# Mathematics Grade 10 P2 Gauteng June 2017 Memo (English)

> Grade 10 Mathematics Paper 2 memorandum from the Gauteng June 2017 (English), 6 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr10-math-p2-english-june-2017-possible-answers/26dc2a6d-2c76-4c1c-a2b8-614d6e862891
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 10
- Language: English
- Document type: Memorandum
- Year: 2017
- Exam period: Gauteng June
- Paper: 2
- Curriculum: CAPS
- Pages: 6

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## About this document

Mathematics · Grade 10 · Gauteng June · 2017 · English. Memorandum, 6 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Matching paper and memo

- [Gr10 Math P2 (English) June 2017 Question Paper](https://www.hlayiso.com/library/gr10-math-p2-english-june-2017-question-paper/b5f34c52-34e2-4402-9d23-ed733526a7b3) — QP · 2017 · Eng

## Document text

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### Page 1

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          GAUTENG DEPARTMENT OF EDUCATION
              PROVINCIAL EXAMINATION
                    JUNE 2017
                     GRADE 10

                   MATHEMATICS
                     PAPER 2

                  MEMORANDUM

6 pages

### Page 2

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                                                            MATHEMATICS
                                           MEMORANDUM
                                                            (Paper 2)         GRADE 11

                                  GAUTENG DEPARTMENT OF EDUCATION
                                       PROVINCIAL EXAMINATION

                                                          MATHEMATICS
                                                          (Paper 2)

                                                          MEMORANDUM

QUESTION 1

      OP 2  4  3  Pythagoras                                 ✓ OP2  4  3
1.1               2           2                                                  2    2

      OP 2  25                                                      ✓ OP2  25
      OP  5                                                         ✓ OP  5
                                                                                     (3)

1.2   1.2.1    sin 
                      3                                              ✓ answer
                                                                                    (1)
                      5

      1.2.2    cot
                      4                                              ✓ answer
                                                                                    (1)
                      3

      1.2.3    sin 2   cos2 
                          2        2
                3  4
                                                               ✓
                                                                          4
                5  5                                                  5
                 9 16
                
                25 25                                                ✓ answer
               1                                                                    (2)
                                                                                     [7]

                                                 2

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                                                         MATHEMATICS
                                        MEMORANDUM
                                                         (Paper 2)         GRADE 11

QUESTION 2

PENALISE JUST ONCE FOR INCORRECT ROUNDING-OFF.

2.1   2.1.1   3 sin 138,7
              1,980                                                   ✓ answer
                                                                                         (1)

      2.1.2   sec 50
              1,556                                                   ✓ answer
                                                                                         (1)

      2.1.3   4 tan 2 288,2. cos164,6
                       sin 199,4
                                                                       ✓✓ answer
              107 ,402                                                                  (2)

2.2   2.2.1   cos 30  tan 60
                  3                                                       3
                    3                                                ✓
                 2                                                       2
                  32 3                                                ✓ 3
              
                    2
                                                                       ✓ answer
                3 3
                                                                                        (3)
                 2
      2.2.2   sin 45
                       5 cos ec 90  3 tan 2 30
              cos 45                                                     sin 45
                                                                       ✓          1
                                                                          cos 45
                  2
                               1 
                                    2
                                                                       ✓ cosec 90 1
                2         
                      5 1  3                                                   1
                  2            3                                     ✓ tan 30 
                 2                                                                  3
                                                                         1
                         1                                           ✓
              1  5  3                                               3
                         3
                                                                       ✓ answer
               3                                                                       (5)
                                                                                        [12]

                                                     3

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                                                            MATHEMATICS
                                            MEMORANDUM
                                                            (Paper 2)         GRADE 11

QUESTION 3

3.1   tan   4,96
          78,60                                                        ✓ answer
                                                                                              (1)

3.2   2 sin 2 10 1

       sin 2 10 
                     1
                                                                          ✓divide by 2
                     2
         2 10 30                                                   ✓ 30
                     2  40
                           20                                          ✓answer
                                                                                              (3)
                                                                                              [4]

QUESTION 4

4.1              y                                                        f ( x)  2 tan x
                                                                          ✓ shape
             2
                                                                          ✓ asymptote
                                                                          ✓ (45 ; 2)
                                                                          g ( x)  cos x 1
                           90     180   270   360   x
                 0
                                                                          ✓ shape
                                                                          ✓ x -intercept
        -2                                                                ✓ y -intercept
                                                                                              (6)

4.2   4.2.1          1                                                    ✓ answer
                                                                                              (1)

      4.2.2          180                                                 ✓ answer
                                                                                              (1)

      4.2.3          y  0 ; 2                                          ✓ critical values
                     OR                                                   ✓ correct brackets/
                     0 y  2                                               inequality
                                                                                            (2)
                                                                                          [10]

                                                    4

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                                                                           MATHEMATICS
                                            MEMORANDUM
                                                                           (Paper 2)        GRADE 11

QUESTION 5

5.1   Both pairs of opposite sides are parallel.
      All sides are equal.
      Diagonals bisect the angles.                                                       ✓✓ any two
      Diagonals bisect at right angles.                                                     answers
      Both pairs of opposite angles are equal.                                                            (2)

5.2                                           A                                 B
                                                  x  20

                                                                2x  60
                                        D                                  C

      Aˆ  Cˆ                                       Opp angles of // m                   ✓ statement
      x  20  2 x  60
      x  80                                                                            ✓ answer
      Cˆ  100                                                                                           (2)
                                                                                                          [4]

QUESTION 6

6.1
                            A       1                               B
                                2

                                                        2
                    D                               1
                                                            C

      Construct diagonal AC
      In  ADC and  ABC                                                                 ✓ construction
       AC  AC                                      Common
                                                                                         ✓alternate angles
        Aˆ  Cˆ
        1       1                                   Alternate angles AB//DC
                                                                                         AB//CD
       Aˆ 2  Cˆ 2                                  Alternate angles AD//BC              ✓alternate angles
       ADC  ABC                                  s<<                                  AD//BC
       AB = DC                                                                          ✓  ADC  ABC
                                                     ADC  ABC
       AD = BC                                                                          ✓ AB = DC AND
                                                     ADC  ABC
                                                                                         AD = BC
                                                                                                         (5)

                                                    5

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                                                                               MATHEMATICS
                                                MEMORANDUM
                                                                               (Paper 2)                 GRADE 11

6.2
                                                  A                                             D
                                                                                        1
                                                                                            2
                                                                                2
                                                                           1        F
                                                                 O

                                                  E 2
                                                   1
                                        1
                                    B       2
                                                                                C

      6.2.1   Fˆ1  Eˆ 2                               Alternate angles AF//EC                      ✓statement and
                                                                                                    reason
              Fˆ  Fˆ  Eˆ  Eˆ 180                  Angles on straight line
                1    2      1   2                                                                   ✓statement and
              Fˆ2  Eˆ 2                                                                           reason
                                                                                                    ✓Fˆ2  Eˆ 2
                                                                                                                     (3)

      6.2.2   In  AFD and  BEC                                                                    ✓identify correct 
              AF = EC                                  Opposite sides //   m
                                                                                                    ✓ AF = EC
               Dˆ 1  Bˆ 2                             Alternate angles AD//BC
               Fˆ  Eˆ
                2     1                                proved                                       ✓ AFD  BEC
               AFD  BEC                            s<<                                          ✓ AD=BC
              AD = BC                                  AFD  BEC
              AB//BC                                                                                ✓ reason
              ABCD is a parallelogram                  One pair of opp sides =//                                    (5)
                                                                                                                   [13]

                                                                                                         TOTAL: 50

                                                        6

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