1.1

a = 9,5
b = 0,909.. = 0,91
y = 9,5 + 0,91x

a = 9,5
b = 0,91
equation (3)

1.2

correct slope going through 2 points:
(50 ; 55) or
(40 ; 46) or
(60 ; 64) or
(0 ; 9,5) or
(45 ; 50)
(2)

1.3

Final exam mark ≈72,22%(calculator)
OR
y = 9,5 + 0,91(69)
≈72,29%

answer(2)
substitution
answer
(2)

1.4

r = 0,95

🗸 answer(A) (1)

1.5

There is a very strong positive correlation between
the Mathematics and Physical Sciences mark.

🗸 strong(1)

1.6

The teacher concludes that the higher the learners' Mathematics
marks, the higher the learners' Physical Sciences marks.

answer (1)

[10]

2 018

2 175

2 182

2 215

2 254

2 263

2 267

2 271

2 293

2 323

2 334

2 346

2.1

July

🗸 answer

2.2

x = 26 941 Answer only: Full marks
         12
= 2 245,083.. ≈ 2 245,08 aircraft landings

🗸 26 941
🗸 answer

2.3

Standard deviation for landings at the King Shaka International airport:
s = 86,30

🗸🗸 answer

2.4

(x - σ ; x + σ) = (2 245,08 – 86,30 ; 2 245,08 + 86,30)
limit = (2 158,78 ; 2 331,38)
There were 9 months when the aircraft arrivals at the King Shaka
International airport were within one standard deviation of the mean.

🗸 x - σ
🗸 x + σ
🗸 answer

2.5

The standard deviation of the number of landings at the Port Elizabeth
Airport will be higher than the standard deviation of the number of arrivals
at the King Shaka International Airport OR C.

🗸 answer

[9]

3.1

m_WP  = 4 - (-4) = 8 
            - 4 - 0    - 4
m_WP = -2

🗸 substitution of W and P
🗸 m_WP
(2)

3.2

m_ST = ½ (given)
(m_WP)(m_ST ) = (- 2)(½)
= -1
ST ⊥ WP

🗸 (m_WP)(m_ST)
🗸 (m_WP)(m_ST) = -1

3.3

5y + 2x + 60 = 0
y = ^{- 2}/₅x -12
-²/₅x -12 = ½x + 6
- 4x -120 = 5x + 60
9x = -180
x = -20
y = ^{- 2}/₅ (- 20)-12
y = -4
S(–20; –4)
OR

🗸 equating
🗸 x value
🗸 substitution
🗸 y value
(4)

5y + 2x + 60 = 0
5(½x + 6) + 2x + 60 = 0
⁵/₂x + 30 + 2x + 60 = 0
⁹/₂x = -90        x = -20
y = -²/₅ (- 20)-12
y = -4
S(–20; –4)

OR
5y + 2x = -60........... (1)
2 y - x = 12............. (2)
(1) + 2(2) : 9 y = -36
y = -4
2(–4) –  x = 12
x = –20

🗸 substitution
🗸 x value
🗸 substitution
🗸 y value
(4)

🗸 adding
🗸 y value
🗸 substitution
🗸 x value
(4)

y = -²/₅ (- 4)-12   OR    5y + 2(-4) + 60 = 0
y = -⁵²/₅
R(- 4; -⁵²/₅) OR   R(- 4 ; - 10,4)
WR = 4 -(-⁵²/₅)  OR WR = √(-4 - (-4))² + (4 - (-⁵²/₅ )²
WR = ⁷²/₅ units or WR = 14²/₅ units

OR
WR = ST - SK
= ½ x + 6 -(-²/₅x - 12)
= ⁹/₁₀ x + 18
=⁹/₁₀ (-4) + 18
= 14,4 units

🗸 substitution
🗸 y value
🗸 method or subst into distance formula
🗸 answer
(4)

🗸 substitution
🗸 simplification
🗸 subst x = –4
🗸 answer  (4)

m_SK = - ²/₅
β = 158,19...°   (Ref. ∠ =21, 801…°)
MNS = 21,80 .. °
m_ST  = ½
NMS = 26,56 .. °
θ = 21,80...° + 26,56...° [ext ∠ of Δ]
θ = 48,366...° = 48,37°

🗸 m_SK
🗸 size of b
🗸  size of  NMS
🗸 method
🗸 answer
(5)

In ΔSRW:
⊥h= –4 – (–20)
⊥h = 16units
Area DSRW = ½(⊥h)(WR )
=½(16)( ⁷²/₅)
=115,2square units
Area SWRL = 2Area ΔSRW
=2(115,2)
= 230,4square units

OR
In DSRW:
⊥h = –4 – (–20)
⊥h = 16units
Area SWRL =16 x ⁷²/₅
= 230,40 square units

OR
SW = √(- 20 + 4)²  +(- 4 - 4)² = 8√5 = 17,89
SR = √(- 20 + 4)² +(- 4 + 10²/₅)² =16√29  = 17,2
                                                          5
Area SWRL = 2 x Area ΔSRW
= 2(½SW x SR sinθ)
=2(½8√5 x 16√29 sin 48,37°)
                     5
=230,41square units

🗸⊥h
🗸 substitution
🗸 area Δ
🗸 answer
(4)

🗸 ⊥h
🗸 🗸 substitution
🗸 answer
(4)

🗸SW =8√5
🗸SR =16√29
               5
🗸substitution
🗸answer
(4)

4.1

x² + y ² = r ²
r ² = (- 3)2  + (4)2  = 25
x² + y ² = 25

🗸 substitution
🗸 answer
(2)

4.2

TM ⊥ TN   [tangent ⊥ radius]
T(–11 ; 4)
r = –3 – (–11) = 8
(x + 3)² + (y - 4)² = 64

🗸 x_T = -11
🗸 LHS 🗸 RHS
(3)

4.3

O (0 ; 0 ) and M(–3; 4)
m_OM = 4 - 0 = - 4   OR    0 - 4   = - 4
          - 3 - 0     3            0 - (-3)     3
m_NM = 3 
            4
y - 4 = ³/₄ (x - (-3))          OR      y = ³/₄x + c
y - 4 =³/₄x + 9                         4 =³/₄ (-3) + c
y = ³/₄x + 25                            c = 25
y =³/₄x + ²⁵/₄

🗸 m_OM = - ⁴/₃
🗸 mNM = ³/₄
🗸 substitution of m and M
🗸 equation
(4)

N(–11 ; p)
y = ³/₄x + ²⁵/₄
p =³/₄(-11)+ ²⁵/₄   OR    4 - p    = ³/₄
                                  - 3 - (-11)
p =- 2
N(-11;- 2)
- 3 + x_S= 0    and 4 + y_S  = 0
    2                        2
S(3;- 4)
SN =√(-11- 3)²  + (- 2 - (-4))²
= 10√2 units or 14,14 units

🗸subst x = –11 into eq or gradient
🗸p =- 2
🗸xS 🗸 yS
🗸answer (CA)
(5)

B(- 2; 5)
BM = √2 units
Radius of circle centred at M = 8 units
k = 8 -    2        or      k = 8 +    2
= 6,59 units               = 9,41 units
= 6,6 units                 = 9,4 units

🗸      2

🗸🗸 k = 6,6

🗸🗸 k = 9,4
(5)

5.1

Period of g = 360°

🗸 answer (1)

5.2

Amplitude of f =½

🗸 answer (A) (1)

5.3

f (180°) – g(180°)
= ½ - (-½)
= 1

🗸1 (1)

5.4.1

x = 140,9°

🗸 x = 140,9° (1)

5.4.2

3 sin x + cos x ≥ 1
√3 sin x +½cos x ≥ ½
 2
sin x cos 30° + cos x sin 30°≥ ½
sin(x + 30°) ≥ ½
sin(x + 30°) = ½ at   x = 0° or x = 120°
x ∈ [0°;120°] OR     0° ≤ x  ≤120°

🗸dividing by 2
🗸cos30°; sin 30°
🗸sin(x + 30°) ≥ ½
interval
(4)

[8]

6.1.1

tanq = - ¹²/₅ or - 2²/₅

🗸answer (1)

6.1.2

(OP)² = (–5)² + (12)²
OP = 13
cosq = - ⁵/₁₃

🗸 Pythagoras
🗸 OP
🗸 answer
(3)

6.1.3

sin(θ + 90°) =  b  
                      6,5
cosθ =  b   
           6,5
- 5 =  b 
 13   6,5
b = - ⁵/₂

OR
cos(90° +θ ) =  a     
                       6,5
- sinθ=  a  
            6,5
- 12 =  a    ∴ a = -6
   13   6,5
b = √(6,5)² - (-6)² = - ⁵/₂

🗸 sin(θ + 90°) =  b  
                          6,5
🗸 cosθ
🗸 - 5 =  b 
     13   6,5
🗸 value of b
(4)

🗸 cos(θ + 90°) = a  
                          6,5
🗸 - sinθ
🗸 value of a
🗸 value of b
(4)

sin 2x. cos(-x) + cos 2x. sin(360° - x)
                sin(180° + x)
= sin 2x cos x + cos 2x( - sin x)
                    - sin x
= sin(2x - x)
     - sin x
= sin x
  -sin x
= -1

🗸 sin(360° - x) = -sin x
🗸sin(180° + x) = -sin x
🗸 numerator = sin x
🗸 answer
(5)

6 sin²x + 7 cos x - 3 = 0
6(1 - cos²x) + 7 cos x - 3 = 0
6 - 6 cos²x + 7 cos x - 3 = 0
6 cos²x - 7 cos x - 3 = 0
(3cos x + 1)(2 cos x - 3) = 0
cos x = - ¹/₃   or    cos x = ³/₂ (N/A)
x = 109,47° + k.360°; k ∈ Z or
x = 250,53° + k.360°; k ∈ Z

🗸 identity
🗸 standard form
🗸 factors
🗸 both solutions of cos x
🗸 x = 109,47° & 250,53°
🗸 + k.360°; k ∈ Z
(6)

x + ¹/_x = 3cos A
(3cosA²)  = (x + ¹/_x)²
9cos²A = x² + 1  + 2
                       x²
9 cos ² A = 2 + 2
cos ² A = ⁴/₉
cos 2A = 2 cos ² A -1
= 2(⁴/₉)- 1
= - ¹/₉

OR
x² - 2 + 1 = 0
            x²
(x - ¹/_x)² = 0
x² = 1
x = ±1

3cosA = 2       or    3cosA = -2
cosA = ²/₃        or     cosA = -²/₃ 
cos2A = 2cos²A - 1
= 2(±²/₃)² - 1
= - ¹/₉

🗸squaring both sides
🗸 9cos²A = x² + 1  + 2
                          x²
🗸cos ² A = ⁴/₉
🗸 cos 2A= 2cos² A-1
🗸 answer
(5)

🗸 x = ±1

🗸 cos A =²/₃ 
🗸 cos A = - ²/₃ 
🗸double angle identity
🗸 answer
(5)

7.1

tan 30° = √3r 
               QS
QS=  √3r   
       tan 30°
= √3r     or  √3r 
    1            √3 
   √3             3
= 3r

🗸🗸 trig ratio
🗸 QS subject
(3)

7.2

Area of flower garden = π (3r )2 - πr ²
= 9πr ² - πr ²
= 8πr ²

🗸 substitution into difference of areas
🗸 answer
(2)

7.3

RS2 = r² + (3r)² - 2(r)(3r) cos 2x
= r² + 9r² - 6r² cos 2x
= 10r² - 6r² cos 2x
= r²(10 - 6 cos 2x)
RS = r √10 - 6 cos 2x

🗸substitution into cosine rule correctly
🗸10r² - 6r² cos 2x
🗸r²(10 - 6 cos 2x)
(3)

7.4

RS = 10√10 - 6 cos 2(56)
= 34,9966...
» 35 m

🗸substitution
🗸 answer
(2)

[10]

8.1.1(a)

O₂   = 64°
[∠ at centre = 2 × ∠ at circumference]

🗸 S 🗸 R
(2)

8.1.1(b)

M₂  = 90°     [Line from centre to midpt of chord]
KON=90° + 26° = 116°  [ext ∠ of D]
O1 =116° - 64° =52°

OR
M₂  = 90° [Line from centre to midpt of chord]
O₃  = 64°  [sum of ∠s in D]
O₁ = 52°  [ ∠s on straight line]

🗸 S 🗸 R
🗸 S
🗸 answer (4)

🗸 S 🗸 R
🗸 S
🗸 answer (4)

8.1.2

PKO + P = 128°  [sum of  ∠s in ∆]
PKO= P [ ∠s opp = sides]
= 64°
K₂  = 32° or  K₂  = K₁
KN bisects OKP

OR
K₂ = KNO  [ ∠s opp = sides]
K₂  + KNO = 64° [sum of ∠s in ∆]
K₂  = 32° or  K₂  = K₁
KN bisects OKP

🗸 S
🗸 S
🗸 S
(3)

🗸 S
🗸 S
🗸 S
(3)

F₁  = D₁ [tan chord theorem]
D₁ = B  [Given]
F₁ = B
FG || BC [corresp ∠s =]

🗸 S 🗸 R
🗸 F₁ =  B
🗸 R
(4)

GC = FB   [line || one side of Δ]
AC    AB
 x + 9  = 5 
2x - 6     7
7x + 63 =10x - 30
3x = 93
x = 31

OR
AG= 2x - 6 -(x + 9) = x -15
AG = AF   [line || one side of Δ]
GC   FB
 x -15 = 2 
x + 9     5
5x - 75 = 2x +18
3x = 93
x = 31

OR
AF = AG [line || one side of Δ]
AB    AC
 2  = x - 15
 7     2x - 6
7x - 105 = 4x -12
3x = 93
x = 31

🗸 S 🗸 R
🗸 substitution
🗸 answer
(4)

🗸 S 🗸 R
🗸 substitution
🗸 answer
(4

🗸 S 🗸 R
🗸 substitution
🗸 answer
(4)

9.1

Construction: Draw diameter KS and draw KR
QST = 90° - TSK  [radius ^ tangent]
SRK =  90°     [∠ in semi circle]
SRT = 90° - KRT
TSˆK = TRK  [∠s same segment]
QST = R

construction
S/R
S/R
S
S/R
(5)

Construction: Draw radii OS and OT
QST = 90° - OST   [radius ⊥ tangent]
OST = STO     [∠s opp = sides]
SOT =180° - 2OST    [∠s of D]
R = 90° - OST  [∠ at centre = 2 × ∠ circumf]
QST = Rˆ

construction
S/R
S/R
S
S/R (5)

Construction: Draw diameter KS and join K to T.
QST = 90° - TSK   [radius ⊥ tangent]
STK = 90°    [∠ in semi circle]
K  = 90° - TSˆK
QSˆT = K
but Rˆ   = K  [∠s same segment]
QST = Rˆ

Construction: Draw radii OT, OR and OS
OST = OTS [∠s opp = radii]
Also: OTR = ORT  and  ORS = OSR
2x + 2 y + 2z = 180° [∠s ofΔ]
x + y + z = 90°
y + z = 90° - x
OSQ = 90° [radius ⊥ tangent]
TSQ = 90° - x
TSQ = y + z = R

Construction: Draw radii OT and OS, tangent QT
OSˆQ = 90°     [radius ⊥ tangent]
TSQ = 90° - TSO
TSO = STO  [∠s opp = radii]
TOS = 180° - 2TSˆO  [∠s of D]
R = 90° - TSO  [∠ at centre = 2 × ∠ circum]
TSQ = R

S
R
(2)

Q₂  = x   [tan chord theorem]
OR
M₂ = x   [tan chord theorem]
Q₂   = x  [∠s in same segment

S
R (2)
S/R
(2)

MN = MS     [QN || PR; Prop Th]
NR    SP
N₁  = N₂  = x [given]
P₃  = x  [∠s  in  same  segment]
P₃ = Q₂  [= x]
SQ = PS [sides opp = ∠]
MN = MS
NR    SQ

S
R
(6)

10.1.1

DBE = 90°   [∠ in semi-circle]
DMA = 90°     [ AM ⊥ DE ]
FBDM is a cyclic quadrilateral
[converse opp ∠s cyclic quad]

OR
DBE = 90°   [∠ in semi-circle]
M₂  = DBE=90°
FBDM is a cyclic quadrilateral
[converse ext ∠ of cyclic quad]

S
R (3)

S
R (3)

B₃  = D₂   [tangent chord th]
F₁  = D₂     [ext ∠ cyc quad]
B₃  = F₁

OR
B₁ = E = x  [tangent chord th]
F₁  = 90° - x [∠ sum in Δ]
D₂  = 90° - x [∠ sum in Δ]
F₁  = D₂
B₃ = D₂  [tangent chord th]
B₃  = F₁

OR
B₁ = E = x  [tangent chord th]
B₃ = 90° - x [straight  line ]
F₁ = 90° - x   [sum of ∠s Δ]
B₃  = F₁

S
R (4)
F₁ = 90° - x
= D₂

R (4)

S
R (4)

In ∆CDB and ∆CBE
C = C [common ∠] CBD = CEB  [tangent chord th]
CDB = CBE  [∠ sum in D]
∆CDB ||| ∆CBE

OR
In ∆CDB and ∆CBE
CBD = CB    [tangent chord th]
C = C    [common ∠]
∆CDB ||| ∆CBE [∠∠∠]

S
S/R
R (3)

S/R
S
R (3)

BC = DC     [||| Δs]
EC    BC
BC² = EC x DC
= 8 x 2
= 16
BC = 4

ratio
substitution
answer (3)

BC = DB    [||| Δs]
EC    BE
DB = 4 = 1 
BE    8    2
BE = 2DB
DB² + BE² = DE²  [Pyth theorem]
DB² + (2DB)² = 36
5DB² = 36
DB² = 36 
            5
DB = 6 = 2,68 units
        √5

BE = 2DB
substitution into Pyth theorem
DB² = 36 
            5
answer (4)