Monthly income (in rands)

9 000

13 500

15 000

16 500

17 000

20 000

Monthly repayment (in rands)

2 000

3 000

3 500

5 200

5 500

6 000

1.1

a = -1946,875... = -1946,88
b = 0,41                                            
Answer only: Full marks
yˆ = -1946,88 + 0,41x

a = -1946,88

b = 0,41

equation

(3)

1.2

Monthly repayment ≈ R3 727,16  (calculator)

OR

y = -1946,88 + 0,41(14000)
= R3 793,12

answer (2)

substitution

answer
(2)

1.3

r = 0,946 …. ≈ 0,95

🗸 answer

(1)

1.4

Not to spend R9 000 per month because the point (18 000 ; 9 000)
lies very far from the least squares regression line. OR

🗸🗸 answer
(2)

[8]

2.1

Number people paid R200 or less = 19

🗸 answer (1)

2.2

7 +12 + a + 35 + b + 6 = 100
a = 40 - b
309 = (50 x 7) + (150 x 12) + (250 x a) + (350 x 35) + (450 x b) + (550 x 6)
                                                       100
309 = (50 x 7) + (150 x 12) + (250 x (40 - b) + (350 x 35) + (450 x b) + (550 x 6)
                                                       100
350 + 1800 + 10000 – 250b + 12250 + 450b +3300 = 30900
200b = 3200
b = 16
a = 24

OR

7 +12 + a + 35 + b + 6 = 100
b = 40 - a

309 = (50 x 7) + (150 x 12) + (250 x a) + (350 x 35) + (450 x b) + (550 x 6)
                                                       100
309 = (50 x 7) + (150 x 12) + (250 x a) + (350 x 35) + (450 x(40 - a)) + (550 x 6)
                                                       100
350 + 1800 + 250a + 12250 + 1800 - 450a = 30900
200a = 3200
b = 16
a = 24

🗸 ∑x = 100

🗸 a = 40 - b

🗸 ∑ fX

🗸 ∑ fX = 309
        n
🗸 200b = 3200 (5)

🗸 ∑x = 100

🗸 a = 40 - a

🗸 ∑ fX

🗸 ∑ fX = 309
        n
🗸 200a = 4800 (5)

(5)

2.3

Modal class: 300 < x ≤ 400

🗸 answer (1)

🗸 grounded at (0 ; 0)
🗸 (600 ; 100)
🗸 cumulative frequencies for y-coordinates
🗸 smooth shape
(4)

3.1

Equation of PR: y = 5

🗸 answer (1)

3.2.1

m_RS  = y₂ - y₁
            x₂ - x₁
m_RS = 5 - (-7) = 12  Answer only: Full marks
            3 - (-3)    6
= 2

🗸 substitution of R & S into gradient formula

🗸 answer
(2)

3.2.2

mRS = mPT [PT || RS]
tanθ = 2
θ= 63,43°

🗸 m_RS = m_PT
🗸 tanθ = 2
🗸 θ = 63,43°
(3)

3.2.3

Equation of RS:

y - 5 = 2(x - 3) or y - (-7) = 2(x - (-3)) or 5 = 2(3) + c
y - 5 = 2x - 6              y + 7 = 2x + 6             c = -1
y = 2x -1                   y = 2x -1             y = 2x -1
 D(0 ; –1)

OR
m_RS = m_RD = mD_S  Answer only: Full marks
2 = 5 - y =  y + 7   
     3 - 0     0 - (-3)
y = -1
D(0 ; –1)

🗸 substitution

🗸 equation of RS

🗸 coordinates of D
(3)

🗸 equating gradients

🗸 value of y

🗸 coordinates of D

(3)

ST = 2√5 = √[-5 - (-3)]² + (k - (-7))²
20 = 4 + (k + 7)²
(k + 7)² = 16
k + 7 = ±4
k = -11 or k = -3
k = –3

OR

ST = 2√5 =√[-5 - (-3)]² + (k - (-7))²
20 = 4 + k ² +14k + 49
k ² +14k + 33 = 0
(k +11)(k + 3) = 0
k = -11 or k = -3
k = –3

🗸 substitute S and T into distance formula

🗸 isolate square

🗸 square root both sides

🗸 answer
(4)

🗸 substitute S and T into distance formula

🗸 standard form

🗸 factors

🗸 answer
(4)

Method: translation
T→S:
(x ; y) → (x + 2 ; y - 4)
by symmetry: D→N:
D(0 ; –1) → N(0 + 2 ; –1 – 4)
N(2 ; –5)
Answer only: Full marks

OR

Midpoint of TN = Midpoint of SD
x + (-5) = - 3 + 0   and    y + (-3) = - - 7 + (-1)
   2               2                      2                    2
x = 2 and  y = -5
N(2 ; – 5)   Answer only: Full marks

🗸 method

🗸 x-coordinate

🗸 y-coordinate

(3)

🗸 method:

midpoint of diagonals

🗸 x-coordinate

🗸 y-coordinate

(3)

β is the inclination of RS β = 63,434...°
OFD = 63,434...°  [vert opp ∠ s]
ODF = 90° - 63,434...° = 26,565...°
RDR^/  = 2(26,565...°) = 53,13°

OR
PEFR is a ||m   [both pairs of opp sides || ]
R  = θ = 63,434...°        [opp ∠s of ||m]
RR'D = 63,434...°            [∠s opp = sides:RD = R'D ]
RDR'  = 180° - (63,43° + 63,43°) [sum of ∠s in ∆]
RDR'  = 53,13°

OR
tan ODF = ³/₆
ODF = 26,565..°
RDR^/  = 2(26,565...°) = 53,13°

OR
R^/(-3; 5)    [reflection of R(3 ; 5) about the y-axis]
RD = √(3 - 0)² + (5 - (-1)²
RD =     √45 = R^//D  or    3√ 5  or  6,71
(RR/ )2 = (√45⁾² + (√45)² - 2(√45)(√45)(cos RDR^/) 

6² = 45 + 45 - 2(45)(cos RDˆ R^/ )
cos RDR^/  = 45 + 45 - 36
                         2(45)
cos RDR^/  = ³/₅
RDR^/  = 53,13°

🗸 b = 63,43°

🗸 ODˆ F = 26,57°

🗸 answer (3)

🗸  Rˆ  = 63,43°

🗸  RR^/D = 63,43°

🗸 answer

(3)

🗸 trig ratio

🗸 ODF = 26,565..°

🗸 answer

(3)

🗸 R/ (-3; 5) OR

RD =    45 = R^//D

🗸 substitution into cosine rule

🗸 answer

(3)

4.1

M(-1;1)
(x +1)2 + ( y -1)2   = 1 
Answer only: Full marks

🗸 M(-1;1)

🗸LHS 🗸 RHS

(3)

4.2

Midpoint of CB, N: (– 0,5 ; 1,5)
x_C + 0 = - ½   and  y_C + 1 = 3 
    2                            2        2       
Answer only: Full marks
C(– 1 ; 2)

OR

B→N:
(x ; y) → (x - 0,5 ; y + 0,5)
N→C:
Answer only: Full marks
(x ; y) → (x - 0,5 ; y + 0,5)
C(– 0,5 – 0,5 ; 1,5 + 0,5)
C(– 1 ; 2)

🗸x value 🗸 y value

(2)

🗸x value 🗸 y value

(2)

m_tangent = 1
y = mx + c
y = x + c
2 = 1(-1) + c
c = 3
y = x + 3
y – x = 3

OR

m_radius = 2 -1 
             -1 - 0
= -1
mtangent = 1
y - y₁ = m(x - x₁)
y - y₁ = 1(x - x₁)
y - 2 = 1(x -(-1))
y - 2 = x +1
y = x + 3
y – x = 3

🗸 mradius

🗸 mtangent 

🗸 substitute (– 1 ; 2) and m

🗸 simplification

(4) 

🗸 mradius

🗸 mtangent 

🗸 substitute (– 1 ; 2) and m

🗸 simplification

(4)

Tangents to circle: y = x + 3 and y = x +1
t > 3 or    t < 1   
Answers only: Full marks

🗸 y = x +1 
🗸 t > 3      🗸 t < 1
(3)

Draw rectangle CNED:

OR
D(– 3 ; 0) C→N:
(x ; y) → (x + 0,5 ; y - 0,5)
D→E:
D (x ; y) →E(x + 0,5 ; y - 0,5) Answer only: Full marks
E(– 3 + 0,5 ; 0 – 0,5)
E(– 2,5 ; – 0,5)

🗸 midpt of DN

🗸x value 🗸 y value
(3) 

🗸 coordinates of D

🗸x value 🗸 y value

(3)

area of trapezium AOBC = ½ (1 + 2)(1)
= 1½ square units

area of DACD = ½(2)(2)
= 2 square units

area of quadrilateral OBCD = 3½ square units
2a² = ⁷/₂
a² = ⁷/₄
a = ^√7/₂

OR

BM produced cuts the tangent at F. area of ΔCFB = ½ (2)(1)
= 1 square unit
area of trapezium BFDO = ½(2 + 3)(1)
= 2½ square units
area of quadrilateral OBCD = 3½ square units
2a² = ⁷/₂
a² = ⁷/₄
a = ^√7/₂

OR

Join DB
area of ΔODB = ½ (3)(1)
= ³/₂ square unit
area of ΔDCB = ½(2√2 )(√2)
= 2 square unit
area of OBCD = ³/₂ + 2 = square units 2
2a² = ⁷/₂
a² = ⁷/₄
a = ^√7/₂

OR 

Let E be the point of intersection of DC with the positive y–axis.
area of DDEO = ½(3)(3)
= ⁹/₂ square unit
area of DECB = ½(2)(1) or ½(2√2 )(√2)
= 1 square unit
area of quadrilateral OBCD = 9 - 1 = 3 1 square units
2a² = ⁷/₂
a² = ⁷/₄
a = ^√7/₂

🗸 substitution into area of trapezium form

🗸 area of trapezium 

🗸 area of triangle

🗸 area of OBCD 

🗸 equating area OBCD to 2a²

(5)

🗸 area of triangle

🗸 substitution into area of trapezium

🗸 area of trapezium

🗸 area of OBCD

🗸 equating area OBCD to 2a² 

(5)

🗸 area of Δ

🗸 subst into area of Δ

🗸 area of Δ

🗸 area of OBCD

🗸 equating area OBCD to 2a²

(5)

🗸 area of Δ

🗸 subst into area of Δ

🗸 area of Δ

🗸area of OBCD

🗸 equating area OBCD to 2a²

(5)

5.1

    sin x     + sin(180° + x) cos(90° - x)
cos x.tan x
=      sin x      + (-sin x) sin x
   cos x. sin x
             cos x
=1- sin ²x
=cos ²x

🗸 –sin x 🗸 sin x
🗸 tan x = sin x
                cos x
🗸 1- sin ²x
🗸 cos²x
(5)

5.2

sin ² 35° -cos ² 35°
  4sin10° cos10°
= - (cos2 35° - sin 2 35° )
        2(2sin10° cos10°)
= - cos 70°
    2sin 20°
=   - cos 70°      OR    =    - sin 20°     
     2 cos 70°                      2 sin 20° 

🗸  - (cos2 35° - sin2 35° )

🗸– cos 70°

🗸 2sin 20°

🗸 answer

(4)

5.3

2sin 2 77° = 2[sin(90° -13°)]²
= 2cos² 13°
= 2cos² 13° -1+1
= cos 26° + 1
= m +1

OR

1- 2sin ² 77° = cos154°
2sin ² 77° = 1- cos154°
= 1- (-cos 26°)
= 1+ m

🗸 using co-ratio
🗸 reduction
🗸 2cos² 13° -1 = cos 26°
🗸 answer
(4)

🗸 1- 2sin ² 77° = cos154°
🗸 2sin ² 77° = 1- cos154°
🗸 reduction
🗸 answer

(4)

5.4.1

sin(x + 25°) cos15° - cos(x + 25°)sin15° = tan165°
sin(x + 25° -15°) = -0,2679... OR - 2 + √3
sin(x + 10°) = -0,2679...OR - 2 + √3
x + 10° = 195,54° + k.360°  or x + 10° = 344,46° + k.360°
x = 185,54° + k.360°; k∈Z or  x = 334,46° + k.360°; k∈Z

OR

sin(x + 25°)sin 75° - cos(x + 25°) cos 75° = tan165°
- (cos(x + 25°) cos 75° - sin(x + 25°)sin 75°) = -0,2679...
cos(x + 100°) = 0,2679...
ref. ∠ = 74.4577…°
x + 100° = 74,46° + k.360°
or x + 100° = 285,54° + k.360°
x = –25,54° + k.360°; k∈Z or x = 185,54° + k.360°; k∈Z

🗸 - 0,2679...
🗸 195,54° & 344,46°
🗸 185,54° & 334,46°
🗸 + k.360°; k∈Z
(6)

🗸🗸 cos(x + 100°)
🗸 - 0,2679...
🗸 74,46° & 285,54°
🗸 - 25,54° & 185,54°
🗸 + k.360°; k∈Z

(6)

f (x) = sin(x +10°)  Answers only: Full marks
For minimum value of sin x : x = 270°
For minimum value of sin(x +10°) : x = 260°

🗸 f (x) = sin(x +10°)
🗸 270°
🗸 answer
(3)

6.1

Range of f: y ∈[- 2 ;0] OR - 2≤ y ≤ 0

🗸 critical values
🗸 notation
(2)

6.2

x∈(90° ; 270°) OR x∈[90° ; 270°]

🗸 critical values
🗸 notation
(2)

6.3

PQ= cos 2x -(sin x -1)
= 1- 2sin²x -sin x + 1
= - 2sin ²x -sin x + 2
sin x = - b
           2a
= - (- 1)
   2(- 2)
sin x = -¼
x = 194,48° or x = 345,52°

🗸 PQ= cos 2x -(sin x -1)
🗸 cos 2x =1- 2sin ²x
🗸substitution into formula
🗸 sin x =-¼
🗸 194,48° 🗸 345,52°

(6)

7.1

sin 60° = AK 
                x
AK= x sin 60° or √3 x or 0,866x
                            2

🗸 trig ratio
🗸 answer
(2)

7.2

KCF= 120°

🗸 answer (1)

7.3

KF² = CF² + CK² - 2CF.CKcos KCF

AKF = y
Area ΔAKF= ½.AK.KFsin AKF
=½. √3x . √7x sin y
        2        2
= x² √21sin y
          8

🗸 correct use of cosine rule
🗸substitution
🗸 cos120° = - ½
🗸 KF = 7x
             2
🗸 correct use of area rule
🗸substitution
🗸 answer in terms of x and y
(7)

[10]

8.1.1

Rˆ  = 80°
[co-int ∠s; QW || RK]

S R (2)

8.1.2

Pˆ  = 100° [opp ∠s of cyclic quad]

S R (2)

8.1.3

PQR = 136° [ext ∠ of cyclic quad]
Q₂ = 36°

OR
PUW =S₂  = 136°  [corresp∠s  QW || RK]
PQˆW + P = PUW   [ext ∠s of Δ QPU]
PQW +100° = 136° PQˆ W = 36°
OR

U₃  = 180° -136° = 44°     [co-int∠s; QW || RK]
U₁  = U₃  = 44°    [vert opp∠s ]
PQW = 180° - (100 + 44°)    
[sum of ∠s in ∆] PQW = 36°

S R
S
(3)

S R
S
(3)

S R
S
(3)

U₂  = S₂  = 136°  [alt ∠s; QW || RK]
OR
U₂  = 100° + 36°  [ext ∠s of ∆QPU]
= 136°

OR
U₂ = PUW = 136° [vert opp ∠s]

OR
U₂  = 180° - U₃  [∠s on a str line]
= 180° - 44°
= 136°

S R
(2)

S R
(2)

S R
(2)

S R
(2)

8.2.1

In ΔEFT and DDCT:
EF = 9 = 1 
CD  18    2
FT = 5 = 1 
TC  10     2
ET = 7 = 1 
TD  14    2
∆EFT ||| ∆DCT [ Sides of ∆ in prop]
EFD = ECD

OR
In DFET:                                      In DTDC:
49 = 25 + 81- 2(5)(9)cosF           196 = 100 + 256 - 2(10)(18)cosC
cosF = 19                                      cosC  = 19 
            30                                                   30
F = 50,7°                                       C = 50,7°

all 3 ratios = ½

∆EFT ||| ∆DCT
R (4)

F = 50,7°

C  = 50,7°
(4)

EFD = ECD      [proved in 8.2.1]
E, F, C and D are concyclic
EFCD is a cyclic quad   [converse ∠s in the same segment]
DFC = DEC [∠s in the same segment]

S R
R
(3)

O₂   =360° - x  [∠s round a pt]
M  =180° - ½x  [∠ at centre = 2 ×∠ at circumference
T₂ + P₁ = ½x   [sum of ∠s in ∆]
T₂ = P₁  = ¼x   [∠s opp equal sides]
STM = P₁  = ¼x     [tan chord theorem]

OR
O₂  =360° - x  [∠s round a pt]
M =½O₂  [∠ at centre = 2 ×∠ at circumference]
T₂ + P₁  =180° - M  [sum of ∠s in ∆]
T₂  = P₁  [∠s opp equal sides]
= 180° - M  = 180° - ½O₂ = 180° - ½(360° - x) = ¼x
        2                    2                           2                   
STˆM =  ¼x [tan chord theorem] 

O₂   =360° - x
M  =180° - ½x

T₂ + P₁ = ½x 
P₁  = ¼x 
R (7)

O₂  =360° - x 
S R
S
R
S
R
(7)

10.1

Constr: Draw h1 from E⊥AD and h₂ from D ⊥ AE
Proof:
area ΔADE =½ ADxh₁= AD
area ΔBDE   ½DBxh₁   DB

area ΔADE =½ AE x h₂ = AE 
area ΔDEC   ½ EC x h₂   EC

But areaΔBDE = area ΔDEC [same base & height or DE || BC]

area ΔADE = area ΔADE
area DBDE    area ΔDEC
AD = AE
DB   EC

🗸 constr OR
reason: common vertex or same height

area ΔADE =½ ADxh₁= AD
area ΔBDE   ½DBxh₁   DB

area ΔADE = AE 
area ΔDEC    EC        

🗸 S 🗸R
🗸 S
(6)

10.2.1

V₃   = x    [Tans from same point]
R  = x     [tan chord theorem]
W₃   = x   [corresp ∠s; WT || RV]

🗸 S 🗸 R
🗸 S 🗸 R
🗸 S 🗸 R (6)

10.2.2(a)

V₃   = W₃   = x  [proved in 10.2.1]
W, S, T and V are concyclic
WSTV is a cyclic quad [converse ∠s in the same segmen

🗸 S
🗸 R
(2)

10.2.2(b)

W₂   = S₂   = x  [∠s in the same segment]
V₁  = W₂   = x [alt ∠s ; WT || RV]
But  R  = x     [proved in 10.2.1]
Rˆ  = V₁  = x
WR = WV  [sides opp equal ∠s]
ΔWRV is isosceles

OR

S₂ =W₂  = x  [ ∠s in the same segment ]
W₂  = W₃ = x
W₂  + W₃ = Rˆ  + V₁   [ext ∠ of Δ]
V₁  = x = R
WR = WV   [sides opp equal ∠s]
ΔWRV is isosceles

S 🗸R
S/ R
S (4)

S 🗸R
S/ R
S
(4)

In ∆WRV and ∆TSV
R  = S₂  = x  [proved OR tan chord theorem]
V₁ = V₃   = x  [proved]
∆WRV ||| ∆TSV        [∠, ∠, ∠ ]

OR
In ∆WRV and ∆TSV
R  = S₂   = x  [proved OR tan chord theorem]
V₁   = V₃   = x  [proved]
W₁  = STV = x   [sum of ∠s in ∆]
∆WRV ||| ∆TSV

S

S

R (3)

S

S

S

(3)

RV = WR     [ DWRV ||| DTSV]
SV     TS
WR x SV = RV ´TS

WR = KV   [prop theorem; WT || RV]
SR     SV

WR x SV = KV x SR

RV x TS = KV x SR
RV = KV
SR    TS

OR

In ∆RVS and ∆VKT
SVR = K₄   [alt ∠s, WT || RV]
SRˆ V = V₃    [proven]
∆RVS ||| ∆VKT [∠, ∠, ∠ ]
RV = KV
SR    VT

but VT = ST  [tans from same point]
RV = KV
SR    TS

correct ratios

WR = KV   R
SR     SV

equating WR x SV

(4)

identifying correct

∆s

proving |||

correct ratio

S

(4)