MARKING CODES

M

Method

A

Accuracy

AO

Answer only

CA

Consistent accuracy

F

Formula

I

Identity

R

Rounding

S

Simplification

ST

Statement

RE

Reason 

ST RE

Statement and correct reason 

SF

Substitution correctly in correct formula

NPU

No penalty for omitting units 

Q

M

1.1

𝑞 = −4

√  A    (1)

1.2

𝐴𝐵² = (−2 + 5)² + (−4 − 0)²
= 9 + 16
𝐴𝐵 = 5 units 

√ M
√ 5 CA   (2)

1.3

𝑚_𝐶𝐷  = −4−0
             3−0
= − 4
      3
tan 𝜃 = − 4
               3
Ref ∠ = tan−1(⁴/₃)
= 53,13°
∴ 𝜃 = 126,87°

1.4

𝐵𝐶 = 3 − (−2) = 3 + 2 = 5 units 
𝐴𝑂 = 0 − (−5) = 5 units 
∴ABCO is a parallelogram (One pair of opp. side = and 
But , AO = AB (from  1.2)
∴ 𝐴𝐵𝐶𝑂 is a rhombus
(parallelogram with all sides = )

√ 𝐵𝐶 & 𝐴𝑂 M
√ parm and  reason 
√ rhombus 
√ reason  (4)

[10]

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2.1.1

𝑥² + 𝑦² = 25 … 1
𝑦 − 𝑥 − 1 = 0 … 2
𝑦 = 𝑥 + 1 … 3
sub 3 in 1:
𝑥² + (𝑥 + 1)² = 25
𝑥² + 𝑥² + 2𝑥 + 1 = 25
2𝑥² + 2𝑥 − 24 = 0
𝑥² + 𝑥 − 12 = 0
(𝑥 + 4)(𝑥 − 3) = 0
𝑥 = −4 or  𝑥 = 3
∴ 𝑦 = −3 or 𝑦 = 4

√ 𝑦 = 𝑥 + 1
√ 𝑥² + (𝑥 + 1)² = 25
√ 𝑥² + 𝑥 − 12 = 0
√ 𝑥 = −4
√ 𝑥 = 3
√ 𝑦 = −3 CA
√ 𝑦 = 4 CA    (7)

2.1.2

(3; 2): (3)² + (2)² = 13
∴ 𝑥² + 𝑦² < 𝑟²
∴ lies inside the circle 

√ 13
√ 𝑥² + 𝑦² < 𝑟² CA
√ conclusion  (3)

2.1.3

𝑚_𝑟𝑎𝑑   =   0−3  
              0−(−4)
= − 3
      4
∴ 𝑚_𝑡𝑎𝑛 × 𝑚_𝑟𝑎𝑑 = −1
∴ 𝑚_𝑡𝑎𝑛   × − ³/₄  = −1
∴ 𝑚_𝑡𝑎𝑛 = −1 ÷ − ³/₄
∴ 𝑚_𝑡𝑎𝑛 = ⁴/₃
𝑦 = 𝑚𝑥 + 𝑐
∴ 𝑦 = ⁴/₃ 𝑥 + 𝑐
(−4; 3): 3 = ⁴/₃ (−4) + 𝑐
3 = −¹⁶/₃ + 𝑐
²⁵/₃ = 𝑐
∴ 𝑦 = ⁴/₃ 𝑥 + ²⁵/₃

√  − ³/₄

√ ⁴/₃

√ ²⁵/₃ = 𝑐

√ 𝑦 = ⁴/₃ 𝑥 + ²⁵/₃ CA   (4)

2.2

36𝑥² + 49𝑦² = 1764
36𝑥²     +  49𝑦²  = 1
1764       1764
𝑥² + 𝑦² = 1
49   36

√ 𝑥² + 𝑦² = 1
  49   36
√ Shape 
√ 𝑥-intercept 
√ y-intercept   (4)

[18]

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3.1.1

√ SF

√ −0,8

(2)

3.1.2

√        1
  𝑠𝑖𝑛(^𝐴_/3 +  2𝐵)

√ SF

√ −1,4

(3)

3.2.1

𝑡𝑎𝑛𝐴̂ = 5 = −5
          12   −12
∴ 𝑐𝑜𝑠𝑒𝑐²𝐴̂ =     1     
                   𝑠𝑖𝑛²𝐴̂
𝑐𝑜𝑠𝑒𝑐²𝐴̂ = 169
                 25

OR

𝑡𝑎𝑛𝐴̂ = 5 = −5
          12   −12
∴ 𝑐𝑜𝑠𝑒𝑐²𝐴̂ = (¹³/₋₅)²
𝑐𝑜𝑠𝑒𝑐²𝐴̂ = 169
                 25

r² = (-12)² + (-5)²
∴r = 13

√ 𝑡𝑎𝑛𝐴 = 5 = −5 A
             12   −12

√ correct quadrant 

√ 𝑟 = 13 A

√  1    or  𝑐𝑜𝑠𝑒𝑐²𝐴̂ = (¹³/_-5)² CA
  𝑠𝑖𝑛²𝐴

√ ¹⁶⁹/₂₅ CA  (5)

3.2.2

𝑠𝑒𝑐𝐴̂ − 𝑠𝑖𝑛𝐴̂ =   1    − 𝑠𝑖𝑛𝐴̂
                      𝑐𝑜𝑠𝐴̂
=   1      −    −5 
 ⁻¹²/₁₃          13
= 13   − −5 
  −12     13
= −109 
     156

OR

𝑠𝑒𝑐𝐴̂ − 𝑠𝑖𝑛𝐴̂ =   1    − 𝑠𝑖𝑛𝐴̂
                     𝑐𝑜𝑠𝐴̂
= 13   −   −5  
  −12       13
= −109
     156

√   1    𝐨𝐫  13 
  𝑐𝑜𝑠𝐴      −12

√    1   
 ⁻¹²/₁₃ 

√ −109 CA   (3)
    156

3.2.3

𝑡𝑎𝑛𝐴̂ = ⁵/₁₂
ref ∠ = 𝑡𝑎𝑛⁻¹ ( ⁵/₁₂)
= 22,62°
∴ 𝐴̂ = 180° + 22,62°

𝐴̂ = 202,62°

√ method 

√ 22,62°

√ 202,62°

[16]

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4.1

𝑠𝑖𝑛(𝜋 − 𝑥). 𝑐𝑜𝑠𝑒𝑐(2𝜋 − 𝑥). 𝑡𝑎𝑛(𝜋 + 𝑥)
        𝑠𝑒𝑐(2𝜋 − 𝑥). 𝑐𝑜𝑠(2𝜋 − 𝑥)
= 𝑠𝑖𝑛(180° − 𝑥). 𝑐𝑜𝑠𝑒𝑐(360° − 𝑥). 𝑡𝑎𝑛(180° + 𝑥)
         𝑠𝑒𝑐(360° − 𝑥). 𝑐𝑜𝑠(360° − 𝑥)
= 𝑠𝑖𝑛(𝑥). −   1/sin (𝑥) . 𝑡𝑎𝑛(𝑥)
          1/cos (𝑥). 𝑐𝑜𝑠(𝑥) 
= −1. 𝑡𝑎𝑛(𝑥)
         1
= −tan (𝑥)

√ correct conversion of rad to degrees 

√ 𝑠𝑖𝑛(𝑥)

√ −     1     
       sin (𝑥)

√ 𝑡𝑎𝑛(𝑥)

√     1     
  cos (𝑥)

√  𝑐𝑜𝑠(𝑥)

√ −1.𝑡𝑎𝑛(𝑥)
        1

√  −tan (𝑥)    (8)

OR

𝑠𝑖𝑛(𝜋 − 𝑥). 𝑐𝑜𝑠𝑒𝑐(2𝜋 − 𝑥). 𝑡𝑎𝑛(𝜋 + 𝑥)
       𝑠𝑒𝑐(2𝜋 − 𝑥). 𝑐𝑜𝑠(2𝜋 − 𝑥)
= 𝑠𝑖𝑛(180° − 𝑥). 𝑐𝑜𝑠𝑒𝑐(360° − 𝑥). 𝑡𝑎𝑛(180° + 𝑥)
              𝑠𝑒𝑐(360° − 𝑥). 𝑐𝑜𝑠(360° − 𝑥)
= 𝑠𝑖𝑛(𝑥). −𝑐𝑜𝑠𝑒𝑐(𝑥). 𝑡𝑎𝑛(𝑥)
         sec (𝑥). 𝑐𝑜𝑠(𝑥)
= −1. 𝑡𝑎𝑛(𝑥)
        1
= −tan (𝑥)

√ correct conversion of rad to degrees
√ 𝑠𝑖𝑛(𝑥)
√ −𝑐𝑜𝑠𝑒𝑐(𝑥)
√ 𝑡𝑎𝑛(𝑥)
√ 𝑠𝑒𝑐(𝑥)
√ 𝑐𝑜𝑠(𝑥)
√ −1.𝑡𝑎𝑛(𝑥)
         1
√  −tan (𝑥)  (8)

4.2

𝐿𝐻𝑆/𝐿𝐾 =     𝑐𝑜𝑠𝜃    − 𝑡𝑎𝑛𝜃
                 1−𝑠𝑖𝑛𝜃
= 𝑐𝑜𝑠𝜃     − 𝑠𝑖𝑛𝜃
  1−𝑠𝑖𝑛𝜃     𝑐𝑜𝑠𝜃
= 𝑐𝑜𝑠𝜃.𝑐𝑜𝑠𝜃−𝑠𝑖𝑛𝜃(1−𝑠𝑖𝑛𝜃)
         𝑐𝑜𝑠𝜃(1−𝑠𝑖𝑛𝜃)
= 𝑐𝑜𝑠²𝜃−𝑠𝑖𝑛𝜃+𝑠𝑖𝑛²𝜃
      𝑐𝑜𝑠𝜃(1−𝑠𝑖𝑛𝜃)
=       1−𝑠𝑖𝑛𝜃     
   𝑐𝑜𝑠𝜃(1−𝑠𝑖𝑛𝜃)
=    1   
   𝑐𝑜𝑠𝜃
= 𝑠𝑒𝑐𝜃 = 𝑅𝐻S/RK

√ 𝑠𝑖𝑛𝜃
   𝑐𝑜𝑠𝜃
√ 𝑐𝑜𝑠𝜃(1 − 𝑠𝑖𝑛𝜃)
√ 𝑐𝑜𝑠²𝜃 − 𝑠𝑖𝑛𝜃 + 𝑠𝑖𝑛²𝜃
√     1−𝑠𝑖𝑛𝜃      
  𝑐𝑜𝑠𝜃(1−𝑠𝑖𝑛𝜃)
√   1          (5)
   𝑐𝑜𝑠𝜃

[13]

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5.1

√ cos start and end point 

√ cos turningpoints 

√ cos 𝑥- intercepts

√ sin start and end point

√ sin turningpoints 

√ sin 𝑥- intercepts  (6)

5.2.1

120°

√ 120°

(1)

5.2.2

(a)

𝑥 = 30° and / en 𝑥 = 120°

√ 𝑥 = 30°

√ 𝑥 = 120°

(2)

(b)

90° ≤ 𝑥 ≤ 150°

√ 90° ≤

√ ≤ 150°

(2)

[11]

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6.1

AB = 8 cm (opp sides of rec = )

√ 8 cm ST
√ RE  (2)

6.2

𝐵𝐸 = tan 30°
 8
BE = 4,62 cm = 5 cm

√ 𝐵𝐸 = tan 30° M
    8
√ 5 cm  (2)

6.3

sin 𝐸𝐵̂𝐶  = sin 30°
   9               4,62
sin EB̂C = sin 30° × 9
                  4,62
EB̂C = 76,91° = 77°

√ SF
√ sin 𝐸𝐵̂𝐶 = sin 30° × 9
                     4,62
√  77°  (3)

6.4

𝐵𝐸̂𝐶 = 180° − 30° − 77°
= 73° (Int. ∠’s of ∆BEC)
𝐴𝑟𝑒𝑎 ∆BCE = ½ 𝑎. 𝑏. 𝑠𝑖𝑛𝐶̂
= ½ (5)(9)𝑠𝑖𝑛73°
= 21,52 cm²

√ ST RE
√ ½ (5)(9)𝑠𝑖𝑛73° SF
√ 21,52 cm² CA   (3)

6.5

𝐶𝐹² = 𝐶𝐸² + 𝐹𝐸² − 2(𝐶𝐸)(𝐹𝐸) cos 𝐶𝐸̂𝐹
𝐶𝐹² = (9)² + (10)² − 2(9)(10) cos(25°)
= 17,86 …
𝐶𝐹 = √17,86 …
𝐶𝐹 = 4,22 … = 4 cm

√ (9)² + (10)² − 2(9)(10) cos(25°) SF
√ 17,86 … S
√ 4 cm CA    (3)

[13]

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M

7.1

double the size of the angle subtended by the same arc at the circumference of a circle

(1)

7.2

7.2.1

𝐴𝑂̂𝐵 = 180° − 48° …(suppl. ∠’s)
= 132°

√ ST RE

(1)

7.2.2

𝐶̂ = ½ (132°)…( ∠ at centre = 2 x ∠ at circum.)
= 66°

√ RE

√ 66° ST

 (2)

7.2.3

𝑂𝐵̂𝐸 = 90° …( tan ⊥ rad)
𝐴𝐸̂𝐷 = 180° − (90° + 48°)
= 42° (int  ∠′s of triangle)

√ ST

√ RE

√ ST RE

(3)

7.2.4

𝐷̂  = 𝐴𝐸̂𝐷 = 42° …(∠′s opp.= sides)
𝐴𝑂̂𝐷 = 84 ….(ext ∠ of ∆)

√ ST RE

√ ST RE

(2)

7.2.5

𝐵𝐸² = 𝑂𝐸² − 𝑂𝐵²    ….(Pythagoras)
= 7² − 5²
= 24
𝐵𝐸 = √24 ≈ 4,9 cm

√ M Pythagoras

√ ST 𝐵𝐸² = 24

√ ST 𝐵𝐸 = √24

(3)

[12]

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8.1

supplementary 

√  (1)

8.2

8.2.1 (a)

𝑄𝑆̂𝑅 = 𝑇̂  = 𝑥 (∠′s subt.chord QR )

√ ST

√ RE  (2)

8.2.2 (b)

𝑃𝑄̂𝑅 = 180° − (58° + 𝑥) (opp. ∠′s quad.)

= 122° − 𝑥

√ ST RE
√ 122° − 𝑥 ST   (3)

8.2.2

𝑈̂ + 𝑈𝑄̂𝑅 + 𝑈𝑅̂𝑄 = 180° (int. ∠′s ∆)
22° + 72° + 𝑥 + 58° = 180°
𝑥 = 28°

√ ST RE
√ 𝑥 = 28° ST  (2)

OR

OR

𝑃̂ + 𝑆̂ + 𝑈̂ = 180° (int. ∠′s ∆)
72° + 58° + 𝑥 + 22° = 180°
𝑥 = 28°

√ ST RE
√ 𝑥 = 28° ST  (2)

[8]

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M

9.1

• All angles of the one triangle is equal to the angles in the other triangle
• Sides of triangles are in proportion 

√ ST
√ ST  (2)

9.2

9.2.1

In ∆𝐶𝐷𝐹 and ∆𝐸𝐶𝐹:
𝐹𝐶̂𝐷 = 𝐶𝐸̂𝐷 = 52° (tanchord)
𝐶𝐹̂𝐷 = 𝐶𝐹̂𝐸 (common ∠)
𝐶𝐷̂𝐹 = 𝐷𝐶̂𝐸 (int. ∠′s ∆)
∴ ∆𝐶𝐷𝐹⦀∆𝐸𝐶𝐹 (∠∠∠)

√ ST RE
√ ST RE
√ ST ∆𝐶𝐷𝐹⦀∆𝐸𝐶𝐹 OR  RE (∠∠∠) (4)

9.2.2

𝐶𝐷 = 𝐷𝐹 = 𝐶𝐹 (∆𝐶𝐷𝐹⦀∆𝐸𝐶𝐹)
𝐸𝐶    𝐶𝐹     𝐸𝐹
∴ 𝐶𝐹² = 𝐸𝐹. 𝐹𝐷

√ 𝐶𝐷 = 𝐷𝐹 = 𝐶𝐹 ST
   𝐸𝐶    𝐶𝐹     𝐸𝐹
√ 𝐶𝐹² = 𝐸𝐹. 𝐹𝐷 ST   (2)

9.2.3

𝐷𝐹 = 15 − 6 = 9
𝐶𝐷 = 𝐷𝐹 = 𝐶𝐹 (from 9.2.2)
𝐸𝐶    𝐶𝐹    𝐸𝐹
∴ 9   = 𝐶𝐹
  𝐶𝐹     15
∴ 𝐶𝐹² = 135
∴ 𝐶𝐹 ≈ 12 cm

√ 𝐷𝐹 = 9 A
√ 9 = 𝐶𝐹 SF
  𝐶𝐹   15
√ 𝐶𝐹² = 135 CA
√ 12 cm CA R   (4)

OR

OR

𝐶𝐹² = 15 × 9 (from 9.2.2)
𝐶𝐹 = √135
𝐶𝐹 = 11,619 …
∴ 𝐶𝐹 = 12 cm

√ 𝐶𝐹² = 15 × 9 A
√ 𝐶𝐹 = √135 S
√ 𝐶𝐹 = 11,619 …  CA
√ 12 cm CA R   (4)

9.2.4

𝐶𝐷 = 12 = 4
𝐸𝐶    15    5

√ 12 SF A
   15
√ 4 A (2)
   5

9.2.5

𝐸𝐶̂𝐹 = 44° + 52° = 96° ≠ 90°
∴ CE not a diameter (Diameter not perpendicular to tangent)

√ ST
√ ST RE   (2)

𝐶𝐷̂𝐸 = 180° − 44° − 52° = 86° ≠ 90°
∴ 𝐶𝐸 𝑛𝑜𝑡 𝑎 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 / 𝐶𝐸 𝑛𝑖𝑒 ′𝑛 𝑚𝑖𝑑𝑑𝑒𝑙𝑙𝑦𝑛 𝑛𝑖𝑒 (not Converse)

√ ST

√ ST RE  (2)

Q

M

10.1

10.1.1

𝐴̂ = 30° ×   𝜋  
              180°
𝐴̂ = 𝜋 radians 
      6

√ ×  𝜋  
     180°
√ 𝜋 radians   (2)
   6

10.1.2

𝑠 = 𝑟𝜃
𝐵𝐺 = (4)  (^𝜋/₆)
𝐵𝐺 = 2𝜋   cm
         3

√ F
√ SF
√ 3𝜋 cm²
    2

(3)

10.1.3

Area of a sector / 𝑂𝑝𝑝 𝑣𝑎𝑛 ′𝑛 𝑠𝑒𝑘𝑡𝑜𝑟 = 𝑟²𝜃
                                                              2
∴ Area of sector AEC / 𝑂𝑝𝑝 𝑣𝑎𝑛 𝑠𝑒𝑘𝑡𝑜𝑟 𝐴𝐸𝐶 =(9)2𝜋/6
                                                                             2
= 3𝜋 cm²
    2

√  F
√ SF
√ 3𝜋 cm²
    2

(3)

10.1.4

Area of a sector / 𝑂𝑝𝑝 𝑣𝑎𝑛 ′𝑛 𝑠𝑒𝑘𝑡𝑜𝑟 = 𝑟²𝜃
                                                              2
∴ Area of sector ABG / 𝑂𝑝𝑝 𝑣𝑎𝑛 𝑠𝑒𝑘𝑡𝑜𝑟 𝐴𝐵𝐺 =(4)2𝜋/6
                                                                             2
= 4𝜋 cm²
    3
∴ Area of BECG  = 27𝜋 − 4𝜋
                                4        3
∴ Area of BECG  = 65𝜋 cm²
                               12

√ SF
√ 4𝜋
    3
√ M
√ 65𝜋
   12

(4)

10.2

10.2.1

𝜔 = 2𝜋𝑛
𝜔 = 2𝜋(120)
𝜔 = 240𝜋 rad

√ F
√ SF
√  240𝜋 rad   (3)

10.2.2

𝑣 = 𝜋𝐷𝑛
𝑣 = 𝜋(3 × 2)(120)
𝑣 = 720𝜋 cm

√ F
√ SF
√ 720𝜋 cm

OR

OR

𝑣 = 𝑟𝜔
𝑣 = (3)(240𝜋)
𝑣 = 720𝜋 cm

√ F
√ SF
√  720𝜋 cm  (3)

10.2.3

Linear speed of large pulley 
= 720𝜋 cm 
𝑣 = 𝜔𝑟
720𝜋 = 𝜔(15)
48𝜋 rad  = 𝜔

√ F
√ SF
√ 48𝜋 rad   (3)

10.3

4ℎ² − 4𝑑ℎ + 𝑥² = 0
4ℎ² − 4(10)ℎ + (8)² = 0
4ℎ² − 40ℎ + 64 = 0
ℎ2 − 10ℎ + 16 = 0
(ℎ − 8)(ℎ − 2) = 0
∴ ℎ = 8 cm or ℎ = 2 cm
∴ ℎ = 2 cm

√ F
√ SF
√ Standard form
√ M
√ ℎ = 2 cm

OR

OR

𝐶𝐸 = 4 (line from centre perpendicular to chord)
𝐹𝑂 = 𝑂𝐺 (radii)
𝑂𝐶² = 𝑂𝐸² + 𝐶𝐸² (Pyth)
∴ 𝑂𝐸² = 𝑂𝐶² − 𝐶𝐸²
∴ 𝑂𝐸² = (5)² − (4)²
∴ 𝑂𝐸² = 25 − 16
∴ 𝑂𝐸2 = 9
∴ 𝑂𝐸 = ±√9
∴ 𝑂𝐸 = ± 3
∴ 𝑂𝐸 = 3 cm
∴ 𝐸𝐺 = 5 − 3
∴ 𝐸𝐺 = 2 cm

√ ST RE

√ ST RE

√ 𝑂𝐸² = (5)² − (4)² SF

√ 𝑂𝐸 = 3 cm   CA

√ 𝐸𝐺 = 2 cm   CA  (5)

[26]

Q

M

11.1

√ F
√ SF
√ S
√ 75,15 cm²

OR

√ F
√ SF
√ S
√ 75,15 cm²   (4)

11.2

√ F
√ SF
√ 17,22 m²

[7]

TOTAL:

150