# Mathematics Grade 12 Paper 2 North West Mock 2022 Memo

> Grade 12 Mathematics Paper 2 memorandum from the North West Mock Exam 2022, 34 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/nw-maths-grade-12-sept-2022-p2-and-memo/1ad68574-d446-4249-ad02-6d932bdafc9d
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2022
- Exam period: North West Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 34

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## About this document

Mathematics · Grade 12 · North West Mock Exam · 2022. Memorandum, 34 pages. Read online or download the PDF.

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### Page 1

. Ff; ‘ (stag cur

education

Lefapha la Thuto la Bokone Bophirima
Noordwes Departement van Onderwys
North West Department of Education
NORTH WEST PROVINCE

NATIONAL i
SENIOR CERTIFICATE
i
i
GRADE 12
MATHEMATICS P2 |
SEPTEMBER 2022
| MARKS: 150 |
H TIME: 3 hours
i

This question paper consists of 14 pages, 1 information sheet
and an answer book of 20 pages.

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{SRE SPA ET

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NSC

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1

2.

This question paper consists of 10 questions.

Answer ALL the questions in the SPECIAL ANSWER BOOK provided.

Clearly show ALL calculations, diagrams, graphs, etc. that you have used in determining
your answers.

Answers only will NOT necessarily be awarded full marks.

You may use an approved scientific calculator (non-programmable anc hon-graphical),
unless stated otherwise. S .

If necessary, round off answers to TWO decimal places, aia otherwise.
Diagrams are NOT necessarily drawn to scale. es
An information sheet with formulae is included at;théend of the question paper.

Write neatly and legibly.

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PAS

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ny

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QUESTION 1
A study was conducted among eight (8) employees in a company to understand the
relationship between the additional number of rest days given in a year and the productivity
of each employee.

The results are shown in the table below.

Additional number of

rest days 5 2 9 1 3 10 4 6
Productivity of the a
employee (in %) 85 63 90 56 70 88 @ 62

a
Ll Calculate the correlation coefficient of the data. \? qd)
1.2 Consider the correlation coefficient calculated in Q AN 1.1 and the four

phrases below. Only write down the letter of the at best describes the
relationship between the variables in the given

A, Fairly strong, negative correlation Ss

Very weak, positive correlation

B.
Cc. Fairly strong, positive cgebition
D.

Perfect, positive a ion (¢3)

1.3 Determine the cao the least squares regression line of the data. QB)

14 Use the cai tained in QUESTION 1.3 to predict an employee’s productivity
if he/she was, iven eight (8) additional rest days. (2)

2)

15 Should the regression line in QUESTION 1.3 be used to predict the productivity
of an employee if he/she was given thirty (30) additional rest days in the year?
Explain your answer. (2)

19]

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QUESTION 2

A survey was conducted among a number of learners who were using the same cellphone
company to establish the number of WhatsApp messages that they sent during a day.
The results were summarised in the partially completed cumulative frequency graph (ogive)

below.

Cumulative Frequency Graph (Ogive)

560

450 at

400 i

we
a
oe

w&
iJ
—)
usd

nN
wa
eo

200

Cumulative frequency
i

150 j

100 ra 7

50

pat
0 oF i
A B Cc D E F G H

Number of WhatsApp messages sent during a day

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The data shown in the cumulative frequency graph is also summarised in the partially completed

frequency table below.

WhatsApp messages sent | Frequency
40 <x <80 20
80 < x <120 30
120 < x < 160 Q
160 < x < 200 R
200 < x < 240 100
240 < x < 280 90
280 < x < 320 60
24 Write down the value of A, a tick mark on the horizontal axis of the cumulative
frequency curve. (1)
2.2 How many learners participated in this survey? ()
2.3 Calculate the values of Q and R, the frequencies in the frequency table above. (3)
2.4 Estimate the mean number of WhatsApp messages sent per day. (2)
2.5 Calculate the interquartile range. GB)
2.6 The cellphone company wants to introduce a new rule where you are not allowed
to send more than 200 WhatsApp messages perdaym[f this rule is applied to this
group of learners on the existing data, how would the standard deviation for the
data be affected? Motivate your answer. (2)
[12]

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QUESTION 3

In the diagram below, AC-1 ; 7), BG ; 1) and C are the vertices of AABC. The angle of
inclination of AC is @. BC || x-axis and BC = 4 units. D and E are the x-intercepts of the
lines AB and AC respectively. B and E are joined.

py
AC1;7)
D \ > g e
BG ; -1) 40 c
3.1 Write down the equation of the line BC. a)
3.2 Write down the x-coordinate of C. qd)
3.3 Determine the equation of the line AC. (4)
34 Calculate the size of ACB. (4)

3.5 If point K is the reflection of E in the line BC:

3.5.1 Write down the coordinates of K. G3)
3.5.2 Calculate the area of quadrilateral BECK. (3)
3.6 F is a point in the first quadrant, such that it forms an equilateral ABFC.
Calculate the coordinates of F. 6)
(21)

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QUESTION 4

In the diagram below, a circle with centre M and a tangent PD to the circle at P(O ; 6) are
drawn. The equation of the circle is given as x’ + 6x+y? —4y -12=0.
PR is a diameter, T is a y-intercept and chord RT is produced to D. RTD |! x-axis.

Ay
PO; 6)
M,
x
oO
R T D
4.1 Determine the coordinates of M. (3)
42 Determine the length of PR. Q)
43 Determine the equation of the tangent PD. GB)
44 Determine the coordinates of D. (4
4.5 The circle, having centre M, is translated to the right such that the point D lies on
the translated circle. Determine the coordinates of the possible centres of the
translated circle. 4
[16]

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Mathematics,

NSC

QUESTION 5

NW/September 2022

In the diagram below, f(x) = sinax and an incompleted g(x) = tan(x + 5) are drawn for the
interval x <¢ [0° ; 135°].

5.1

5.2

5.3

5.4

5.5

5.6

we

‘45° 90 135°

Write down the values of a and b.

Write down the range of f

Write down the period of g.

Write down the equation of the asymptote of g for the given interval.
Write down TWO values of x, in the given interval, where f(x) = 1 — g(x)

For which value(s) of x, in the given interval, is f’(x).g'(x) > 0?

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(2)
(2)
(1)
(i)
(2)

(2)
[10]

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QUESTION 6

6.1 In the diagram, K( P3 3 ) is a point in the 2"! quadrant. T is a point on the

positive x-axis and obtuse KOT = 0.

K(p 3 ty
6
ob X
(e) T
6.1.1 Write down the value of tan @ in terms of p. qd)

6.1.2 If 6 =120°, WITHOUT using a calculator, calculate the value ofp. (3)

6.2 If cos 42° = ¢, WITHOUT using a calculator, write the following expressions in

terms of t:
6.2.1 sin 48° 2)
6.2.2 cos 84° (3)
6.2.3 cos 72° (5)
6.3 Simplify the following expression to ONE trigonometric ratio:
sinx ‘ sin (180° ~ x) | sin” x + cos’ x ©
sin 2x cos x 2sin(90° + x)
6.4 Consider f(x) = sin(2@ - 15°).cos(@ — 30°) +cos(2@ — 15°).sin(@ — 30°)
Determine the general solution of f(x) = 0,8 (7)
6.5 Itis given that 48°* = 6, 6°°* =8 and 8 = 32!~2sin’s
WITHOUT solving for x, determine the value of tan 2x. (5)
[32]

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QUESTION 7

In the diagram below, AD is a vertical pole having height # metres. B, D and C are three
points in the same horizontal plane. AB and AC are cables and the angle of depression from

Ato Cis 30°, AB=3h and BAC = 2x.

7A Write down the size of ACD . (i)
7.2 Determine the distance AC in terms of h. (2)
7.3 Calculate the size of ABD. (2)
74 Caloulate the size of x if BC = V7h. (5)
{10}

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Give reasons for your statements in QUESTIONS 8, 9 and 16.
QUESTION 8

In the diagram, ABCD is a cyclic quadrilateral. G is a point on AD such that BG || CD:
ECF is a tangent to the circle at C. BD is a chord of the circle.

GBD = 30° and DCE = 60°.

A

8.1 Calculate, with reasons, the size of:
8.1.1 3B, _ 7 (1)
8.12 B, ; . (2)
813 é, Me a
8.1.4 DAB . a . 2)
8.2 Is BDa diameter of the circle? Motivate your answer. i a Q)

[8]

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MathelD rH

QUESTION 9

9.1 In the diagram below, AMVT and AAKF are drawn such that M=A, W=K and T=F.

Use the diagram in the ANSWER BOOK to prove the theorem which states that
if two triangles are equiangular, then the corresponding sides are in proportion,

that is ——-= ——
i)

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9.2 In the diagram below, cyclic quadrilateral PORS is drawn. Chord SR produced
and chord PQ produced meet at T. KPL is a tangent to the circle at P. U and V

are points on PS and PQ respectively such that UV || SQ || KL. Let KPS = x

L

9.2.1 Write down, with reasons, FIVE other angles EACH equal to x. (5)

9.2.2 Prove that:

(a) PQ=PS qd)
(b) KP is a tangent to the circle that passes through the points P, U and
V, at P. @)
9.2.3. Give a reason why = = 5 @
9.2.4 Prove that:
(a) APTS ||| ARTQ 4)
(b) PT.RQ=PS.RT @)
(c) PQ=/PSPT-TSROQ (4)

[25]

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QUESTION 10

In the diagram below, the circle with centre O, passes through the points C, D, and E,
OT bisects chotd CD. DOE.= 4x

. . oO
Prove, with reasons, that sin x.cosx = = [7]

TOTAL: 150

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Mathematics/P2

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INFORMATION SHEET: MATHEMATICS

~b Vb? —4ac

2a

x=
A= P(tni) A= P(\-ni)

T, =a+(n~-Vd

T,=ar S,

a

i i
d= (x, =x)? +0, =») ml

yr MX+e y-y, = m(x-x,)

(e-a)' +(y-b) =”?
a 5b c
sind sinB  sinC

In SABC:

a? =b* +02 —2be.cos A

area AABC = + ab,sin Cc

sina + Ay= sin a.cos B + cosa@,sin B
cosa + 8) =cosa.cos f ~ sin a.sin 2
cos’ @~sin’ @
cos 2a =41-2sin’ a
2c0s? a -1
x

Le

Pay =A

n(S)

poathx

pa ee 20-7)
Se-37

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mt alr” -

roi

; pats)" -1] poxc a+")

A=P0-i" A=P0+i)"

S,= 502 +(n-Nd]

Ss =); -1ere<l

sre 0
, l-r

He) = tim LTD LO)
Fo

A +%y Vit 2
;

pee)

y27¥
ma 2k
x2 7%

m=tand

sin(a — 8) = sin a.cos 8 —cosa.sin 8

cos(a ~ 8) = cosa@.cos f +sin a.sin 8

sin 2a = 2sin a.cosa

” 2

; XG -3)

P(A or B) = P(A) + P(B) — P(A and B)

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education

Lefapha la Thuto la Bokone Bophirima
Noordwes Departement van Onderwys
North West Department of Education
NORTH WEST PROVINCE

NATIONAL
SENIOR CERTIFICATE/
NASIONALE
SENIOR SERTIFIKAAT

GRADEI/GRAAD 12

MATHEMATICS P2/WISKUNDE V2
SEPTEMBER 2022

MARKING GUIDELINES/NASIENRIGLYNE

MARKSI/PUNTE: 150

These marking guidelines consist of 18 pages with 3 pages containing the cognitive grid/
Hierdie nasienriglyne bestaan uit 18 bladsye met 3 bladsye wat die kognitiewe tabel bevat.

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NSC/NSS — Marking Guidelines/Nasienriglyne

NOTE:
e If acandidate answers a question TWICE, only mark the FIRST attempt.
e If acandidate has crossed out an attempt of a question and has not redone the question,
mark the crossed out version.
e Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at
the second calculation error.
e Assuming answers/values in order to solve a problem is NOT acceptable.

NOTA:
e As ‘n kandidaat ‘n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na.
e As ‘n kandidaat ‘n antwoord van ‘n vraag doodtrek en nie oordoen nie, merk die
doodgetrekte poging.
e Volgehoue akkuraatheid word in ALLE aspekte van die nasienriglyne toegepas.
Hou op nasien by die tweede berekeningsfout.
e Om antwoorde/waardes te aanvaar om ‘n probleem op te los, word NIE toegelaat nie.

QUESTIONIVRAAG 1

Additional rest days given/
Addisionele rusdae gegee

Productivity of the employee(in %)/

Produktiwiteit van werknemer (in %) 85 63 90 56 70 88 72 62

11 |r=0,81 ¥r=081 (1)
1.2 |[C ¥C (1)
13 | j=a+bx ¥ a=56,79
a = 56,79 and/en b = 3,29 v b=3,29
a ¥ — equation/
“. Y = 56,79+3,29x vergelyking
(3)
14 | y =56,79+3,29(8) ¥ substitute 8 into eq.
¥ = 83,11% productivity / produktiwiteit /vervang 8 in vgl.
y 4 productivity / produktiwitei ¥83,11
OR/OF
y ~ 83,13% (calculator / sakrekenaar) vv answer/antwoord
(2)
1.5 | No/Nee v¥ No/Nee
The line of regression is only valid for /Die regressielyn is slegs | Y explanation/
geldig vir 1<x<10 verduideliking
(2)
[9]

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Cumulative frequency/ Kumulatiewe frekwensie

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QUESTIONIVRAAG 2

Cumulative Frequency Graph (Ogive)/

ng Guidelines/Nasienriglyne

Kumulatiewefrekwensie-grafiek (Ogief)

NW/September 2022

500 WhatsApp messages sent / | Frequency!
WhatsApp-boodskappe Frekwensie
450 a4 gestuur
40 < x < 80 20
400
Fo 80 < x <120 30
350 120 < x <160 Q
ae 160 < x < 200 R
300 $+: 200 < x < 240 100
AH 240 < x < 280 90
250 280 < x < 320 60
200 #$:
150
a
100 an’
tH t
50 t
0 eed
A B Cc D E F G H
Number of WhatsApp messages sent/
Aantal WhatsApp-boodskappe gestuur
2.1 40 yv_ 40 (1)
2.2 | 450 Y¥ 450 (1)
2.3 Q = 100 — 50 = 50 (calculating from/bereken vanaf interval) ¥VQ=50
R= 100 vy R=100 (3)
2.4 209 vv 209 (2)
25 Q, = 168 Also accept/Aanvaar ook vy Q, = 168
Q3 = 255 Q, € [167;169] YQ, = 255
“ IQR / IKO = 255 - 168
Q, € [254; 256]
= 87 ¥ — answer/antwoord
(3)
2.6 The standard deviation will be smaller/Die standaardafwyking sal v¥smaller/kleiner
kleiner wees.
Because three intervals are taken away, the data becomes more ¥ explanation/
closely grouped together/Omdat drie intervalle verwyder word, sal verduideliking
die data nader aan mekaar gegroepeer wees. (2)
[12]

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QUESTIONIVRAAG 3
Ay
A(-1; 7)
x
31 |y=-1 (1)
3.2 Xo =3+45=7 (1)
3.3 C(7;-1)
¥ method
= 7-(-) gradient/metode
ac -1-7 gradiént
=-1 ¥Me =-1
Through / Deur(-1;7) OR/OF (7;-1)
y-7 = -1(x-(-1)) y- (-1) = -1(x-7) v Subst./Vervang

y-7=-1(x +) y+l=-x+7 (-1;7) OR/OF (7;-1)
“ny=-xX+6 Vy=-X+6
(4)
3.4 tan = Myo v¥ tang =-1
tand=-1
tan“ (-1) =@
0 = — 45° + 180°
“8 = 135° ¥ 135°
AED = 45° [4s on straight line / Ze op reguit lyn] v AED = 45°
. ACB = 45° [corresp./s / oreenk.Ze; DE || BC] Y ACB= 45°
(4)

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3.5.1 |E(6;0) vx, =6
K(6 ; -2
( ) VX =6
YY = 72
(3)
3.5.2 | EK =2 units/eenhede v¥ EK=2
BC x EK
Area BECK = v¥ method/metode
_4x2
2
= 4units’ / eenhede? Y answer/antwoord
ORIOF (3)
1
Area ABEC = =(4)(1)
, v Area ABEC = 2
1
Area ABKC = 3 (4)
-2 v Area ABKC = 2
Area BECK = 2 + 2 = 4 units’ / eenhede” ¥ answer/antwoord
(3)
3.6 % = Xp + Xo
2
_—3+7
-5 2 ¥ method/metode
“FG: y) vx, =5

BF = (5-3) +(y +1) OR/JOF FC = (5-7) +(y +1)
BF = FC = BC =4
16=4+(y +1)

v distance form./
afstandsformule

(y +1)? =12 Y(y+2) =12

y=t N12 -1

ays V12 -1 (first quadr. / eerste kwadr.) ¥ antwoord/answer
(5)
[21]

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QUESTIONIVRAAG 4 Ay
P(O; 6)
M
oO
R T
4.1 | x? + 6x + (3)? + y? —4y + (-2)? =124+94+4 ¥ method/metode
(x +3) +(y-2) = 25
. M(-3; 2) v¥x=-3
Answer only: Full Marks vy=2
(3)
4.2 | r? =12 + (3)? + (-2)? = 25 vr? = 25 (can also be shown in
“r=5 4.1/ kan ook in 4.1 gewys word)
“PR =10 ¥ PR=10
(2)
43) _ 6-2
MP 0-(-3)
4 4
“3 “Tae 3
3 3
Meo =F [rad L tangent/rad 1 raaklyn] Ym, =- 4
y=-—x+c
y= _ 3, +6 ¥ equation / vergelyking
4 (3)

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Yo + Y,
44 ae You ¥ method/metode
6+ Ye _ 2
2
Ye =-2 YVR = 2
D(x ; — 2)
3 .
2 = - 7x46 ¥ subst. y = —2 into
32 equation/vervang y = —2 in
X= 3 vergelyking
32 32
“ D(—;-2 ¥ D(—=;-
( 3 ) D(Fi-2)
(4)
45 |T, >D,= = units / eenhede v =
new centre / nuwe middelpunt (F ; 2) v & ; 2|
R, 2D, = ° units / eenhede v0
3

new centre / nuwe middelpunt (F ; 2]

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QUESTIONIVRAAG 5
4y
x
5° 90) 135° .
5.1 | a=2 and/en b =— 45° Ya=2
vb=- 45°
(2)
5.2) -1<y<1 ORIOF ye [-1 : 1 ¥ critical values/kritiese
waardes
¥ notation/notasie
(2)
5.3 | 180° vy 180° (1)
5.4 | x= 135° v 135° (1)
5.5 | x=0° orlof x = 45° or/of x = 90° v vany 2 correct solutions/
enige 2 korrekte oplossings
(one mark per solution/ een
punt per oplossing)
(2)
5.6 | O° < x <45° ORIOF- x e[0%;A45°) ¥ 0° and/en 45°
¥ notation/notasie
(2)
[10]

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QUESTIONI/VRAAG 6
K(p 33 y
o> X
0 T
6.1.1 tang - 8 v x3
p p
(1)
6.1.2 | tan120° = ¥3
p
tan (180° = 60°) = aft
= Pp
—tan 60° = RE v —tan 60°
3
—J3 = 3 ¥—\3 = 3
p p
p=-1 vp=-l
(3)
6.2.1 cos 42° ¥ sin 48° = cos 42°
= sin 48° vt
=t
Answer only: Full Marks
(2)
6.2.2 | cos2(42°) ¥c0s2(42°)
= 2cos’ 42° -1 v 2cos’ 42° -1
= 2t?-1 ¥ 2? -1
ORIOF
cos 2(42°) ¥cos2(42°)
=1- 2sin? 42° ¥1- 2sin’ 42°
2
=1- 2( 1-1")
= 2t? -1 2? 1

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622 ORIOF ¥c0s2(42")
cos 2(42°) Y cos? 42° — sin? 42°
= cos’ 42° — sin? 42°
2

=? -(i-")
v2? -1

= 2t?-1 (3)

6.2.3 cos (42° + 30°) ¥ manipulation /

manipulasie

= COS 42°.cos30° — sin 42°.sin 30° ¥ expansion of identity/
uitbreiding van identiteit

-.8_ foe vB gh

ato yin ts 2 2
vt&/jl-t?

_ Bt fl -0?

2 2

2
_ vat~ yi-0 ¥ answer/antwoord
2 (5)
6.3 sinx | sin(180° - x) _ sin’ x + cos’ x

sin 2x COS X ~ 2sin (90° + x)

sin x . sin x 1
2sinx.cosx cOosSx 2CcOSXx

1 sin x 1
= +—-
2cosx cosx 2cosx

sin x
cos X

=  tanx

¥ sin 2x = 2sin x.cos X

¥ sin (180° — x) = sinx

v sin? x + cos? x =1

¥ 2sin(90° + x) = 2cosx

y Sinx

COS X

v tan x

(6)

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i}

ng Guidelines/Nasienriglyne

NSC/NSS — Mar!

NW/September 2022

6.4 | f(x) = sin[(20 - 15°) + (@ - 30°) ]
f (x) = sin[3@ — 45°]
.. sin[30 — 45°] = 0,8

30 — 45° = 53,1301...° + k.360°, ke Z
30 = 98,1301...° + k.360°
6 = 32,71° + k.120°

or / of
30 — 45° = 180° — 53,1301...° + k.360°, keZ

30 = 171,8699...° + k.360°
6 = 57,29° + k.120°

Y £(x) = sin[30 — 45°]
Y sin[30 — 45°] = 0,8

¥ 30 — 45° = 53,1301...°

¥6@= 32,71° + k.120°

v
30 — 45° = 180° — 53,1301...°

¥ 0 = 57,29° + k.120°
¥+k.120°,keZ

0)
6.5 4sinx < 92sinx _ Y 2esinx _
geex = (22x ye = 2esinxcosx _ g ¥ 2esinx.cosx _ g
Q = 22sinxc0sx _ gol - 2sin? x
sin 2x = 5cos2x ¥ sin2x = 5cos2x
tan2x =5 ¥tan2x=5
(5)
[32]

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NSC/NSS — Marking Guidelines/Nasienriglyne
A
QUESTIONIVRAAG 7

7.1 | ACD = 30° ¥30°
(1)
. h . h
7.2 30° = —~ v sin30° = —
sn AC AC
“. AC = 2h ¥ AC = 2h
(2)
73 | sinaBp - t= Y sinABD = +
3h 3 3
ABD = sin
3
ABD = 19,47° ¥ ABD =19,47°
(2)
7.4 | BC? = AB? + AC? - 2AB.AC.cos 2x ¥ correct use of cosine-
rule/korrekte gebruik van
cos-reél
2 2 2 .
(¥7h) = (3h) + (2h) — 2(3h)(2h)cos2x ¥ subst/vervanging
Th? = Qh? + 4h? — 12h? cos2x
—6h? = —12h’ cos 2x
1
cos 2x = — Y cos2x = +
2 2
2x = 60° v 2x = 60°
x = 30° vx = 30°
(5)
[10]

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NSC/NSS — Marking Guidelines/Nasienriglyne
GEOMETRY/MEETKUNDE

Please read carefully through the following table before marking QUESTION 8-10 /
Lees asseblief sorgvuldig deur die volgende tabel alvorens VRAAG 8-10 nagesien word.

The order in which the candidate answers a geometry question must follow
logically/ Die volgorde waarin ‘n kandidaat ‘n meetkundevraag beantwoord moet
logies volg.

Example/Voorbeeld

Given/Gegee AB || CD and/en EFD = 115°
A D

65

B c

The candidate first need to calculate x BEFORE he/she can calculate y/Die
kandidaat moet eerste vir x bereken VOORDAT hy/sy vir y kan bereken.

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)

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)
Award a mark if the statement AND reason are both correct
(Both MUST be correct to get one mark)

Ken ‘n punt toe as die bewering EN rede beide korrek is

(Beide MOET korrek wees om een punt te kry)

S/R

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NSC/NSS — Marking Guidelines/Nasienriglyne

QUESTIONIVRAAG 8 A

8.1.1] D, = 30° [alt 4s/ verwiss Ze; BG || DC] vSIR

(4)
8.1.2 | B, = 60° [tan-chord th/raaklyn - koordst. ] vSYR

(2)
8.1.3 | C, = 90° [Zs of A/binne.Ze van A] vSIR

(1)
8.1.4 | DAB = 90° [opp.4 of cq / teenoorst.Ze van kvh] v¥SYR

(2)
8.2 | Yes/Ja v yes/ja

BD is adiameter/BD is 'n middellyn [converse  omgekeerde Zin su | vSIR
(2)
[8]

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NSC/NSS — Marking Guidelines/Nasienriglyne

QUESTIONIVRAAG 9
M

9.1 Construction: Mark P and Q such that MP = AK and MQ = AF. Y constr/
Draw PQ konstr
Konstruksie: Merk P en Q sodat MP = AK en MQ = AF.

Trek PQ

In AMPQ and/en AAKF
M = A [given / gegee]
MP = AK [construction / konstruksie]

MQ = AF [construction / konstruksie]

“. AMPQ = AAKF[S, Z,S] Yv SIR
P, = K [from / vanuit = ] YS
but /maar V = K [given / gege]

P= V vs

.. PQ|IVT [corresp.2s = /ooreenk.Ze =] v¥ SIR
MV MT we

—= rop.th / eweredigheidst.; PQ|| VT

MP” MO [prop 9 QIIVT] Vv SYR
MP = AK and/en MQ = AF

_MV _MT

“AK AF (7)

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NSC/NSS — Marking Guidelines/Nasienriglyne

9.2 L

9.2.1 KPS = PQS =x _ [tan-chord th./raaklyn - koordst. | VSIR
KPS = PSQ=x [alt.Zs/verwissel.Ze;KL||SQ | VSIR
KPS = PUV=x [alt. Zs/verwissel. Ze; KL||UV ] YSIR
PVU = PQS = x [corresp. Zs/ooreenk. Ze; UV||SQ] VSIR
LPQ = PSQ = x [tan-chord th./raaklyn - koordst.] YSIR
(5)
9.2.2 (a) | PSQ =PQS= x
-. PQ = PS [sides opp. = /s/ sye teenoor ="Ze } YR
(4)
9.2.2(b) | KPU=PWU =x
-. KP is a tangent /'n raaklyn
[converse tan-chord th. / omgekeerde raaklyn — koordst. | YR
(4)
9.2.3 Prop.th / eweredigheidst. ; UV || SQ YR
(1)

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NSC/NSS — Marking Guidelines/Nasienriglyne
9.2.4 (a) In APTS and/en ARTQ
T =T [common / gemeenskaplik ] YSIR
TPS=TRQ [ext.Z of cq/buitez van kvh] YSYR
PST = TOR [sum of “s of A/som van Ze van A] VSIR
J ATPS ||| ATRQ [2 2 4] (4)
9.2.4 (b
(0) | TP _ PS [from / vanuit ||| A] vSvYR
TR RO
. PT.RQ = PS.TR (2)
9.2.4 (Cc vs
() PS _ 1S [from / vanuit ||| A]
RQ TQ
PS.TQ =TS.RQ vs
but /maar TQ = PT — PQ
PS(PT — PQ) =TS.RQ vs
but/maar PQ = PS
PS.PT — PQ’ = TS.RQ
PQ? = PS.PT — TS.RQ vs
PQ = JPS.PT —TS.RQ (4)
[25]

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NSC/NSS — Marking Guidelines/Nasienriglyne

QUESTIONIVRAAG 10
D
4x
Cc [e) E
7 Z at centre = 2x /at circumference/

DCE = 2x . vSVR

midpts.7 = 2xomtreks.Z

A line from centre to midpt of chord/

OTC =90°| . P Y¥SYR

midpt.0 ; midpt.koord
In A OTC
sin2x = oT vs

oc

2sin XCOs X = oc ¥ 2sinxcosx
sin Xcos x = OT

2x OC v axOG

OT *

sin Xcos x = —

CE

(7)
[7]

TOTALITOTAAL: 150

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