# Technical Mathematics Grade 12 P2 2026 Memo (Eng)

> Grade 12 Technical Mathematics Paper 2 memorandum from the NSC November Prediction Exam 2026 (English), 13 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-12-technical-mathematics-2026-nsc-november-prediction-exam-memorandum-paper-2/dbc9742c-a38b-4b69-a882-f688106881aa
- Updated: 2026-10-29
- Subject: Technical Mathematics
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Prediction Exam
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 13

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## About this document

Technical Mathematics · Grade 12 · NSC November Prediction Exam · 2026 · English. Memorandum, 13 pages. Read online or download the PDF.

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## Document text

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### Page 1

hlayiso mock exams
                           Practice Series:
                           Grade 12 NSC Preparation
                           WWW.HLAYISO.COM

                                NATIONAL
                           SENIOR CERTIFICATE

                                    GRADE 12

                        TECHNICAL MATHEMATICS P2
                      PREDICTION EXAMINATION 2026
                            MARKING GUIDELINES

MARKS: 150

                     These marking guidelines consist of 13 pages.

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### Page 2

Technical Mathematics/P2                       2                              Hlayiso/Prediction 2026
                                     NSC – Marking Guidelines

MARKING CODES
 A   Accuracy                        CA     Consistent accuracy
 M   Method                          R      Rounding
 NPR No penalty for rounding         NPU    No penalty for units omitted
 S   Simplification                  SF     Substitution in correct formula
 ST  Statement                       RE     Reason
 S/R Statement with reason           AO     Answer only
NOTE:

    • If a candidate answers a question TWICE, only mark the FIRST attempt.
    • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines where indicated.
    • No penalty for rounding (NPR) for ALL questions, unless stated otherwise. A tolerance range is
      given where rounding of an earlier answer changes the final answer.
    • In QUESTIONS 7, 8 and 9 (Euclidean geometry) a statement without the correct reason earns
      no mark for the reason. Accept any valid alternative proof.

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### Page 3

Technical Mathematics/P2                     3                        Hlayiso/Prediction 2026
                                   NSC – Marking Guidelines

QUESTION 1

                     1−5      −4
 1.1         mAB =          =                                 ✓ SF                        A
                   2 − (−4)    6
                     2                                        ✓ answer                   CA
             mAB = −                                          AO: full marks
                     3
                                                                                          (2)

                       2
 1.2         tan θ = −                                        ✓ tan θ = mAB              CA
                       3
             reference ∠ = 33,69◦
             θ = 180◦ − 33,69◦ = 146,31◦                      ✓ answer                   CA
                                                                                         (2)

                 √
 1.3         AB√= (2 −
                     √(−4)) + (1 − 5)
                            2        2                        ✓ SF                        A
             = 52 = 2 13 units                                ✓ answer in surd form      CA
                                                                                         (2)
                (              )
                −4 + 2 5 + 1
 1.4         M         ;
                   2     2
                                                              ✓ x = −1                     A
             M(−1 ; 3)
                                                              ✓y=3                         A
                                                                                          (2)

 1.5         mAB × m⊥ = −1                                    ✓ m1 × m2 = −1              A
                    3                                         ✓ m⊥ = 23                  CA
             m⊥ =
                    2                                         ✓ substitution of M        CA
                      3
             y − 3 = (x − (−1))                               ✓ equation                 CA
                      2                                                                  (4)
                  3     9
             y = x+       OR y = 1,5x + 4,5
                  2     2
                3     9
 1.6         t = (1) + = 6                                    ✓t =6                      CA
                2     2
                                                                                         (1)

                                                                                         [13]

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### Page 4

Technical Mathematics/P2                           4                        Hlayiso/Prediction 2026
                                         NSC – Marking Guidelines

QUESTION 2

 2.1.1       r2 = (−3)2 + (6)2 = 45                                 ✓ substitution of T        A
             x2 + y2 = 45                                           ✓ equation                CA
                                                                                              (2)

                   √         √
 2.1.2       r=        45 = 3 5 units                               ✓ answer                  CA
                                                                                              (1)

                      6−0
 2.1.3       mOT =           = −2                                   ✓ mOT                      A
                      −3 − 0
                         1                                          ✓ mtangent                CA
             mtangent =     (tangent ⊥ radius)                      ✓ substitution of T       CA
                         2
                      1                                             ✓ equation                CA
             y − 6 = (x − (−3))
                      2                                                                       (4)
                  1     15
             y = x+          OR y = 0,5x + 7,5
                  2     2

 2.1.4       K(0 ; 7,5)                                             ✓ answer                  CA
                                                                                              (1)

              25x2 4y2    100
 2.2.1            +     =                                           ✓ dividing by 100           A
              100 100 100
              x2 y2           x2 y2
                +   = 1   OR    +   =1                              ✓ standard form           CA
              22 52           4 25
                                                                                              (2)

 2.2.2                  y                                           ✓ x-intercepts ±2         CA
                            5                                       ✓ y-intercepts ±5 and
                                                                    elliptical shape          CA
                                                                                              (2)

                                     x
              −2                 2

                            −5

                                                                                              [12]

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### Page 5

Technical Mathematics/P2                         5                          Hlayiso/Prediction 2026
                                       NSC – Marking Guidelines

QUESTION 3

                     180◦
 3.1.1       2,5 ×        = 143,24◦                               ✓ answer                     A
                      π
                                                                                              (1)

                                            1
 3.1.2       cot(143,24◦ − 41,3◦ ) =                              ✓ tan 101,94
                                                                         1
                                                                               ◦              CA
                                       tan 101,94◦
             = −0,21                                              ✓ answer                    CA
                                                                                              (2)

              sin2 41,3◦     0,4357 . . .
 3.1.3                  ◦
                          =                                       ✓ substitution              CA
             cos 143,24     −0,8011 . . .
             = −0,54                                              ✓ answer                    CA
                                                                                              (2)

 3.2.1       k2 + (−5)2 = 62 (Pythagoras)                         ✓ Pythagoras                 A
             k2 =√11                                                      √
             k = 11 (k > 0, R in quadrant IV)                     ✓k=         11              CA
                                                                                              (2)

                       √
 3.2.2       6 sin(θ + )11 tan θ (    )
                      5    √       5                              ✓ sin θ = − 65              CA
             =6 −        + 11 − √
                      6            11                             ✓ tan θ = − √511            CA
             = −5 − 5 = −10
                                                                  ✓ answer                    CA
                                                                                              (3)

 3.3         5 cos x = −2
             cos x = −0,4                                         ✓ cos x = −0,4               A
             reference ∠ = 66,42◦                                 ✓ reference angle           CA
             x = 180◦ − 66,42◦ = 113,58◦                          ✓ 113,58◦                   CA
             or x = 180◦ + 66,42◦ = 246,42◦                       ✓ 246,42◦                   CA
                                                                                              (4)

                                                                                             [14]

QUESTION 4

 4.1.1       − cos x                                              ✓ answer                     A
                                                                                              (1)

 4.1.2       − tan x                                              ✓ answer                     A
                                                                                              (1)

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### Page 6

Technical Mathematics/P2                                      6                         Hlayiso/Prediction 2026
                                                    NSC – Marking Guidelines

             (− cos x)(− tan x)
 4.1.3                          + (cos x)2 (tan x)2                            ✓ − cos x                      A
                  − sin x
                         sin x                                                 ✓ − tan x                      A
                 cos x ·                     2                                 ✓ − sin x                      A
             =−          cos x + cos2 x · sin x
                     sin x                cos2 x                               ✓ cos2 x · tan2 x              A
             = −1 + sin x2                                                     ✓ −1 + sin2 x                 CA
             = − cos2 x                                                        ✓ answer                      CA
                                                                                                             (6)

 4.2.1       1 + tan2 θ = sec2 θ                                               ✓ answer                        A
                                                                                                              (1)

                      sin θ (1 + tan2 θ )
 4.2.2       LHS =                                                             ✓ common factor                A
                             sec θ
               sin θ · sec2 θ                                                  ✓ sec2 θ                       A
             =                                                                 ✓ sin θ · sec θ               CA
                    sec θ                                                        sin θ
                                          1                                    ✓ cos θ = tan θ               CA
             = sin θ · sec θ = sin θ ·                                                                       (4)
                                       cos θ
             = tan θ = RHS

                                                                                                             [13]

QUESTION 5

 5.1               y                                                           f:
               3                              x = 90◦
                                                                               ✓ intercepts (0 ; 1), 45◦ and
                                                                               135◦
                                 g
               2
                                                                                                           A
                                                                               ✓ turning point (90◦ ; −1),
                       f
               1
                                                                               end point (180◦ ; 1)        A
                           45◦          90◦             135◦   180◦
                                                                      x        ✓ shape                     A
              −1                                                               g:
              −2
                                     (90◦ ; −1)
                                                                               ✓ asymptote x = 90◦         A
                                                                               ✓ intercepts (0◦ ; 0) and
              −3
                                                                               (180◦ ; 0)                  A
                                                                               ✓ shape (through (45◦ ; 1)
                                                                               and (135◦ ; −1))            A
                                                                                                          (6)

 5.2.1       180◦                                                              ✓ answer                        A
                                                                                                              (1)

 5.2.2       x = 90◦                                                           ✓ answer                        A
                                                                                                              (1)

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### Page 7

Technical Mathematics/P2                      7                         Hlayiso/Prediction 2026
                                    NSC – Marking Guidelines

 5.3         45◦ < x < 90◦ or 135◦ < x < 180◦                  ✓ 45◦ < x < 90◦            CA
             OR x ∈ (45◦ ; 90◦ ) ∪ (135◦ ; 180◦ )              ✓ 135◦ < x < 180◦          CA
                                                               Penalise 1 mark if inclusive
                                                               notation is used
                                                                                          (2)

 5.4         −3 ≤ y ≤ 1    OR   y ∈ [−3 ; 1]                   ✓ −3                       CA
                                                               ✓ 1 and notation           CA
                                                                                          (2)

                                                                                          [12]

QUESTION 6

                        522 + 382 − 702
 6.1         cos BĈD =                                        ✓ SF in cosine rule         A
                          2(52)(38)
                                                               ✓ −0,19                    CA
             cos BĈD = −0,1903 . . .
                                                               ✓ answer                   CA
             BĈD = 100,97◦
                                                                                          (3)

                           1
 6.2         Area △BCD = (BC)(CD) sin BĈD                     ✓ area rule                 A
                           2
               1                                               ✓ SF                       CA
             = (52)(38) sin 100,97◦                            ✓ answer                   CA
               2
             = 969,95 m2                                       Accept 969,9 – 970,0
                                                                                           (3)

                       AB
 6.3         tan 24◦ =                                         ✓ trig ratio                A
                       70
             AB = 70 tan 24◦ = 31,17 m                         ✓ answer                   CA
                                                                                          (2)

                        AB 31,17
 6.4         tan AĈB =     =                                  ✓ trig ratio                A
                        BC    52
             AĈB = 30,94√◦                                    ✓ substitution             CA
             OR AC = 522 + 31,172 = 60,63 m;                   ✓ answer                   CA
                        31,17                                                             (3)
             sin AĈB =
                        60,63

                                                                                          [11]

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### Page 8

Technical Mathematics/P2                        8                       Hlayiso/Prediction 2026
                                      NSC – Marking Guidelines

QUESTION 7

 7.1         …bisects the chord.                                 ✓ answer                  A
                                                                                          (1)

 7.2.1       AM = 12 cm                                          ✓S                        A
             (line from centre ⊥ to chord)                       ✓R                        A
                                                                                          (2)

 7.2.2       Let the radius be r: OA = OC = r (radii)
             OM = r − 8                                          ✓ OM = r − 8              A
             OA2 = AM 2 + OM 2 (Pythagoras)                      ✓ Pythagoras (R)          A
             r2 = 122 + (r − 8)2                                 ✓ substitution           CA
             r2 = 144 + r2 − 16r + 64
             16r = 208
             r = 13 cm                                           ✓ answer                 CA
                                                                                          (4)

 7.2.3       ∠ in semicircle   (CD is a diameter)                ✓R                        A
                                                                                          (1)

 7.2.4       CD = 26 cm and AC2 = 122 + 82 = 208                 ✓ AC2 = 208 or MD = 18CA
             AD2 = CD2 − AC2 = 676 − 208 = 468
             AD = 21,63 cm                                       ✓ answer                 CA
             OR MD = 5 + 13 = 18 cm;                                                      (2)
             AD2 = 122 + 182 = 468
                               √
             AD = 21,63 cm (6 13)

                                                                                         [10]

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### Page 9

Technical Mathematics/P2                       9                           Hlayiso/Prediction 2026
                                     NSC – Marking Guidelines

QUESTION 8

 8.1         …the angle in the alternate segment.                  ✓ answer                   A
                                                                                             (1)

 8.2.1       AD̂B = T ÂB = 40◦                                    ✓S                         A
             (tan-chord theorem)                                   ✓R                         A
                                                                                             (2)

 8.2.2       AĈB = AD̂B = 40◦                                     ✓S                        CA
             (∠s in the same segment)                              ✓R                         A
             OR (tan-chord theorem)                                                          (2)

 8.2.3       BÂC = BD̂C = 58◦ (∠s in the same segment)            ✓S                         A
             AB̂C = 180◦ − 58◦ − 40◦ (sum of ∠s in △)              ✓R                         A
             AB̂C = 82◦
             OR AD̂C = 40◦ + 58◦ = 98◦ (S)                         ✓ answer                  CA
             AB̂C = 180◦ − 98◦ = 82◦ (opp ∠s of cyclic quad) (R)                             (3)

 8.2.4       V ÂD = AB̂D (tan-chord theorem)                      ✓ S/R                      A
             AB̂D = AD̂B (∠s opp equal sides; AB = AD)             ✓ S/R                      A
             ∴ V ÂD = AD̂B                                        ✓S                         A
             ∴ TV ∥ BD (alt ∠s equal)                              ✓ conclusion with R        A
             OR                                                                              (4)
             AB̂D = AD̂B = 40◦ (∠s opp equal sides)
             T ÂB = 40◦ = AB̂D (given / proven)
             ∴ TV ∥ BD (alt ∠s equal)

 8.3.1       BÔC = 2 × 64◦ = 128◦                                 ✓S                         A
             (∠ at centre = 2 × ∠ at circumference)                ✓R                         A
                                                                                             (2)

 8.3.2       OB̂T = 90◦ (tan ⊥ radius)                             ✓ S/R                      A
             OĈT = 90◦ (tan ⊥ radius)                             ✓ S/R                      A
             OB̂T + OĈT = 180◦
             ∴ OBTC is a cyclic quadrilateral                      ✓ conclusion with R        A
             (opp ∠s supplementary / converse opp ∠s of cyclic                               (3)
             quad)

                                                                                            [17]

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### Page 10

Technical Mathematics/P2                       10                       Hlayiso/Prediction 2026
                                     NSC – Marking Guidelines

QUESTION 9

 9.1         …in the same proportion.                           ✓ answer                   A
                                                                                          (1)

             PS PT
 9.2.1           =       (line ∥ one side of △; ST ∥ QR)        ✓S                         A
             SQ T R
             6     8                                            ✓R                         A
               =
             9 TR
             T R = 12 units                                     ✓ answer                  CA
                                                                                          (3)

             QU   PT     8
 9.2.2          =     =      (line ∥ one side of △; TU ∥ PQ)    ✓ ratio                   CA
             UR T R 12
                   8                                            ✓ answer                  CA
             QU =    × 30 = 12 units                                                      (2)
                  20

 9.2.3       In △PST and △PQR:
             P̂ = P̂ (common)                                   ✓S                         A
             PŜT = Q̂ (corresp ∠s; ST ∥ QR)                    ✓ S/R                      A
             PT̂ S = R̂ (corresp ∠s; ST ∥ QR)                   ✓ S/R or conclusion        A
             ∴ △PST ||| △PQR (∠∠∠)                                                        (3)

             ST     PS
 9.2.4           =       (||| △s)                               ✓ ratio                    A
             QR PQ
                   6                                            ✓ answer                  CA
             ST =     × 30 = 12 units                                                     (2)
                   15
             OR SQUT is a parallelogram (ST ∥ QU, TU ∥ SQ),
             so ST = QU = 12 units

                                                                                         [11]

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### Page 11

Technical Mathematics/P2                            11                       Hlayiso/Prediction 2026
                                          NSC – Marking Guidelines

QUESTION 10

                     3 000
 10.1.1      n=            = 50 rev/s                                ✓ answer                    A
                       60
                                                                                                (1)

 10.1.2      ω = 2π n = 2π (50)                                      ✓ SF                       A
             = 100π ≈ 314,16 rad/s                                   ✓ answer                  CA
                                                                                               (2)

 10.1.3      D = 250 mm = 0,25 m                                     ✓ conversion to m          A
             v = π Dn = π (0,25)(50)                                 ✓ SF                      CA
             v = 39,27 m/s                                           ✓ answer                  CA
             OR v = ω r = 314,16 × 0,125 = 39,27 m/s                                           (3)

 10.1.4      4h2 − 4dh + x2 = 0                                      ✓ formula                   A
             4(60)2 − 4(250)(60) + x2 = 0                            ✓ SF                        A
             14 400 − 60 000 + x2 = 0
             x2 = 45 600                                             ✓ x2 = 45 600             CA
             x = 213,54 mm                                           ✓ answer                  CA
                                                                                               (4)

                         π
 10.2.1      220◦ ×          = 3,84 rad                              ✓ answer                    A
                        180◦
                                                                                                (1)

 10.2.2      s = rθ = 15 × 3,84                                      ✓ SF                      CA
             s = 57,60 cm                                            ✓ answer                  CA
                                                                     Accept 57,6
                                                                                                (2)

 10.2.3      140◦ = 2,4435 . . . rad                                 ✓ 2,44 rad                 A
             s = 6 × 2,4435 = 14,66 cm                               ✓ answer                  CA
                                                                     Accept 14,64 – 14,66
                                                                                                (2)

 10.2.4      Total length = 57,60 + 14,66 + 2(24,73)                 ✓ adding arcs and 2 × 24,73
             = 121,72 cm                                             CA
                                                                     ✓ answer                 CA
                                                                                              (2)

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### Page 12

Technical Mathematics/P2                        12                        Hlayiso/Prediction 2026
                                      NSC – Marking Guidelines

 10.2.5      Minor sector angle = 360◦ − 220◦ = 140◦ = 2,4435     ✓ 140◦ / 2,44 rad          A
             rad                                                  ✓ SF                      CA
                    r2 θ    (15)2 (2,4435)                        ✓ answer                  CA
             Area =      =
                     2             2                              Accept 274,5 – 274,9
             = 274,89 cm2                                                                   (3)

                                                                                           [20]

QUESTION 11

                     18
 11.1.1      a=         =3m                                       ✓ answer                   A
                      6
                                                                                            (1)

                  4,2 + 5,1
 11.1.2      m1 =           = 4,65; m2 = 5,45; m3 = 5,9;          ✓ mid-ordinates m1 –m3  A
                      2
             m4 = 5,7; m5 = 5,0; m6 = 4,25                        ✓ mid-ordinates m4 –m6  A
             AT = 3(4,65 + 5,45 + 5,9 + 5,7 + 5,0 + 4,25)         ✓ SF                   CA
             = 3(30,95) = 92,85 m2                                ✓ answer               CA
             OR (                                             )                          (4)
                      4,2 + 3,9
             AT = 3             + 5,1 + 5,8 + 6,0 + 5,4 + 4,6
                          2

 11.1.3      Turf needed = 92,85 × 1,1 = 102,14 m2                ✓ ×1,1                     M
             Cost = 102,135 × R245 = R25 023,08                   ✓ cost                   CA
             R25 023,08 > R24 000                                 ✓ comparison             CA
             The budget is NOT sufficient.                        ✓ conclusion             CA
                                                                  No marks for the conclusion
                                                                  without the 10% extra.
                                                                                            (4)

                         1 4
 11.2.1      V = π r2 h + × π r3                                  ✓ volume of cylinder       A
                         2 3
                             2                                    ✓ volume of hemisphere     A
             = π (15) (80) + π (15)3
                      2
                             3
             = 18 000π + 2 250π = 20 250π                         ✓ answer                  CA
             ≈ 63 617,25 cm3                                                                (3)

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### Page 13

Technical Mathematics/P2                      13                       Hlayiso/Prediction 2026
                                    NSC – Marking Guidelines

                         1
 11.2.2      A = 2π rh + × 4π r2                               ✓ curved surface of cylinder
                         2
             = 2π (15)(80) + 2π (15)2                          A
             = 2 400π + 450π = 2 850π                          ✓ surface of hemisphere A
             ≈ 8 953,54 cm2
                                                               ✓ answer                  CA
                                                                                         (3)

 11.2.3      30 × 63 617,25 = 1 908 517,5 cm3                  ✓ conversion to m3 (or 2 m3
             = 1,91 m3                                         to cm3 )                 CA
             1,91 m3 < 2 m3 ∴ there is enough concrete.        ✓ conclusion             CA
                                                                                        (2)

                                                                                        [17]

                                                                               TOTAL: 150

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