# Technical Mathematics Grade 10 P2 November Mock 2026 Memo

> Grade 10 Technical Mathematics Paper 2 memorandum from the November Mock Exam 2026 (English), 9 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-10-technical-mathematics-2026-november-mock-exam-memorandum-paper-2/cecb0e32-80d5-4367-be7b-c9bb63365cfe
- Updated: 2026-10-29
- Subject: Technical Mathematics
- Grade: Grade 10
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 9

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## About this document

Technical Mathematics · Grade 10 · November Mock Exam · 2026 · English. Memorandum, 9 pages. Read online or download the PDF.

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## Matching paper and memo

- [Grade 10 Technical Mathematics P2 November 2026 Mock](https://www.hlayiso.com/library/grade-10-technical-mathematics-2026-november-mock-exam-question-paper-paper-2/769ed4e7-7ee1-44e3-a30e-181922777291) — QP · 2026 · Eng

## Document text

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### Page 1

hlayiso mock exams
                             Practice Series:
                             Grade 10 Exam Preparation
                             WWW.HLAYISO.COM

                                  NATIONAL
                             SENIOR CERTIFICATE

                                     GRADE 10

                         TECHNICAL MATHEMATICS P2
                     NOVEMBER EXAMINATION 2026 (MOCK)
                              MARKING GUIDELINES

  MARKS: 100

                       These marking guidelines consist of 9 pages.

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### Page 2

Technical Mathematics/P2                       2                                  Hlayiso/Mock 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
 S/R Statement/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.
    • In QUESTIONS 5 and 6 (Euclidean geometry) a statement without the correct reason earns no
      mark for the reason. Accept any valid alternative method.

QUESTION 1

                  √
 1.1         AB = (2 − (−2))2 + (4 − 1)2                            ✓ SF                        A
               √        √
             = 16 + 9 = 25                                          ✓ answer                   CA
             = 5 units                                                                         (2)

                       4−1     3
 1.2         mAB =           =                                      ✓ SF                        A
                     2 − (−2) 4
                                                                    ✓ answer                   CA
                                                                                               (2)

                   0−4     4
 1.3         mBC =      =−                                          ✓ SF                        A
                   5−2     3( )
                                                                    ✓ mBC = − 34               CA
                        3     4
             mAB × mBC = × −    = −1                                ✓ product = −1             CA
                        4     3                                                                (3)
             ∴ AB ⊥ BC
                (             )
                 −2 + 5 1 + 0
 1.4         M          ;                                           ✓ x-value                    A
                    2      2
               (      )                                             ✓ y-value                    A
                 3 1                                                                            (2)
             M    ;       OR M(1,5 ; 0,5)
                 2 2
                    −3 − 4
 1.5         mBD =          =7                                      ✓ mBD = 7                   A
                      1−2
                                                                    ✓ substitution of point    CA
             y − 4 = 7(x − 2)
                                                                    ✓ answer                   CA
             y = 7x − 10                                                                       (3)

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### Page 3

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

 1.6         y = 7(1,5) − 10 = 0,5                                      ✓ substitution of M          CA
             ∴ M(1,5 ; 0,5) lies on BD                                                               (1)

 1.7         Square.                                                    ✓ square                      A
             All four sides are equal (5 units) AND AB̂C = 90◦ .        ✓ reason                     CA
             (A rhombus with a right angle is a square.)                                             (2)

                                                                                                     [15]

QUESTION 2

 2.1.1       sin(37,4◦ + 18,6◦ ) = sin 56◦                              ✓ 56◦                         A
             = 0,83                                                     ✓ answer                     CA
                                                                                                     (2)

                 1
 2.1.2                 − tan 18,6◦                                      ✓ sec A = cos1 A               A
             cos 37,4◦
             = 1,2587 . . . − 0,3365 . . .
                                                                        ✓ answer                     CA
             = 0,92                                                                                  (2)

                          12
 2.2.1       tan θ =         , θ in the 3rd quadrant                    ✓ 3rd quadrant                 A
                           5
                            y                                           ✓ x = −5 and y = −12           A
                     −5         θ
                                    x

              −12
                                                                        ✓ r = 13                     CA
                          13
                                                                        ✓ answer                     CA
                                                                                                     (4)

             r2 = (−5)2 + (−12)2 = 169, r = 13
                       12
             sin θ = −
                       13
                                        13   −5
 2.2.2       cosec θ + cot θ = −           +                            ✓ cosec θ = − 13             CA
                                        12 −12                                        12
               13   5                                                   ✓ cot θ = 12
                                                                                   5
                                                                                                     CA
             =−   +                                                     ✓ answer                     CA
               12 12
               8      2                                                                              (3)
             =− =−
               12     3

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### Page 4

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

 2.3         cos x = 0,65                                        ✓ cos x = 0,65               A
             x = cos−1 (0,65)                                    ✓ inverse                   CA
             x = 49,46◦                                          ✓ answer                    CA
                                                                                             (3)

                                                                                             [14]

QUESTION 3

                    RQ
 3.1.1       sin 6◦ =                                            ✓ correct ratio               A
                    4,5
             RQ = 4,5 sin 6◦
                                                                 ✓ answer                    CA
             RQ = 0,47 m                                                                     (2)

                     PQ
 3.1.2       cos 6◦ =                                            ✓ correct ratio               A
                     4,5
             PQ = 4,5 cos 6◦
                                                                 ✓ answer                    CA
             PQ = 4,48 m                                                                     (2)
             (Pythagoras also accepted)

                       30
 3.2.1       tan 65◦ =                                           ✓ correct ratio              A
                      BC
                    30                                           ✓ BC as subject             CA
             BC =                                                ✓ answer                    CA
                  tan 65◦
             BC = 13,99 m                                                                    (3)

                       30
 3.2.2       sin 65◦ =                                           ✓ correct ratio              A
                      AC
                    30                                           ✓ answer                    CA
             AC =         = 33,10 m                                                          (2)
                  sin 65◦

                       30
 3.2.3       tan 40◦ =                                           ✓ correct ratio              A
                      BD
                    30                                           ✓ BD = 35,75                CA
             BD =         = 35,75 m
                  tan 40◦
             CD = BD − BC = 35,75 − 13,99                        ✓ answer                    CA
             CD = 21,76 m                                                                    (3)

                                                                                             [12]

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### Page 5

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

QUESTION 4

 4.1                 y                                                                      f:
                         (0◦ ; 2)                               (270◦ ; 2)                  ✓ x-intercepts              A
                                                                                            ✓ turning points (90◦ ; −2) &
                 2
                                                                             g

                 1                                                                          (270◦ ; 2)                  A
                                                                                            ✓ shape                     A
                                       90◦            180◦        270◦           360◦
                                                                                        x   g:
                                                                                            ✓ end points (0◦ ; 2),
              −1
                                                                                            (360◦ ; 2)                  A
                                                 f                                                             ◦
                                                                                            ✓ min. point (180 ; 0)      A
              −2
                                    (90◦ ; −2)                                              ✓ shape                     A
                                                                                                                       (6)

 4.2         2                                                                              ✓ answer                    A
                                                                                                                       (1)

 4.3         360◦                                                                           ✓ answer                    A
                                                                                                                       (1)

 4.4         0 ≤ y ≤ 2 OR                        y ∈ [0 ; 2]                                ✓ critical values         CA
                                                                                            ✓ notation                CA
                                                                                                                      (2)

 4.5         0◦ < x < 180◦                       OR     x ∈ (0◦ ; 180◦ )                    ✓ critical values         CA
                                                                                            ✓ notation                CA
                                                                                                                      (2)

                                                                                                                      [12]

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### Page 6

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

QUESTION 5

 5.1         …equal to half of the third side.                        ✓ answer                   A
                                                                                                (1)

 5.2.1       QR = 2 × ST = 14 cm                                      ✓S                         A
             (midpoint theorem)                                       ✓R                         A
                                                                                                (2)

 5.2.2       PŜT = Q̂ = 68◦                                          ✓S                         A
             (corresponding ∠s; ST ∥ QR)                              ✓R                         A
                                                                                                (2)

 5.2.3       P̂ = 180◦ − 68◦ − 52◦ = 60◦                              ✓S                         A
             (sum of ∠s of △)                                         ✓R                         A
                                                                                                (2)

 5.3.1       In △ADE and △ABC:
             Â = Â                                      (common)    ✓ Â common                S
             AD̂E = AB̂C                     (corresp. ∠s; DE ∥ BC)   ✓ S/R                    S/R
             AÊD = AĈB                     (corresp. ∠s; DE ∥ BC)
                                                                      ✓ conclusion with reason R
             ∴ △ADE ||| △ABC                                 (∠∠∠)
                                                                                              (3)

             DE AD
 5.3.2          =                                          (||| △s)   ✓ ratio with AB = 10       A
             BC    AB
              5    4
                =                                                     ✓ answer                 CA
             BC 10
             BC = 12,5 units                                                                   (2)

 5.4         Any TWO of:
             SSS (three sides)                                        ✓ first condition          A
             SAS (two sides and included angle)                       ✓ second condition         A
             AAS (∠∠S)                                                                          (2)
             RHS (90◦ , hypotenuse, side)

                                                                                               [14]

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### Page 7

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

QUESTION 6

 6.1.1       Â = Ĉ                               (opp. ∠s of ∥m)    ✓ S/R                     S/R
             3x + 10◦ = 5x − 30◦                                      ✓ equation                  A
             2x = 40◦
             x = 20◦                                                  ✓ answer                  CA
                                                                                                (3)

 6.1.2       Â = 3(20◦ ) + 10◦ = 70◦                                 ✓ Â = 70◦                CA
             B̂ = 180◦ − 70◦ = 110◦                                   ✓ answer with reason      CA
             (co-interior ∠s; AD ∥ BC)                                                          (2)

 6.2.1       PT̂ Q = 90◦                                              ✓S                          A
             (diagonals of a rhombus are perpendicular)               ✓R                          A
                                                                                                 (2)

 6.2.2       PT = 8 cm and QT = 6 cm             (diagonals bisect)   ✓ PT = 8 and QT = 6        A
             PQ2 = 82 + 62                            (Pythagoras)    ✓ Pythagoras              CA
             PQ2 = 100
                                                                      ✓ answer                  CA
             PQ = 10 cm
                                                                                                (3)

 6.2.3       4 × 10 = 40 cm                                           ✓ answer                  CA
                                                                                                (1)

 6.3         …rectangle.                                              ✓ answer                    A
                                                                                                 (1)

                                                                                                [12]

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### Page 8

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

QUESTION 7

                1,4
 7.1.1       r=      = 0,7 m                                        ✓ r = 0,7                  A
                 2
                                                                    ✓ SF                      CA
             V = π (0,7)2 (1,8)
                                                                    ✓ answer                  CA
             V = 2,77 m3                                                                      (3)

 7.1.2       A = 2π (0,7)2 + 2π (0,7)(1,8)                          ✓ 2π r2 substituted       CA
             = 3,08 + 7,92                                          ✓ 2π rh substituted       CA
             A = 11,00 m2                                           ✓ answer                  CA
                                                                                              (3)

                 1
 7.2.1       A=    × 2,4 × 0,6                                      ✓ SF                       A
                 2
                                                                    ✓ answer                  CA
             A = 0,72 m2
                                                                                              (2)

 7.2.2       V = 0,72 × 1,5                                         ✓ SF                      CA
             V = 1,08 m3                                            ✓ answer                  CA
                                                                                              (2)

 7.2.3       AC2 = 2,42 + 0,62 = 6,12                (Pythagoras)   ✓ Pythagoras               A
             AC = 2,47 m                                            ✓ answer                  CA
                                                                                              (2)

                                                                                              [12]

QUESTION 8

 8.1.1       0,385 × 60′ = 23,1′                                    ✓ 23′                      A
             0,1 × 60′′ = 6′′                                       ✓ answer                  CA
             47◦ 23′ 6′′                                                                      (2)

                   36   45
 8.1.2       18◦ +    +                                             ✓ method                   M
                   60 3 600
             = 18,61◦ (18,6125◦ )                                   ✓ answer                  CA
                                                                                              (2)

                       π                                                 π
 8.2         150◦ ×                                                 ✓ × 180 ◦                  M
                      180◦
                 5π                                                 ✓ answer                  CA
             =      rad                                                                       (2)
                  6

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### Page 9

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

                            π
 8.3         θ = 75◦ ×          = 1,309 rad                       ✓ conversion to radians    A
                           180◦
             s = rθ
                                                                  ✓ SF                      CA
             s = 45 × 1,309
                                                                  ✓ answer                  CA
             s = 58,90 cm                                                                   (3)

                                                                                            [9]

                                                                                  TOTAL: 100

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