# Mathematics Grade 12 Paper 2 NSC November 2020 Memo

> Grade 12 Mathematics Paper 2 memorandum from the NSC November Exam 2020, 27 pages. Read online or download the free PDF.

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- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2020
- Exam period: NSC November Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 27

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Mathematics · Grade 12 · NSC November Exam · 2020. Memorandum, 27 pages. Read online or download the PDF.

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### Page 1

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                 SENIOR CERTIFICATE/SENIOR SERTIFIKAAT
                     NATIONAL SENIOR CERTIFICATE/
                     NASIONALE SENIOR SERTIFIKAAT

                                        GRADE/GRAAD 12

                             MATHEMATICS P2/WISKUNDE V2

                                        NOVEMBER 2020

                       MARKING GUIDELINES/NASIENRIGLYNE

MARKS/PUNTE: 150

                            These marking guidelines consist of 27 pages.
                             Hierdie nasienriglyne bestaan uit 27 bladsye.

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Mathematics P2/Wiskunde V2                         2                                       DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

NOTE:
 If a candidate answers a question TWICE, only mark the FIRST attempt.
 If a candidate has crossed out an attempt of a question and not redone the question, mark the
  crossed out version.
 Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at the
  second calculation error.
 Assuming answers/values in order to solve a problem is NOT acceptable.

LET WEL:
 As 'n kandidaat 'n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na.
 As 'n kandidaat 'n antwoord van 'n vraag doodtrek en nie oordoen nie, sien die doodgetrekte
   poging na.
 Volgehoue akkuraatheid word in ALLE aspekte van die memorandum toegepas. Hou op nasien
   by die tweede berekeningsfout.
 Aanvaar van antwoorde/waardes om 'n probleem op te los, word NIE toegelaat nie.

      GEOMETRY
                A mark for a correct statement
                (A statement mark is independent of a reason)
         S
                'n Punt vir 'n korrekte bewering
                ('n Punt vir 'n bewering is onafhanklik van die rede)
                A mark for the correct reason
                (A reason mark may only be awarded if the statement is correct)
         R
                'n Punt vir 'n korrekte rede
                ('n Punt word slegs vir die rede toegeken as die bewering korrek is)

                Award a mark if statement AND reason are both correct
        S/R
                Ken 'n punt toe as die bewering EN rede beide korrek is

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Mathematics P2/Wiskunde V2                         3                                                 DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 1

1.1   a  9,5                                                                                            a  9,5
      b  0,909..  0,91                                                                                 b  0,91
      yˆ  9,5  0,91x                                                                                   equation
                                                                                                                         (3)
1.2
                                  100

                                   90
                                                                                                         correct
                                   80
                                                                                                         slope going

         PHYSICAL SCIENCES (% )
                                   70                                                                    through 2
                                                                                                          points:
                                   60
                                                                                                         (50 ; 55) or
                                   50                                                                    (40 ; 46) or
                                                                                                         (60 ; 64) or
                                   40                                                                    (0 ; 9,5) or
                                   30                                                                    (45 ; 50)

                                   20
                                  9,5
                                   10

                                    0
                                        0    10   20   30      40    50      60   70      80       90
                                                            MATHEMATICS (% )

                                                                                                                         (2)
1.3 Final exam mark  72,22%                                  (calculator)                               answer
                                                                                                                         (2)
      OR

      yˆ  9,5  0,9169                                                                                substitution
          72,29%                                                                                        answer
                                                                                                                      (2)
1.4 r = 0,95                                                                                             answer(A)
                                                                                                                      (1)
1.5 There is a very strong positive correlation between the Mathematics and                              strong/ sterk
    Physical Sciences mark.
    Daar is ‘n baie sterk positiewe korrelasie tussen die Wiskunde en Fisiese
    Wetenskappunte.                                                                                                      (1)
1.6 The teacher concludes that the higher the learners' Mathematics marks, the                           answer
    higher the learners' Physical Sciences marks.
    Die onderwyser het waargeneem dat hoe hoër die wiskunde punte is, hoe hoër
    is die Fisiese Wetenskappunte.                                                                                       (1)
                                                                                                                        [10]

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Mathematics P2/Wiskunde V2                         4                                       DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 2

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 / Julie                                                                     answer
                                                                                                              (1)
2.2         26 941                                                                       26 941
        x                                        Answer only: Full marks
              12
          = 2 245,083..  2 245,08 aircraft landings                                    answer
                                                                                                              (2)
2.3     Standard deviation for landings at the King Shaka International airport:         answer
          86,30                                                                                             (2)
2.4     ( x   ; x   ) = (2 245,08 – 86,30 ; 2 245,08 + 86,30)                        x 
                   limit = (2 158,78 ; 2 331,38)                                         x 
        There were 9 months when the aircraft arrivals at the King Shaka
        International airport were within one standard deviation of the mean.            answer
                                                                                                              (3)
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
                                                                                                              (1)

                                                                                                              [9]

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Mathematics P2/Wiskunde V2                         5                                       DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 3
                                                                        y
                                                                                            T

                                                     W(–4 ; 4)

      N                                   M                       V
                                                                        O                                 x

                                                                         P(0; –4)
                            S

                                                              R

                                                                                     K
3.1             4  (4)   8
          m WP                                                               substitution of W and P
                 40 4
          mWP  2                                                             mWP
                                                                                                              (2)
3.2                1
          mST       (given)
                   2
                                                                               mWP mST 
          mWP mST    2 1                                           mWP mST  = –1
                                  2
                        1                                                                                   (2)
          ST  WP
3.3       5 y  2 x  60  0
                    2
           y   x  12
                    5
             2          1
           x  12  x  6                                                     equating
             5          2
           4 x  120  5x  60
          9 x  180                                                            x value
          x  20
           y    20  12
                    2                                                          substitution
                    5
           y  4                                                             y value
           S(–20; –4)                                                                                        (4)

          OR

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Mathematics P2/Wiskunde V2                         6                                       DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

                                                                        y
                                                                                            T

                                                      W(–4 ; 4)

    N                                         M                   V
                                                                        O                                 x

                                                                        P(0; –4)
                          S

                                                              R

                                                                                     K

        5 y  2 x  60  0
         1                                                                     substitution
        5 x  6   2 x  60  0
         2         
           5
             x  30  2 x  60  0
           2
                           9
                             x  90           x  20                          x value
                           2
         y    20  12
                  2
                                                                                 substitution
                  5
         y  4
                                                                                 y value
         S(–20; –4)
                                                                                                              (4)

        OR
               5 y  2 x  60 ........(1)
                2 y  x  12    ........(2)
        (1)  2(2) : 9 y  36                                                   adding
                       y  4
                                                                                 y value
             2(–4) – x = 12                                                      substitution
                        x = –20                                                  x value
                                                                                                              (4)

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Mathematics P2/Wiskunde V2                         7                                       DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

3.4
        y
              2
                 4  12 OR 5 y  2(4)  60  0                               substitution
              5
              52
       y                                                                       y value
               5
                   52 
        R   4;   OR R(–4 ; –10,4)
                    5
                   52                                 52                     method or subst into
        WR  4     OR WR = ( 4  ( 4)) 2  (4    ) 2                  distance formula
                   5                                  5
              72                  2                                              answer
        WR      units or WR  14 units
               5                  5                                                                             (4)
       OR

        WR  ST  SK
               1           2        
                 x  6    x  12                                            substitution
               2            5       
                9                                                                simplification
                  x  18
               10
                9
              ( 4)  18
               10                                                                subst x = –4
              14,4 units
                                                                                 answer                        (4)
3.5             2
        mSK                                                                    mSK
                5
          158,19... (Ref.  =21, 801…°)
                                                                                 size of 
        MN̂S  21,80...
              1
        mST 
              2
        NM̂S  26,56...                                                         size of NM̂S
          21,80...  26,56... [ext  of ]                                   method
                                                                                 answer
          48,366... = 48,37°
                                                                                                               (5)
3.6    In SRW:
        h = –4 – (–20)
        h = 16units                                                             h
       Area SRW   h WR 
                      1
                      2
                            72 
                    16 
                     1
                                                                                 substitution
                     2      5
                    115,2 square units                                          area 

       Area SWRL = 2Area SRW
                 =2(115,2)
                 = 230,4square units                                             answer
                                                                                                              (4)
       OR

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Mathematics P2/Wiskunde V2                         8                                       DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

       In SRW:
        h = –4 – (–20)
        h = 16units                                                             h
                           72
       Area SWRL = 16                                                            substitution
                            5
                      = 230,40 square units                                      answer
                                                                                                               (4)
       OR
                                                                                SW = 8 5
        SW      20  42   4  42  8 5  17,89
                                          2
                                  2
       SR   20  4    4  10  
                                        16 29                                           16 29
                                               17,23
                          2
                                                                                SR=
                                  5     5                                               5
       Area SWRL = 2  Area SRW
          1              
        2 SW  SR sin  
          2              
            1      16 29             
         2 8 5        sin 48,37                                          substitution
            2        5               
        =230,41square units                                                     answer
                                                                                                               (4)
                                                                                                              [21]

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Mathematics P2/Wiskunde V2                         9                                        DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 4
                                                                     y

                                    T            M(–3 ; 4)

                                                                 O                                               x

                                    N(–11 ; p)
                                                                             S

4.1     x2  y2  r 2
                                                                                substitution
        r 2   3  4  25
                     2      2

        x 2  y 2  25                                                          answer
                                                                                                               (2)
4.2    TM  TN        [tangent  radius]
       T(–11 ; 4)
       r = –3 – (–11) = 8                                                       xT  11
        x  32   y  42  64                                               LHS  RHS
                                                                                                               (3)
4.3    O (0 ; 0 ) and M(–3; 4)
                  40     4         04        4                                                4
       mOM                  OR                                             mOM  
                 30     3       0  ( 3)    3                                                3
                 3                                                                          3
        m NM                                                                   m NM 
                 4                                                                          4
                3                                3
       y  4  ( x  (3))           OR     y     xc                          substitution of m and M
                4                                4
               3   9                            3
       y 4  x                            4  (3)  c
               4   4                            4
              3    25                          25
       y x                               c
              4    4                            4
                                                                                equation
                                               3    25
                                           y  x                                                              (4)
                                               4     4

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Mathematics P2/Wiskunde V2                         10                                      DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

4.4    N(–11 ; p)

           3    25
        y   x
           4     4
                                           4 p
        p   11 
           3         25                               3
                                OR                                           subst x = –11 into eq or
           4         4                   3  ( 11) 4                         gradient
        p2                                                                  p2

        N 11;  2

         3  xS               4  yS
                 0      and          0
            2                     2
       S3;  4                                                             xS  yS

        SN      11  32   2  (4)2
              10 2 units or 14,14 units                                      answer (CA)
                                                                                                               (5)
4.5     B 2 ; 5
        BM  2 units                                                             2

       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                        k  6,6
                                                                              k  9,4
                                                                                                               (5)
                                                                                                              [19]

                                                         y

                                                         6,6

                                              (–2 ; 5)

                                           9,4                                                  x

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Mathematics P2/Wiskunde V2                         11                                        DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 5

          y

                                                    g

                                                                  A(130,9° ; 0,33)
                                                                                         x
              0°                              90°                150°           180°

                          f

5.1     Period of g = 360                                                     answer
                                                                                                              (1)
5.2                           1                                                answer (A)
        Amplitude of f =                                                                                      (1)
                              2
5.3     f (180) – g(180)
           1  1
            
           2  2
                                                                               1
        =1
                                                                                                              (1)
5.4.1   x  140,9                                                             x  140,9
                                                                                                              (1)
5.4.2     3 sin x  cos x  1
          3        1        1
            sin x  cos x                                                     dividing by 2
         2         2        2
        sin x cos 30  cos x sin 30 
                                          1                                    cos 30; sin 30
                                          2
                        1                                                                           1
        sin( x  30)                                                         sin( x  30) 
                        2                                                                           2
                        1
        sin( x  30)    at x  0 or x  120
                        2
         x [0 ;120] OR 0  x  120                                       interval
                                                                                                              (4)
                                                                                                              [8]

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Mathematics P2/Wiskunde V2                         12                                             DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 6
                                                           y

                                          P(5 ; 12)

                                                       <
                                                               
                                                                                             x

                                                                               T

6.1.1               12                2
        tan               or  2                                              answer
                     5                5
                                                                                                                     (1)
             2           2       2
6.1.2 (OP) = (–5) + (12)                                                          Pythagoras
        OP = 13                                                                   OP
                 5
      cos                                                                      answer
                13
                                                                                                                     (3)
6.1.3                    b
        sin(  90)                                                             sin(  90) 
                                                                                                      b
                        6,5                                        y                                 6,5
                 b                       P(–5 ; 12)
        cos                                                                     cos 
                6,5
        5 b                                     13
            
        13 6,5                                                                    5 b
                                                                                     
              5                                                                    13 6,5
        b                                                                x      value of b
              2                                                    O
                                                                                                                     (4)
        OR                                      6,5
                                     S(a ; b)                                                          a
        cos(90   ) 
                          a                                                       cos(  90) 
                         6,5                                                                          6,5
                          a                                                        sin 
               sin  
                         6,5
                  12      a                                                       value of a
                            a  6
                  13 6,5
                                          5
        b  (6,5) 2  ( 6) 2                                                   value of b
                                          2
                                                                                                                     (4)

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Mathematics P2/Wiskunde V2                         13                                      DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

6.2     sin 2 x. cos( x)  cos 2 x. sin(360  x)
                      sin(180  x)
           sin 2 x cos x  cos 2 x  sin x 
                                                                          cos( x)  cos x
                          sin x                                           sin(360  x)   sin x
           sin(2 x  x)
                                                                          sin(180  x)   sin x
               sin x
            sin x
                                                                          numerator = sin x
            sin x
         1                                                               answer
                                                                                                              (5)
6.3                 2
              6 sin x  7 cos x  3  0
                                                                           identity
        6(1  cos 2 x )  7 cos x  3  0
         6  6 cos 2 x  7 cos x  3  0
                                                                           standard form
              6 cos 2 x  7 cos x  3  0
                                                                           factors
          (3 cos x  1)( 2 cos x  3)  0
                1                     3                                    both solutions of cos x
        cos x        or      cos x  (N/A)
                3                     2
        x  109,47  k.360; k  Z or                                    x  109,47 & 250,53
           x  250,53  k.360; k  Z                                     + k.360; k  Z
                                                                                                              (6)
6.4               1
             x      3 cos A
                  x
                                    2

        3 cos A 2   x  1                                          squaring both sides
                         x
                           1                                                                    1
         9 cos 2 A  x 2  2  2                                           9 cos 2 A  x 2       2
                          x                                                                     x2
         9 cos A  2  2
               2

                     4
          cos 2 A                                                                  4
                     9                                                     cos 2 A 
                                                                                    9
           cos 2A  2 cos 2 A  1                                          cos 2A  2 cos 2 A 1
                       4
                     2   1
                       9
                        1
                    
                        9                                                  answer
       OR                                                                                                     (5)

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Mathematics P2/Wiskunde V2                         14                                      DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

                     1
        x2  2          0
                   x2
                     2
              1
           x    0
              x
                x2  1
                  x  1
        3cosA  2    or 3cosA  2
                 2                           2
        cosA            or       cosA  
                 3                           3
        cos2A  2cos 2 A  1                                               x  1
                              2                                                      2
                    2                                                    cos A 
                 2    1                                                         3
                    3
                                                                                      2
                    1                                                      cos A  
                                                                                    3
                    9                                                     double angle identity

                                                                           answer
                                                                                                               (5)
                                                                                                              [24]

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Mathematics P2/Wiskunde V2                         15                                      DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 7
                                                             P

                                                         r   Q
                                                             2x          30°
                                                  R
                                                                                  S

7.1                    3r
        tan 30                            OR                                  trig ratio
                       QS
                   3r                                                           QS subject
        QS 
                tan 30
                  3r               3r
                         or
                  1                 3
                   3               3
             3r                                                                                               (3)
7.2     Area of flower garden   3r   r 2                                  substitution into
                                                  2

                                                                                 difference of areas
                                         9r 2  r 2
                                         8r 2                                 answer
                                                                                                               (2)
7.3        2       2           2
        RS  r  (3r )  2( r )(3r ) cos 2 x                                   substitution into cosine rule
                                                                                correctly
                 r 2  9r 2  6r 2 cos 2 x
                 10r 2  6r 2 cos 2 x                                          10r 2  6r 2 cos 2 x
                 r 2 (10  6 cos 2 x )                                         r 2 (10  6 cos 2 x )
                                                                                                               (3)
         RS  r 10  6 cos 2 x
7.4     RS  10 10  6 cos 2(56)                                               substitution
             34,9966...                                                        answer
             35 m                                                                                             (2)
                                                                                                              [10]

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Mathematics P2/Wiskunde V2                         16                                        DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 8

                                                                             L

                           K                             O            1
                                 1      2
                                                     1       3        2 M
                                     32°                 2

                                                                    26°

                                                                       N
                                            P

8.1.1(a)   Ô 2  64     [ at centre = 2 × at circumference/                  S R
                           Middelpts  = 2 × omtreks]
                                                                                                                (2)
8.1.1(b)   M̂ 2  90    [Line from centre to midpt of chord/lyn v midpt         SR
                         na midpt v koord]
           KÔN  90  26  116 [ext  of /buite  van ]                    S
           Ô1 116  64  52                                                  answer
           OR                                                                                                   (4)
           M̂ 2  90    [Line from centre to midpt of chord/lyn v midpt         SR
                         na midpt v koord]
           Ô3  64     [sum of s in ]                                        S
           Ô1  52     [s on straight line/op ‘n reguitlyn]                    answer
                                                                                                                (4)
8.1.2      PK̂O  P̂  128 [sum of s in ∆/som e van ∆]
            PK̂O  P̂ [s opp = sides/e teenoor = sye]                          S
                 64                                                            S
            K̂ 2  32 or K̂ 2  K̂1                                            S
                                                                                                                (3)
            KN bisects/halveer OK̂P
           OR
                   K̂ 2  KN̂O [s opp = sides/e teenoor = sye]
                                                                                 S
           K̂ 2  KN̂O  64 [sum of s in ∆/som e van ∆]                       S
            K̂ 2  32 or K̂ 2  K̂1
                                                                                 S
            KN bisects/halveer OK̂P                                                                            (3)

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Mathematics P2/Wiskunde V2                         17                                            DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

8.2

                                                  A

                                                               G
                                        1
                                 F          2

                                                                      1
                                                                              D
                                                                          2

                                                                                     C

             B

8.2.1   F̂1  D̂1    [tan chord theorem/raaklyn koordst]                            S R
        D̂1  B̂     [Given/Gegee]
         F̂1  B̂                                                                   F̂1  B̂
         FG || BC [corresp s =/Ooreenkomstigee =]                                 R
                                                                                                                    (4)
8.2.2      GC FB
                              [line || one side of /lyn || een sy v ]            S R
           AC AB
          x9      5
                                                                                    substitution
         2x  6 7
        7 x  63  10 x  30
              3x = 93
               x = 31                                                                answer
                                                                                                                    (4)
        OR
        AG  2 x  6  x  9  x  15
           AG AF                                                                    S R
                         [line || one side of /lyn || een sy v ]
           GC FB
         x 15 2                                                                     substitution
               
          x9 5
        5x  75  2 x 18
             3x = 93                                                                 answer
              x = 31                                                                                                (4)

        OR

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Mathematics P2/Wiskunde V2                         18                                      DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

        AF AG                                                                 S R
                     [line || one side of  /lyn || een sy v ]
        AB AC
        2 x  15                                                               substitution
           
        7 2x  6
        7 x  105  4 x  12
        3x  93                                                                answer
        x  31                                                                                                 (4)

                                                                                                              [17]

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Mathematics P2/Wiskunde V2                         19                                      DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 9

9.1

                                                          K

                                                                             R
                                    T

                                                          O

                            Q
                                                      S

9.1           Construction: Draw diameter KS and draw KR                           construction
              Konstruksie: Trek middellyn KS en verbind KR
              QŜT  90  TŜK     [radius  tangent/raaklyn]
                                                                                   S/R
              SR̂K  90                [ in semi circle/halfsirkel]              S/R
              SR̂T  90  KR̂T                                                  S
              TŜK  TR̂K               [s same segment/e dieselfde
                                         segment]                                  S/R
                                                                                                       (5)
              QŜT  R̂

                                                   OR

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Mathematics P2/Wiskunde V2                         20                                       DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

                                                                     R
                               T

                                                     O

                       Q
                                                 S

9.1           Construction: Draw radii OS and OT                                    construction
              Konstruksie: Trek radii OS en OT
              QŜT  90  OŜT        [radius  tangent/raaklyn]                   S/R
              OŜT  ST̂O               [s opp = sides/e teenoor = sye]           S/R
               SÔT  180  2OŜT      [s of /e van ]                        S
               R̂  90  OŜT          [ at centre = 2× circumf/                 S/R
                                          midpts = 2× omtreks]
               QŜT  R̂
                                                                                                        (5)

                                                   OR

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Mathematics P2/Wiskunde V2                         21                                       DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

                                                     K

                                                                     R
                               T

                                                     O

                       Q
                                                 S

9.1           Construction: Draw diameter KS and join K to T.                       construction
              Konstruksie: Trek middellyn KS en verbind K tot T K
              QŜT  90  TŜK       [radius  tangent/raaklyn]                    S/R
              ST̂K  90 [ in semi circle/halfsirkel]                              S/R
                K̂  90  TŜK                                                   S
               QŜT  K̂
              but R̂  K̂     [s same segment/e dieselfde segment]                S/R
               QŜT  R̂
                                                                                                         (5)

                                                   OR

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Mathematics P2/Wiskunde V2                         22                                        DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

                                                                      R
                                                              y
                               T            y                     z
                                        x
                                                     O

                                                x z
                       Q
                                                 S

9.1           Construction: Draw radii OT, OR and OS                                 construction
              Konstruksie: Trek radiuse OT, OR en OS
              OŜT  OT̂S         [s opp = radii/e teenoor = radiuse]              S/R
              Also: OT̂R  OR̂T and OR̂S  OŜR
              2 x  2 y  2 z  180 [s of ]
              x  y  z  90
                   y  z  90  x                                                  S
              OŜQ  90     [radius  tangent/raaklyn]
                                                                                     S/R
               TŜQ  90  x
                                                                                    S
               TŜQ  y  z  R̂
                                                                                                          (5)

                                                   OR

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Mathematics P2/Wiskunde V2                         23                                       DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

                                                                     R
                               T

                                                     O

                       Q
                                                 S

9.1           Construction: Draw radii OT and OS, tangent QT                        construction
              Konstruksie: Trek radiuse OT en OS, raaklyn QT
              OŜQ  90 [radius  tangent/raaklyn]                                 S/R
               TŜQ  90  TŜO
               TŜO  ST̂O [s opp = radii/e teenoor = radiuse]                  S
              TÔS  180  2TŜO [s of ]                                        S
              R̂  90  TŜO [ at centre = 2× circumf/
                                midpts = 2× omtreks]
                                                                                    S/R
               TŜQ  R̂
                                                                                                         (5)

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Mathematics P2/Wiskunde V2                         24                                        DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

9.2

                                                                                         R

                                                          N
                                                      1           3
                                                          2

           M        2
                1

                                     S                                x

                                                                      1
                                                                  2
                                                              3
                                                                      P
                            1 2

                        Q

9.2.1(a)       N̂ 2  x           [alt s; PR || NQ/verw. e; PR || NQ]             SR
                                                                                                          (2)
9.2.1(b)       Q̂ 2  x           [tan chord theorem/raaklyn koordstelling]         SR
               OR                                                                                         (2)
               M2  x             [tan chord theorem/raaklyn koordstelling]
                                                                                     S/R
               Q̂ 2  x           [s in same segment/e in dieselfde segm]          S/R
                                                                                                          (2)
9.2.2          MN MS                                                                S R
                             [QN || PR; Prop Th]
               NR SP
               N̂1  N̂ 2  x [given]                                               S
                                                                                    SR
               P̂3  x            [s in same segment/e in dieselfde segm]
               P̂3  Q̂ 2         [= x]
                                                                                    R
               SQ = PS            [sides opp = /sye teenoor = e]
               MN MS
                                                                                                         (6)
                NR SQ
                                                                                                         [15]

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Mathematics P2/Wiskunde V2                         25                                       DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 10

            C                   D                                        M                       E
                                1       2                            1    2

                                                                     4 1
                                                                          F
                                                                     3       2

                                            1
                                                2
                                            B       3

                                                                         A

10.1.1        DB̂E  90     [ in semi-circle/ in halfsirkel]                    S R
               DM̂A  90 [ AM  DE ]
               FBDM is a cyclic quadrilateral/koordevh
                                                                                    R
               [converse opp s cyclic quad/omgek teenoorst e kvh ]                                     (3)
              OR
              DB̂E  90        [ in semi-circle/ in halfsirkel]                 S R
                 M̂ 2  DB̂E  90
               FBDM is a cyclic quadrilateral/koordevh
              [converse ext of cyclic quad/omgek buite van kvh ]                 R
                                                                                                         (3)

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Mathematics P2/Wiskunde V2                         26                                       DBE/November 2020
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

10.1.2        B̂3  D̂ 2 [tangent chord th/raaklyn koordst]                       S R
              F̂1  D̂ 2 [ext  cyc quad/buite  koordevh]                        S R

               B̂3  F̂1                                                                               (4)

              OR
              B̂1  Ê  x [tangent chord th/raaklyn koordst]                     S R
              F̂1  90  x [ sum in / van ]                                    F̂1  90  x
                                                                                  
               D̂ 2  90  x [ sum in / van ]                                   D̂ 2
               F̂1  D 2
              B̂3  D̂ 2    [tangent chord th/raaklyn koordst]                    R
               B̂3  F̂1                                                                               (4)

              OR
              B̂1  Ê  x [tangent chord th/raaklyn koordst]                     S R
               B̂3  90  x [straight line/ reguitlyn ]                          S
              F̂1  90  x [sum of s  / som van e van ]
                                                                                  S
               B̂3  F̂1                                                                               (4)
10.1.3        In ∆CDB and ∆CBE
              Ĉ  Ĉ         [common /gemeenskaplike  ]                        S
                                                                                   S/R
              CB̂D  CÊB [tangent chord th/raaklyn koordst]
               CD̂B  CB̂E [ sum in / van ]                                   R
              ∆CDB ||| ∆CBE                                                                             (3)

              OR
              In ∆CDB and ∆CBE
              CB̂D  CÊB [tangent chord th/raaklyn koordst]                       S/R
               Ĉ  Ĉ    [common /gemeenskaplike  ]                            S
              ∆CDB ||| ∆CBE [, , ]                                             R
                                                                                                        (3)
10.2.1        BC DC
                              [||| s]
              EC   BC                                                              ratio
              BC 2  EC  DC
                    8 2                                                          substitution
                    16
               BC  4                                                             answer
                                                                                                        (3)

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