are the minimum and maximum values of y = sin x ?
(2)
1.2
What is the domain and range of y = sin x
(4)
1.3
Write down the x-intercepts of y = sin x
(2)
1.4
What is the amplitude and period of y = sin x ?
(2)
38
Solutions
1.1
Minimum value = -1 üand maximum value = 1 ü
1.2
Domain : x Î - 3600 ;3600 , x Î R üü
1.3
x-intercepts: - 3600 ;-1800 ;0 0 ;1800 ;3600. üü
(2)
1.4
Amplitude is1 and period 3600 .üü
(2)
[
]
(2)
Range: [- 1;1] y Î R üü
(4)
39
Worked-out Example 2
2.
Consider a function g ( x) = - cos x + 1
[
]
2.1
sketch the graph of g for x Î - 3600 ;3600
2.2
write down the period and amplitude of g
2.3
write down the range of g
Solutions
2.1
2.2
period: 3600 amplitude: 1
2.3
Range: y Î [0;2] OR 0 £ y £ 2
40
Worked-out Example 3
3.1
[
Sketch a graph of y = tan( x + 60 0 ) - 1; x Î - 1500 ;1800
]
Solution
3.1
41
Worked-out Example 4
4.1
Sketch the graph of f ( x ) = sin 3x, x Î éë00 ; 3600 ùû
Some thought process:
•
This is a sine graph with k = 3
•
360 0
= 120 0
The period will be
3
•
For drawing our graph, we can divide all the x-values on the standard
sin graph by 3. That will mean our standard x-values when using table
{
}
now be: {0 ;30 ;60 ;90 ;120 ;...} x Î[0 ;360 ]
0
0
0
0
0
0
0
method: 0 ;90 ;180 ;270 ;360 x Î [0 ;360 ] will for the new graph
0
•
0
0
0
0
0
0
Finding table using Casio fx calculator:
Step 1: click on mode
Step 2: select table
Step 3: type sin 3x
Step 4: Start at 00and end at 3600 since x Î éë00 ;3600 ùû
Step 5: Step by period divide by number of quadrants
Final sketch
4.1
42
Worked-out Example 5
e
•
•
The graph has been reflected, \ - sin e graph, thus a = -1
The graph has shifted 20 0 to the right, p = -20 0
•
Middle y-value is - 1
1
Amplitude is [0 - (-2)] = 1
2
The equation is therefore: y = - sin( x - 20 0 ) - 1
•
•
43
GRAPHICAL INTERPRETATION
Worked-out Example 1
1.
Sketch the graphs of: y = 2 sin x and y = cos 2 x if -1800 £ x £ 1800 on the same system of axes.
1.1
1.2
For which value(s) of x is 2 sin x > 0 ?
For which value(s) of x is
1
cos 2 x - sin x = 0
2
Solutions
1.
1.1
00 < x < 1800
44
1.2
1
cos 2 x - sin x = 0
2
cos 2 x = 2sin x
1 - 2sin 2 x = 2sin x
2sin 2 x + 2sin x - 1 = 0
sin x =
sin x ¹
-2 ± (2) 2 - 4 ( 2 )( -1)
2 ( 2)
-2 - 12
4
or
-2 + 12
4
ref Ð = 21, 47 0
sin x =
x = 21, 27 0 or x = 158, 730
SOLVING 2D AND 3D PROBLEMS
In any triangle:
a
b
c
=
=
sin A sin B sin C
sin A sin B sin C
=
=
a
b
c
2
2
2
Cosine rule : a = b + c - 2bc cos A
Sine rule :
A
A
b 2 = a 2 + c 2 - 2ac cos B
bb
cc
c 2 = a 2 + b 2 - 2ab cos C
Area rule : A. DABC =
C
C
aa
B
B
1
ab sin C
2
1
ac sin B
2
1
= bc sin A
2
=
45
SINE RULE
•
Sine rule is applicable when given two sides and an angle in any triangle, then you can be able to calculate
the 2nd angle.
Worked-out Example:
ˆ
In DPQR, PQ=12cm, QR=10cm and R=80
. Determine the:
0
a) size P̂
b) length of PR
Solutions
a)
P
sin P sin 800
=
10
12
10 ´ sin 80
sin P =
12
0
-1 æ 10 ´ sin 80 ö
ˆ
P = sin ç
÷
12
è
ø
q
r = 12
800
R
p = 10 cm
Q
sin P sin R
=
p
r
Pˆ = 55.150
b)
For you to be able to get the length of PR you
will need to know Q̂ . Now you know two
angles in DPQR then you can get the 3rd one
by applying sum of angles in a D.
Qˆ = 44,850 [ sum of Ðs in a D ]
q
r
=
sin Q sin R
q
12
=
0
sin 44,85
sin 800
12 ´ sin 44,850
q=
sin 800
= 8,59 cm
•
Sine rule is also applicable when given two angles and a side, then you will be able to use it to calculate the
other sides as well as the 3rd angle
46
Worked-out Example 2
0
0
ˆ
ˆ
In DABC, A=50
, C=32
and AB=5cm . Determine:
a) the value of a length of BC.
Solution
b) the size of B̂
c) the value b length of AC
a)
a
c
=
sin A sin C
a
5
=
0
sin 50
sin 320
a = 7, 23
b)
Bˆ = 980
c)
b
c
=
sin B sin C
b
5
=
0
sin 98
sin 320
b = 9,34
C
320
a
b
B
c = 5 cm
A
[ sum of Ðs in a D ]
Then we know now all the angles
and lengths of the sides in this
triangle. You cannot use trig ratios
solving this triangle, as it is not a
right-angled triangle.
47
COSINE RULE
•
Cosine rule is applicable when given length of all the 3 sides of a triangle, you can be able
to calculate any angle in the triangle. The 2nd angle can be calculated by applying cosine
rule or sine rule it will depend on you.
Worked-out Example 1
In DDEF, DE = 7 cm, FE = 9 cm and Eˆ = 550 .
Determine the:
Solutions
a) length of DF
Applying cosine with sides and included
angle
b) size of F̂
a)
DF 2 = DE 2 + EF 2 - 2 DE.EF .cos Eˆ
= ( 7 ) + ( 9 ) - 2 ( 7 )( 9 ) cos 550
2
a)
DF =
2
( 7 ) + ( 9 ) - 2 ( 7 )( 9 ) cos 550
2
2
= 7, 60
b) b) To get the 2nd angle you can apply sine
D
e
f = 7cm
F
550
E
d = 9cm
rule as well but for now we are going to
apply cosine rule when having all the 3
sides. Since we are looking for F̂ , then
the side opposite to F̂ will be the subject
of the formula in this way.
DE 2 = EF 2 + DF 2 - 2.EF .DF cos Fˆ
2
7 2 = 92 + ( 7, 60 ) - 2.9.7, 60.cos Fˆ
9 + ( 7, 60 ) - 7
cos Fˆ =
(2)(9)(7, 60)
2
2
2
2
2 ö
æ 2
-1 9 + ( 7, 60 ) - 7
ˆ
÷
F = cos ç
ç 2(9)(7, 60) ÷
è
ø
0
Fˆ = 48,99
48
AREA RULE
•
Area rule is applicable when you are given two sides and included angle, then you can
calculate the area of the triangle.
Worked-out Example 1
In DABC, Â=500 , AC = 9,34 and AB= 5cm .
a) Determine the area of DABC
C
a
a)
b = 9, 34
B
500
c = 5 cm
A
49
3.4 TYPICAL EXAM QUESTIONS
Example 1
QUESTION 1
1.1
1.2
1.3
50
Solutions
1.1
7.1
h
In D ABC :sin q =
AB
AB =
7.2
1.2
ü
AB =
h
sin q
(1)
h
sin q
ˆ = Dˆ = 900 - q (sides opp = Ðs )
In D ABD : BAD
ˆ = 1800 ( Ðs of D )
900 - q + 900 - q + ABD
ˆ =D
ˆ = 900 - q üReason
ü BAD
(3)
ˆ = 2q
ABD
1.3
7.3
ˆ = 2q
ü ABD
AD
AB
=
sin 2q sin(900 - q )
h
´ sin 2q
AD = sin q
cos q
h
´ 2sin q cos q
sin
q
AD =
cos q
AD = 2h
ü
AD
AB
=
sin 2q sin(900 - q )
(3)
ü cos q
ü 2 sin q cos q
51
WORKSHEETS, QUESTION 2-5
QUESTION 2
2.1
2.2
2.3
2.3.1
2.3.2
2.3.3
2.4
2.4.1
2.4.2
52
QUESTION 3
3.1
3.1.1
3.1.2
3.1.3
3.2
3.2.
1
3.2.2
53
QUESTION 4
4.1
4.1.
1
4.1.2
4.2
4.3
4.4
4.4.1
4.4.2
54
55
QUESTION 2
2.1
2.2
56
2.3.1
2.3.2
57
QUESTION 3
3.1.
1
3.1.2
3.1.
3
58
3.1.3
3.2.1
3.2.2
59
QUESTION 4
4.1.
1
4.1.
2
4.2
60
4.2
4.2
61
4.3
62
4.4.1
4.4.2
63
64
65
66
WORKSHEETS, QUESTION 6-11
67
QUESTION 7
7.1
7.2
7.3
68
69
QUESTION7
7.1
7.2
7.3
70
QUESTION 8
8.1
8.1.1
8.1.2
71
8.2
8.2.1
8.2.2
8.2.3
72
QUESTION 9
9.1
9.1.1
9.1.2
9.1.3
73
9.2
74
QUESTION 10
10.1
10.2
10.3
75
QUESTION 11
11.1
11.2
11.3
76
QUESTION 8
8.1
8.1.1
77
8.1.2
8.1.2
78
8.2
8.2.1
79
8.2.2
8.2.2
8.2.3
8.2.3
80
QESTION 9
9.1.1
1.1.1
1.1.2
1.1.
3
9.1.2
9.1.2
81
9.1.2
9.1.3
9.1.2
9.2
9.2
82
9.2
9.2
83
QUESTION 10
10.1
10.2
84
10.3
10.3
85
QUESTION 11
11.1.1
11.1.1
11.1.2
11.1.2
86
11.2
11.2
11.3
11.3
87
2. EUCLIDEAN GEOMETRY
2.1
2.2
WORK COVERED
•
All theorems on straight lines, triangles and parallel lines
•
Theorem of Pythagoras
•
Similarity and Congruency
•
Midpoint theorem
•
Properties of quadrilaterals
•
Circle Geometry.
•
Proportionality theorems
•
Similar triangles
•
Theorem of Pythagoras (proof by similar triangles)
OVERVIEW OF TOPICS
GRADE 10
GRADE 11
GRADE 12
Ø Revise basic results
established in earlier
grades regarding lines,
angles and triangles,
especially the similarity
and congruence of
triangles.
Ø Investigate line
segments joining the
midpoints of two sides of
a triangle.
Ø Define the following
special quadrilaterals:
the kite, parallelogram,
rectangle, rhombus,
square and trapezium.
Investigate and make
conjectures about the
properties of the sides,
angles, diagonals and
areas.
Ø Investigate and
prove theorems of
the geometry of
circles assuming
results from earlier
grades, together
with one other result
concerning tangents
and radii of circles.
Ø Solve circle
geometry problems,
providing reasons
for statements when
required.
Ø Prove riders.
Ø Revise earlier (Grade 911) work on similar
polygons.
Ø Prove (accepting results
established in earlier
grades):
• that a line drawn parallel
to one side of a triangle
divides the other two
sides proportionally (and
the Mid-point theorem as
a special case of this
theorem);
• that equiangular triangles
are similar;
• that triangles with sides in
proportion are similar;
• the Pythagorean Theorem
by similar triangles and
riders
88
According to the National Diagnostic Reports, the previous learners had challenges
to:
•
Make assumptions on cyclic quadrilaterals as equal, right angles where there is none,
angles as equal, lines as parallel, etc.
• When proving cyclic quad or that a line is a tangent they use that as a reason in their
proof.
• State incomplete or incorrect reasons for statements
• Identify correct sides that are in proportion
• State proportions without reasons
• Write proof of a theorem without making the necessary construction
• Differentiate when to use similarity or congruency when solving riders
• Understand properties of quadrilaterals and also the connections between shapes, eg
(1) all squares are rectangles (2) all squares are rhombi (3) etc.
• Solve problems that integrates topics e.g. Trigonometry and Euclidean Geometry and
Analytical Geometry
• Prove cyclic quad or // lines or tangents
SUGGESTIONS TO ADDRESS THE CHALLENGES:
•
•
•
•
•
•
•
•
•
•
Scrutinise the given information and the diagram for clues about which theorems could
be used in answering the question.
Differentiate between proving a theorem and applying a theorem
Use the list of reasons provided in the Examination Guidelines.
Identify the correct sides that are in proportion
All statements must be accompanied by reasons. It is essential that the parallel lines be
mentioned when stating that corresponding angles are equal, alternate angles are
equal, the sum of the co-interior angles is 180° or when stating the proportional
intercept theorem.
Note that construction is necessary when proving theorems
Understand the difference between the concepts “similarity” and “congruency”
Revise properties of quadrilaterals done in earlier grades
Practise more exercises where the converses of the theorems are used in solving
questions
Practice solving problems that integrate topics e.g. Trigonometry and Euclidean
Geometry
89
Revision of earlier (Grade 9-10) Geometry
Note:
•
•
You must be able to identify, visualise theorems, axioms to apply in every situation.
When presented with a diagram they should be able to write the theorem in words.
Straight Lines
The sum of angles around a
Adjacent angles on a straight
Vertically opposite angles
point is 360 °
line are supplementary
are equal.
2
1
a b
c
1 O 3
4
2
B
In the diagram, Bˆ1 + Bˆ 2 = 180 °
Oˆ1 = Oˆ 3 and Oˆ 2 = Oˆ 4
Corresponding angles are
equal (F-shape).
Alternate angles are equal (z or
N-shape)
Co-interior angles are
supplementary (U-shape)
If AB//CD, then the
corresponding angles are
equal.
If AB//CD, then alternate angles
are equal.
If AB//CD, then co-interior
angles are supplementary.
In the diagram, a + b + c =
360°
Parallel Lines
A
C
1
B
1
A
D
C
Aˆ1 = Cˆ1
B
1
!!
Aˆ1 = D
1
1
D
1
1
Aˆ1A+ Bˆ1 = 180 °
90
B
Triangles
The interior angles of a triangle are
TheThe
interior
exterior
angles
angle
of aistriangle
equal toare
the sum
The ex
supplementary
supplementary
of the interior opposite angles
the inte
A
A
C
B
1
B
ˆ + Bˆ + Cˆ = 180°
A
In an equilateral triangle all sides are equal
and all angles are equal to 60°
ˆ ++ BB
ˆ!! = Cˆ
Cˆ = AˆA
2
Aˆ + Bˆ
In an equilateral triangle all sides are equal
Angles opposite equal sides are equal
and all angles are equal to 60°
Sides opposite equal angles are equal
Angles
Sides
A
A
B
2
C
C
ˆ = Bˆ = Cˆ = 60° , and AB = BC = AC
A
B
C
If AB =
!ˆ! = Cˆ
= AC , then B
ˆ =IfBˆ AB
A
= Cˆ = 60° , and AB = BC = AC
Conve
!ˆ! = Cˆ , then AB = AC
Conversely, if B
Congruency
Congruency of triangles (four conditions)
Condition 1
Two triangles are congruent if three sides
of one triangle are equal in length to the
three sides of the other triangle. (SSS)
91
Condition 2
Two triangles are congruent if two sides
and the included angle are equal to two
sides and the included angle of the other
triangle. (SAS)
Condition 3
Two triangles are congruent if two angles
and one side of a triangle are equal to two
angles and a corresponding side of the
other triangle. (AAS)
Condition 4
Two right-angled triangles are congruent if
the hypotenuse and a side of the one
triangle is equal to the hypotenuse and a
side of the other triangle. (RHS)
The Midpoint Theorem
A
D
B
A
E
D
C
If AD = DB and AE = EC , then DE // BC
1
and DE = BC
2
B
E
C
If AD = DB and DE // BC , then AE = EC and
1
DE = BC
2
92
Properties of Quadrilaterals: (Properties of quadrilaterals and their application are important
in solving Euclidean Geometry problems).
Parallelogram (Parm)
•
•
•
•
Opposite sides are parallel and equal in length
Opposite angles are equal
Diagonals bisect each other
Area = base x perpendicular height
Rhombus
..
x x
Square
x
..
.. . .
.. . .
x
•
•
•
•
•
All sides are equal in length
Opposite sides are parallel
Opposite angles are equal
Diagonals bisect each other at 90°
Diagonals bisect the corner angles
1
• Area =
(diagonal 1 x diagonal 2)
2
•
•
•
•
All sides are equal in length
Opposite sides are parallel
Corner angles equal 90°
Diagonals are equal and bisect each other at
90°
•
•
Diagonals bisect the corner angles
Area = side x side
•
•
•
•
Opposite sides are parallel and equal in length
Corner angles equal 90°
Diagonals are equal and bisect each other
Area = base x height
Rectangles
93
• Two pairs of adjacent sides are equal
Kite
• A single pair of opposite angles are equal
• Diagonals intersect each other at 90°
X
X
• One diagonal bisects the corner angle
• Shorter diagonal is bisected by the longer
diagonal
• Area =
1
(diagonal1 x diagonal2)
2
Trapezium
• At least one pair of opposite sides are parallel
• Area =
1
(sum of 2 //sides) x height
2
Area of Triangle
(a) The height or altitude of a triangle is always relative to the chosen base.
C
B
se
Ba
Height (h)
Base
t (h
igh
He
Bas
e
B
A
A
)
He
igh
A
t (h
C
B
Height (h)
A
)
C
B
Base
In all cases, the area of the triangles can be calculated by using the formula
Area of DABC =
1
(base) ´ (height )
2
94
C
(b) Two triangles which share a common vertex have a common height.
A
F
h
h
G
B
C
(c)
D
D
E
Triangles with equal or common bases lying between parallel lines have the same area
A
D
E
H
h
A
h
h
A
h
B
Base
C
F
Base
G
95
GRADE 11-12 EUCLIDEAN GEOMETRY
NOTE: Grade 11 Geometry is very important as it is examinable in full with the Grade 12
Geometry. The nine circle geometry theorems must be understood and mastered in order to
achieve success in solving riders.
KEY CONCEPTS
Proofs of the following theorems are examinable:
Ø The line drawn from the centre of a circle perpendicular to a chord bisects the
chord;
Ø The line drawn from the centre of a circle to the meet point of the chord is perpendicular
to the chord;
Ø The angle subtended by an arc at the centre of a circle is double the size of the
angle subtended by the same arc at the circle (on the same side of the chord as
the centre);
Ø The opposite angles of a cyclic quadrilateral are supplementary;
Ø The angle between the tangent to a circle and the chord drawn from the point of
contact is equal to the angle in the alternate segment;
Ø A line drawn parallel to one side of a triangle divides the other two sides
proportionally;
Ø Equiangular triangles are similar
96
The line drawn from the centre of a circle perpendicular to a chord bisects the chord
GIVEN: The line drawn from the centre
of a circle perpendicular to a chord
R.T.P. bisects the chord
GIVEN: Circle O and OM ^ AB
R.T.P. : AM = MB
O
Construction: Draw radii OA and OB
Proof:
In D OAM and D OBM,
OA = OB . . .
A
M
B
( Radii)
OM = OM . . . (Common)
OM̂A = OM̂B . . . (Each = 90° )
\DOAM º DOBM . . . (RHS)
\ AM = MB
. . . (From congruency)
97
NOTE: Conversely, a line segment drawn from the centre of a circle to the midpoint of a
chord, is perpendicular to the chord
Example:
O
Given: Circle with centre O and chord AB. OC ^ AB, cutting AB at D,
with C on the circumference. OB = 13 units and
AB = 24 units. Calculate the length of CD.
A
D
B
C
AD = DB . . . (Line from centre ^ chord)
But AB = 24 units . . . . (Given)
∴ DB = 12 units
In D ODB,
𝑂𝐵! = 𝑂𝐷! + 𝐷𝐵! . . .(Pythagoras)
13! = 𝑂𝐷! + 12!
𝑂𝐷! = 13! − 12!
OD = √169 − 144
= 5 units
But OB = OC = 13 units . . . . (Radii)
And 𝐶𝐷 = 𝑂𝐶 − 𝑂𝐷 = 13 – 5
= 8 units
98
The angle subtended by an arc at the centre of a circle is double the size of the angle
subtended by the same arc at the circle (on the same side of the chord as the centre)
GIVEN: The angle subtended by
an arc at the centre of a circle
R.T.P. is double the size of the angle subtended
by the same arc at the circle
FORMAL PROOF:
Given
: Circle with centre O and arc AB subtending AÔB at the centre and AĈB at the circle
R.T.P
: AÔB = 2 ´ AĈB
Construction : Draw CO and produce
Diagram 1
Diagram 2
C
1
O
A
C
B
2
1
Diagram 3
1
2
2
O
1
B
O
2
A
1 2
A
1
C
2
B
Proof:
Oˆ1 = Cˆ1 + Aˆ . . . . . . . (Ext Ðs of D = sum of int. opp Ðs )
ˆ . . . . . . . ( Ðs opp = sides OA and OC radii)
But Cˆ 1 = A
\ Oˆ1 = 2 Cˆ1
ˆ = 2 Cˆ
similarly O
2
2
(
ˆ + Oˆ = 2 Cˆ + Cˆ
In Diagram 1 & 2: O
1
2
1
2
In Diagram 3:
(
)
\ Oˆ 2 - Oˆ1 = 2 Cˆ 2 - Cˆ1
)
\ AOˆ B = 2 ´ ACˆ B
\ AOˆ B = 2 ´ ACˆ B
99
The inscribed angle subtended by the diameter of a circle at the circumference is a right angle.
(∠ in a semi-circle).
In the diagram alongside, PT is a diameter of the
circle with centre O. M and S are points on
the circle on either side of PT. MP, MT, MS
and OS are drawn. M̂ = 37° .
Calculate, with reasons, the size of:
a) M̂ 1
b) Ô1
SOLUTION:
a) PMˆ T = 90° . . . . ( Ðs in a semi - circle )
Mˆ 1 = 90° - 37°
C
Mˆ 1 = 53 °
b) Oˆ 1 = 2(53°) = 106 ° . . . . ( Ð at centre = 2 ´ Ð at circumference)
A
M
B
NOTES:
a). If, for any circle with centre M, point B moves in an
anticlockwise direction, it reaches a point where arc AB
becomes a diameter of the circle. In that case, arc AB subtends ∠AMB
at the centre and ∠ACB at the circumference. Using the above theorem
and the fact that ∠AMB is a straight angle, it can be deduced that ∠ACB = 90° .
b). Equal chords subtend equal angles at
the centre and at the circumference.
100
Example:
In the accompanying diagram, PR and PQ are equal chords of the circle
with centre M. QS ^ PR at S. PS = x units and MR is drawn.
a). Express, with reasons, QS in terms of x.
(5)
S
x
P
1
b). If x = √12 units and MS = 1 unit, calculate the
length of the radius of the circle.
c). Calculate, giving reasons, the size of ∠P.
M 2
(2)
(5)
SOLUTION:
a). PS = SR = x
R
Q
………. (Line from centre ^ chord)
∴ PR = PQ = 2x ………. (Equal chords, given)
In D PQS,
PQ 2 = QS 2 + PS 2 …….. (Pythagoras)
QS 2 = (2 x) 2 - x 2
= 3x 2
QS =
3 x units
b). Radius = QS – SM
3 x -1
=
=
3 12 - 1
= 6–1
= 5 units
c). tan P =
QS
6
=
PS
2 3
=
3
3
=
3
\ P̂ = 60°
101
The opposite angles of a cyclic quadrilateral are supplementary
Note that all 4 vertices of a quadrilateral must lie on the same circle for the quadrilateral to be cyclic.
GIVEN: The opposite angles of a cyclic
quadrilateral
R.T.P. are supplementary
Given any circle with centre O, passing through the vertices of cyclic quadrilateral ABCD
R.T.P.: Â + Ĉ = 180° and B̂ + D̂ = 180°
Construction
: Draw BO and OD
Proof:
A
D
O
Ô 2 = 2Â ( Ð at the centre = 2 ´ Ð at circle)
1
2
Ô 1 = 2 Ĉ ( Ð at the centre = 2 ´ Ð at circle)
Ô 1 + Ô 2 = 2 (Â + Ĉ)
but
Ô 1 + Ô 2 = 360° (Ð' s around a point)
hence
 + Ĉ = 180°
also
B̂ + D̂ = 180° (sum of int Ð' s of quad)
C
B
102
Example
D, E, F, G and H are points on the circumference of a circle.
Gˆ1 = x + 20 ° and Hˆ = 2x + 10° DE || FG.
a) Determine the size of DÊG in terms of x
b) Calculate the size of DĤG
SOLUTION
L
a) D E G = 180 - (2 x + 10) (opposite angles of a cyclic quadrilateral)
180 - 2 x - 10
= 170 - 2 x
L
L
b) G1 = E ( alt <s DE // FG )
x + 20 = 170 - 2 x
3x = 150
x = 50 0
Ù
\ D H G = 70 0
103
Angle between a tangent and a chord is equal to the angle in the alternate segment.
GIVEN: Angle between a tangent and the chord
R.T.P. is equal to the angle in the
alternate segment
GIVEN: MKˆ L OR Kˆ 1 with chord KL and tangent MK
RTP: MKˆ L = Nˆ
D
N
N
D
N
1
.O
O
1
2
. 1
L
2
2
1
K
M
Construction: Draw diameter KOD
and join DL
Proof:
Kˆ 1 + Kˆ 2 = 90° …..tan ^ rad
Kˆ 1 = 90° - Kˆ 2
Dˆ = 90° - Kˆ 2
Dˆ = Kˆ ……. Both = 90° - Kˆ
2
ˆ = Nˆ . . . Ðs in same segment
But D
\ Kˆ 1 = Nˆ …..both = D̂
1
M
Construction: Join OL and OK
Proof:
Kˆ 1 + Kˆ 2 = 90° . . . tan ^ rad
Kˆ = 90° - Kˆ
2
Oˆ1 + Kˆ 2 + Lˆ1 = 180 ° . . . sum of Ðs
in a D
Kˆ 2 + Lˆ1 = 180° - Oˆ1
Oˆ = 2 Nˆ . . . Ð at centre = 2 Ð at
1
circumf.
Kˆ 2 + Lˆ1 = 180 ° - 2 Nˆ
But Kˆ = Lˆ . . . Ðs opp = sides
2
1
Kˆ 2 = 90° - Nˆ
Nˆ = 90° - Kˆ ; \ Kˆ = Nˆ
2
.O
L
2
1
K
1
DLˆ K = 90° …. Ðs at semi-circle
Kˆ 2 + Dˆ = 90° ….sum of Ðs in a D
1
2
L
1
M
K
Construction: Join OK and
extend to D on the
circumference.
Join ND.
Kˆ 1 + Kˆ 2 = 90° . . .
tan ^ rad
Nˆ 1 + Nˆ 2 = 90° . . . Ðs Ð in
the semi-circle
Kˆ 1 + Kˆ 2 = Nˆ 1 + Nˆ 2 . . both
= 90°
But Nˆ 2 = Kˆ 2 . . . Ðs in
same segment
\ Kˆ 1 = Nˆ 1
1
104
PROPORTIONALITY
•
It is important to stress to learners that proportion gives no indication of actual length. It
only indicates the ratio between lengths.
•
Make sure that you know the meaning of ratios. For example, the ratio
AB 2
does not
=
BC 3
necessarily mean that the length of AB is 2 and the length of BC is 3.
The line drawn parallel to one side of a triangle divides the other two sides proportionally
GIVEN: The line drawn parallel
to one side of a triangle
R.T.P. divides the other two sides proportionally
105
Given : D ABC, D lies on AB and E lies on AC. And DE // BC.
R.T.P.:
AD AE
A
=
DB EC
A
h
h
E
B
D
D
0
E
k
B
A
E
C
C
B
1
AD ´ h
Area DADE 2
AD
=
=
( same height )
Area DBDE 1 DB ´ h DB
2
1
AE ´ k
Area DADE 2
AE
=
=
( same height )
Area DCED 1
EC
EC ´ k
2
Make sure the height used corresponds with the correct base as indicated in the construction
But Area DBDE = Area DCED (same base and between // lines)
\
Area D ADE Area D ADE
=
Area D BDE
Area DCED
\
D
k
k
C
h
AD AE
=
DB EC
106
SIMILARITY THEOREM:
•
Know the conditions under which two triangles are similar
•
Note that to prove that sides are in proportion, similarity of triangles is proved and not
congruency
When proving that the two triangles are similar, make sure that the equal angles
correspond: i.e. if given that DABC /// DBDA then you cannot say that DABC /// DABD
•
•
•
To prove triangles are similar, we need to show that two angles (AAA) are equal OR
three
sides in proportion (SSS).
The examples on similar triangles illustrate a highly systematic and effective strategy
which has been used in the teaching of triangle geometry.
Equiangular triangles are similar
GIVEN: Equiangular triangles
R.T.P. are similar
In DAXY and DDEF
AX = DE (constructi on)
AY = DF (constructi on)
\ AXˆY = Bˆ
Þ XY // BC (corresponding Ðs =)
AB AC
=
(line // one side of D)
AX AY
but AX = DE and AY = DF (constructi on)
AB AC
\
=
B
DE DF
D
A
 = D̂ ( given )
\ DAXY º DDEF
(SAS )
AXˆY = Eˆ but Eˆ = Bˆ (given)
X
Y
C
Similarly by marking off equal lengths on BA and BC, it can be shown that:
\
F
E
AB BC
=
DE EF
AB AC
BC
=
=
DE DF
EF
107
THEOREMS AND THEIR CONVERSES
(Opp. Sides parm =)
(Conv. Opp. Sides parm)
If given a parallelogram, then the opposite sides
of the parm are equal
If opposite sides of a quadrilateral are equal,
(diags. parm)
If given a parallelogram, then the diagonals bisect
each other
(opp. Ð' s parm)
then the quadrilateral is a parallelogram
(Conv. diags. parm)
If diagonals of a quadrilateral bisect each other,
then the quadrilateral is a parallelogram
(conv. opp. Ð' s parm)
108
(Sides of rhomb.)
(diags rhomb.)
(conv. sides of rhomb.)
(conv. diags rhomb.)
(conv. diags rhomb.)
(diags rhomb.)
Conv. rectangle
Conv. Ð' s in the same segment
109
Conv. opp. Ð' s cyclic quad
(conv. ext, Ð' s cyclic quad.)
110
NOTE: success in answering Euclidean Geometry comes from regular practice,
starting off with the easy and progressing to the difficult.
Important points about solving riders in Geometry
1
Read the problem carefully for understanding. You may need to underline important
points and make sure you understand each term in the given and conclusion. Highlight
key word like centre, diameter, tangent, because they are linked to theorems you would
need to solve riders.
2
Draw the sketch if it is not already drawn. The sketch need not be accurately drawn
but must as close as possible to what is given i.e. lines and angles which are equal
must look equal or must appear parallel etc. Also indicate further observations based
on previous theorems.
3
Indicate on the figure drawn or given all the equal lines and angles, lines which are
parallel, drawing in circles, measures of angles given if not already indicated in the
question. Put in the diagram answers that you get as you work along the question; you
may need to use them as you work along the question. It might be more helpful to have
a variety of colour pens or highlighters for this purpose.
4
Usually you can see the conclusion before you actually start your formal proof of a
rider. Always write the reason for each important statement you make, quoting in brief
the theorem or another result as you proceed.
5
When proving similar triangles, the triangles are already similar, you just need to
provide reasons for similarity. It helps to highlight the two triangles so that it will be
easy to see why corresponding angles are equal. Do not forget to indicate the reason
for similarity that is AAA or ÐÐÐ .
Answers must be worked out sequentially, there’s always a way out.
6
7
9
Sometimes you may need to work backwards, asking yourself what I need to show to
prove this conclusion (required to be proved) and then see if you can prove that as
you reverse. NB, do not use answer/ what is supposed to prove in the proof.
WRITE GEOMETRY REASONS CORRECTLY. Refer to acceptable reasons as
reflected in the Examination Guidelines.
111
2.1 PRACTICE EXERCISES
•
Diagrams are not drawn to scale
•
Refrain from making assumptions. (For example, if a line looks like a tangent, but
no tangent is mentioned in the description statement, the three theorems
associated with a tangent cannot be applied. Sometimes the examiner may want
you to prove that it is a tangent)
QUESTION 1
Are the following pairs of triangles similar? Give a reason for your answer.
(a)
P
(b)
A
10
8
L
8
B
6
C
M
4
5
4
3
A
N
Q
1,5
C
R
3
B
QUESTION 2
In the accompanying figure, AOB is a diameter
B
of the circle AECB with centre O.
OE // BC and OE meets AC at D.
O
B and E are joined.
A
D
C
E
112
2.1
Prove that AD = DC
2.2
Prove that EB bisects ABˆ C
2.3.
If EBˆ C = x , express BAˆ C in terms of x.
QUESTION 3
In the diagram alongside, BC and CAE are tangents to circle DAB and BD = BA.
E
D 2
3
1
1 2
B
3.1
A
2 1
C
Prove that
3.1.1
Dˆ 2 = Aˆ 2 + Aˆ 3
3.1.2
DA // BC
ED EA
=
AB AC
3.2
Hence, deduce that
3.3
Calculate the length of AB, if it is further given that EC : EA = 5 : 2 and ED = 18 units.
113
3.4
Prove that ∆ EDA ||| ∆ EAB.
QUESTION 4
In the diagram, BC = 17 units, where BC is a diameter of the circle. The length of the
chord BD is 8 units.
The tangent at B meets CD produced at A.
B
8
E
C
F
A
D
4.1 Calculate, with reasons, the length of DC
4.2 E is a point on BC such that BE : EC = 3 : 1. EF is parallel to BD with F on DC.
4.2.1 Calculate, with reasons, the length of CF
4.2.2 Prove that ΔBAC /// ΔFEC
4.2.3 Calculate the length of AC
114
QUESTION 5
In DADC , E is a point on AD and B is a point on AC such that EB//DC.
F is a point on AD such that FB//EC.
It is also given that AB = 2BC
5.1
Determine the value of AF: FE
5.2
Calculate the length of ED if AF = 8cm
115
QUESTION 6
In the accompanying figure, PQRS is a cyclic
quadrilateral with RS = QR. A straight
line (not given as a tangent) through R,
parallel to QS, meets PS produced at T.
T
S
1
P
3
2
1
2
P and R are joined.
V
2
If Rˆ3 = x
1
Q
4
6.1
Prove giving reasons that RT is a tangent to the circle at R
6.2
Prove that Rˆ1 = Tˆ
6.3
Prove that DRST /// DPQR
6.4
If PQ = 4cm and ST = 9cm, Calculate the length of QR
3
R
116
QUESTION 7
In the figure PY is a diameter of the circle
and X is on YP produced. XT is a tangent to
the circle at T and XB is
perpendicular to YT produced.
7.1
7.2
Prove that BX // TP
Prove that
XB XT
=
YB YX
117
QUESTION 8 (WC 2016 Trial)
In the diagram, O is the centre of the circle. A, B, C, D and E are points on the
circumference of the circle. Chords BE and CD produced meet at F. Ĉ = 100°, F̂ = 35° and
AEˆB = 55° .
8.1
8.2
Calculate, giving reasons, each of the following angles:
8.1.1
Â
(2)
8.1.2
Ê1
(2)
8.1.3
D̂1
(2)
Prove, giving reasons, that 𝐴𝐵 ∥ 𝐶𝐹.
(4)
118
QUESTION 9 (GDE, 2017 Trial)
In the diagram below, O is the centre of the circle. C is the midpoint of chord BD.
Point A lies within the circle such that BA ^ AOD .
9.1
Show that DA.OD = OD 2 + OD.OA .
(1)
9.2
Prove that 2 DC 2 = OD 2 + OD.OA
(7)
[8]
119
QUESTION 10
Determine, with reasons, y in terms of x.
[6]
120
QUESTION 11 (GDE, 2018 TRIAL)
In the diagram below, O is the centre of the circle. ABCD is a cyclic quadrilateral. BA
and
CD are produced to intersect at E such that AB = AE = AC.
11.1
11.2
Determine each of the following angles in terms of x:
11.1.1
B̂ 2
(2)
11.1.2
Ê
(5)
11.1.3
C2
(3)
If Eˆ = Cˆ 2 = x , prove that ED is a diameter of circle AED.
(4)
[14]
121
QUESTION 12 (WC, Sept. 2015)
12.1
Complete the following statement:
If two triangles are equiangular, then the corresponding sides are …
12.2
In the diagram, DGFC is a cyclic quadrilateral and AB is a tangent to the circle
at B. Chords DB and BC are drawn. DG and CF produced meet at E and DC is
produced to A. EA | | GF.
E
1 2
G 2
1
1
D
2
1
1
2
2
2
F
C
1
2
A
1
B
12.2.1 Give a reason why Bˆ = Dˆ .
1
1
(1)
12.2.2 Prove that DABC /// DADB.
(3)
12.2.3 Prove that Eˆ = Dˆ .
2
2
(4)
12.2.4 Prove that AE 2 = AD ´ AC.
(4)
12.2.5 Hence, deduce that AE = AB.
(3)
122
QUESTION 13 (GDE, 2016 Trial)
In the diagram below NE is a common tangent to the two circles. NCK and NGM are
double
chords. Chord LM of the larger circle is a tangent to the smaller circle at point C. KL,
KM and
CG are drawn.
Prove that:
13.1
KC MG
=
KN MN
(4)
13.2
KMGC s a cyclic quadrilateral if CN = NG.
(3)
13.3
DMCG /// DMNC.
(3)
13.4
MC 2 KC
=
MN 2 KN
(4)
123
QUESTION 14 (WC, 2016 Trial)
In the diagram, P, S, G, B and D are points on the circumference of the circle such
that PS || DG || AC. ABC is a tangent to the circle at B. GBˆ C = 𝑥.
14.1
Give a reason why Gˆ 1 = x.
14.2
Prove that:
14.2.1
BE =
BP.BF
BS
(1)
(2)
14.2.2
DBGP /// DBEG
(4)
14.2.3
BG 2 BF
=
BP 2 BS
(3)
[10]
124
QUESTION 15 (WC, 2016 Trial)
15.1
15.2
Calculate, giving reasons, the length of:
15.1.1
FC
(3)
15.1.2
BD
(4)
Determine the following ratio:
Area DECF
Area DABC
(4)
[11]
125
QUESTION 16 (DBE, Nov. 2017)
16.1 Give a reason why:
16.1.1
Dˆ 3 = 90°
16.1.2 ABDE is a cyclic quadrilateral.
16.1.3
Dˆ 2 = x
(1)
(1)
(1)
16.2 Prove that:
16.2.1 AD = AE
16.2.2
DADB /// DACD
(3)
(3)
16.3 It is further given that BC = 2AB = 2r.
16.3.1 Prove that AD 2 = 3r 2 .
(2)
16.3.2 Hence, prove that DADE is equilateral.
(4)
126
QUESTION 17 (GDE, 2018 Trial)
17.1
Prove that
BC DA
=
YZ DX
(5)
PRACTICE EXERCISE SOLUTIONS
QUESTION 1
1.1
1.2
YES,
LN LM MN 1
=
=
=
AB AC BC 2
NO
127
QUESTION 2
2.1
Ĉ = 90° ….. (Ð at semi-circle)
ˆ A = 90°..…(corr. Ðs = (OE // BC))
OD
AD = DC…..(line segment from centre ^ to
chord
EBˆ C = BEˆO ……(alt. OE // BC )
but EBˆ O = BEˆ O ……(Ðs opp = sidesradii (OE = OB))
\ EBˆ C = EBˆ O
bisects the chord)
2.3
2.2
\ EB bisects ABˆ C
AOˆ E = 2 ABˆ E ……. (at centre = 2 ´ at
circumf.)
but ABˆ E = EBˆ C = x...( EB bisects ABˆ C )
AOˆ E = 2 x
In DABC, BAˆ C + 90° + 2 x = 180°
BAˆ C = 90° - 2 x
QUESTIONS 3
3.1.1
3.1.2
Aˆ3 = Bˆ1
(tan chord)
ˆ
Bˆ 2 = D
1
ˆ + Bˆ
ˆ =A
D
2
2
1
(exterior Ð of a ∆)
ˆ
ˆ =A
D
1
2
\ Dˆ 2 = Aˆ 2 + Aˆ3
(tan chord)
( Ðs opposite
= sides)
\ Bˆ 2 = Aˆ 2
\DA // BC
(alternate Ðs =)
128
3.2
(proportionality theorem)
ED
EA
=
DB
AC
But DB = AB
(given)
EC
5
=
EA
2
EA + AC
5
=
EA
2
\ ED = EA
AB
3.3 EC:EA = 5:2
1+
AC
AC
5
=
EA
2
AC
3
=
EA
2
3.4
EA
2
=
AC
3
In ∆ EDA and ∆ EAB
Dˆ 2 = Aˆ 2 + Aˆ3
(proved)
Aˆ3 = Bˆ1
(tan chord)
Eˆ = Eˆ
(common)
\∆ EDA ||| ∆ EAB
\
ED
EA
=
AB
AC
18
2
=
AB
3
AB = 27 units
(Ð Ð Ð)
QUESTION 4
BDˆ C = 90 o.....(Ð in semi - circle)
2
2
2
4.1 BC = DC + DB ( Pythagoras theorem)
DC 2 = 17 2 - 82
DC = 15
4.2.1
CF CE
=
(line // one side of D)
CD CB
CF 1
=
15 4
4CF = 15
\ CF = 3,75
129
4.2.2
4.2.3
ABˆ C = 90 o.....(tan ^ rad )
EC =
BDˆ C = 90 o.....(Ð in semi - circle )
EFˆC = BDˆ C = 90 o.....(Corr Ðs, EF // BD)
In ΔBAC and ΔFEC
1
´ 17 = 4,25
4
AC BC
=
(DBAC /// DFEC )
EC FC
AC
17
=
4,25 3,75
AC = 19,27
Cˆ = Cˆ ...... (common)
ABˆ C = EFˆC = 90 o ( proven above)
BAˆ C = FEˆ C (3rd Ð of D)
\ DBAC /// DFEC ...( AAA)
QUESTION 5
5.2
AF 2
=
FE
1
5.1
AF 8
FE =
= = 4cm
2
2
AE = 12cm
QUESTION 7
7.1
Tˆ3 = 90° [Ðs in semi circle ]
XBˆ Y = 90° [ given]
\ Tˆ3 = XBˆ Y [both = 90°]
BX // TP [corresp Ðs =]
ED 1
=
[ BE // DC ; prop theorem ]
AE 2
ED 1
=
12 2
ED = 6cm
In DXBT and DXBY
XBˆ T = XBˆ Y [common Ð]
Xˆ 2 = Tˆ2 [alt Ðs ; BX // TP ]
7.2 Tˆ = Yˆ [tan chord theorem]
2
Xˆ 2 = Yˆ [both = Tˆ2 ]
Tˆ = BXˆY [3rd Ð of the D]
1
\ DXBT /// DXBY [ÐÐÐ]
XB XT
=
[/// Ds]
YB YX
130
QUESTION 8
8.1.1
BÂE = 90°
Ð semi circle
üS üR
(2)
8.1.2
Ê 1 = 80°
opp angles cyclic quad
üS üR
(2)
8.1.3
Dˆ 1 = 45°
ext Ð of D FED
üS üR
8.2
Bˆ1 = 35°
Interior Ð of D
üS üR
Fˆ = 35°
given
\ AB || CF
(4)
üS
üR
Altternate Ðs =
[10]
QUESTION 9
9.1
DO. OD = OD(OD + OA)
ü OD(OD + OA)
= OD 2 + OD.OA
9.2
(1)
In ΔDAB and ΔDCO
D̂ = D̂
Ĉ 2 = 90°
(common )
(line from centre to midpt of a chord/Midp t thm )
üS
D̂ = D̂
üS Ĉ 2 = 90° üR
Ĉ 2 = Â
B̂ = Ô3
(3 Ð of a D)
\ ΔDAB||| ΔDCO
\
rd
(ÐÐÐ)
DA AB DB
=
=
DC CO DO
üS Ĉ 2 = Â
üS B̂ = Ô3
DA AB DB
=
=
DC CO DO
DA.DO = DC.DB
ü
OD 2 + OD.OA = DC.2DC
ü
OD 2 + OD.OA = 2DC 2
OD 2 + OD.OA = 2DC 2
(7)
131
[8]
QUESTION 10
10
PT̂R = 90°
(Ð in semi - circle)
üS/R
x = 900 + R̂
(ext Ðof D)
üS/R
\ R̂ = x - 90°
ST̂P = x - 90°
(tan chord theorem)
x + x - 90° + y = 180°
(sum of Ðs in D)
\ y = 270° - 2 x
üS üR
üS
üanswer
(6)
QUESTION 11
11.1.1 In DOBC
B̂2 = Ĉ3
(Ðs opposite = radii)
B̂ 2 = 90° - 2 x
(sum of Ð' s of a D)
üS üR
ü B̂ 2 = 90° - 2 x
(2)
11.1.2
üS üR
 3 = 2 x
(Ð at centre = 2 ´ Ðat circumference)
Â3 = Ĉ1 + Ê
(ext Ðof D)
But AB = AC = AE
(given)
Ĉ1 = Ê
(Ðs opposite = sides)
\ Ê = x
11.1.3
B̂1 + B̂2 = Cˆ 2 + Ĉ3
üS
üS
ü Ê = x
(5)
(Ðs opposite = sides)
B̂1 = Cˆ 2 = 180 ° - (2 x + 90° - 2 x + 90° - 2 x )
üS
(sum of Ð' s of a D)
üS
\ Ĉ 2 = x
üS
(3)
132
11.2
(ext.Ðs of a cyclic quadrilateral) üS üR
Â1 = Ĉ
Â1 = 90° - 2 x + x + x
Â1 = 90°
(line subtends90° Ð)/
ED is a diameter of circle
(converseof Ð in a semi - circle)
ü Â1 = 90°
üR
(4)
QUESTION 12
12.1
tangent-chord theorem
12.1.2
In DABC and DADB :
Â1 = Â1
(1)
(common)
B̂1 = D̂1
(provedin10.2.1)
\ DABC ||| DADB
(ÐÐÐ)
üS
üS
üR
OR
Â1 = Â1
(common)
B̂1 = D̂1
(provedin10.2.1)
BĈA = B̂ 2
(Ðs of a Δ = 180°)
\ DABC ||| DADB
12.1.3
Ê 2 = F̂1
(alternateÐs ; EA || GF)
F̂1 = D̂ 2
(ext.Ðs of a cyclic quad DGFC)
\ Eˆ 2 = D̂ 2
üS
üS
üR
(3)
üSüR
üSüR
(4)
133
12.1.4
In DAEC and DADE :
(common)
 2 =  2
(provedin10.2.3)
Ê 2 = D̂ 2
(ÐÐÐ)
\ DAEC ||| DADE
\
AE AC
=
AD AE
üS
üS
üR
üS
\ AE 2 = AD ´ AC
OR
In DAEC and DADE :
 2 =  2
(common)
Ê 2 = D̂ 2
(provedin10.2.3)
AĈE = Ĝ 1
(Ðs of a Δ = 180° OR ext Ð of cyclicquad DGFE)
\ DAEC ||| DADE
\
üS
üS
üR
AE AC
=
AD AE
üS
\ AE = AD ´ AC
2
(4)
12.1.5
AB AC
=
AD AB
(DABC ||| DADB)
üS
AB = AD ´ AC
2
= AE2
\ AB = AE
(from10.2.4)
üS
üS
(3)
[16]
134
QUESTION 13
13.1
N̂1 = Ĉ 4
(tan chord theorem)
N̂1 = K̂ 2
(tan chord theorem)
üS/R
üS/R
\ Ĉ 4 = K̂ 2
CG || KM
(corresp Ðs =)
KC MG
=
KN MN
(line|| to one sideof D OR prop theorem)
üS/R
üR
(4)
13.2
Ĉ 4 = K̂ 2
Ĉ 4 = Ĝ 2
(proved)
(Ðs opposite = sides)
ü Ĉ 4 = Ĝ 2 üR
\ Ĝ 2 = K̂ 2
\ KMGC is a cyclic quad
13.3
(ext Ð = int opp Ð)
üR
(3)
In D MCG and D MNC :
M̂ 2 = M̂ 2
(common)
Ĉ3 = N̂ 2
(tan chord theorem)
Ĝ1 = Ĉ3 + Ĉ4
(sum of Ðs in D)
\ Δ MCG ||| Δ MNC (ÐÐÐ)
üS
üS/R
üR
(4)
135
13.4
MC MN
=
MG MC
(D |||)
s
üS/R
MC2 = MG . MN
MC2 MG . MN
=
MN 2
MN 2
=
üS
MG
MN
KC MG
=
KN MN
(proved)
MC 2 KC
=
MN 2 KN
üS
üS
(4)
[19]
QUESTION 14
14.1
alt Ðs , YT || RQ
14.2.1
BP BS
=
BE BF
BE 2 =
üR
(Prop theorem,EF || PS)
BP . BF
BS
üSüR
(2)
14.1.2 In D BGP and D BEG :
1) Ĝ 1 = P̂1
(tan chord theorem)
2) B̂ = B̂
(common)
\ Δ BGP ||| Δ BEG (ÐÐÐ)
üSüR
üS/R
136
OR
üS/R
In D BGP and D BEG :
(tan chord theorem)
1) Ĝ 1 = P̂1
2) B̂ = B̂
(common)
3) BĜP = BÊG (sum of Ðs in D)
üSüR
üS/R
\ Δ BGP ||| Δ BEG
üS
(4)
14.1.3
BG BP
=
BE BG
Δ BGP ||| Δ BEG
\BG 2 = BP . BE
BG 2 = BP .
BG 2 =
\
üS
BP . BF
BS
üS
2
BP . BF
BS
üSubst
2
BG
BF
=
2
BP
BS
(4)
[10]
QUESTION 15
15.1.1
FC 4
=
20 5
(EF || AD)
\FC = 16
15.1.2
36 4
=
DB 5
üS üR
üanswer
(3)
(DE || AB)
ü DC = 16
üS üR
\DB = 45
137
üanswer
(4)
15.2
1
. 4k . 8 . sin C
Area of D ECF
2
=
Area of D ABC 1 . 9k . 81. sin C
2
1
ü .4k .8. sin C
2
1
ü .9k .40.5. sin C
2
Area of D ECF 32
=
Area of D ABC 81
üüanswer
QUESTION 16
16.1.1 Angles in a semi-circle
üR
(1)
16.1.2
Exterior Ð of a quad = opp interior Ð
OR
Opp Ð s of a quad supplement ary
üR
(1)
16.1.3 tangent chord theorem
üR
(1)
16.2.1 In Δ AEC
Ê = 180° - (90° + x )
(sum of Ðs in D)
Ê = 90° - x
D̂1 = 180 ° - (90° + x )
(Ðs on a straight line)
üS
üS
D̂1 = 90° - x
\AD = AE
(sides opp = Ðs)
üR
(3)
138
16.2.2 In Δ ADB and Δ ACD
üS
 2 =  2
(common)
D̂ 2 = C
(proven)
B̂2 = D̂ 2 + D̂3
(sum of Ðs in D)
üS
üS
\ D ADB ||| D ACD
OR
In Δ ADB and Δ ACD
 2 =  2
(common)
D̂ 2 = C
(proven)
\ D ADB ||| D ACD
16.3.1
AD AB
=
AC AD
(|||Ds)
üS
üS
üR
(3)
(ÐÐÐ)
üratio
AD = AC . AB
2
= 3r . r
= 3r 2
üsubstitution
(2)
139
16.3.2
AD = AE = 3 r
(from11.2.2(a)) &11.2.3(a)
AB = r and BC = 2 r \AC = 3 r
üAC ito r
In Δ ACE :
tan Ê =
=
AC
AE
ütrig ratio
3r
= 3
3r
üsimplification
\Ê = 60°
\D̂1 = 60°
(from11.2.2(a))
\Â1 = 60°
( Ðs of D = 180°)
üall 3 Ðs = 60°
\ D ADE is a cyclic quad
OR
AD DB
=
AC CD
( ||| Ds )
3 r DB
=
3r
CD
tan x =
ü
1
3
3 r DB
=
3r
CD
ü tan x =
\In D BDC : x = 30°
1
3
ü x = 30°
\Ê = 60°
\D̂1 = 60°
(from11.2.2(a))
\Â1 = 60°
( Ðs of D = 180°)
üall 3 Ðs = 60°
\ D ADE is a cyclic quad
140
OR
AD DB
=
AC CD
( ||| Ds )
3 r DB
CD
=
\ BD =
3r
CD
3
DC 2 = BC 2 - DB2
DC 2 = 4 r 2 -
CD 2
3
3DC 2 = 12r 2 - CD 2
ü BD =
CD
3
4 DC 2 = 12r 2
DC =
3r
EC 2 = EA 2 + AC2
= 3r 2 + 9 r 2
ü DC = 3r
EC = 2 3r
\ED = EC - DC
=
3r
\ED = EA = AD
ü EC = 2 3r
\ADE is equilateral
ü ED = EA = AD
141
ACCEPTABLE REASONS: EUCLIDEAN GEOMETRY
In order to have some kind of uniformity, the use of the following shortened versions of the theorem
statements is encouraged.
Acceptable reasons: Euclidean Geometry (English)
THEOREM STATEMENT
ACCEPTABLE REASON(S)
LINES
The adjacent angles on a straight line are supplementary.
If the adjacent angles are supplementary, the outer arms of these angles
form a straight line.
The adjacent angles in a revolution add up to 360°..
Vertically opposite angles are equal.
If AB || CD, then the alternate angles are equal.
If AB || CD, then the corresponding angles are equal.
If AB || CD, then the co-interior angles are supplementary.
If the alternate angles between two lines are equal, then the lines are
parallel.
Ðs on a str line
adj Ðs supp
Ðs round a pt OR Ðs in a rev
vert opp Ðs =
alt Ðs; AB || CD
corresp Ðs; AB || CD
co-int Ðs; AB || CD
alt Ðs =
If the corresponding angles between two lines are equal, then the lines
are parallel.
corresp Ðs =
If the co-interior angles between two lines are supplementary,
then the lines are parallel.
coint Ðs supp
TRIANGLES
The interior angles of a triangle are supplementary.
The exterior angle of a triangle is equal to the sum of the interior opposite
angles.
Ð sum in D OR sum of Ðs in ∆
OR Int Ðs D
ext Ð of D
The angles opposite the equal sides in an isosceles triangle are equal.
Ðs opp equal sides
The sides opposite the equal angles in an isosceles triangle are equal.
sides opp equal Ðs
In a right-angled triangle, the square of the hypotenuse is equal to the
sum of the squares of the other two sides.
Pythagoras OR
Theorem of Pythagoras
Converse Pythagoras
OR
Converse Theorem of Pythagoras
SSS
If the square of the longest side in a triangle is equal to the sum of the
squares of the other two sides then the triangle is right-angled.
If three sides of one triangle are respectively equal to three sides of another
triangle, the triangles are congruent.
If two sides and an included angle of one triangle are respectively equal to
two sides and an included angle of another triangle, the triangles are
congruent.
SAS OR SÐS
142
THEOREM STATEMENT
ACCEPTABLE REASON(S)
If two angles and one side of one triangle are respectively equal to
two angles and the corresponding side in another triangle, the
triangles are congruent.
AAS OR ÐÐS
If in two right-angled triangles, the hypotenuse and one side of one
triangle are respectively equal to the hypotenuse and one side of the
other, the triangles are congruent
RHS OR 90°HS
The line segment joining the midpoints of two sides of a triangle is
parallel to the third side and equal to half the length of the third side
Midpt Theorem
The line drawn from the midpoint of one side of a triangle, parallel
to another side, bisects the third side.
line through midpt || to 2nd side
A line drawn parallel to one side of a triangle divides the other
two sides proportionally.
line || one side of D
OR prop theorem; name || lines
If a line divides two sides of a triangle in the same proportion,
then the line is parallel to the third side.
line divides two sides of ∆ in prop
If two triangles are equiangular, then the corresponding sides are in
proportion (and consequently the triangles are similar).
||| Ds OR equiangular ∆s
If the corresponding sides of two triangles are proportional, then the
triangles are equiangular (and consequently the triangles are
similar).
If triangles (or parallelograms) are on the same base (or on bases
of equal length) and between the same parallel lines, then the
triangles (or parallelograms) have equal areas.
Sides of ∆ in prop
CIRCLES
The tangent to a circle is perpendicular to the radius/diameter of the
circle at the point of contact.
If a line is drawn perpendicular to a radius/diameter at the point
where the radius/diameter meets the circle, then the line is a tangent
to the circle.
The line drawn from the centre of a circle to the midpoint of a chord
is perpendicular to the chord.
same base; same height OR
equal bases; equal height
tan ^
radius tan
^ diameter
line ^ radius OR
converse tan ^ radius
OR converse tan ^
diameter
line from centre to midpt of
chord
The line drawn from the centre of a circle perpendicular to a chord
bisects the chord.
line from centre ^ to chord
The perpendicular bisector of a chord passes through the centre
of the circle;
perp bisector of chord
The angle subtended by an arc at the centre of a circle is double the
size of the angle subtended by the same arc at the circle (on the
same side of the chord as the centre)
Ð at centre = 2 ×Ð at
circumference
The angle subtended by the diameter at the circumference of the
circle is 90°.
Ðs in semi- circle OR
diameter subtends right angle
If the angle subtended by a chord at the circumference of the
circle is 90°. then the chord is a diameter.
chord subtends 90° OR
converse Ðs in semi -circle
143
THEOREM STATEMENT
ACCEPTABLE REASON(S)
Angles subtended by a chord of the circle, on the same side of the
chord, are equal
Ðs in the same seg.
If a line segment joining two points subtends equal angles at two
points on the same side of the line segment, then the four points are
concyclic.
Equal chords subtend equal angles at the circumference of the
circle.
Equal chords subtend equal angles at the centre of the circle.
line subtends equal Ðs OR
converse Ðs in the same seg.
Equal chords in equal circles subtend equal angles at the
circumference of the circles.
Equal chords in equal circles subtend equal angles at the centre
of the circles.
The opposite angles of a cyclic quadrilateral are supplementary
If the opposite angles of a quadrilateral are supplementary then the
quadrilateral is cyclic.
The exterior angle of a cyclic quadrilateral is equal to the interior
opposite angle.
If the exterior angle of a quadrilateral is equal to the interior
opposite angle of the quadrilateral, then the quadrilateral is cyclic.
Two tangents drawn to a circle from the same point outside the
circle are equal in length
The angle between the tangent to a circle and the chord drawn from
the point of contact is equal to the angle in the alternate segment.
If a line is drawn through the end-point of a chord, making with
the chord an angle equal to an angle in the alternate segment,
then the line is a tangent to the circle.
equal chords; equal Ðs
equal chords; equal Ðs
equal circles; equal chords; equal
Ðs
equal circles; equal chords; equal
Ðs
opp Ðs of cyclic quad
opp Ðs quad supp OR
converse opp Ðs of cyclic quad
ext Ð of cyclic quad
ext Ð = int opp Ð OR
converse ext Ð of cyclic quad
Tans from common pt OR
Tans from same pt
tan chord theorem
converse tan chord theorem OR
Ð between line and chord
QUADRILATERALS
The interior angles of a quadrilateral add up to 360.
The opposite sides of a parallelogram are parallel.
If the opposite sides of a quadrilateral are parallel, then the
quadrilateral is a parallelogram.
sum of Ðs in quad
opp sides of ||m
opp sides of quad are ||
The opposite sides of a parallelogram are equal in length.
If the opposite sides of a quadrilateral are equal , then the
quadrilateral is a parallelogram.
opp sides of ||m
opp sides of quad are =
OR converse opp sides of a parm
opp Ðs of ||m
opp Ðs of quad are = OR
converse opp angles of a parm
diag of ||m
diags of quad bisect each other
OR converse diags of a parm
The opposite angles of a parallelogram are equal.
If the opposite angles of a quadrilateral are equal then the
quadrilateral is a parallelogram.
The diagonals of a parallelogram bisect each other.
If the diagonals of a quadrilateral bisect each other, then the
quadrilateral is a parallelogram.
If one pair of opposite sides of a quadrilateral are equal and parallel,
then the quadrilateral is a parallelogram.
pair of opp sides = and ||
The diagonals of a parallelogram bisect its area.
diag bisect area of ||m
The diagonals of a rhombus bisect at right angles.
diags of rhombus
144
The diagonals of a rhombus bisect the interior angles.
diags of rhombus
All four sides of a rhombus are equal in length.
All four sides of a square are equal in length.
The diagonals of a rectangle are equal in length.
The diagonals of a kite intersect at right-angles.
A diagonal of a kite bisects the other diagonal.
A diagonal of a kite bisects the opposite angles
sides of rhombus
sides of square
diags of rect
diags of kite
diag of kite
diag of kite
TERMINOLOGY
Term
Explanation
Euclidean Geometry
Geometry based on the postulates of Euclid. Euclidean geometry
deals with space and shape using a system of logical deductions
A statement that has been proved based on previously established
statements
A statement formed by interchanging what is given in a theorem and what is
to be proved
A problem of more than usual difficulty added to another on an examination
paper
Straight line from the centre to the circumference of a circle or sphere. It is
half of the circle’s diameter
Straight line going through the centre of a circle connecting two points on the
circumference
Line segment connecting two points on a curve. When the chord passes
through the centre of a circle it is called the diameter
A 4-sided closed shape (polygon)
A quadrilateral whose vertices all lie on a single circle. This circle is called
the circumcircle or circumscribed circle, and the vertices are said to be
concyclic
A straight line joining two opposite vertices (corners) of a straight sided
shape. It goes from one corner to another but is not an edge
The distance around the edge of a circle (or any curved shape).
It is a type of perimeter
The area bound by a chord and an arc
Part of the circumference of a circle
The area bound by two radii and an arc
A statement that follows with little or no proof required from an
already proven statement. For example, it is a theorem in geometry that the
angles opposite two congruent sides of a triangle are also congruent
(isosceles triangle). A corollary to that statement is that an equilateral
triangle is also equiangular.
In any right-angled triangle, the square on the hypotenuse is equal to the
sum of the squares on the other two sides.
The longest side in a right-angled triangle. It is opposite the right
angle.
theorem
converse
rider
radius
diameter
chord
quadrilateral
cyclic quadrilateral
diagonal
circumference
segment
arc
sector
Corollary (Theorem that
follows on from another
theorem)
Theorem of Pythagoras
hypotenuse
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Complementary angles
Supplementary angles
Vertically opposite angles
Intersecting lines
Perpendicular lines
parallel lines
transversal
Corresponding angles
alternate angles
co-interior angles
congruent
similar
proportion
ratio
area
tangent
Point of tangency
exterior angle
subtend
polygon
Radii (plural of radius)
Angles that add up to 90º.
Angles that add up to 180º.
Non-adjacent opposite angles formed by intersecting lines.
Lines that cross each other.
Lines that intersect each other at a right angle.
Lines the same distance apart at all points. Two or more lines are
parallel if they have the same slope (gradient).
A line that cuts across a set of lines (usually parallel).
Angles that sit in the same position on each of the parallel lines in the
position where the transversal crosses each line.
Angles that lie on different parallel lines and on opposite sides of the
transversal.
Angles that lie on different parallel lines and on the same side of the
transversal.
The same. Identical.
Looks the same. Equal angles and sides in proportion.
A part, share, or number considered in comparative relation to a
whole. The equality of two ratios. An equation that can be solved.
The comparison of sizes of two quantities of the same unit. An
expression.
The space taken up by a two-dimensional polygon.
Line that intersects with a circle at only one point (the point of
tangency)
The point of intersection between a circle and its tangent line
The angle between any side of a shape, and a line extended from the next
side
The angle made by a line or arc
A closed 2D shape in which all the sides are made up of line
segments. A polygon is given a name depending on the number of sides it
has. A circle is not a polygon as although it is a closed 2D shape it is not
made up of line segments
This is common when triangles are drawn inside circles –
look out for lines drawn from the centre. Remember that all
radii are equal in length in a circle
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Acknowledgement
The Department of Basic Education (DBE) gratefully acknowledges the following officials
for giving up their valuable time and families and for contributing their knowledge and
expertise to develop this resource booklet for the children of our country, under very
stringent conditions of COVID-19:
Writers: Mrs Nontobeko Gabelana, Mr Tshokolo Mphahama, Ms Thandi Mgwenya, Mrs
Nomathamsanqa Princess Joy Khala, Mr Eric Makgubje Maserumule, Mr Avhafarei
Edward Thavhanyedza and Mrs Thavha Hilda Mudau.
DBE Subject Specialist: Mr Leonard Gumani Mudau
The development of the Study Guide was managed and coordinated by Ms Cheryl Weston
and Dr Sandy Malapile.
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