# Mathematics Grade 12 Paper 2 NSC June 2019 Memo

> Grade 12 Mathematics Paper 2 memorandum from the NSC June Exam 2019, 20 pages. Read online or get the free PDF download, question paper linked when available.

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- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2019
- Exam period: NSC June Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 20

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Mathematics · Grade 12 · NSC June Exam · 2019. Memorandum, 20 pages. Read online or download the PDF.

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### Page 1

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            SENIOR CERTIFICATE EXAMINATIONS/
                SENIORSERTIFIKAAT-EKSAMEN
        NATIONAL SENIOR CERTIFICATE EXAMINATIONS/
           NASIONALE SENIORSERTIFIKAAT-EKSAMEN

                                        MATHEMATICS P2/
                                          WISKUNDE V2

                         MARKING GUIDELINES/NASIENRIGLYNE

                                                 2019

                                                                                     MARKS: 150
                                                                                     PUNTE: 150

                           These marking guidelines consist of 20 pages.
                           Hierdie nasienriglyne bestaan uit 20 bladsye..

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Mathematics P2/Wiskunde V2                         2                                                   DBE/2019
                          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.

NOTA:
• 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 nasienriglyne toegepas. Hou op
   nasien by die tweede berekeningsfout.
• Om antwoorde/waardes te aanvaar om 'n probleem op te los, word NIE toegelaat NIE.

                                GEOMETRY • MEETKUNDE
                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/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 1
1.1 45 children                                                                                                    answer
                                                                                                                                       (1)
1.2
      x=
                              ∑ fx = (4 × 2) + (8 × 10) + (12 × 9) + (16 × 7) + ( 20 × 8) + (24 × 7) + (28 × 2)
           n                                                               45
         692                                                                                                       692
      x=     OR x = 15,38 minutes                                 Answer only: full marks
          45                                                                                                       answer
                                                                                                                                       (2)

1.3
                                         Time taken (t)         Number of         Cumulative
                                          (in minutes)           children         frequency
                                            2<t≤6                     2               2
                                                                                                                   first 4 cum freq
                                            6 < t ≤ 10                10               12                           correct
                                                                                                                   last 3 cum freq
                                           10 < t ≤ 14                9                21                           correct
                                           14 < t ≤ 18                7                28
                                          18 < t ≤ 22                 8                36
                                          22 < t ≤ 26                 7                43
                                           26 < t ≤ 30                2                45
                                                                                                                                       (2)
1.4
                                        CUMULATIVE FREQUENCY GRAPH
                                                  (OGIVE)
                               50

                               40                                                                                  plotting cum

       Cumulative Frequency
                                                                                                                    freq at upper
                               30
                                                                                                                    limits correctly
                                                                                                                    (all points)
                                                                                                                   shape (smooth)
                               20                                                                                  grounding (2;0)

                               10

                                0
                                    0        5           10      15         20       25         30
                                                          Time in minutes
                                                                                                                                       (3)

1.5 On graph at the y-value of 22,5 or 23                                                                          graph
    Median = ± 15 minutes.                                            Answer only: full marks                      answer
                                                                                                                                    (2)
                                                                                                                                   [10]

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Mathematics P2/Wiskunde V2                         4                                                   DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 2

2.1      a = 12,44                                                   value of a
         b = 0,98                   Answer only: full marks          value of b
          y = 12,44 + 0 ,98 x                                        equation
                                                                                                     (3)
2.2.1                   15
         Percentage =      × 100
                        50
                      = 30%                                          answer
                                                                                                     (1)
2.2.2    yˆ = 12,44 + 0 ,98 x
         yˆ = 12,4 4 + 0 ,98(30)                                     substitution of 30
         yˆ = 41,84
            = 42                 Answer only: full marks             answer as integer
         OR
                                                                                                     (2)
         ŷ = 41,87 (if using calculator)                            value of y
         ŷ = 42                                                     answer as integer
                                                                                                     (2)

         OR
                21
         yˆ =                                                         answer
                50
                                                                                                     (2)

2.3.1    standard deviation =13,88                                    answer
                                                                                                     (2)
2.3.2    x = 50,67 − 45,67                                           50,67 − 45,67
           = 5%                    Answer only: full marks           answer
                                                                                                    (2)
                                                                                                   [10]

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Mathematics P2/Wiskunde V2                         5                                                    DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 3
                                                     y
                                                          K

                                                          B
                                                          1

                                                                                                                    x
                                        E(–2; 0)      0

  A

                                                                             C(2 ; –3)

                                        D(–2 ; –5)

3.1.1 Midpoint of EC:
         − 2 + 2 0 + (−3)        −3
      =         ;          = 0 ;                                              x value  y value
         2           2           2 
                                                                                                               (2)
3.1.2         − 3 − (−5)    − 5 − (−3)
        mDC =            OR                                                       substitution
               2 − (−2)      −2−2
              2 1                   Answer only: full marks
             = =                                                                  answer
              4 2                                                                                              (2)
3.1.3            1                                                                          1
         mAB =             [AB || DC]                                             mAB =
                 2                                                                          2
            1                                      1
        y=    x+c                         y − y1 =   ( x − x1 )
            2                                      2
            1                                      1
        0 = (−2) + c         OR             y − 0 = ( x − (−2))                   substitution of ( − 2 ;0)
            2                                      2
        c =1
              1                                                                  equation
        ∴ y = x +1
              2                                                                                                (3)
3.1.4                  1                                                                   1
        tan α = mAB =                                                             tan α =
                       2                                                                   2
            α = 26,57°                                                            value of α
            θ = 90° + 26,57°       [ext ∠of Δ]
              =116,57°                                                            value of θ
                                                                                                               (3)

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Mathematics P2/Wiskunde V2                         6                                                                        DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

3.2
        B(0 ; 1)                                                                                      coordinates of B
                1 − (−3)                               (−3) − 1
        mBC =                           OR mBC =
                  0− 2                                  2 −0
             = −2                               = −2                                                  mBC = − 2
                       1
             mAB × mBC = × −2
                       2                                                                              product of gradients = –1
                     = −1
             ∴ AB ⊥ BC
                                                                                                                                   (3)
3.3.1   AB̂C = 90°
        ∴ EC is diameter [converse : ∠ in semi circle]
                                   3
        ∴ centre of circle =  0 ; −                                                                 answer
                                   2
                                                                                                                                   (1)
                                    2
3.3.2
        (x − 0)2 +  y + 3  = r 2                                                                  substitution of centre
                              2
                                        2                                              2

        (− 2 − 0)2 +  0 + 3  = r 2 OR              (2 − 0)2 +  − 3 −  − 3   = r 2       correct substitution of
                               2                                          2                      E(–1 ; 0), B(0 ; 1) or
                         −3 
                                            2                                                          C(2 ; –3) to calculate
        OR (0 − 0 ) + 1 − 
                       2               2
                                = r                                                                 r 2 or r
                         2 
                  EC            (−2 − 2) 2 + (0 − (−3))2
        OR r =       =
                   2                       2
                      3
        OR r = 1 − (− )                                                                               value of r 2 or r
                      2
               25           5
        ∴ r2 =       or r =
                4           2
                           2
         2   3   25
        x +y+  =                                                                                    equation
             2   4
                                                                                                                                   (4)
                                                                                                                                  [18]

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Mathematics P2/Wiskunde V2                         7                                                   DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 4

                                                      y
                                                               L

                                        T(–2 ; b)

                                                                M(2 ; 1)
                                                                                                                    x

                                                               K

                P

4.1      ( x − 2) 2 + ( y −1) 2 = 25                                             equation of the circle
                                                                                substitution of point T
        (−2 − 2) 2 + (b − 1) 2 = 25
                    (b − 1) 2 = 9   OF                                          simplification
                        b −1 = ± 3
            ∴ b = 4 or b ≠ − 2                ∴                                  answer
                                                                                                              (4)
4.2.1 K(2 ; 1 – 5)
      ∴ K(2 ; –4)                         Answer only: full marks                x value  y value
                                                                                                              (2)
4.2.2            4 −1       3
        mMT =           =−                                                       mMT
               −2−2         4
               4                                                                           4
        m PL =        [radius ⊥ tangent]                                         mPL =
               3                                                                           3
            4
        y= x+ c
            3
            4
        4 = (−2) + c                                                             substitution of mPL
            3                                                                     and the point T
            20
        c=
             3
            4    20
        y= x+
            3     3                                                              equation
                                                                                                              (4)

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Mathematics P2/Wiskunde V2                         8                                                    DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

        OR
                  4 −1       3
        mMT =            =−                                                       mMT
                −2−2         4
               4                                                                            4
        m PL =         [radius ⊥ tangent]                                         mPL =
               3                                                                            3
                 4
        y − y1 = ( x − x1)
                 3
                4
        y − 4 = ( x + 2)
                3                                                                 substitution of mPL and
            4     20                                                                the point T
        y= x+
            3      3
                                                                                  equation
        OR                                                                                                     (4)

        P(–11 ; –8)
                  4 − (−8)                                                        coordinates of P
        mPL =
                − 2 − (−11)
                4
              =
                3                                                                           4
            4                                                                     mPL =
        y= x+ c                                                                             3
            3
               4
        − 8 = (−11) + c
               3
              20                                                                  substitution of mPL and
          c=
               3                                                                    the point P or T
            4     20
        y= x+
            3      3
                                                                                  equation
                                                                                                               (4)
4.2.3          4       20 28                                                              28
        yL =     (2) +     =                                                      yL =
               3        3     3                                                           3
               28                          28          40
        L 2 ;      and K (2 ; − 4 ) : LK =    − (−4) =                          length of LK
                3                          3           3
        Coordinates of P:
        x+2          1          y−4                              L(2 ;   )
               = −4       and        = −6
          2          2           2

        ∴ x = −11               y = −8                                            xP  y P
        ∴ P( −11; − 8)
        ⊥ height (PH) = 2 − (−11) = 13                                            length of ⊥ height
                       1                                           K(2 ; –4)
        Area ∆ PKL = (LK)(PH)
                       2
                       1  40       P(–11 ; –8)
                     =  (13)                                    H(2 ; –8)       substitution into
                       2 3 
                                                                                   the area formula
                       260
                     =        OR 86,67 square units                               answer
                         3
                                                                                                               (7)

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Mathematics P2/Wiskunde V2                         9                                                     DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

4.2.3 OR
              4       20 28                                                                28
        yL =    (2) +     =                                                        yL =
              3        3     3                                                             3
              28                          28          40
       L 2 ;      and K (2 ; − 4 ) : LK =    − (−4) =                            length of LK
               3                          3           3
       Coordinates of P:
                                                                 L(2;      )
       x+2          1          y−4
              = −4       and        = −6
         2          2           2

       ∴ x = −11                    y = −8
                                                                                   xP  y P
       ∴ P( −11; − 8)

                                                                    K(2 ; –4)
        PK 2 = (− 11 − 2 ) + (− 8 − (−4) )
                          2                  2

       PK = 185 units                                   θ
                                      P(–11 ; –8)
               − 8 − (−4) 4                                        H(2 ; –8)
       mPK =             =
                − 11 − 2    13
                4
       tan θ =       ∴ θ = 17,1027...°
               13
       ∴ PK̂L = 90° + 17,1027...° = 107,1°                                         PK̂L

                    1
       Area ∆ PKL = (PK)(LK).sin PK̂L
                    2
                  =
                    1
                    2
                          (     )
                             40 
                        185  sin 107,10°                                         substitution into
                             3                                                    the area rule
                  = 86,67 square units                                             answer
                                                                                                                (7)
4.3    The centres of the two circles lie on the same vertical line                correct method
       x = 2. and the sum of the radii = 10                                        sum of radii = 10
       n −1 = 10                   1 − n = 10
                         or                                                        n =11  n = − 9
          n =11                        n = −9
                                                 Answer only: full marks                                        (4)
                                                                                                               [21]

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Mathematics P2/Wiskunde V2                         10                                                  DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 5

5.1.1   sin 191°
        = − sin 11°                                                                       − sin 11°
                                                                                                                    (1)
5.1.2   cos 22°
        = cos(2 × 11°)
        = 1 − 2sin 2 11°                                                                  answer
                                                                                                                    (1)
5.2     cos( x − 180°) +      2 sin ( x + 45°)
        = − cos x + 2 (sin x cos 45° + cos x sin 45°)                                     − cos x  expansion
                               1             1 
        = − cos x + 2  sin x     + cos x                                      special angle ratios
                                2             2  
        = − cos x + sin x + cos x                                                        simplification of last
                                                                                          2 terms
        = sin x                                                                          answer
                                                                                                                    (5)
        OR

        cos( x − 180°) +      2 sin ( x + 45°)
        = − cos x + 2 (sin x cos 45° + cos x sin 45°)                                     − cos x  expansion
                              2           2 
        = − cos x + 2  sin x     + cos x                                           special angle ratios
                                2          2 
                                            
        = − cos x + sin x + cos x                                                        simplification of last
                                                                                          2 terms
        = sin x                                                                          answer
                                                                                                                    (5)
5.3     sin P + sin Q = sin P + cos P                                                     sin Q = cos P
                                      2

        (sin P + cos P )2 =  7                                                        squaring
                             5
                                                 49                                       expansion
        sin 2 P + 2 sin P cos P + cos 2 P =
                                                 25
                     49
         2 sin P cos P =−1                                                                sin 2 P + cos 2 P = 1
                     25
                  49 25 
        sin 2P =    −   
                  25 25 
                  24                                                                     answer
               =
                  25                                                                                                (5)
                                                                                                                   [12]

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Mathematics P2/Wiskunde V2                         11                                                  DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 6
6.1    cos( x − 30°) = 2 sin x
       cos x cos 30° + sin x sin 30° = 2 sin x                                               expansion
               3          1
                 cos x + sin x = 2 sin x                                                     special ∠ s
              2           2
               3          3
                 cos x = sin x
              2           2                                                                  simplification
                       3
             tan x =                                                                         equation in tan
                      3
             x = 30° + k .180° ; k ∈ Z                                                       30°
                                                                                             k.180°; k ∈ Z
            OR                                                                              OR
            x = 30° + k .360° or x = 210° + k .360° ; k ∈ Z                                  30° and 210°
                                                                                             k .360° ; k ∈ Z
                                                                                                             (6)

                                                   y

                                                                                      g
                                                        D

           –180°                                                                                   x
                                        B         0°                         A            180°
                                                                                  f
                   C

6.2.1(a)     A(120° ; 0 )                                                                    answer
                                                                                                                   (1)
6.2.1(b)     C(− 150° ; − 1)                                                                 x value  y value
                                                                                                              (2)
6.2.2(a)     x ∈ (−90° ; 30°) OR − 90° < x < 30°                                             endpoints
                                                                                             correct interval
                                                                                                              (2)
6.2.2(b)     x ∈ (− 160° ; 20°) OR − 160° < x < 20°                                         endpoints
                                                                                            correct interval
                                                                                                              (2)
6.2.3       y = 2 2 sin x +3
            Range of y = 2 sin x : y ∈ [− 2 ; 2] OR − 2 ≤ y ≤ 2
            Range of y = 2 sin x + 3 : y ∈ [1 ; 5] OR 1 ≤ y ≤ 5                              1 5
            Range: y = 2 2 sin x + 3 : y ∈ [2 ; 32] OR 2 ≤ y ≤ 32                            2  32
                                                                                             correct interval
                                                    Answer only: full marks                                  (5)
                                                                                                            [18]

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Mathematics P2/Wiskunde V2                         12                                                  DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 7

                                                                                            E

                 A

                       θ
                                                                                           D

                 B                               2θ         60°      x+2
                                x
                                                 C
                                                                                           H

                 F

                                                     G
7.1.1             x
           sin θ =                                                               trig ratio
                AC           OR
                  x
           AC =                                                                  simplification
                sin θ                                                                                     (2)
7.1.2               x +2
           cos 60° =                                                             trig ratio
                     CE                OR
                      x+2
               CE =
                    cos 60°                                                      making CE the
                    x+2                                                           subject
                  =       = 2( x + 2 )
                       1
                       2                                                                                  (2)
7.2                    1                                                         use area rule
          Area ∆ ACE = AC.EC. sin AĈE                                            correctly
                       2
                        1 x                                                    substitution of
                     =            (2(x + 2 ))sin 2θ                               x
                        2  sin θ                                                      (2( x + 2))
                        x( x + 2) × 2 sin θ cos θ                                 sin θ
                     =                                                           substitution of
                                   sin θ                                          sin 2θ
                     = 2 x(x + 2 ) cos θ                                                                  (3)

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Mathematics P2/Wiskunde V2                         13                                                  DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

7.3       EC = 2(12 + 2) = 28                                                    EC
          AE 2 = AC 2 + EC 2 − 2(AC)(EC )cosAĈE                                 use cosine rule
                             2
                                                                                  correctly
                 12                 12 
                                               (28) cos110°
                                2
               =          + 28 − 2                                            substitution
                 sin 55°            sin 55° 
            AE = 35,77 m                                                         answer
                                                                                                          (4)
                                                                                                         [11]

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Mathematics P2/Wiskunde V2                         14                                                  DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 8

8.1

                                                                       P

                                                N                                   M
                                                     2
                                                 1

                                                         O

                            L
                                                                                    S

                                                               31°

                                                         T

8.1.1(a)   MÔS = 62° [∠ at centre = 2× ∠ at circumf/middelpts∠ =2omtreks∠]                 SR
                                                                                                                   (2)
8.1.1(b)   L̂ = 31°        [equal chords; equal ∠s /= koorde; = ∠e]                         SR
                                                                                                                   (2)
8.1.2      LN = NP and LO = OM                                                              LO = OM
                  1
           ∴ ON = 2 PM       [ midpoint theorem/middelpuntstelling]                         SR
                  1
           ∴ ON = 2 MS       [PM = MS]                                                     S
                                                                                                                   (4)
           OR
           N̂1 = 90° [line from centre to midpt chord/lyn v midpt na midpt kd]             SR
            P̂ = 90°      [∠ in semi-circle/∠ in halfsirkel]
           L̂ is common/gemeen
           ∴ ∆NLO ||| ∆ PLM (∠∠∠)                                                           S/R
               NL NO 1
                   =    =                                                                   S
               PL PM 2
                     1
           ∴ ON = 2 PM
                     1
           ∴ ON = 2 MS        [PM = MS]                                                    S
                                                                                                                   (4)

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Mathematics P2/Wiskunde V2                         15                                                  DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

8.2
                                                     B
                                            2y

                                        K
                                                 L

                           3y

                      A          N          M                                                     C
                                10x                          23x

8.2.1      AN AK
             =   [line || one side of ∆ OR prop theorem; KN ||BM/                          R
           AM AB
                  lyn || sy van ∆ OR eweredigheidst; KN||BM]

           AN 3 y 3
             =   =
           AM 5 y 5                                                                        S
                                                                                                                    (2)
8.2.2      AM 10 x
              =       [given]
           MC 23 x
           AM = 5 y =10 x ∴ y = 2 x                                                         S
           LC MC
             =        [line || one side of ∆ OR prop theorem; KN ||LM/                      R
           KL NM
                       lyn || sy van ∆ OR eweredigheidst; KN||BM]

                    23 x 23 x 23                                                            S
                =       =    =
                    2y    4x   4                                                                                    (3)

           OR

           AM 10 x
              =      [given]
           MC 23 x
           AN 3 y 6 x
              =    =                                                                        S
           MN 2 y 4 x
           LC MC
             =       [line || one side of ∆ OR prop theorem; KN ||LM/                       R
           KL NM
                      lyn || sy van ∆ OR eweredigheidst; KN||BM]

                    23 x 23 x 23                                                            S
                =       =    =
                    2y    4x   4                                                                                    (3)
                                                                                                                   [13]

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Mathematics P2/Wiskunde V2                         16                                                  DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 9

          K
                   x
                                            M
                                    1       2
                                                            N
                                                        1   2

                                        1
                                                                                                 A
                                                2
                                                    3
                                        B

9.1      B̂1 = x            [ ∠' s opp = sides/∠e teenoor = sye]                    S
         M̂ 2 = 2 x          [ext ∠ of ∆] OR M̂1 = 180° − 2 x [∠s of ∆]             S R
         BM = MN             [ 2 tans from a common point/raaklyne vanuit
                              dieselfde punt]                                       S R
                180° − 2 x
         N̂ 1 =               = 90° − x [ ∠' s opp = sides/∠e teenoor = sye]        answer
                      2                                                                                     (6)
         OR
         NM = BM [ 2 tans from a common point/raaklyne vanuit                       S R
                              dieselfde punt]
          B̂ 2 = N̂ 1 [ ∠' s opp = sides/∠e teenoor = sye]                          S R
          B̂1 = x [ ∠' s opp = sides/∠e teenoor = sye]                              S
         In ∆ KBN:
          x + x + B̂ 2 + N̂ 1 =180° [sum of ∠' s of ∆]
         2 x + 2 N̂ 1 =180°
           x + N̂ 1 = 90°
                N̂ 1 = 90° − x                                                      answer
                                                                                                            (6)
9.2       MB̂A = B̂ 2 + B̂ 3 = 90° [tangent⊥diameter/raaklyn⊥middellyn]             S  R
              B̂3 = 90° − B̂2
                   = 90° − (90° − x ) = x                                           S
            B̂ 3 = K̂ = x                                                           S
         ∴ AB is a tangent/raaklyn converse tan-chord theorem/                      R
                                   omgekeerde raakl koordst]]                                               (5)

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Mathematics P2/Wiskunde V2                         17                                                  DBE/2019
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         OR
         B̂ 2 = N̂ 1
         B̂1 + B̂2 = x + (90° − x) = 90°
                                                                                    S
         ∴ KN is diameter/middellyn [converse ∠ in semi-circle/
                                           omgekeerde ∠ in halfsirkel]              R
         MB̂A = B̂ 2 + B̂ 3 = 90°        [tangent ⊥ diameter]
         ∴ AB is a tangent/raaklyn converse tan-chord theorem/                      S R
                                       omgekeerde raakl koordst]]                   R

                                                                                                            (5)
                                                                                                           [11]

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Mathematics P2/Wiskunde V2                         18                                                  DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 10

10.1

                                   A
                                                                               D

                       M                      N                          
                                                                     E                  F

                   
               B                                       C

10.1       Constr: Let M and N lie on AB and AC respectively such that                   Constr / Konstr
                    AM = DE and AN = DF. Draw MN.
           Konst: Merk M en N op AB en AC onderskeidelik af sodanig dat
                    AM = DE en AN = DF. Verbind MN.
           Proof:
           In ∆ AMN and ∆ DEF
           AM = DE [Constr]
           AN = DF [Constr]
             Â = D̂ [Given]                                                              Δ AMN ≡ Δ DEF
           ∴ Δ AMN ≡ Δ DEF (SAS)
                                                                                            SAS
           ∴AM̂N = Ê = B̂
           MN || BC [corresp ∠' s are equal/ooreenkomstige ∠e =]                         MN || BC and R
             AB AC
                  =        [line || one side of ∆ OR prop theorem; MN ||BC]                  AB AC
             AM AN                                                                            =   R
                                                                                             AM AN
             AB AC
           ∴      =        [AM = DE and AN = DF]
             DE DF

                                                                                                                   (6)

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Mathematics P2/Wiskunde V2                         19                                                         DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

10.2

                                                        D

                                                                      4       G
                                                                          3
                                                                              2 1

                                                                2 1                  2 1
                               A                      O           F                     B          C

                                                       E

10.2.1(a)    DÔB = 90°
             DĜF = Ĝ 3 + Ĝ 4 = 90°          [∠ in semi-circle/∠ in halfsirkel]                 S R
            DÔB + DĜF = 180°
            ∴ DGFO is a cyclic quad.           [converse: opp ∠s of cyclic quad/                 R
                                             omgekeerde teenoorst ∠e v koordevh]
                                                     OR
                                     ∠s of quad = 180°/∠e van koordevh = 180°]                                    (3)
            OR
            EÔB = 90°
             DĜF = Ĝ 3 + Ĝ 4 = 90°          [∠ in semi-circle/∠ in halfsirkel]
                                                                                                  S R
            EÔB = DĜF
            ∴ DGFO is a cyclic quad.      . [converse: ext ∠ = opp int ∠/                        R
                                       omgekeerde buite∠ = teenoorst ∠]
                                                OR
                      ext∠ of quad = opp int ∠/buite∠ v vh = teenoorst ∠]                                         (3)

10.2.1(b)          F̂1 = D̂                                                                      SR
                                        [ext ∠of cyclic quad/buite∠ v koordevh]
                                                                                                 SR
            Ĝ 1 + Ĝ 2 = D̂            [tan-chord theorem/raakl koordst]
               ∴ F̂1 = Ĝ 1 + Ĝ 2
                                                                                                 R
              ∴ GC = CF                 [ sides opp equal ∠s/sye teenoor = ∠e]                   (5)

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Mathematics P2/Wiskunde V2                         20                                            DBE/2019
                          SC/SS/NSC/NSS – Marking Guidelines/Nasienriglyne

10.2.2(a) AB = DE = 14                        [diameters/middellyne]              S
          ∴ OB = 7 units                                                          S
          ∴ BC = OC – OB = 11 – 7                Answer only: full marks
               = 4 units                                                          S
                                                                                                   (3)
10.2.2(b)   In Δ CGB and Δ CAG
            Ĝ 1 = Â = x        [tan-chord theorem/raakl koordst]                 S/R
            Ĉ = Ĉ              [common]
            ∆CGB ||| ∆CAG        [∠, ∠, ∠ ]                                       S
             CG CB
                   =                                                              S
             CA CG
             CG      4
                   =
              18 CG                                                                CA = 18
            CG 2 = 72
              CG = 72 or 6 2 or 8,49 units                                        answer
                                                                                                   (5)
10.2.2(c) OF = OC – FC
              = 11 – 72                                                            OF
                   OF
          tan E =                                                                  trig ratio
                   OE
                   11 − 72
                 =         = 0,36
                       7                                                           substitution
              Ê = 19,76°                                                          answer
                                                                                                   (4)
            OR
            OF = OC – FC
               = 11 – 72                                                           OF
            FE = OE 2 + OF 2
               2

                 = 7 2 + (11 − 72 ) 2
             FE = 7,437.. = 7,44
                    OE                                    OF
            cos E =                      OR        sin E =                         trig ratio
                     FE                                   FE
                       7                                 11 − 72
                  =        = 0,94                      =         = 0,338           substitution
                    7,44                                   7,44
                Ê = 19,76°                          Ê = 19,76°                   answer
                                                                                                   (4)
                                                                                                  [26]

                                                                             TOTAL/TOTAAL: 150

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