# Mathematics Grade 10 Paper 2 KwaZulu Natal June 2019 Memo

> Grade 10 Mathematics Paper 2 memorandum from the KwaZulu Natal June 2019, 20 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/maths-p2-and-memo-june-2019-gr-10/279beb3f-6369-4cd3-9892-544f2a80be2d
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 10
- Document type: Memorandum
- Year: 2019
- Exam period: KwaZulu Natal June
- Paper: 2
- Curriculum: CAPS
- Pages: 20

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Mathematics/P2 Common Test June 2019

NSC- GRADE 10

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

Le This question paper consists of 3 questions.
2 Answer ALL the questions.
ch Clearly show ALL calculations, diagrams, graphs, etc. which you have used in

determining your answers.
4. Answers only will NOT necessarily be awarded full marks.

> You may use an approved scientific calculator (non-programmable and
non-graphical), unless stated otherwise.

6. If necessary, round off answers correct to TWO decimal places, unless stated
otherwise.

c Diagrams are NOT necessarily drawn to scale.

8. Write neatly and legibly.

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Mathematics/P2 Common Test June 2019
NSC- GRADE 10

QUESTION 2

21 Given that a@ = 24,6° and £ =132,7° calculate the value of the following

(correct to TWO decimal places):

i
21.1 cosa (1)
.
2 led cosec2 8 (2)
22 Various options are provided as possible answers to the following questions.

Choose the correct answer and write only the letter (A — D) next to the question number, for
example: 2.2.4 A

221 In the diagram below of right angle triangle EFG, EF = 17cm and FG = 15cm.

E 17cm F
5
8cem 15cm
G
Which trigonometric equation could be used to determine the value of angle £?
: 17 15
A. sin E =— B. cos E = —
15 7.
15 : 8
(e, tan F = — D. sin E =—
8 17 (1)
2.2.2 In the diagram below scalene ALMN shown, M =90°.
-
M o N
Which of the following statements is always true?
A. sin L =cosL B. sin L =cosN (1)
(o cos L =cos M D. sin =cosM

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Mathematics/P2 Common Test June 2019
NSC- GRADE 10
QUESTION 3
3.1 Complete the following:
The line joining the mid-points of two sides of a triangle is ....... to the third side and equal
to ee the length of the third side. (2)
3.2 In the diagram below, APOR has E, F and G the midpoints of PQ, QR and PR

respectively. QG // FH .

Prove:
3.2.1 QOGHF isa parallelogram G3)
822 BG GH. (3)
3.2.3 ER // PH (3)
(1)
TOTAL: 50

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                                1                                  Common Test June 2019
Mathematics P2                          NSC-Marking Guideline

                                 MATHEMATICS P2

                             MARKING GUIDELINE

                                   COMMON TEST

                                      JUNE 2019

                                   NATIONAL
                              SENIOR CERTIFICATE

                                      GRADE 12

 MARKS: 150

                     N.B. This marking guidelines consists of 14 pages.

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                                                     2                           Common Test June 2019
Mathematics P2                                 NSC-Marking Guideline

QUESTION 1

 1.1          220                                             A 220
         𝑥=
         ̅        = 22
              10                                               CA answer                         (2)
                                                              Answer only full marks
 1.2     σ = 3,95                                              AA answer                        (2)
                                                              If formula is used 1CA mark for
                                                              substitution and 1CA mark for
                                                              answer.
 1.3

                                                               A minimum & maximum value
                                                               A quartile 1 value
                                                               A median value
                                                              A quartile 3 value

                                                              If No Diagram No marks       (4)
 1.4     ( x − σ; x + σ)
         (22 – 3,95; 22 + 3,95)
         (18,05; 25, 95)                                     CA 18,05
         6 runners                                            CA 25,95
                                  (answer only – full marks)  CA answer                          (3)
                                                                                                  [11]

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                                3                                                            Common Test June 2019
Mathematics P2                                                      NSC-Marking Guideline

QUESTION 2

 2.1     A = 250                                                                                 (A)A
         B = 502                                                                                 (A) B           (2)

 2.2
                                    2000 + 5000 + 600 + 1280 + 1150 + 18000 + 33250             CA sum
         𝑥̅ =
                                                          1300
              61280                                                                             CA 61280
         𝑥̅ =
              1300
         𝑥̅ = 47,14                                                                              CA
         (answer only – full marks)                                                             answer            (3)
 2.3     65 < 𝑥 ≤ 75                                                                            AA
                                                                                                answer            (2)
 2.4                                                                                            CA upper
                                           AGES OF CONSUMERS                                    limits
                                    1400                                                        CA
                                                                                                grounding

             CUMULATIVE FREQUENCY
                                    1200
                                                                                                (5; 0)
                                    1000
                                                                                                CA joining
                                     800                                                        points with a
                                     600                                                        smooth curve
                                     400
                                     200
                                      0
                                           5   15   25   35         45    55     65     75
                                                              AGE
                                                                                                                  (3)

 2.5     Not a normal distribution. Highest frequency is found between the ages                 A No             (2)
         55 to 75. Mean < median, therefore skewed to the left.                                 A Reason
                                                                                                                  [12]

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                                                      4                              Common Test June 2019
Mathematics P2                                  NSC-Marking Guideline

QUESTION 3

 3.1                            2           2                           A substitution
           BC = √(𝑥2 − 𝑥1 ) + (𝑦2 − 𝑦1 )

           = √(−2 − 1)2 + (−2 − 4)2

                 =    9  36
                                                                        CA answer              (2)
                 =     45

                 = 3 5
 3.2                                                                         1
              1−2 4−2                                                   A
           M(     ;    )                                                     2
               2     2
               1                                                        A 1
           M (− ; 1)
               2                                                                                 (2)

 3.3                  2 4    6
           m AB =            =
                       2 1   3

                                    = 2                                  A M AB
                       1
           mMD = –              (DM ⊥ AB)                               CA gradient of MD
                       2

                 y = mx + c

                1  1                                                                      1
           1 = − (− ) + 𝑐                                               CA subst. (− 2 ; 1) into
                2  2
                                                                        eq.
                                1
                     c = 1–
                                4

                            3
                       =
                            4

              1   3
           𝑦=− 𝑥+
              2   4
                                                                        CA answer
                                                                                                 (4)

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                                5                                                      Common Test June 2019
Mathematics P2                                  NSC-Marking Guideline

 3.4       E is the midpoint since ME║BC.                                A S/R

             1+4 4+1                                                    A substitution
           E(    ;    )
               2    2
               5 5
           = E( ; )                                                     CA      answer      (provided
               2 2
                                                                        coordinates are positive)
                                                                                                    (3)
 3.5                  5
                 −2 − 2
           𝑚𝐵𝐸 =
                      5
                 −2 − 2
           𝑚𝐵𝐸 = 1                                                      CA m BE (must be positive)

           𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )        OR    y = mx + c
           𝑦 − (−2) = 1(𝑥 − (−2))           −2 = 1(−2) + c              CA substitution
           𝑦+2=𝑥+2                           0=c
           𝑦=𝑥                               y=x                        CA answer (must be
                                                                        positive)         (3)

 3.6                  2 1     1                                        A Substitution
           mBC =            =
                     24       2                                       CA mBC =
                                                                                      1
                                                                                       2
                           1
           mBC  tan  
                           2                                                          1
                                                                        CA tan  
                                                                                      2
                       𝜃 = 26,57°                                       CA answer
                                                                                                   (4)

                                                                                                  [18]
           QUESTION 4

 4.1                                        2                           A subst. into dist. formula
           𝐴𝑄 = √(−6 − 2)2 + (−7 − (−1))
           𝐴𝑄 = √(−8)2 + (−6)2
           𝐴𝑄 = √64 + 36
           𝐴𝑄 = √100

            AQ  10                                                    CA answer
                                                                                                   (2)

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                                                             6                               Common Test June 2019
Mathematics P2                                         NSC-Marking Guideline

 4.2
           ( x  a) 2  ( y  b) 2  r 2
                                                                               A (𝑥 − 2)2 + (𝑦 + 1)2
            (𝑥 − 2)2 + (𝑦 + 1)2 = 100                                          CA 100
                                                                                                         (2)

 4.3.1                 7 1 3                                                       3
           m AQ                                                              A
                       6 2 4                                                       4

                              4
            mQP                            𝑟𝑎𝑑. ⊥ 𝑡𝑎𝑛                       CA − 3
                                                                                         4
                              3
                                                                                                         (2)

 4.3.2            7 1      3                                                      3
           m AR                                                             A
                  10  2     4                                                      4
                       4
             m PR            𝑟𝑎𝑑. ⊥ 𝑡𝑎𝑛                                            4
                       3                                                       CA
                                                                                     3
                                                                                                         (2)
 4.4.1                    4
           mQP  
                          3

                      y  y1  m ( x  x1 )

                                   4
                      y+7 =         ( x  6)                                  CA substitution
                                   3

                               4                                               CA answer
                      y        x  15
                               3                                                                         (2)

 4.4.2                4
           mPR 
                      3

                 y  y1  m( x  x1 )

                              4
                 y7           ( x  10)                                      CA substitution
                              3

                          4 x 40
                     y         7
                           3   3

                          4    61                                              CA answer
                      =     x
                          3    3                                                                         (2)

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                                7                                                              Common Test June 2019
Mathematics P2                                         NSC-Marking Guideline

 4.5.1     4    61    4                                                        CA Equating
             𝑥−    = − 𝑥 − 15
           3     3    3
           8    16
             𝑥=
           3     3

           x =2                                                                CA x value
                53
           y=− 3                                                               CA y value

                      53                                                       CA both co-ordinates
           P(2; = − 3 )

           OR

           The x – co-ordinate of P is 2 (ARPQ is a kite)                      A x = 2

                                            4
                     Subst x  2 in y       x  15
                                            3
                               4
                     y         (2)  15
                               3                                               CA substitution

                               8
                       = − − 15
                               3

                                8  45
                           =
                                  3
                                                                               CA y value
                                53
                           =
                                3
                                                                               CA both co-ordinates
                        53                                                                                 (4)
             𝑃 (2; − 3 )

 4.5.2     In  SPR

                       P̂         (ext  of SPR)                          A S/R

                                                                               A P̂    
                      P̂    
                                                                                                           (2)

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                                                         8                                Common Test June 2019
Mathematics P2                                     NSC-Marking Guideline

 4.5.3                        sin(    )
           tan (   ) =
                              cos(   )

                 sin  cos   cos  sin 
           =                                                               A expansion
                  cos  cos   sin  sin 

              4 3        3 4
             ( ) ( ) − (− ) ( )
           =  5   5      5 5                                               A numerator
                3 3      4 4                                               A denominator
             (− ) ( ) + ( ) ( )
                5 5      5 5
             12 12
               +
           = 25 25                                                         A Simplification
             −9 16
               +
             25 25
               24 25
           =     ×
               25 7
               24
           =
                7

               OR
                                                                                                      (4)
                    4                                                      A answer
           𝑡𝑎𝑛𝛽 =        ∴ 𝛽 = 53.13°
                    3
                       4
            𝑡𝑎𝑛𝛼 = − 3 ∴ 𝛼 = 126,87°                                       A answer
                             sin(126.87  53.13)
           tan (   ) =                                                  A simplification
                             cos(126.87  53.13)
                         = 3, 42857…….
                           24
                         ≅ 7                                               A Answer                  (4)

                                                                                                     [22]

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                                9                                             Common Test June 2019
Mathematics P2                                  NSC-Marking Guideline

QUESTION 5

 5.1                 3               1
         tan A =       ;   sin B =
                     4               3

                 Diagram 1                               Diagram 2

                                                                                  A diagram 1

         𝑟2 = (−4)2 + (−3)2 … .          𝑃𝑦𝑡ℎ           𝑥2 = 32 − 12 … 𝑃𝑦𝑡ℎ
         𝑟2 = 25                                         𝑥2 = 8
         𝑟=5                                             𝑥 = 2√2

 5.1.1 cos 2𝐴 2                                                                   A identity
       = 2𝑐𝑜𝑠 𝐴 − 1
              −4 2
       = 2( ) −1
               5
           7                                                                      CA answer          (3)
       =
          25
 5.1.2 sin(𝐴 + 𝐵)
       = sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵                                                A Diagram 2
           −3 2√2           −4 1                                                  A expansion
       = ( )(         ) + ( )( )
             5     3        5    3
             4 + 6√2                                                              CA answer          (3)
       =−
                15
 5.2 sin 20o cos 320o + cos (–20o) sin 400o
       = sin 20o cos 40o + cos 20o sin 40o                                        A reduction
       = sin (20o + 40o)                                                          CA simplify
       = sin 60o
            3
       =                                                                          CA answer          (3)
           2
                                                                                  (provided
                                                                                  special angle
                                                                                  ratio)
                                                                                  ANSWER
                                                                                  ONLY = 0

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                                                     10                        Common Test June 2019
Mathematics P2                                  NSC-Marking Guideline

 5.3                       
                cos 2 90 o  θ       
         cos (θ)  sin (90 o θ) cos θ

               =       sin 2 θ                                                      A numerator
                   cos θ  cos2 θ                                                   A denominator
                                                                                    A common
                            sin 2 θ
                                                                                   factor
                       cos θ (1  cos θ)                                             A difference
                            1  cos 2 θ                                             of squares
                     
                         cos θ (1  cos θ)
                                                                                    A
                         1  cos θ                                                  simplification
                                                                                                        (5)
                           cos θ

                         1
                            1
                       cos θ
                      RHS

 5.4.1 cos 2  cos2   sin 2 
              = cos   sin   cos   sin                                     A expansion
              = p.q
                                                                                     A answer           (2)
 5.4.2 1  sin 2 
         cos 2
           sin 2   2 sin  cos   cos 2                                         A numerator
         =
                     cos 2   sin 2                                               A denominator
                                                                                    A factorise
                     sin   cos  
                                         2

         =                                                                          A factorise
             (cos   sin  )(cos   sin  )
             p                                                                      CA answer           (5)
         =
             q
 5.5     6 cos 2 x  sin x  5  0
                                                                                    A identity
         6 (1  sin 2 x )  sin x  5  0
         6  6 sin 2 x  sin x  5  0
                                     −6𝑠𝑖𝑛2 𝑥 + sin 𝑥 + 1 = 0
                                  (3 sin 𝑥 + 1)(−2 sin 𝑥 + 1) = 0                   CA factors
                                                1             1                     CA both ratios
                                     sin 𝑥 = − ; sin 𝑥 =
                                                3             2                     A 𝑘 ∈ ℤ
          𝑥 = 199,47° + 𝑘. 360°; 𝑘 ∈ ℤ               𝑂𝑅 𝑥 = 30° + 𝑘. 360°; 𝑘 ∈ ℤ    CA both
          𝑥 = 340,53° + 𝑘. 360°; 𝑘 ∈ ℤ               𝑂𝑅 𝑥 = 150° + 𝑘. 360°; 𝑘 ∈ ℤ   solutions of
                                                                                               1
                                                                                    sin 𝑥 = − 3
                                                                                    CA both
                                                                                    solutions of
                                                                                             1      (6)
                                                                                    sin 𝑥 = 2

                                                                                                         [27]
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                                11                                  Common Test June 2019
Mathematics P2                        NSC-Marking Guideline

QUESTION 6

 6.1                                                                 A x-intercept

                                                                     A t. pt.
                                                                     (600; 1)

                                                                     A shape

                                                                                           (3)
 6.2     60° < 𝑥 < 90°                                               CA end
                                                                     values                (2)
                                                                     CA notation
 6.3     Graph of f moves 600 left.
                                                                     A shifts 600
                                                                     A to the left

                                                                                           (2)
                                                                                           [7]

QUESTION 7

   7            DA                                            A trig ratio
         tan 𝜃 =
                AB
         ∴ AD = AB tan 𝜃                                       AAD value
         Also
           AB              𝑘
                  =
          sin 𝛼       sin(180°−2𝛼)                            A substitute into sine
         AB sin 2𝛼 = 𝑘 sin 𝛼                                  rule
                 𝑘 sin 𝛼                                      A
         AB =
              2 sin 𝛼 cos 𝛼                                    sin 2𝛼 = 2 sin 𝛼 cos 𝛼
                 𝑘
         AB =
              cos 𝛼                                           A
                 𝑘. tan 𝜃                                                 𝑘. sin 𝛼
         ∴ AD =                                                 𝐴𝐵 =
                 2 cos 𝛼                                               2 sin 𝛼 cos 𝛼

                                                              A making AB value -
                                                              simplified

                                                                                            [6]

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                                                      12                 Common Test June 2019
Mathematics P2                                   NSC-Marking Guideline

QUESTION 8

 8.1     Qˆ 4  Wˆ 2  x … (tan chord theorem)
                                                                          AS/R
         Wˆ  Wˆ  x … (WQ bisects P𝑊R.)
            2        1

         Qˆ1  Wˆ1  x … (tan – chord theorem)                            AS/R
                                                                          AS/R
         Tˆ  Wˆ  x … ( s in same segment)
           2         2                                                    AS/R
       Sˆ2  Wˆ1  x … s in same segment)                                AS/R                (5)
       ̂2 = Q
 8.2.1 T     ̂1 = 𝑥                                                       AS
        TS // PR … (alternate s equal)                                  AR                  (2)
 8.2.2 Tˆ  Pˆ … (corresponding ∠𝑠 ; TS//PR)                              AASR
           3

         Tˆ3  Qˆ 3 … ( s in same segment)                               AS/R
          P̂  Q̂ 3
                                                                                               (3)
 8.2.3 In Δ TQS
       Tˆ2  x                                                            AS
       Ŝ 2  x                                                           AS
        Tˆ2  Sˆ2  x                                                    AS                  (4)
        Δ TQS isosceles … ( s opposite equal sides)                     AR
 8.2.4 WQ  ̂ P = WŜQ. . . (tan − chord theorem)                          AASR
       T1 = WŜQ … (ext  of cyclic quad)
       ̂
                                                                          AS/R
                                                                                               (3)

                                                                                               [17]

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                                13                                            Common Test June 2019
Mathematics P2                               NSC-Marking Guideline

QUESTION 9

       P1 = ̂
 9.1.1 ̂    Q1              given
       ̂    ̂
       P1 = R               tan-chord theorem                              AASR
         ̂
       ∴ Q1 = R̂
        ∴ TQ ∥ SR (corr ∠s 𝑎𝑟𝑒 𝑒𝑞𝑢𝑎𝑙)                                      AR                        (3)

 9.1.2 ̂
       P1 = Q ̂1              given
       TS = TP                tan from same point                          AS/R
       ̂
       P1 = S1̂               equal ∠𝑠 opp equal sides                     AS/R
       ∴Q ̂ 1 = Ŝ1
       ∴QPTS is a cyclic quad        converse equal ∠𝑠 subtended by same   AR
                                     chord                                                            (3)
 9.1.3 QPTS is a cyclic quad
       ∴̂ P1 = ̂Q2                   ∠𝑠 in same ⊙ segm                     AASR
       but ̂P1 = Q̂1                 given
          ̂
       ∴ Q1 = Q 2̂                                                         AS
       ∴TQ bisect SQ  ̂P                                                                              (3)
 9.2.1 In Δ LPK and Δ NPL
       KL̂P = LN  ̂ P … tan chord theorem                                  AASR
       ̂ 2 = 90° … …
       P                     ∠ in semi ⊙                                   AS/R
       P1 = ̂
       ̂      P2 … both = 90o
       PK̂ L = NL̂P … remaining angle
       ∴ ΔLPK///ΔNPL . . . ∠∠∠                                             AR
                                                                                                      (4)
 9.2.2   PL      KL   PK                                                   AASR
              = NL = PL… Δ LPK /// Δ NPL
         NP
         PL      PK
            = PL ……. ΔLPK///ΔNPL                                           A proportionality
         NP
         ∴ PL2 = NP. PK
                                                                                                      (3)
 9.2.3 Δ NLK                                                               A answer                  (1)
 9.2.4
       ΔNLK///ΔNPL
         𝐾𝑁 𝐿𝑁
       ∴     =      ⌈///∆′𝑠⌉
         𝐿𝑁 𝑁𝑃                                                             A S/R
       𝐿𝑁 2 = 𝐾𝑁. 𝑁𝑃                                                       A  Substitution
            = 16 x 10
            = 160
        LN = √160                                                          CA NL value
                1
       Radius = 2 √160                                                     CA radius =
                                                                           1
       Area of Circle = 𝜋𝑟 2                                                 160
                           1                                               2
                      = 𝜋(2 √160)2
                      = 125.66cm2

                                                                           CA Substitution
         OR                                                                CA Answer
                                                                                                      (6)

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                                                         14                     Common Test June 2019
Mathematics P2                                      NSC-Marking Guideline

         From Question No. 9.2.2: PL2  NP.PK
                                      = 10 cm x 6 cm
                                      = 60 cm2

         NL2   =     PL2 + PN2 … Pythagoras                                  A PL2 = 60 cm2
               =     60 + 100     (Δ LPN)
               =     160 cm2
                                                                             A Pythagoras
         ∴ NL = √160𝑐𝑚
         1       1           1
           NL  diameter      160 cm
         2       2           2                                               CA NL value
                                                                             CA radius =
         Area of circle = r 2                                               1
                                                                               160
                                          2
                                                                             2
                                1
                         = 𝜋 × (2 √160)       𝑐𝑚2
                                                                             CA substitution
                         = 125,66 cm2 OR        40𝜋 𝑐𝑚2

                                                                             CA answer
                                                                                                        [23]

QUESTION 10
 10.1      AE 4                                                                A S/R
               =    … … prop theorem ; EB//FC
           EF 6
           AE   AC
           EF
              = CD … … prop theorem; EC//FD                                    A S/R
           4 10
              =
           6 CD
           CD = 15 𝑢𝑛𝑖𝑡𝑠                                                       CA answer            (3)

 10.2      ∆FEC      3
                  =5                same height
           ∆CFA                                                                A S/R
           Area ∆CFA   10   2
           Area ∆FAD
                     = 25 = 5       same height                                A S/R

           Area ∆FEC Area ∆FEC Area ∆CFA
                    =         ×
           Area ∆FAD Area ∆CFA Area ∆FAD
             3 2                                                               CA simplify
           =   ×
             5 5
              6
           =
             25                                                                CA answer
                                                                                                     (4)
                                                                                                     [7]

                                                                            TOTAL MARKS: 150

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