1.1

a = 62
b = 51
c = 42
d = 35
e = 31

value of a 
value of b 
value of c 
value of d 
value of e 

(5)

1.2

Skewed to the right 

OR

Positively skewed 

answer 

(1)

1.3

Yes 
Q₃ = 51 and the upper 25% is from Q₃ above..

Yes
Reason

(2)

[8]

Positive impact 

The number of learners obtaining lower marks decreased while those obtaining higher marks increased in the Post Test.

Frequency
Pre-test
Post test

Cummulative frequency
Pre-test
Post test

2.6

Pre: 90 – 78 = 12 learners obtained 60% and more
Post: 90 – 48 = 42 learners obtained 60% and more
Therefore, the teacher achieved the target.

12
42
conclusion

[14]

3.1

t = 2

value of t 

(1)

3.2.1

3.2.2

mPR =y2 - y1 = 0 - 6
           x2 - x1    -5 - 2
= 6
   7

substitution

gradient of PR       (2)

3.3

tan RPS = ^6/₇
RPS = 40, 6º
40, 6º = PRQ = 90º (ext. ∠ of Δ)
∴ PRQ = 49, 4º

tanθ = ^6/₇
value of PRQ
method
reason 
value of  RPS (5)

3.4

m_PR = ^6/₇
m_PQ = -4 - 0  = -4
           2-(-5)      7
m_PR  × m_PQ = ^6/₇ × [-⁴/₇] = - ²⁴/₄₉ # - 1

∴ ΔPRQ is not right angled at R

OR 

RQ² =100
PR² = 85
PQ² = 65
∴ RQ² #  PR² + PQ²
∴PQR is not right angled at P

substitution 
gradient of PR 
method
conclusion 

OR 

substitution 
squares 
RQ² #  PR² + PQ²
conclusion   (4)

3.5

m_newline = m_PQ = - ⁴/₇
y - 0 =  - ⁴/₇  x (x - 0)
∴ y =  - ⁴/₇ x

gradient of new line
substitution 
equation 

3.6

SP = 7 units
SR = 6 units
RQ = 10 units
Area of ΔSPR
= ½ × 7 × 6
=  21 units² 
Area of ΔRPQ
=  ½ × 7 × 10
= 35 units²
= Area of ΔSPR
   Area of ΔPRQ
= 21
   35
= 3 
   5

length of SP 
length of RQ 
Area of ΔSPR
Area of ΔRPQ
answer

(5)

[22]

4.1.1

x² + y² + 6x - 6y + 9 = 0
x² + 6x + 9 + y² - 6 y + 9 = -9 + 9 + 9
∴ ( x + 3)² + ( y - 3)² = 9
L(-3;3) and r  3 units

method 
( x + 3)² + ( y - 3)² = 9
coordinates of L
value of r

(4)

4.1.2

method 
value of x 
value of y 

4.1.3

m_LM OR m_KM
m_tangent
substitution 
equation 

4.1.4

( x + 3)² + ( y - 3)² = 9
( 0 + 3)² + ( y - 3)² = 9
( y - 3)² = 0
y = 3
N(0;3)

value of x
value of y 

4.2.1

L ( – 3 ; 3)
L^/ (2  ; –4)

value of x 
value of y   (2)

4.2.2

m_ML/ = -4 - 1 = - 5 
            2 + 1      3
y - (1) = -5 (x + 1)
              3
∴ y = - 5 x - 2
           3      3

Not passing through the origin

m_ML/
substitution 
equation 
conclusion     (4)

[19]

5.1.1

value of x
value of r 
expansion 
substitution
answer   (5)

5.1.2

expansion
substitution 
answer        (3)

5.1.3

reduction
substitution      (2)

5.2.1

1- cosθ = 0 or sinθ = 0
cosθ =1 or sinθ = 0
∴θ = 180.k     (k∈θ)

OR 

θ = 360º.k     or θ =180º + 360º.k            (k∈θ)

method 
answer    (2)

5.2.2

common denominator
simplification
identity 
simplification   (4)

LHS/LK = sin (x - y)
              cos x.cos y
= sin x cos y - cos x sin y
               cos x.cos y
= sin x cos y - cos x sin y
   cos x.cos y  cos x.cos y
= tan x - tan y
= 3k - 2k
= k

identity
method 
simplification 
substitution   (4)

6.1.1

1

1 (1)

6.1.2

120°

120° (1)

6.2

f (x) = g(x)
cos (x - 60º) = sin 3x
cos (x - 60º) =cos (90º - 3x)
± (x - 60º) = 90º - 3x + k.360º  (k∈)
x - 60º = 90º - 3x + k.360º or  - x + 60º = 90º - 3x + k.360º
4x = 150º + k.360º or  2x = 30º + k.360º
x = 37, 5º + k.90º or  x = 15º + k.180º
x∈{37, 5, - 52, 5º;15º; 127, 5º}

cos(x - 60º) - cos(90º - 3x)
without cos
simplification
x = 37, 5º + k.90º or  x = 15º + k.180º
two values
two remaining values (6)

6.3

f:
endpoints
both intercepts
shape 

g:
both intercepts
shape (5)

6.4

x = -30º    or   x = 150º

both values of x beide waardes van x

(1)

6.5

f (x) = cos(x - 60º + 15º)
h(x) = cos (x – 45°)

h(x) = cos (x – 45°)

(1)

[15]

5² = 4² + 5² - 2(4)(5) cos A
cos A = ²/₅
4² = 5² + 5² - 2(5)(5) cos B
cos B= ¹⁷/₂₅
cos A - cos B = 2 - 17
                         5   25
= -7 ≈ - 0, 28
   25

substitution into cosine rule
value of cos A 
substitution into cosine rule
value of cos B 
value of cos A – cos B

[5]

OM ⊥ CD         (line from centre which bisects the chord)
AO = OB = OD =11cm
In Δ OMD:
MD² + 7² = 11²         (Pythagoras theorem)
MD = 6√2
∴ CD =12√2

S/R
S
use of Pythagoras theorem 
length of MD
length of CD (5)

[5]

9.1

R₂ =  Q₂         ( ∠s in the same segment)
U₄ =  U₁         ( vert.opp. ∠s )
S₂ =  T₂          ( ∠s in the same segment)
∴ Δ RUS ||| Δ QUT (∠∠∠)

S/R
S/R
S/R   (3)

9.2

S₂ = T₂ = x     (∠s in the same seg)
U₃ =S₂ = x     (tan-chord theorem)
U₆ = U₃ = x    (vertically opp. ∠s)

S/R
S
R
S/R  (4)

9.3.1

R₁ + x + 90º - x = 180º ( sum of ∠s of Δ)
∴ R₁ = 90º
QTis a diameter (QT subtends a right angle)

S
value of R₁    
QT is a diameter
R   (4)

9.3.2

P = 90º        (∠in thesemicircle)
OR 
P + 90º = 180º (opp.∠sof a cyclicquad)
∴P =  90º

S
R   (2)

10.1

Construction: Draw diameter GOE. Join EH
Proof:
G₁ + G2 = 90º (tangent ⊥ diameter)
EHG = 90º    (∠in the semi circle)
G₂ + E = 90º    (sum of ∠s of Δ) 
∴ G₁ + G₂ = G₂ + E
∴ G₁ =  E
But: E = F (∠sin the same segment)
∴ DGH = F

construction
S/R
S/R
S
S/R

(5)

10.2

10.2.1

Kite

answer   (1)

10.2.2

KLO = 90º (tan ⊥ rad.)
KPO = 90º (tan ⊥ rad.)
MLP = 90º (∠sin the semi circle)

S/R
S
S/R        (5)

10.2.3

KLO + KPO = 90º + 90º
=180º
∴KLOP is a cyclic quad.         (Opp.∠s are supp.)

S
R  (2)

10.2.4

K + LOP =180º        (Opp.∠sof cyclicquad.)
But M = 67º  (sopp.equal sides)
∴ LOP = 67º  +  67º   (Ext.∠ of Δ)
=134º
∴ K + 134º = 180º
∴ K = 46º

OR

M = 67º         ( ∠s opp. = sides)
LOM = 46º    ( ∠s of Δ)
∴ K = 46º      (ext. ∠of cyclic quad.)

S/R
S/R
value of LOP
value of K
van K

OR

S/R
value of LOM
value of K
reason     (5)

[18]

AP = 3 (Prop. theorem; PQ||BC)
PB   5
       AP       = 3
  PR + BR     5
∴ 5AP = 3PR + 3BR
BR = 1            (Prop. theorem; RS||AC)
RA    3
     AP       = 1 
AP + PR      3
3BR = AP + PR
∴ 5AP = 3PR + AP + PR
4AP = 4PR
∴ AP = PR

S/R
S
simplification 
S/R
S
simplification 
substitution               (7)

[7]