# Mathematics Grade 12 Paper 2 Limpopo Mock 2022 Memo

> Grade 12 Mathematics Paper 2 memorandum from the Limpopo Mock Exam 2022, 41 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/limpopo-mathematics-grade-12-pretrial-2022-p2-and-memo/084f9683-b9d7-451f-b7e4-add3a2982a5c
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2022
- Exam period: Limpopo Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 41

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## About this document

Mathematics · Grade 12 · Limpopo Mock Exam · 2022. Memorandum, 41 pages. Read online or download the PDF.

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                                    NATIONAL
                               SENIOR CERTIFICATE

                                          GRADE 12

                                    MATHEMATICS P2

                                       AUGUST 2022

                                PRETRIAL EXAMINATION

MARKS: 150

TIME: 3 HOUR

                This question paper consists of 15 pages and 1 information sheet.

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                                                      NSC

 INSTRUCTIONS AND INFORMATION

 Read the following instructions carefully before answering the questions.

 1.         This question paper consists of 10 questions.

 2.         Answer ALL the questions in the SPECIAL ANSWER BOOK provided.

 3.
                                                                                 s.co
            Clearly show ALL calculations, diagrams, graphs, etc. which you have used in
            determining your answers.                                                m
                                                                               si c
 4.                                                 st
            Answers only will NOT necessarily be awarded full marks.
                                                       an
 5.         You      may use        an   approved        m
                                                     scientific
                                                          or      calculator   (non-programmable     and
            non-graphical), unless stated otherwise.
                                                             ep
 6.
                                                               hy
            If necessary, round off answers correct to TWO decimal places, unless stated otherwise.
                                           om
            Diagrams are NOT necessarily drawn to scale.
                                         fr
 7.
                        oa
            An information sheet with formulae is included at the end of the question paper.
                          de
 8.                         d
            do
            Write neatly and legibly.
 9.           wn
                l

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                                                   NSC

QUESTION 1

 Each child in a group of four-year-old children was given the same puzzle to complete.
 The time taken (in minutes) by each child to complete the puzzle is shown in the table below.

                               TIME TAKEN (t)            NUMBER OF
                                (IN MINUTES)              CHILDREN
                                     2t6                      2
                                                                          s.co
                                    6  t  10                 10
                                                                              m
                                    10  t  14                 9
                                                                     si c
                                    14  t  18   st            7
                                    18  t  22
                                                     an
                                                       mor
                                                                8
                                    22  t  26                 7
                                    26  t  30
                                                           ep   2
                                                             hy
                                           om
 1.1        How many children completed the puzzle?
                                       fr                                                                 (1)

 1.2
                        oa
            Calculate the estimated mean time taken to complete the puzzle.                               (2)
                          de
                            d
 1.3        do
            Complete the cumulative frequency column in the table given in the ANSWER BOOK.               (2)
              wn
 1.4            l
            Draw a cumulative frequency graph (ogive) to represent the data on the grid provided
            in the ANSWER BOOK.                                                                           (3)

 1.5        Use the graph to determine the median time taken to complete the puzzle.                      (2)
                                                                                                          [10]

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                                                     NSC

QUESTION 2

 Learners who scored a mark below 50% in a Mathematics test were selected to use a computer-
 based programme as part of an intervention strategy. On completing the programme, these
 learners wrote a second test to determine the effectiveness of the intervention strategy. The mark
 (as a percentage) scored by 15 of these learners in both tests is given in the table below.

  LEARNER            L1     L2      L3   L4     L5   L6    L7     L8     L9   L10 L11 L12 L13 L14 L15

    TEST 1
                     10     18      23   24     27   34    34     36     37   39    40     44    45     48     49
       (%)
    TEST 2
                     33     21      32   20     58   43    49     48     41   55    50     45    62     68     60
       (%)

 2.1         Determine the equation of the least squares regression line.                                     (3)

 2.2         A learner's mark in the first test was 15 out of a maximum of 50 marks.

             2.2.1         Write down the learner’s mark for this test as a percentage.                       (1)

             2.2.2         Predict the learner's mark for the second test. Give your answer to the
                           nearest integer.                                                                   (2)

 2.3         For the 15 learners above, the mean mark of the second test is 45,67% and the standard
             deviation is 13,88%. The teacher discovered that he forgot to add the marks of the last
             question to the total mark of each of these learners. All the learners scored full marks
             in the last question. When the marks of the last question are added, the new mean mark
             is 50,67%.

             2.3.1        What is the standard deviation after the marks for the last question are added
                          to each learner's total?                                                            (2)

             2.3.2        What is the total mark of the last question?                                        (2)
                                                                                                              [10]
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                                                     NSC

QUESTION 3

 In the diagram, A, B, C(2 ; –3) and D(–2 ; –5) are vertices of a trapezium with AB || DC.
 E(–2 ; 0) is the x-intercept of AB. The inclination of AB is  . K lies on the y-axis
 and KB̂E  .

                                                      y
                                                           K

                                                       B

                                                                                                      x
                                        E(–2 ; 0)     0

  A

                                                                    C(2 ; –3)

                                        D(–2 ; –5)
 3.1        Determine:

            3.1.1       The midpoint of EC                                                             (2)

            3.1.2       The gradient of DC                                                             (2)

            3.1.3       The equation of AB in the form y  mx  c                                      (3)

            3.1.4       The size of                                                                   (3)

 3.2        Prove that AB  BC.                                                                        (3)

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                                                   NSC
 3.3        The points E, B and C lie on the circumference of a circle. Determine:

            3.3.1       The centre of the circle                                                              (1)

                        The equation of the circle in the form ( x  a)  ( y  b)  r
                                                                       2          2    2
            3.3.2                                                                                             (4)
                                                                                                              [18]

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                                                   NSC

QUESTION 4
 In the diagram, the circle is centred at M(2 ; 1). Radius KM is produced to L, a point outside
 the circle, such that       KML || y-axis.      LTP       is a tangent to the circle at    T(–2 ; b).
       1      
 S   4 ;  6  is the midpoint of PK.
       2      

                                                       y
                                                                 L

                                      T(–2 ; b)

                                                                  M(2 ; 1)
                                                                                                                    x

                                                                 K

                P

 4.1        Given that the radius of the circle is 5 units, show that b = 4.                                  (4)

 4.2        Determine:

            4.2.1         The coordinates of K                                                                (2)

            4.2.2         The equation of the tangent LTP in the form y = mx + c                              (4)

            4.2.3         The area of  LPK                                                                   (7)

 4.3        Another circle with equation ( x  2) 2  ( y  n) 2  25 is drawn. Determine, with an
            explanation, the value(s) of n for which the two circles will touch each other
            externally.                                                                                       (4)
                                                                                                              [21]

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                                                 NSC

QUESTION 5

 5.1        Without using a calculator, write the following expressions in terms of sin 11 :

            5.1.1         sin 191                                                                         (1)

            5.1.2         cos 22                                                                          (1)

 5.2        Simplify cos(x  180)  2 sin( x  45) to a single trigonometric ratio.                      (5)

                                     7
            Given: sin P  sin Q      and P̂  Q̂  90
 5.3                                 5
            Without using a calculator, determine the value of sin 2P .
                                                                                                           (5)
                                                                                                           [12]

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                                                      NSC

QUESTION 6

 6.1        Determine the general solution of cos( x  30)  2 sin x .                                          (6)

 6.2        In the diagram, the graphs of f ( x)  cos( x  30) and g ( x)  2 sin x are drawn for
            the interval x  [180 ;180] . A and B are the x–intercepts of f. The two graphs
            intersect at C and D, the minimum and maximum turning points respectively of f.

                                                     y

                                                                  D                        g

  –180°                                 B                0°                    A               180°          x

                                                                                      f
              C

            6.2.1         Write down the coordinates of:

                          (a)       A                                                                            (1)

                          (b)       C                                                                            (2)

            6.2.2         Determine the values of x in the interval x  [180 ; 180] , for which:

                          (a)       Both graphs are increasing                                                   (2)

                          (b)       f ( x  10)  g ( x  10)                                                  (2)

            6.2.3         Determine the range of y = 22sin x + 3                                                 (5)
                                                                                                                 [18]

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                                                      NSC

QUESTION 7

 In the diagram below, CGFB and CGHD are fixed walls that are rectangular in shape and vertical
 to the horizontal plane FGH. Steel poles erected along FB and HD extend to A                  and E
 respectively. ∆ACE forms the roof of an entertainment centre.

 BC = x, CD = x + 2, BÂC   , AĈE  2 and EĈD  60

                                                                                      E

                     A
                         
                                                                                     D

                     B                            2
                                                            60°   x+2
                                    x
                                                 C
                                                                                     H
                     F

                                                      G

 7.1        Calculate the length of:

            7.1.1            AC in terms of x and                                                          (2)

            7.1.2            CE in terms of x                                                               (2)

 7.2        Show that the area of the roof ACE is given by 2 x ( x  2) cos  .                            (3)

 7.3        If   55 and BC = 12 metres, calculate the length of AE.                                      (4)
                                                                                                            [11]

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                                                     NSC

QUESTION 8

 8.1        In the diagram, O is the centre of the circle and LOM is a diameter of the circle.
            ON bisects chord LP at N. T and S are points on the circle on the other side of
            LM with respect to P. Chords PM, MS, MT and ST are drawn. PM = MS and
             MT̂S  31

                                                                      P

                                                N                               M
                                                     2
                                                 1

                                                         O

                                L
                                                                                S

                                                              31°

                                                         T

            8.1.1         Determine, with reasons, the size of each of the following angles:

                          (a)       MÔS                                                                   (2)

                          (b)       L̂                                                                     (2)

                                            1
            8.1.2         Prove that ON =     MS.                                                          (4)
                                            2

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                                                    NSC

 8.2        In ABC in the diagram, K is a point on AB such that AK : KB = 3 : 2.               N
            and M are points on AC such that KN || BM. BM intersects KC at L.                AM
            : MC = 10 : 23.

                                                    B

                                        K
                                                L

                     A              N       M                                          C

            Determine, with reasons, the ratio of:

                          AN
            8.2.1                                                                                      (2)
                          AM

                          CL
            8.2.2                                                                                      (3)
                          LK
                                                                                                       [13]

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                                                  NSC

QUESTION 9

 In the diagram, tangents are drawn from point M outside the circle, to touch the circle at B and
 N. The straight line from B passing through the centre of the circle meets MN produced in A.
 NM is produced to K such that BM = MK. BK and BN are drawn.

 Let K̂  x .

          K
                 x
                                    M
                                    1 2

                                                          N
                                                      1   2

                                                                                       A
                                    1 2
                                          3
                                    B

 9.1          Determine, with reasons, the size of N̂1 in terms of x.                                   (6)

 9.2          Prove that BA is a tangent to the circle passing through K, B and N.                      (5)
                                                                                                        [11]

        QUESTION 10

 10.1         In the diagram, ∆ABC and ∆DEF are drawn such that Â  D̂ , B̂  Ê and Ĉ  F̂ .

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                                                NSC

                                    A
                                                                         D

                                                                     
                                                                 E              F

                      
               B                                   C

            Use the diagram in the ANSWER BOOK to prove the theorem which states that
            if two triangles are equiangular, then the corresponding sides are in proportion,
                      AB AC
            that is        .
                      DE DF                                                                             (6)

 10.2       In the diagram, O is the centre of the circle and CG is a tangent to the circle at G.
            The straight line from C passing through O cuts the circle at A and B.      Diameter
            DOE is perpendicular to CA. GE and CA intersect at F. Chords             DG, BG and
            AG are drawn.

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                                                      NSC

                                                       D

                                                                     4       G
                                                                         3
                                                                             2 1

                                                               2 1                 2 1
                              A                        O         F                   B      C

                                                       E

            10.2.1      Prove that:

                        (a)         DGFO is a cyclic quadrilateral                                                 (3)

                        (b)         GC = CF                                                                        (5)

            10.2.2      If it is further given that CO = 11 units and DE = 14 units, calculate:

                        (a)         The length of BC                                                               (3)

                        (b)         The length of CG                                                               (5)

                        (c)         The size of Ê.                                                                (4)
                                                                                                                   [26]

                                                                                                  TOTAL:           150

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                                                                NSC

                                 INFORMATION SHEET: MATHEMATICS

           b  b 2  4ac
    x
                2a
    A  P(1  ni )             A  P(1  ni )                A  P(1  i) n                              A  P(1  i) n

    Tn  a  (n  1)d                   Sn 
                                                n
                                                  2a  (n  1)d 
                                                2

    Tn  ar n1                         Sn 
                                                 
                                                a r n 1  ; r 1                             S 
                                                                                                        a
                                                                                                            ; 1  r  1
                                                  r 1                                                 1 r

    F
            
            x 1  i   1
                     n
                                                 P
                                                         
                                                       x 1  1  i 
                                                                      n
                                                                           
                   i                                          i
                       f ( x  h)  f ( x )
     f ' ( x)  lim
                h 0            h
                                                               x  x 2 y1  y 2 
     d  ( x 2  x1 ) 2  ( y 2  y1 ) 2                     M 1      ;         
                                                               2          2 
                                                                                           y 2  y1
      y  mx  c                        y  y1  m( x  x1 )                          m                         m  tan
                                                                                           x 2  x1
     x  a 2   y  b2  r 2
                           a        b        c
     In ABC:                           
                        sin A sin B sin C
                         a 2  b 2  c 2  2bc. cos A
                                          1
                        area ABC  ab. sin C
                                          2
     sin(   )  sin  cos   cos  sin                             sin(   )  sin  cos   cos  sin 
     cos(   )  cos  cos   sin  sin                             cos(   )  cos  cos   sin  sin 
              cos 2   sin 2 
              
     cos 2  1  2 sin 2                                             sin 2  2 sin  . sin 
              2 cos2   1
              
                                                                                  n            2

          x                                                                     xi  x 
     x                                                                2  i 1
            n                                                                          n
                n( A )
     P(A)                                                            P(A or B) = P(A) + P(B) – P(A and B)
                nS

     yˆ  a  bx                                                        b
                                                                                x  x ( y  y )
                                                                                 (x  x)          2

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                                NATIONAL
                           SENIOR CERTIFICATE

                                     GRADE 12

                       MATHEMATICS P2 MARKING GUIDE
                        WISKUNDE V2 NASIENRIGLYNE

                                       AUGUST 2022

MARKS/PUNTE: 150

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

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                                  Marking Guidelines/Nasienriglyne

NOTE:
• If a candidate answers a question TWICE, only mark the FIRST attempt.
• If a candidate has crossed out an attempt of a question and not redone the question, mark
   the crossed out version.
• Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking
   at the second calculation error.
• Assuming answers/values in order to solve a problem is NOT acceptable.

NOTA:
• As 'n kandidaat 'n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na.
• As 'n kandidaat 'n antwoord van 'n vraag doodtrek en nie oordoen nie, sien die
   doodgetrekte poging na.
• Volgehoue akkuraatheid word in ALLE aspekte van die nasienriglyne toegepas. Hou op
   nasien by die tweede berekeningsfout.
• Om antwoorde/waardes te aanvaar om 'n probleem op te los, word NIE toegelaat NIE.

                                 GEOMETRY • MEETKUNDE

                A mark for a correct statement
                (A statement mark is independent of a reason)
         S
                'n Punt vir 'n korrekte bewering
                ('n Punt vir 'n bewering is onafhanklik van die rede)

                A mark for the correct reason
                (A reason mark may only be awarded if the statement is correct)
         R
                'n Punt vir 'n korrekte rede
                ('n Punt word slegs vir die rede toegeken as die bewering korrek is)

                Award a mark if statement AND reason are both correct
        S/R
                Ken 'n punt toe as die bewering EN rede beide korrek is

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                             Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 1
1.1 45 children                                                                             answer
                                                                                                                (1)
1.2

        ∑ fx (4×2)+ (8×10) + (12×9)+ (16×7)+ (20×8)+(24×7)+ (28×2) x
                                                                                            692
      x=             OR x = 15,38 minutes      Answer only: full marks                      answer
                                                                                                                (2)
1.3
                        Time taken (t)        Number of      Cumulative
                         (in minutes)          children       frequency
                          2<t≤6                    2                2                            first 4 cum
                          6< t ≤ 10
                                                                                           freq correct
                                                   10           12                               last 3 cum
                         10 < t ≤ 14
                                                                                           freq
                                                   9            21                            correct
                         14 < t ≤ 18               7            28

                         18 < t ≤ 22               8            36

                         22 < t ≤ 26               7            43

                          26< t ≤ 30               2            45                                              (2)
1.4
                      CUMULATIVE FREQUENCY GRAPH
                                (OGIVE)
            50

            40
                                                                                                 plotting
                                                                                           cum freq at
            30                                                                             upper limits
                                                                                           correctly (all
                                                                                           points)
            20                                                                                   shape
                                                                                           (smooth)
            10
                                                                                                 grounding
                                                                                           (2;0)

             0
                 0         5          10      15        20     25        30
                                        Time in minutes

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                                  Marking Guidelines/Nasienriglyne

                                                                                                             (3)

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

QUESTION/VRAAG 2

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

         OR
                                                                                               (2)

          yˆ =

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

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QUESTION/ VRAAG 3
                                                     y
                                                         K

                                                         B
                                                         1

                                                                                                         x
                                        E(–2; 0)     0

  A

                                                                  C(2 ; –3)

                                        D(–2 ; –5)

3.1.1 Midpoint of EC:
          12  2 0  (3)         3
               ;            0;                                 x value  y value
          2         2             2

                                                                      substitution
3.1.2
        mDC =              OR                                         answer
                                                                                                    (2)
              =
3.1.3       mAB       =[AB||DC]
                                                                      mAB

        y         c                        y− y1 = (x−x1)
                                                                      substitution of (−2;0)
                      c    OR               y− 0 = (x− (−2))
            c=1
                                                                     equation
                                                                                                    (3)

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### Page 22

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          tanα=
3.1.4          1
                 mAB
               2                                                            
          α=26,57°                                                           value of α
        θ=90°+ 26,57°
            =116,57°                                                         value of θ
                                       [ext ∠of Δ]                                                         (3)

3.2
        B(0 ; 1)
                                                                              coordinates of B
                1− (−3)
        mBC =                      OR mBC
                 0− 2
              = −2                      = −2                                  mBC =−2

            mAB ×m BC
                                                                              product of gradients = –1
                   =−1
            ∴AB⊥ BC
                                                                                                            (3)
3.3.1      ˆ
         AB C= 90°
        ∴ EC is diameter [converse: ∠ in semi circle]
                                 3
         ∴centre of circle=  0;                                            answer
                                 2                                                                        (1)
                               2

         x  0    y 
                 2           3
                               r
                                   2

3.3.2                       2                                                    substitution of centre
                               2                                2
                                                            3
         0  0    0    r 2 or  2  0    3  ( )   r 2
                 2        3                     2

                         2                               2                     correct substitution of
                EC        (2  2)  (1  (3))
                                   2             2
                                                                             E(–1 ; 0), B(0 ; 1) or     C(2
        or R                                                               ; –3) to calculate r or r
                                                                                                     2
                 2                2
                    3 
        or r  1   
                    2 
              25            5                                                      value of r2 or r
        r2        or r 
              4             2
                         2
                 3   25
        x2   y   
                 2   4
                                                                                   equation
                                                                                                            (4)

                                                                                                          [18]

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QUESTION/VRAAG 4

                                                     y
                                                             L

                                        T( –2 ; b)

                                                              M(2 ; 1)
                                                                                                                  x

                                                             K

               P

4.1      (x−2)2 +(y−1)2 =25                                                  equation of the circle
                                                                            substitution of point T
                   2         2
        (−2− 2) + (b−1) = 25
                                                                            simplification
                       (b −1)2 =9
        OF
                                               ∴
                                                                             answer
                         b −1=±3                                                                          (4)

           ∴b=4           or b≠− 2

4.2.1 K(2 ; 1 – 5)
      ∴ K(2 ; –4)
                                                                             x value  y value
                                           Answer only: full marks
                                                                                                           (2)

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4.2.2
                                                                           mMT
        mMT mPL               [radius ⊥ tangent]
                                                                           mPL =
              y       c

                          c
                                                                           substitution of mPL
                                                                            and the point T

        c         y

                                                                           equation
                                                                                                        (4)

        OR

                                                                            mMT
        mMT

                                                                            mPL =
        mPL           [radius ⊥ tangent]       y

                                                   y
                                                                            substitution of mPL and
                                               y                             the point T

                                                                            equation
                                                                                                             (4)
        OR

        P(11; 8)
                                                                            coordinates of P
                 4  (8)
        mPL 
                2  (11
               4
            
               3
            4                                                               mPL =
        y  xc
            3
              4
        8  x  c
              3
                                                                            substitution of mPL and
            20
        c                                                                   the point P or T
             3
            4       20
        y  x
            3       3                                                       equation
                                                                                                             (4)
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4.2.3
              4        20 28                                                   y=
         yL    (2)      
              3        3    3                                                  length of LK
           28 
        L  2;  and K  2; 4 
           3 
               28
        LK        (4)
                3
               40
             
                3
        Coordinates of P:
        x+ 2         1        y− 4
               =−4
                                                       L(2 ;
                                        =−6
                         and
          2          2          2

        ∴ x=−11             y =−8                                              xP  yP
        ∴P(−11;−8)
        ⊥height (PH) = 2 − (−11) = 13                   K(2 ; –4)              length of ⊥ height

                                         P(–11 ; –8)
                         1                             H(2– 8
        Area    PKL=      LK  PH
                         2                                                    substitution into
           1  40 
            13                                                   the area formula
           2 3 
          269                                                                  answer
              OR 86,67
           3                                                                                              (7)

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4.2.3 OR

             4       20 28                                                     yL =
       yL     (2)     
             3       3    3
          28 
       L  2;  and K  2; 4                                                         length of LK
          3 
              28
       LK        (4)
               3
              40
            
               3

        Coordinates of P:                                                             xP  yP
        x+ 2       1       y− 4
              =−4      and
          2        2        2

        ∴ x=−11                     y =−8
        ∴P(−11;− 8)
        PK2 = (−11− 2)2 +(−8− (−4) )
        PK = 185 units                                                                   ˆ
                                                                                      PK L

        mPK

                                            °
           ˆ
        ∴PK L = 90°+17,1027...° = 107,1°                                             substitution into        the
                                                                               area rule
                           1                                                         answer
        Area     PKL=                    ˆ
                             LK  LK sin PKL                                                                  (7)
                           2
        
           1  40 
                       
                185 sin107,1
           2 3 
                                
          269
              OR 86,67
           3

4.3     The centres of the two circles lie on the same vertical line            correct method
        x = 2. and the sum of the radii = 10                                    sum of radii = 10
        n−1 = 10                            1 −n= 10
                           or                                                   n=11  n= −9
             n=11                               n = −9
                                                                                                              (4)
                                                   Answer only: full marks

                                                                                                             [21]

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QUESTION/VRAAG 6
6.1         cos(x−30°) = 2sin x cosxcos30°+sin
                                                                                            expansion
              3 xsin30°= 2sin x cosx+ sin x=
             2 2sin x                                                                       special ∠s
              3
             2 cosx= sin x                                                                  simplification
                    3
            tan x= 3                                                                        equation in
            x=30° +k.180°; k∈Z                                                        tan

           OR
                                                                                            30°
           x = 30° +k.360° or x = 210° +k.360°;                                             k.180°; k∈Z
           k∈Z                                                                        OR
                                                                                            30° and 210°
                                                                                            k.360°; k∈Z
                                                                                                       (6)

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7.1.1               x
          sinθ=                                                            trig ratio
                  AC             OR
                   x
           AC =                                                            simplification
                  sinθ
                                                                                                    (2)
7.1.2                   x+2
          cos60°=                                                          trig ratio
                        CE                OR x+2
               CE=
                        cos60° x + 2                                       making CE the
                                                                            subject
                   =             =2(x + 2)
                                                                                                    (2)
7.2                                                                              use area rule
                                                  ˆ
          Area ∆ACE = AC.EC.sin AC E                                      correctly
                                 1    x                                          substitution
                                                                          of
                             =               (2(x+2))sin2θ                     x
                                 2 sinθ x(x+                                      (2(x+2))
                                 2)×2sinθcosθ                                sinθ
                             =                                                   substitution
                                           sinθ                           of    sin2θ               (3)

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                          = 2x(x+2)cosθ

                                                                                         E

                 A

                      θ
                                                                                        D

                 B                                2θ         60 °   x+2
                                x
                                                  C
                                                                                        H

                 F

                                                      G

7.3        EC = 2(12 + 2) = 28                                                      EC
                                                                                    use cosine rule
             AE 2  AC 2  EC  2 AC.EC cos ACE
                                                                             correctly
                                      12  
                          2
                12 
                         28  2  
                               2
                                                 .28  cos110                      substitution
                sin 55             sin 55  
             so                                                                      answer
             AE  35, 77m                                                                              (4)
                                                                                                      [11]

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