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PREPARATORY EXAMINATION
GRADE 12
MATHEMATICS P2
SEPTEMBER 2020
MARKS: 150
TIME: 3 HOURS
This question paper consists of 13 pages, an information sheet and
an answer book of 20 pages.
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Mathematics P2 2 FS/September 2020
Grade 12 Prep. Exam.
Copyright reserved Please turn over
INSTRUCTIONS AND INFORMATION
Read the following instructions carefully before answering the questions.
1.
2.
3.
4.
5.
6.
7.
8.
9.
This question paper consists of 11 questions.
Answer ALL the questions in the SPECIAL ANSWER BOOK provided.
Clearly show ALL calculations, diagrams, graphs, etc. which you have used in
determining your answers.
You may use an approved scientific calculator (non-programmable and non-
graphical), unless stated otherwise.
If necessary, round off answers to TWO decimal places, unless stated otherwise.
Diagrams are NOT necessarily drawn to scale.
Answers only will NOT necessarily be awarded full marks.
An information sheet with formulae is included at the end of the question paper.
Write neatly and legibly.
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Mathematics P2 3 FS/September 2020
Grade 12 Prep. Exam.
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QUESTION 1
The table below gives the average exchange rate and the average monthly oil price for the
year 2010.
Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec
Exchange
rate in R/S 7.5 7.7 7.2 7.4 7.7 7.7 7.6 7.3 7.1 7.0 6.9 6.8
Oil price
in $ 69.9 68.0 72.9 70.3 66.3 67.1 67.9 68.3 71.3 73.6 76.0 81.0
1.1 Draw a scatterplot to represent the exchange rate (in R/S) versus the oil price
(in $).
(3)
1.2 Determine the equation of the least square regression line. (3)
1.3 Calculate the value of the correlation coefficient. (1)
1.4 Comment on the strength of the relationship between the exchange rate
(in R/S) and the oil price (in $).
(2)
1.5 Determine the mean oil price. (1)
1.6 Determine the standard deviation of the oil price. (1)
1.7 Generally there is a concern from the public when the oil price is higher than
two standard deviations from the mean.
In which months would the public have been concerned?
(2)
[13]
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Mathematics P2 4 FS/September 2020
Grade 12 Prep. Exam.
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QUESTION 2
The average percentage of 150 learners for all their subjects is summarised in the cumulative
frequency below.
PERCENTAGE
INTERVAL
CUMULATIVE
FREQUENCY
π < π β€ ππ 5
ππ < π β€ ππ 21
ππ < π β€ ππ 50
ππ < π β€ ππ 70
ππ < π β€ ππ 88
ππ < π β€ ππ 110
ππ < π β€ ππ 135
ππ < π β€ ππ 142
ππ < π β€ ππ 147
ππ < π β€ πππ 150
2.1 Draw the ogive (cumulative frequency graph) to represent the above data on
the grid provided in the ANSWER BOOK.
(4)
2.2 Use the ogive to approximate the following:
2.2.1 The number of learners who scored less than 85% (2)
2.2.2 The median (1)
[7]
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Mathematics P2 5 FS/September 2020
Grade 12 Prep. Exam.
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QUESTION 3
In the diagram below shows, quadrilateral ABCD with AD // BC. The coordinates of the
vertices are A(1; 7); B(π; π); C(β2; β8) and D(β4; β3). BC intersects the x-axis at F.
DοΏ½οΏ½π΅ = πΌ.
3.1 Calculate the gradient of AD. (2)
3.2 Determine the equation of BC in the form π¦ = ππ₯ + π. (3)
3.3 Determine the coordinates of point F. (2)
3.4 ABβ²CD is a parallelogram with π΅β² on BC. Determine the coordinates of π΅β², using a transformation (π₯; π¦) β (π₯ + π; π¦ + π) that sends A to π΅β².
(2)
3.5 Show that πΌ = 48,37Β°. (4)
3.6 Calculate the area of βDCF. (6)
[19]
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Mathematics P2 6 FS/September 2020
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QUESTION 4
Circle C1 and C2 in the figure below have the same centre M. P and A are points on
C2.
PM intersects C1 at D. The tangent BD to C1 intersects C2 at B and E. The equation
of circle C1 is given by π₯2 + 2π₯ + π¦2 + 6π¦ + 2 = 0 and the equation of line PM is
π¦ = π₯ β 2.
4.1 Calculate the coordinates of centre M. (3)
4.2 Determine the radius of circle C1. (1)
4.3 Determine the coordinates of D1 the point where line PM and circle C1
intersects.
(5)
4.4 Give a reason why MD B = 90Β°. (1)
4.5 If is given that DB = 4β2, determine MB, the radius of circle C2. (4)
4.6 Write down the equation of C2 in the form (π₯ β π)2 + (π¦ β π)2 = π2. (1)
4.7 Is the point πΉ(2β5 ; 0) inside circle C2? Support your answer with
calculations.
(4)
4.8 Determine the gradient of the tangent to circle πΆ2 at point P. (2)
[21]
π¦
π₯
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QUESTION 5
5.1 In the diagram, P is the point (π; β21) such that OP = 5 units.
BOP = π as indicated.
5.1.1 Calculate the numerical value of π. (2)
5.1.2 Determine (without the use of a calculator), the numerical value
of the following:
a) cos π (1)
b) tan π + π ππ2π (2)
c) sin 2π (2)
5.2 Simplify (without the use of a calculator):
)90(cos
690cos.tan).180sin(2 β
β
x
xx
(5)
[12]
QUESTION 6
6.1 Prove the identity: x
x
xx
x
sin
cos
2sintan2
sin2 2
=β
(5)
6.2 Show that: πππ220Β° + πππ240Β° + πππ280Β° =3
2
(Hint: 40Β° = 60Β° β 20Β° and 80Β° = 60Β° + 20Β°)
(7)
[12]
π
O
P(π; ββ21 )
B
y
x
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Mathematics P2 8 FS/September 2020
Grade 12 Prep. Exam.
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QUESTION 7
In the figure below Thabo is standing at a point A on top of building AB that is h (m) high.
He observes two cars, C and D that are in the same horizontal plane as B. The angle of
elevation from C to A is π and the angle of elevation from D to A is πΌ. CοΏ½οΏ½π·=π½.
7.1 Calculate the length of AC in terms of h and π. (2)
7.2 Calculate the length of AD in terms of h and πΌ. (2)
7.3 Determine the distance between the two cars, which is the length of CD in
terms of πΌ, π, π½ and h.
(3)
[7]
π
πΌ
π½
β
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Mathematics P2 9 FS/September 2020
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QUESTION 8
In the diagram are the graphs of the functions π(π₯) = cosπ₯
2 and
π(π₯) = sin(π₯ β 30Β°) for π₯ π[β180Β°; 180Β°]. The curves intersect at A and B.
.
8.1 Calculate the x-coordinates of the points A and B. (6)
8.2 Determine the values of π₯ π[β90Β°; 180Β°], for which π(π₯).π(π₯) < 0. (3)
[9]
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Mathematics P2 10 FS/September 2020
Grade 12 Prep. Exam.
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D
QUESTION 9
A, B and C are three points on the circle with centre O such that AB = BC = 3
2 AO.
AO = π₯.
9.1 Calculate the size of οΏ½οΏ½1 rounded off to the nearest degree. (5)
9.2 If οΏ½οΏ½1 = 97Β° and π₯ = 10 cm, calculate the length of AC correct to two decimal
places.
(4)
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Mathematics P2 11 FS/September 2020
Grade 12 Prep. Exam.
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9.3
In the diagram, TP is a tangent to the circle with centre M at P. QS is a
diameter of the circle and R is on the circumference of the circle. TοΏ½οΏ½π = 32Β°.
Calculate the following, giving reasons:
9.3.1 οΏ½οΏ½1
(2)
9.3.2 οΏ½οΏ½4
(2)
9.3.3 οΏ½οΏ½1 (2)
9.3.4 οΏ½οΏ½ (4)
[19]
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Mathematics P2 12 FS/September 2020
Grade 12 Prep. Exam.
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QUESTION 10
10.1 Complete the following statement:
If two triangles are equiangular, then the corresponding sides are β¦
(1)
10.2 Use the diagram to prove the theorem which states that a line drawn parallel
to one side of a triangle divides the other two sides proportionally, that is
prove that .LZ
XL
KY
XK=
(6)
10.3 In the figure, D is a point on side BC of ABC such that BD = 6 cm and
DC = 9 cm. T and E are points on AC and DC respectively such that TE||AD
and AT : TC = 2 : 1
A
B D
E
C
T
F
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Mathematics P2 13 FS/September 2020
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10.3.3 Calculate the numerical value of:
(a)
ABD of Area
ADC of Area
(3)
(b)
ΞABC of Area
ΞTEC of Area
(3)
[19]
QUESTION 11
In the diagram, TPR is a triangle with TP = 4,5 units. Points Q and S are on TR and PR
respectively such that QR = 9,6 units, QS = 4 units, TS = 3,6 units, PS = 1,5 units and
SR = 12 units.
11.1 Prove that PT is a tangent to the circle which passes through the points T,S
and R.
(7)
11.2 Calculate the length of TQ. (5)
[12]
TOTAL: 150
10.3.1 Show that D is the midpoint of BE. (3)
10.3.2 If FD = 2 cm, calculate the length of TE. (3)
P R S
Q
T
9, 6
12 1, 5
4, 5 3, 6
4
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Mathematics P2 14 FS/September 2020
Grade 12 Prep. Exam.
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INFORMATION SHEET: MATHEMATICS
a
acbbx
2
42 ββ=
( )inPA += 1 ( )inPA β= 1 ( )niPA β= 1 ( )niPA += 1
( )dnaTn 1β+= ( ) dnan
Sn 122
β+=
1β= n
n arT ( )
1
1
β
β=
r
raS
n
n ; 1r
r
aS
β=
1; 11 β r
( ) i
ixF
n11 β+
= [1 (1 ) ]nx iP
i
ββ +=
h
xfhxfxf
h
)()(lim)('
0
β+=
β
22 )()( 1212 yyxxd β+β= M
++
2;
2
2121 yyxx
cmxy += )( 11 xxmyy β=β 12
12
xx
yym
β
β= tan=m
( ) ( ) 222rbyax =β+β
In ABC: C
c
B
b
A
a
sinsinsin== Abccba cos.2222 β+=
CabABCarea sin.2
1=
( ) sin.coscos.sinsin +=+ ( ) sin.coscos.sinsin β=β
( ) sin.sincos.coscos β=+ ( ) sin.sincos.coscos +=β
β
β
β
=
1cos2
sin21
sincos
2cos
2
2
22
cos.sin22sin =
n
fxx
=
( )
n
xxn
i
i2
2
1
=
β
=
( )Sn
AnAP
)()( = P (A or B) = P (A) + P (B) β P (A and B)
bxay +=Λ ( )
β
ββ=
2)(
)(
xx
yyxxb
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PREPARATORY EXAMINATION/
VOORBEREIDENDE EKSAMEN
GRADE/GRAAD 12
MATHEMATICS P2/
WISKUNDE V2
SEPTEMBER 2020
MARKS/PUNTE: 150
MARKING GUIDELINES/
NASIENRIGLYNE
This marking guidelines consists of 13 pages./
Hierdie nasienriglyne bestaan uit 13 bladsye.
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Mathematics P2/Wiskunde V2 2 FS/VS/September 2020
Grade/Graad 12 Prep. Exam./Voorb. Eksam.Marking Guidelines/Nasienriglyne
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
NOTE:
β’ If a candidate answers a question TWICE, only mark the FIRST attempt.
β’ If a candidate has crossed out an attempt of a question and not redone the question,
mark the crossed out version.
β’ Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking
at the second calculation error.
β’ Assuming answers/values in order to solve a problem is NOT acceptable.
NOTA:
β’ As 'n kandidaat 'n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na.
β’ As 'n kandidaat 'n antwoord van 'n vraag doodtrek en nie oordoen nie, sien die
doodgetrekte poging na.
β’ Volgehoue akkuraatheid word in ALLE aspekte van die nasienriglyne toegepas. Hou op
nasien by die tweede berekeningsfout.
β’ Om antwoorde/waardes te aanvaar om 'n probleem op te los, word NIE toegelaat NIE.
GEOMETRY/MEETKUNDE
S
A mark for a correct statement
(A statement mark is independent of a reason)
'n Punt vir 'n korrekte bewering
('n Punt vir 'n bewering is onafhanklik van die rede)
R
A mark for the correct reason
(A reason mark may only be awarded if the statement is correct)
'n Punt vir 'n korrekte rede
('n Punt word slegs vir die rede toegeken as die bewering korrek is)
S/R
Award a mark if statement AND reason are both correct
Ken 'n punt toe as die bewering EN die rede beide korrek is
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Grade/Graad 12 Prep. Exam./Voorb. Eksam.Marking Guidelines/Nasienriglyne
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QUESTION/VRAAG 1
1.1
β 4 points
correctly
plotted
β 9 points
correctly
plotted
β All points
(3)
1.2 π¦ = 158,67 β 11,96π₯ βββ Correct
equation
(3)
1.3 π = β0,91 β value of r
(1)
1.4 Exchange rate increase,oil price decrease/Wisselkoers
verhoog, olieprys verlaag
OR/OF
Strong Negative correlation/Sterk negatiewe korrelasie
β β reason
(2)
1.5 π¦ =71,05 β π¦ =71,05
(1)
1.6 Standard deviation/Standaard afwyking : π = 4,09 β π = 4,09 (1)
1.7 71,05+2(4,09) = 79,23 (With calculator 79,22)
December
β 79,23
β December/
Desember
(2)
[13]
0
10
20
30
40
50
60
70
80
90
6.6 6.8 7 7.2 7.4 7.6 7.8
Oil
Pri
ce
Exchange Rate in R/S
Scatter PlotScatter Plot/Spreidiagram O
il P
rice
/Olie
pry
s
Exchange rate/Wisselkoers in R/$
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QUESTION/VRAAG 2
2.1
βgrouded
βCf βupper limit
β curve
(4)
2.2.1 (8;144)
144 below 85 % (accept/aanvaar 144-146)
β(8;144)
β 144 (2)
2.2.2 π2 = 42,77 (accept/aanvaar 41-43) βπ2 = 42,77
(1)
[7]
0
20
40
60
80
100
120
140
160
0 20 40 60 80 100 120
OGIVEOGIVE/OGIEF
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QUESTION/VRAAG 3
3.1 ππ΄π· =7β(β3)
1β(β4)
= 2
β Substitution into the
correct formula
β Answer (2)
3.2 AD// BC
ππ΄π· = ππ΅πΆ = 2
β8 = 2(2) + π
β΄ π¦ = 2π₯ β 4
β ππ΅πΆ = 2
β sub
β Answer
(3)
3.3 At F π¦ = 0
0 = 2π₯ β 4
β΄ π₯ = 2
β΄ πΉ(2; 0)
β 0 = 2π₯ β 4
β π₯ = 2 (2)
3.4 π΅(π₯; π¦) β 7 πΆ(π₯ + 2; π¦ β 5)
π΄(1; 7) β π΅/(3; 2)
OR
π₯π΅ = β2 + (1 + 4) = 3
π¦π΅ = β8 + (7 + 3) = 2
β β π΅/(3; 2)
(2)
3.5 ππ΅πΆ = tan π = 2
π = 63,43Β°
ππ·πΆ =β8 β (β3)
β2 β (β4)=
β5
2
tan π½ =β5
2
π½ = 180Β° β 68,20Β° = 111,80Β°
πΌ = 111.80Β° β 63,43Β° = 48,37Β°
β π = 63,43Β°
β β5
2
βπ½ = 111,80Β°
β πΌ = 48,37
(4)
3.6 DC= β(β4 + 2)2 + (β3 + 8)2
= β29
CF= β(β2 β 2)2 + (β8 β 0)2
= β80
Area/Oppvl βπ·πΆπΉ =1
2DC.CF sin πΌ
= 1
2β29. β80 sin 48,37Β°
= 18 units/πππβπππ
β sub correct formula
β β29
β sub correct formula
β β80
β sub correct formula
β18 π’πππ‘π
(6)
[19]
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QUESTION/VRAAG 4
4.1 π₯2 + π¦2 + 2π₯ + 6π¦ + 2 = 0
π₯2 + 2π₯ + 1 + π¦2 + 6π¦ + 9 = β2 + 10
(π₯ + 1)2 + (π¦ + 3)2 = 8
π(β1; β3)
β 8
ββπ(β1; β3)
(3)
4.2 π = β8 β π = β8
(1)
4.3 π₯2 + (π₯ β 2)2 + 2π₯ + 6(π₯ β 2) + 2 = 0
2π₯2 + 4π₯ β 6 = 0
π₯2 + 2π₯ + 3 = 0 (π₯ + 3)(π₯ β 1) = 0
π₯ = β3 or π₯ β 1
π¦ = β3 β 2 = β5
β΄ π·(β3; β5)
βsubstitution
βπ₯2 + 2π₯ + 3 = 0
β(π₯ + 3)(π₯ β 1) = 0
ββ π·(β3; β5)
(5)
4.4 Angle between radius/diameter and tangent/ Hoek
tussen radius/deursnee en raaklyn β R (1)
4.5 ππ΅2 = ππ·2 + π·π΅2 Pyth
= (β8)2
+ 4(β2)2
= 40
MB= β40 radius of Circle/radius van sirkel πΆ2
β S
β (β8)2
+ 4(β2)2
β 40
β MB= β40
(4)
4.6 (π₯ + 1)2 + (π¦ + 3)2 = 40 βEquation ofπΆ2
(1)
4.7 Distance from/Afstand van (2β5; 0) to centre/tot
middelpunt
= β(2β5 + 1)2
+ (0 + 3)2
= 6,24
6,24< 6,32 (β40)
Distance to centre < radius of circle/Afstand tot
middelpunt < radius van die sirkel
β΄ (2β5; 0) lies inside/lΓͺ binne
β correct sub
β 6,24
β6,24< 6,32 (β40)
β lies inside
(4)
4.8 πππ·π =β3β(β5)
β1β(β3)=
2
2= 1
β΄ M tangent = β1
β MMDP = 1
β M tangent = β1
(2)
[21]
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Grade/Graad 12 Prep. Exam./Voorb. Eksam.Marking Guidelines/Nasienriglyne
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QUESTION/VRAAG 5
5.1.1
( ) 22
2 521 =+c
2
4
2125
2
2
=
=
β=
c
c
c
β Subst. into
pyth
β Answer
(2)
5.1.2
a)
5
2cos =
β Answer
(1)
b)
ββ21
2+ (
ββ21
5)
2
β5β21 + 4250
β
β ββ21
2+ (
ββ21
5)
2
β Answer
(2)
c) cossin2
= 2 (ββ21
5) (
2
5)
=β4β21
25
β 2 (ββ21
5) (
2
5)
β Answer
(2)
5.2
2)sin(
)30360cos(.tan).sin(
x
xx
β
ββ
.sin
)30(costan.sin2 x
xx β=
= ββ3
2tan π₯ Γ· sin π₯
=ββ3
2 cos π₯
β β sin π₯
β (β sin π₯)2
β ββ3
2
ββ
ββ3
2 cos π₯
(5)
[ππ]
π O
P(π; ββ21 )
B
y
x
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QUESTION/VRAAG 6
6.1
x
x
x
xx
x
x
x
xx
x
x
xxx
x
xLHS
sin
cos
sin
cossin
cos
cos1
sin
coscos
1sin2
sin2
cossin2cos
sin2
sin2
2
2
2
2
=
=
β=
β
=
β
=
β΄ πΏπ»π = π π»π
β x
x
cos
sin
β 2π ππ π₯πππ π₯
β
2 sin π₯ (1
cos π₯β
cos π₯)
β x
x
x
cos
cos1
sin2β
β π ππ2π₯
(5)
6.2
πππ220Β° + πππ240Β° + πππ280Β°
= πππ220Β° + [πππ(60Β° β 20Β°)]2 + [πππ(60Β° + 20Β°)]2
= πππ220Β° + [πππ60Β°πΆππ 20Β° β πΆππ 60Β°πππ20Β°]2
+ [πππ60Β°πΆππ 20Β° + πΆππ 60Β°πππ20Β°]2
= πππ220Β° + [β3
2πΆππ 20Β° β
1
2πππ20Β°]
2
+ [β3
2πΆππ 20Β° +
1
2πππ20Β°]
2
=3
2πππ220Β° +
3
2πΆππ 220Β°
=3
2(πππ220Β° + πΆππ 220Β°)
=3
2(1)
=3
2
ββcompound
βsub of special
angle
βsimplification
βcommon factor
βsquare identity
βanswer
(7)
[12]
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QUESTION/VRAAG 7
7.1 In β ABC
ππππ =β
π΄πΆ
β΄ π΄πΆ =β
ππππ
β trig ratio
β answer
(2)
7.2 In β ABD
ππππΌ =β
π΄π·
β΄ π΄π· =β
ππππΌ
β trig ratio
β answer (2)
7.3 πΆπ·2 = π΄πΆ2 + π΄π·2 β 2π΄πΆ. π΄π· cos π½
= (β
ππππ)
2
+ (β
ππππΌ)
2
β 2.β
ππππ.
β
ππππΌπΆππ π½
=β2
πππ2π+
β2
πππ2πΌβ
2β2πΆππ π½
ππππππππΌ
β cosine rule
β substitution
β answer
(3)
[7]
QUESTION/VRAAG 8
8.1 cosπ₯
2= sin(π₯ β 30Β°)
= cos[(90Β° β (π₯ β 30Β°))]
= cos(120Β° β π₯)
β΄π₯
2= 120Β° β π₯ + 360Β°. π
π₯ = 240Β° β 2π₯ + 720Β°. π
π₯ = 80Β° + 240Β°. π
OR π₯
2= 360Β° β (120Β° β π₯) + 360Β°. π
π₯
2= 360Β° β 120Β° + π₯ + 360Β°π
π₯
2= 240Β° + π₯ + 360Β°π
π₯ = 480Β° + 2π₯ + 720π
βπ₯ = 480Β° + 720π
β΄ π₯ = β480Β° β 720Β°π where/π€πππ π β π§
π₯π΄ = β160Β° and/en π₯π΅ = 80Β°
β cos(120Β° β π₯)
β80Β° + 240Β°. π
β240Β° + π₯ + 360Β°π
β β 480Β° β 720Β°π
βπ₯π΄ = β160Β°
βπ₯π΅ = 80Β°
(6)
8.2 β150Β° < π₯ < 30Β° β β Critical values
β Notation
(3)
[9]
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Mathematics P2/Wiskunde V2 10 FS/VS/September 2020
Grade/Graad 12 Prep. Exam./Voorb. Eksam.Marking Guidelines/Nasienriglyne
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
QUESTION/VRAAG 9
9.1 π΄π΅2 = π΄π2 + π΅π2 β 2(π΄π)(π΅π) cos οΏ½οΏ½1
(3
2π₯)
2
= π₯2 + π₯2 β 2π₯2 cos οΏ½οΏ½1
9
4π₯2 β 2π₯2 = β2π₯2 cos οΏ½οΏ½1
cos οΏ½οΏ½1 =1
4π₯2 Γ· 2π₯2
=1
8
β΄ οΏ½οΏ½1 = 97Β°
βcosine rule
β substitution
β simplification
β
1
8
βοΏ½οΏ½1 = 97Β°
(5)
9.2 οΏ½οΏ½ = 48,5Β° angle at centre/hoek by middelpunt =twice angle at circ/twee keer hoek by sirk
οΏ½οΏ½1 + οΏ½οΏ½2 = 83Β° int angles of/binne hoeke van β π΄πΆ
sin 83Β°=
15
sin 48,5Β°
β΄ π΄πΆ = 19,88
β S
β S
β sine rule
β π΄πΆ = 19,88
(4)
9.3.1 οΏ½οΏ½1 = 32Β° tan chord theorem/tan koordstelling
β β S/R
(2)
9.3.2 οΏ½οΏ½1 = οΏ½οΏ½4 = 32Β° angle opp = sides/hoeke teenoor = sye β β S/R
(2)
9.3.3 οΏ½οΏ½1 = οΏ½οΏ½1 + οΏ½οΏ½4 ext angle of β = op pint angles/
Verl hoek van β = opp hoeke
= 32Β° + 32Β°
β΄ οΏ½οΏ½1 = 64Β°
β S
β answer
(2)
9.3.4 οΏ½οΏ½ = 180Β° β 122Β° sum of angles of β = 180Β°/
Som van hoeke van β = 180Β°
β΄ οΏ½οΏ½ = 58Β°
But/Maar οΏ½οΏ½ = οΏ½οΏ½ = 58Β° angle sub by same chord/
hoek verv by dieselfde koord
β β S/R
β β S/R
(4)
[19]
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Mathematics P2/Wiskunde V2 11 FS/VS/September 2020
Grade/Graad 12 Prep. Exam./Voorb. Eksam.Marking Guidelines/Nasienriglyne
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
QUESTION/VRAAG 10
10.1 In the same proportion/
In dieselfde verhouding β same proportion
(1)
10.2 Const: Join KZ & LY & draw β1 from Kβ₯ XL & β2/
Konst: Verbind KZ & LY & skets β1 van Kβ₯ XL &
β2
Proof/Bewys
Area/ππππ£π βππΎπΏ
Area/ππππ£π βπΏππΎ =
1
2ππΎΓβ1
1
2πΎπΓβ2
=ππΎ
πΎπ
Area/ππππ£π βππΎπΏ
Area/ππππ£π βπΎπΏπ=
12 ππΏ Γ β2
12 πΏπ Γ β2
=ππΏ
πΏπ
Area of βππΎπΏ = π΄πππ β ππΎπΏ common/
Oppvl van βππΎπΏ = ππππ£π β ππΎπΏ algemeen
But Area β πΏπΎπ = π΄πππ β πΎπΏπ same base &
height; LK// YZ/
Maar Oppvl β πΏπΎπ = ππππ£π β πΎπΏπ selfde basis &
hoogte; LK// YZ
Area/ππππ£π β ππΎπΏ
Area/ππππ£π βπΏππΎ=
Area/ππππ£π βππΎπΏ
Area/ππππ£π βπΎπΏπ
β΄ππΎ
πΎπ=
ππΏ
πΏπ
β const
βπ΄πππ βππΎπΏ
π΄πππ βπΏππΎ =
ππΎ
πΎπ
β
π΄πππβππΎπΏ
π΄πππβπΎπΏπ=
ππΏ
πΏπ
β S β R
β S
(6)
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Mathematics P2/Wiskunde V2 12 FS/VS/September 2020
Grade/Graad 12 Prep. Exam./Voorb. Eksam.Marking Guidelines/Nasienriglyne
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
10.3.1 πΈπ·
π·πΆ=
π΄π
π΄πΆ=
2
3 line // to one side of /
DE =2
3Γ 9 lyn // aan die eenkant van
= 6 and/en BD = 6 given/gegee
β΄ π· is the midpoint of BE/is the middelpunt van BE
β β S/R
β answer
(3)
10.3.2
BF= FT conv of midpoint theorem
FD=1
2TE midpoint theorem/middelpunt bewys
β΄ ππΈ = 4 cm
β S/R
β R
β ππΈ = 4
(3)
10.3.3
(a)
π΄πππ ππ ADC
π΄πππ ππABD
=1
2Γ π1 Γ β Γ·
1
2Γ π2 Γ β same height/selfde hoogte
=3π¦
2π¦
=3
2
ββ S/R
β 3
2
(3)
(b) 12 Γ ππΆ Γ πΈπΆ Γ sin πΆ
12 Γ π΄πΆ Γ π΅πΆ Γ sin πΆ
=(π₯)(π¦)
(3π₯)(5π¦)
=1
15
ββ
12
Γ ππΆ Γ πΈπΆ Γ sin πΆ
12 Γ π΄πΆ Γ π΅πΆ Γ sin πΆ
β
1
15
(3)
[19]
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Mathematics P2/Wiskunde V2 13 FS/VS/September 2020
Grade/Graad 12 Prep. Exam./Voorb. Eksam.Marking Guidelines/Nasienriglyne
Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief
QUESTION/VRAAG 11
11.1 In TPS and QSR
ππ
ππ=
1,5
4=
3
8
ππ
ππ =
4,5
12=
3
8
ππ
ππ =
3,6
9,6=
3
8
β΄ TPS/// QSR sides of triangles in proportion
sye van die driehoeke is in verhouding
β΄ ποΏ½οΏ½S= οΏ½οΏ½
β΄ ππ is a tangent converse of tan chord theorem/
is 'n raaklyn teenoorg. van tan koordstelling
β
ππ
ππ=
1,5
4
β
ππ
ππ =
4,5
12
β
ππ
ππ =
3,6
9,6
ββ S/R
β ποΏ½οΏ½S= οΏ½οΏ½
βR
(7)
11.2 οΏ½οΏ½ = ποΏ½οΏ½π s ///
ππ|| TP corr angles =/korr hoeke
ππ
9,6=
1,5
12 proportional theorem/verhoudingstelling
β΄ ππ = 1,2
β S
β R
ββ S/R
β ππ = 1,2
(5)
[12]
TOTAL/TOTAAL: 150
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