# Mathematics Grade 12 Paper 2 NSC June 2025 Memo (Afrikaans)

> Grade 12 Mathematics Paper 2 memorandum from the NSC June Exam 2025 (Afrikaans), 23 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/mathematics-p2-may-june-2025-mg-afr-eng/65a3255a-9416-4208-a023-419ec400e4fc
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Language: Afrikaans
- Document type: Memorandum
- Year: 2025
- Exam period: NSC June Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 23

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## About this document

Mathematics · Grade 12 · NSC June Exam · 2025 · Afrikaans. Memorandum, 23 pages. Read online or download the PDF.

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### Page 1

SENIOR CERTIFICATE EXAMINATIONS/
         NATIONAL SENIOR CERTIFICATE EXAMINATIONS

                               MATHEMATICS P2/WISKUNDE V2

                                MAY/JUNE 2025/MEI/JUNIE 2025

                         MARKING GUIDELINES/NASIENRIGLYNE

                                                                             MARKS/PUNTE: 150

                            These marking guidelines consist of 23 pages.
                             Hierdie nasienriglyne bestaan uit 23 bladsye.

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### Page 2

Mathematics P2/Wiskunde V2                          2                         DBE/May/June/Mei/Junie 2025
                              SC/NSC/SS/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 guidelines. 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 nasienriglyne toegepas. Hou op nasien
   by die tweede berekeningsfout.
• Aanvaar van antwoorde/waardes 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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### Page 3

Mathematics P2/Wiskunde V2                          3                         DBE/May/June/Mei/Junie 2025
                              SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 1

    134 215 325 326 362 429 515 531 598 610 624 728 923 1 034                                     1 200

           8554                                                8554
 1.1    x=
            15
          = 570,27                                             answer
                                                                                                     (2)
 1.2    σ = 291, 03                                            291,03
                                                                                                     (1)
 1.3   (279,24 ; 861,3)                                        (x − σ ; x + σ )
       ∴10 premiums                                            answer
                                                                                                     (2)
              118            k + 100                              118
        1791×      +  6763 ×                               1791×
 1.4          100              100                                100
                                          = 686, 44
                      15                                               k + 100
                                                               6763 ×
                                                                         100
                                                                sum of new premiums
                                                                                   = 686,44
               k + 100                                                    15
        6763 ×          = 8183, 22
                 100
        k + 100
                = 1, 209...
          100
        k + 100 = 120,999...
        k = 21%                                                answer
                                                                                                     (4)
                                                                                                     [9]

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### Page 4

Mathematics P2/Wiskunde V2                                           4                         DBE/May/June/Mei/Junie 2025
                                               SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 2

     Number of
    items/Aantal                          10          3    20       14        17        9        12     18        15        19
       items (x)
        Time
  (in minutes)/Tyd                        5           5     9        7        6         6        8      11        10        12
    in minute) (y)

 2.1
                                               Scatter plot / Spreidiagram
                                 14
                                                                                                       3 points correct
                                 12

           Time (in minutes) /
                                 10                                                                    6 points correct
                                 8

             Tyd (in minute)
                                 6                                                                     all points
                                 4                                                                      correct
                                 2
                                 0
                                      0           5        10            15        20       25
                                                  Number of items / Aantal items
                                                                                                                         (3)
 2.2   a = 3,079…                                                                                      a = 3,08
       b = 0,351…                                                                                      b = 0,35
       =yˆ 3, 08 + 0,35 x                                                                              equation
                                                                                                                         (3)
 2.3    r = 0,74                                                                                       0,74
                                                                                                                     (1)
 2.4    y = 3,08 + 0,35(13)                                                                            substitute x=13
        y = 7,63                                                                                       answer
                                                                                                                      (2)
        OR

        y = 7,65 (calculator)                                                                          7,65
                                                                                                                         (2)
 2.5    It does not make sense to pack 0 items in 3,08 minutes./                                       answer
        Dit maak nie sin dat 0 items in 3,08 minute gepak kan word nie.
                                                                                                                         (1)
                                                                                                                        [10]

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### Page 5

Mathematics P2/Wiskunde V2                           5                         DBE/May/June/Mei/Junie 2025
                               SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 3

                                                                    y

                                                                    T

                                  V

         S(m ; 1)
                                                                                                      x
             W                                    R                 O                     R/

 3.1      2 x − y + 10 = 0
          2 x − 0 + 10 = 0                                                       y=0
          x = −5                                                                 x = –5
          R(–5 ; 0)
                                                                                                             (2)
 3.2      R(–5 ; 0)
          T(0 ; 10)                                                              T(0 ; 10)

          (RT) 2 = (−5 − 0) 2 + (0 − 10) 2 OR (RT )2 = 5 2 + 10 2 (Pythag)       subst of R & T into
                                                                                  distance formula or
                                                                                  Pythagoras
          RT = 125 units                 OR/OF          RT = 5 5 units           answer
                                                                                                             (3)
 3.3           2
          2RT = 5SR     2

          2(125) = 5[(m − (−5)) 2 + (1 − 0) 2 ]                                  length of 2RT2
          5[(m + 5) 2 + (1) 2 ] = 250                                            length of 5SR2

          (m + 5) 2 + 1 = 50
          m 2 + 10m − 24 = 0              OR/OF          (m + 5) 2 = 49          standard form or
          (m − 2)(m + 12) = 0                           m + 5 = ±7                isolating square
                                                           m = −5 ± 7
          m = 2 or m = – 12                           m = 2 or m = – 12
          N/A                                         N/A
          ∴m = – 12                                   ∴m = – 12                  negative answer
                                                                                                             (4)

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### Page 6

Mathematics P2/Wiskunde V2                          6                         DBE/May/June/Mei/Junie 2025
                              SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

                                                                   y

                                                                   T

                                 V

         S(m ; 1)
                                                                                                   x
             W                                R                    O                   R/

                    1 − 10
 3.4      mST =                                                               substitution of S & T
                   − 12 − 0                                                    into gradient formula
                   3                                                          mST
          mST =
                   4
                        4                                                                    1
          ∴ mVR = −                                                           mVR = −
                        3                                                                   mST
                 4                                        4
          y = − x+c              OR/OF          y − y1 = − ( x − x1 )
                 3                                        3
                 4                                    4
          0 = − (−5) + c                    y − 0 = − ( x − (−5))             substitution of R
                 3                                    3
                 20                                   4
          c=−                                   y = − ( x + 5)
                  3                                   3
                  4      20                           4    20                 equation
          y =− x−                               y =− x−
                  3       3                           3     3                                           (5)
 3.5                     4  20
          VR: y = − x −
                         3   3
                      3
          ST: y = x + 10
                      4
          3               4  20
             x + 10 = − x −                                                   equating VR and ST
          4               3   3
          9 x + 120 = −16 x − 80

          25 x = −200                                                         simplification leading
                                                                               to x = −8
          x = −8

          y=4

          ∴ V(−8 ; 4)
                                                                                                        (2)
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### Page 7

Mathematics P2/Wiskunde V2                             7                         DBE/May/June/Mei/Junie 2025
                                 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
                                                              y

                                                              T

                           V

 S(m ; 1)
                                                                                             x
                                         R                    O                   R/

 3.6        R ′ (5 ; 0 )

            VR = [−8 − (−5)]2 + (4 − 0) 2 = 5                                    length of VR
                                                                                 length of VT
            VT = (−8 − 0) 2 + (4 − 10) 2 =10

                          1
            Area of RVTR ′ =(VR )(VT ) + 1 (RR ′)(OT )
                          2              2
                          1          1                                           area ∆VRT
                       =    (5)(10) + (10 )(10 )
                          2          2                                           area ∆ RTR ′
                        = 25 + 50
                               = 75 units 2
                                                                                 answer
          OR/OF                                                                                            (5)
                  3
         ST: y = x + 10
                  4
            3
         0
         =    x + 10
            4
              40         1
         x= −     or −13                                                         x-intercept of ST
               3         3
            40       
         W−      ; 0
            3        
                    40  55      1
         WR' = 5 −  −  =   = 18                                                length of WR ′
                    3  3        3
         Area of RVTR' = Area of ∆TWR' − Area of ∆WVR
                                  1    40       1     40                     area ∆ TWR ′
                            =       5 +  (10 ) −  −5 +  ( 4 )
                                  2     3       2      3                     area ∆ WVR
                               = 75 units 2                                      answer
                                                                                                          (5)
                                                                                                         [21]
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### Page 8

Mathematics P2/Wiskunde V2                           8                         DBE/May/June/Mei/Junie 2025
                                     SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
                                                                 y
      QUESTION/VRAAG 4

                                                E(–1 ; 3)

                                              D              O                M                                                 x
S(–11 ; –2)

                                                     T

               C(–7 ; p)

       4.1     CÊM = 90°                                                                  answer
                                                                                                                          (1)
                       0−3
       4.2     mME =
                     3 − ( −1)
                       3
               mME = −                                                                                   3
                       4                                                                   mME = −
                                                                                                         4
                      4
              ∴ mED =                                                                      mED
                      3
                  4                                               4
              3 = ( −1) + c
                  3
                                               OR/OF             3
                                                              y −=
                                                                  3
                                                                    ( x − ( −1) )          substitution of E(–1 ; 3)
                  4    13                                       4     13
              y
              =     x+                                        y
                                                              =   x+                       equation
                  3     3                                       3      3
                                                                                                                          (4)
              OR/OF
                                         2
                                  15 
                         ( 5) + 
                            2
              DM
              =                                            [Pythagoras]
                                  4
                     25
               DM =      or 6,25 units
                      4
                   13                                                                    coordinates of D
              ∴ D  − ;0 
                   4 
                         3−0
              mED =
                           13 
                     −1 −  − 
                           4
                       4                                                                   mED
              ∴ mED =
                       3
                  4                                               4
              3 = ( −1) + c
                  3
                                               OR/OF             3
                                                              y −=
                                                                  3
                                                                    ( x − ( −1) )          substitution of E(–1 ; 3)

                  4     13                                      4     13                   equation
              y
              =     x+                                        y
                                                              =   x+
                  3      3                                      3      3                                                  (4)
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### Page 9

Mathematics P2/Wiskunde V2                          9                         DBE/May/June/Mei/Junie 2025
                              SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

             4     13
 4.3    y
        =      x+
             3      3
             4     13
         0= x+
             3      3
                13
         xD = −                                                                  xD
                 4

                     13                                                        xM − xD
         ∴ DM = 3 −  − 
                     4
                25
          DM =      or 6,25 units                                                answer
                 4                                                                                              (3)

         OR/OF

         EM = 5 units                                                            EM = 5 units
              15
         ED =     units
               4

                               2
                         15
               ( 5) +                                                        substitution of EM & ED
                      2
        DM
        =                                         [Pythagoras]
                        4
              25                                                                 answer
        DM =       or 6,25 units
               4                                                                                                (3)
                 4    13
 4.4    EC: =
            y      x+
                 3      3
                 4         13
            p = ( −7 ) +                                                         substitution of C(–7 ; p)
                 3          3                                                     into equation of EC
            p = −5                                                                                       (1)

         OR/OF

                4
         mEC =
                3
           p −3      4
                   =                                                             substitution of C(–7 ; p)
         −7 − ( −1) 3                                                             into gradient of EC
                  4
         p −3=      ( −6 )
                  3
         p = −5
                                                                                                                (1)

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### Page 10

Mathematics P2/Wiskunde V2                         10                         DBE/May/June/Mei/Junie 2025
                              SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

 4.5     M → E: (x ; y) → (x – 4 ; y + 3)           [translation]                method: translation
         C → S: (–7 ; –5) → (–7 – 4 ; –5 + 3)
         ∴S(–11 ; –2)                                                            xS = −11  yS = −2
                                                                                                                (3)
         OR/OF

         M → C: (x ; y) →(x – 10 ; y – 5)           [translation]                method: translation
         E → S: (–1 ; 3) → (–1 – 10 ; 3 –5)
         ∴S(–11 ; –2)                                                            xS = −11  yS = −2
                                                                                                                (3)
         OR/OF

         E(–1 ; 3) and C(–7 ; –5)
          − 1 + (− 7 ) 3 + (− 5) 
                      ;                           [Midpoint of EC]             method: midpoint
               2           2     
         = (− 4 ; − 1)

         S(x ; y) and M(3 ; 0)
          xS + 3                      yS + 0
                 = −4                        = −1
            2                           2
         xS = −11                     yS = −2                                    xS = −11  yS = −2
         ∴S(–11 ; –2)
                                                                                                                (3)
 4.6     rNEW = 5 + 7
         rNEW = 12                                                               rNEW = 12

         MS =    (3 − (− 11))2 + (0 − (− 2))2
         MS = 200 or 10 2 or 14,14 units                                         MS
         14,14 > 12
         ∴ S(− 11 ; − 2 ) lies outside the circle                                conclusion
                                                                                                                (3)

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### Page 11

Mathematics P2/Wiskunde V2                         11                         DBE/May/June/Mei/Junie 2025
                                     SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

                                                           y

                                           E(–1 ; 3)

                                          D            O                M                                              x
S(–11 ; –2)

                                                 T

              C(–7 ; p)

        4.7      Inclination of EM: tan M ˆ = m = −3
                                                 ME
                                                      4
                                     ref. ∠ = 36,87°
                                     Inclination of EM = 143,13°                        inclination of EM
                 ∴ EM̂D = 36,87°                                                        EM̂D
                                        ˆ = m = 5 or 1
                 Inclination of CM: tan M     CM
                                                   10 2
                                     ∴ M̂ = 26,57°                                      inclination of CM
                 ∴ EM̂T = 26,57° + 36,87°
                        = 63,44°                                                        EM̂T

                 But EM = MT         [radii]
                          180° − 63,44°
                 ∴ ET̂M =
                                2

                 ∴ ET̂M = 58,28°                                                        answer
                                                                                                                        (5)
                                                                                                                       [20]

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### Page 12

Mathematics P2/Wiskunde V2                              12                         DBE/May/June/Mei/Junie 2025
                                   SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 5

          y 2 = 13 2 − (− 5)
                               2
 5.1.1                                    [Pythagoras]
         y = –12                                                                             y = –12
                                                                    θ
         sin 2 θ
                    2
            12                                                    13
         = −                                                                               substitution
            13                                               •
                                                         (–5 ; y)
           144
         =                                                                                   answer
           169                                                                                                       (3)

         OR/OF

         sin 2 θ = 1 − cos 2 θ                                                               square identity
                                   2
                        5                                                                  substitution
         sin 2 θ = 1 −  − 
                        13 
                   144
         sin 2 θ =                                                                           answer
                   169                                                                                               (3)
 5.1.2   tan (360° − θ )
         = − tan θ                                                                           − tan θ
               − 12 
         = −        
               −5 
              12                                                                             answer
         =−
               5                                                                                                     (2)
 5.1.3   cos(θ −135°)

         = cos θ cos 135° + sin θ sin 135°                                                   cpd. ∠ expansion

         = cos θ (− cos 45°) + sin θ (sin 45°)                                               reduction

            5     2   12  2          5  1   12  1 
         =  −  −    +  −     OR  −  −    +  −                             substitution
            13  2   13  2       13    2   13  2 

              7 2                                            7
         =−                                            =−                                    answer
               26                                           13 2

                                                                                                                     (4)

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### Page 13

Mathematics P2/Wiskunde V2                          13                         DBE/May/June/Mei/Junie 2025
                               SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

 5.2      2 cos(180° − x )sin (− x )
            1 − 2 cos 2 (90° − x )
                                                                               cos(180° − x) = − cos x
           2(− cos x )(− sin x )                                               sin( − x) = − sin x
         =
              1 − 2 sin 2 x                                                    cos 2 (90° − x) = sin 2 x
             2 sin x cos x
         =
                cos 2 x                                                        1 − 2 sin 2 x = cos 2 x

             sin 2 x                                                           2 sin x cos x = sin 2 x
         =
             cos 2 x
         = tan 2 x                                                             answer
                                                                                                             (6)
 5.3     (tan 92°)(tan 94°)(tan 96°)...(tan 176°)(tan 178°)
           sin 92°  sin 94°  sin 96°   sin 176°  sin 178° 
         =                           ...                             quotient identity
           cos 92°  cos 94°  cos 96°   cos176°  cos178° 

           cos 2°  cos 4°  cos 6°   sin 4°  sin 2°                   co-ratios
         =                              ...                  
           − sin 2°  − sin 4°  − sin 6°   − cos 4°  − cos 2°         reduction

         =1                                                                    answer
                                                                                                             (4)
         OR

         (tan 92°)(tan 94°)(tan 96°)...(tan 176°)(tan 178°)
         = (tan 92°)(tan 94°)(tan 96°)....(− tan 4°)(− tan 2°)                 reduction

           sin 92°  sin 94°  sin 96°   sin 4°  sin 2°                quotient identity
         =                           ... −     −      
           cos 92°  cos 94°  cos 96°   cos 4°  cos 2° 

           cos 2°  cos 4°  cos 6°   sin 4°  sin 2° 
         =                              ... −     −                  co-ratios
           − sin 2°  − sin 4°  − sin 6°   cos 4°  cos 2° 

         =1                                                                    answer
                                                                                                             (4)
                                                                                                            [19]

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### Page 14

Mathematics P2/Wiskunde V2                           14                         DBE/May/June/Mei/Junie 2025
                                SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 6

 6.1      LHS = 2 cos 2 (45° + x )
          = 2 cos 2 ( 45° + x ) + 1 − 1                                              +1–1
          = cos  2 ( 45° + x )  + 1                                              double angle
          = cos ( 90° + 2 x ) + 1                                                    simplification
                   (−
                   =  sin 2 x ) + 1                                                  reduction
                   = 1 − sin 2 x
                   = RHS
                                                                                                                    (4)
            OR/OF

             LHS = 2 cos 2 (45° + x )
           = 2 ( cos ( 45° + x ) )
                                           2

                   = 2(cos 45° cos x − sin 45° sin x )
                                                            2
                                                                                     compound ∠ expansion
                                                     2
                      2             2        
                 = 2    cos x −      sin x                                       subst. special ∠ values
                      2            2         
                    1                          1     
                 = 2 cos 2 x − sin x cos x + sin 2 x                               simplification
                    2                          2     
                       2                         2
                 = cos x − 2 sin x cos x + sin x                                     sin 2 x + cos 2 x = 1
                 = 1 − sin 2 x
                 = RHS                                                                                              (4)
 6.2.1         = sin ( A − B ) − sin ( A + B )
             LHS
                   = sin A cos B − cos A sin B − ( sin A cos B + cos A sin B )       sin A cos B − cos A sin B
                   = sin A cos B − cos A sin B − sin A cos B − cos A sin B           − sin A cos B − cos A sin B
                  = −2cos A sin B
                  = RHS                                                                                             (2)
 6.2.2     sin 4 x − sin10 x
          = sin ( 7 x − 3 x ) − sin ( 7 x + 3 x )                                    4x =7x–3x & 10x =7x+3x
             = −2cos 7 x sin 3 x                                                     answer
                                                                                                                    (2)
 6.2.3       sin 4 x − sin10 x sin
                                =     3x
             −2 cos 7 x sin 3 x sin 3 x
                                =                                                    substitution
             2 cos 7 x sin 3 x + sin 3 x 0
                                        =
             sin 3 x(2 cos 7 x + 1) = 0                                              factorisation
                                                               1
             sin 3 x = 0                  or        cos 7 x = −                      both equations
                                                               2
             3 x = 0°                               7 x 120
                                                       =     °   or   7 x 240
                                                                         =   °
              x = 0°                                 x = 17,14°       x = 34,29°     answer
                                                                        N/A          answer
                                                                                                                    (5)
                                                                                                                   [13]

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### Page 15

Mathematics P2/Wiskunde V2                               15                         DBE/May/June/Mei/Junie 2025
                                    SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 7                                             y

                   g

            –180°                                                0                                       180°          x

               f

 7.1       Range of f: y ∈ [ −1 ; 3]                                            y ∈ [ −1 ; 3]
                                                                                                                      (1)
           OR/OF

            −1 ≤ y ≤ 3                                                          −1 ≤ y ≤ 3
                                                                                                                      (1)
 7.2       Period of g: 180°                                                    180°
                                                                                                                      (1)
 7.3       f increasing: x ∈ (− 180° ; 0°)                                      x ∈ (− 180° ; 0°)
                                                                                                                      (1)
           OR/OF

           –180° < x < 0°                                                       –180° < x < 0°
                                                                                                                      (1)
 7.4.1      g ( x ). f '( x ) < 0

            x ∈ (− 90° ; 0°) ∪ (0° ; 90°)                                       x ∈ ( −90° ; 0° )  x ∈ ( 0° ; 90° )
                                                                                                                      (2)
           OR/OF

           –90° < x < 0° or 0° < x < 90°                                        –90° < x < 0°  0° < x < 90°
                                                                                                           (2)

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### Page 16

Mathematics P2/Wiskunde V2                           16                         DBE/May/June/Mei/Junie 2025
                                SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

                       1
 7.4.2      cos x ≤ −
                       2
            2 cos x + 1 ≤ 0                                        2 cos x + 1 ≤ 0
            x ∈ [− 180° ; − 120°] ∪ [120° ; 180°]                  x ∈ [ −180° ; − 120°]  x ∈ [120° ; 180°]
                                                                                                                   (3)
           OR/OF

            2 cos x + 1 ≤ 0                                        2 cos x + 1 ≤ 0
           –180° ≤ x ≤ –120° or 120° ≤ x ≤ 180°                    –180° ≤ x ≤ –120°  120° ≤ x ≤ 180°
                                                                                                      (3)
 7.5        g (x ) = sin 2 x
            h(x ) = sin 2(x − 45°)
                   = sin (2 x − 90°)                               sin ( 2 x − 90° )
                   = − sin (90° − 2 x )
                   = − cos 2 x                                     equation of h
                                                                                                                   (2)
                                                                                                                  [10]

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### Page 17

Mathematics P2/Wiskunde V2                             17                         DBE/May/June/Mei/Junie 2025
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QUESTION/VRAAG 8
                                                         D

                                                     p

                                                         A

                                  y

                              x
                 B                                                                       C
 8.1                 p
              tan y =                                                            correct trig ratio
                    AB
                    p                                                            answer
             AB =
                  tan y                                                                                             (2)
 8.2         In ∆BAC:

                   ˆ
              sin BAC        ˆ
                        sin ACB
                      =                                                          correct use of sine-rule
                 BC        AB

              sin (180° − 2 x )         sin x                                    substitute BC & AB
                                  =
                     2p                p 
                                       tan y 
                                             

              sin 2 x       tan y 
                 = sin x ×                                                     sin (180° − 2 x ) = sin 2 x
                2p          p 

              2sin x cos x       tan y                                         sin 2 x = 2sin x cos x
                      = sin x ×        
                  2p             p 

                        tan y 
              2cos x =        (2 p)
                        p 

              cos x = tan y                                                                                         (4)

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Mathematics P2/Wiskunde V2                            18                         DBE/May/June/Mei/Junie 2025
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                                                           D

                                                       p

                                                           A

                                     y

                                 x
                     B                                                                        C

 8.2     OR/OF

         In ∆BAC:

                                               ˆ
         BC2 = AB2 + AC2 − 2 ( AB )( AC ) cos BAC                                         correct use of cos-rule
                             2              2
                       p              p          p  p 
         (2 p) =                                                      cos (180° − 2 x )  substitute AB & BC
                 2
                                +             − 2       
                       tan y          tan y      tan y   tan y 
                     2 p2      2 p 2 ( − cos 2 x )
         4 p2 =            −                                                               cos (180° − 2 x ) = − cos 2 x
                   tan 2 y           tan 2 y
         4 p 2 tan 2 y = 2 p 2 (1 + cos 2 x )
             2   1 + 2 cos 2 x − 1                                                        cos
                                                                                           =   2 x 2 cos 2 x − 1
         tan y =
                        2
            2        2
         tan y = cos x
         tan y = cos x
                                                                                                                       (4)
 8.3     cos x = tan y
         tan y = cos 60°                                                                  substitution of 60°
         tan y = 0,5
         y = 26,57°                                                                       answer

                                                                                                                       (2)
                                                                                                                       [8]

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### Page 19

Mathematics P2/Wiskunde V2                         19                         DBE/May/June/Mei/Junie 2025
                              SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 9

9.1                                                D

                                                                       G

                        E                                 H                                F

 9.1.1       HF GF 2
             = =                            [line || to one side of ∆/lyn || een sy van ∆]  S  R
             EH GD 5
                                                    OR
                                [prop theorem; GH||DE/eweredigheidst.; GH || DE]                                   (2)

 9.1.2       EH DG 5
               =  =                         [line || to one side of ∆/lyn || een sy van ∆]
             EF DF 7
                                                    OR
                                [prop theorem; GH||DE/eweredigheidst.; GH || DE]
            EH 5
               =                                                                                S
            21 7
            EH = 15cm                                                                           answer
                                                                                                                   (2)
            OR/OF

             HF 2
               =                          [line || to one side of ∆/lyn || een sy van ∆]
             EF 7
                                                    OR
                                [prop theorem; GH||DE/eweredigheidst.; GH || DE]
            HF 2
               =                                                                                S
            21 7
            HF = 6cm
            EH
             = 21 − 6
            EH = 15cm                                                                           answer
                                                                                                                   (2)
 9.1.3      ∆FGH ||| ∆FDE                     [∠∠∠]                                             S
                                                                                                                   (1)

 9.1.4       GH FH                                      GH FG
               =               [|||∆'s]       OR/OF       =                     [|||∆'s]        S
             DE FE                                      DE FD
             GH 2
               =                                                                                S
             DE 7
                                                                                                                   (2)

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Mathematics P2/Wiskunde V2                         20                         DBE/May/June/Mei/Junie 2025
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                                                                                    K

9.2

                                                                                             T
                                                               L        1
                                                                    2
                                                              4 3

                                                                                2
                                                      2   1
                                        O       1                           1       M
                                            2         N

                                                                        1
                                                                2
                                                                    R
               P

 9.2.1      L̂ 2 = 90°    [∠ in semi-circle/ OR L̂1 = R̂ 1 [tan chord theorem/                    S  R
                           ∠ in halwe sirkel]               raaklyn-koordst.]
            R̂ 2 = 90°    [∠ in semi-circle]    R̂ 1 = P̂ [tan chord theorem]                     S/R
                           ∠ in halwe sirkel]               raaklyn-koordst.]
            ∴ L̂ 2 = R̂ 2                       ∴ L̂1 = P̂
            ∴ LT || PR [alt ∠s =/verw. ∠e = ] ∴ LT || PR [corresp. ∠s =/                          R
                                                               ooreenk. ∠e =]                                          (4)

 9.2.2       L̂1 = R̂ 1                 [tan chord theorem/raaklyn-koordst.]                      S R
             L̂1 = Ô1                  [corresp. ∠s; LT || OM/ooreenk. ∠e; LT ||OM]              S/R
            ∴ R̂ 1 = Ô1                                                        S
            ∴ L, O, R and M are concyclic.
            ∴ LORM is a cyclic quadrilateral [converse ∠s in the same seg/      R
                                           omgekeerde ∠e in dies. sirkel segm]                                         (5)
 9.2.3       O L̂R = M̂ 1         [∠s in the same seg/∠e in dieselfde segment]                    S/R
             2L̂ 3 = M̂ 1        [∠ at centre = 2 ×∠ at circumference/                            S R
                                  midpt. ∠ = 2× omtreks ∠ ]
            ∴ O L̂R = 2L̂ 3
            ∴L̂ 4 = L̂ 3                                                                          S
            ∴ LN bisects OL̂R
                                                                                                                       (4)
                                                                                                                      [20]

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### Page 21

Mathematics P2/Wiskunde V2                         21                         DBE/May/June/Mei/Junie 2025
                              SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 10

10.1
                                        A                                                P

                          M    1                N                              ♠               ●
                                                                        K                            L

                ♠                                     ●
           B                                              C

 10.1       Construction: Draw line MN, where M and N are points on AB and
            AC respectively such that AM = PK and AN = PL.                                    construction

            In ∆ AMN and ∆PKL
            Aˆ = ˆP                         [given]
            AM = PK                         [construction]
            AN = PL                         [construction]
            ∆AMN ≡ ∆PKL                      [s∠s]                                            S/R
            ∴M ˆ =  ˆ
                    K
                 1

            But Bˆ = Kˆ                     [given]
            ∴Mˆ =   ˆ
                    B                                                                         S
                1

            ∴MN || BC                       [corresp ∠s =/ooreenk. ∠e =]                      S/R
              AB    AC
            ∴     =                         [line || one side of ∆/lyn || een sy v ∆]         SR
              AM AN

            But AM = PK and AN = PL

                AB AC
            ∴      =
                PK   PL                                                                                      (6)

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Mathematics P2/Wiskunde V2                            22                         DBE/May/June/Mei/Junie 2025
                                 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

10.2
       A                                   B           F                                   C

                    R

                             O

 D

 10.2.1     In ∆ CFO and ΔCBD
            Cˆ =C
                ˆ           [common]                                                     S
             CF̂O = CB̂D            [corresp ∠s; BD||FO/ooreenk. ∠e; BD||FO]             S/R
             CÔF = CD̂B            [sum of ∠s in ∆/binne ∠e v ∆]
                                                    OR
                                    [corresp ∠s; BD||FO/ ooreenk. ∠e; BD||FO]
                                                                                         S/R OR R
             ∆ CFO |||ΔCBD          [∠∠∠ ]
                                                                                                               (3)
 10.2.2      FO CO
                =                     [||| ∆s]                                           S/R
            BD CD
            FO.CD = CO.BD
            But RD̂O = FĈO          [given]
            ∴ BD = BC                [sides opp equal ∠s/sye teenoor gelyke ∠e ]         S/R

             ∴ FO.CD = CO.BC                                                                                   (2)

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Mathematics P2/Wiskunde V2                            23                         DBE/May/June/Mei/Junie 2025
                                 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne

 10.2.3     RD = 6 units                 [DR = RB]
            RO 3
                =
            RD 4
                    3
            ∴ RO = (6)
                    4
            RO = 4,5 units                                                                 S
            OR ⊥ BD                      [line from centre to midpt of chord/              S/R
                                          midpt. sirkel; midpt koord]
            ∴ DO = 6 2 + 4,5 2            [Pythagoras]
            DO = 7,5 units                                                                 S

             BF DO
               =                          [line || to one side of ∆/lyn || een sy v ∆]     S/R
             BC DC
                                             OR/OF
                                 [prop theorem; BD||FO/eweredigheidst.;BD||FO]

            BC = BD = 12 units [sides opp equal ∠s/sye teenoor gelyke ∠e]  S
              BF 7,5                                                       S
            ∴    =
              12 19,2
                 7,5 × 12
            BF =
                  19,2
                   75
             BF =     units
                   16                                                                                           (6)
                        RO 4,5 3
 10.2.4      tan RD̂O = = =                                                                ratio
                        RD    6 4
                RD̂O = 36,87°                                                                ˆ
                                                                                           RDO
             FĈO = RD̂O = 36,87°             [given]
            ∴ AB̂D = 73,74°                   [ext ∠ of ∆/buite ∠ v. ∆]                    answer
                                                                                                                (3)

            OR/OF

                                                     ˆ
             CD 2 = BC2 + BD 2 − 2 ( BC )( BD ) cos DBC
                             2       2        2
                  ˆ = 12 + 12 − 19, 2
             cos DBC                                                                       substitution into
                         2 (12 )(12 )                                                       cosine-rule
                  ˆ = − 7
             cos DBC
                        25
               ˆ
             DBC = 106,26°                                                                    ˆ
                                                                                           DBC
            ∴ AB̂D = 73,74°              [∠s on a straight line/∠e op ’n reguitlyn]        answer
                                                                                                                (3)
                                                                                                            [20]
                                                                                         TOTAL/TOTAAL: 150
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