Technical Mathematics Grade 10 Paper 2 North West June 2019 Memo

Downloaded from hlayiso.com PROVINCIAL ASSESSMENT GRADE 10 TECHNICAL MATHEMATICS P2 JUNE 2019 MEMORANDUM MARKS: 75 MEMORANDUM This memorandum consists of 7 pages. Demo NW/JUNE/TEC-MATH/ EMIS/6*******

Downloaded from hlayiso.com Technical Mathematics/P2 2 NW/June 2019 Memorandum – Grade 10 NOTE:  If the learner answered the question TWICE, mark the FIRST attempt ONLY.  If the learner crossed out an attempt of a question and did not REDO the question, mark the crossed out question.  Consistent Accuracy (CA) applies in all aspects of memorandum QUESTION 1 1.1 1.1.1 AB  AC sin C  (1) AC 1.1.2 AB  cot A cot A  BC (1) 1.2.1 sin x  2 cos 3 y  sin 38  2 cos 72  Substitution  Answer  1,23 (2) 1.2.2 3 tan x  y  2   3 tan 2 38  72   Substitution  Answer  22,65 (2) 1.3 1.3.1 x  15  x  15 r  17  r =17 x2  y2  r 2  Substitution into Pythagoras  15  y  17 2 2 2  Simplification 225  y 2  289 y 2  64  y  8  y  8 (5) Demo NW/JUNE/TEC-MATH/ EMIS/6*******

Downloaded from hlayiso.com Technical Mathematics/P2 3 NW/June 2019 Memorandum – Grade 10 1.3.2 8 8 (a) sin    17 17 (1) 1.3.2 cos   sin  2 2 (b) 2 2  15   8   Substitution        17   17  225 64    Simplification 289 289 289  289  Answer 1 (3) [15] QUESTION 2 2.1 a2  a2 (1) 2.2 360  360 (1) 2.3 0 y2  End points OR  Notation y  0;2 (2) 2.4 0   x  180  End points  Notation (2) 2.5 h( x)  (cos x  1)  2  Multiplying by negative   cos x  1  2  +2 Answer only (Full marks)   cos x  1 (2) [8] QUESTION 3 3.1 Demo NW/JUNE/TEC-MATH/ EMIS/6*******

Downloaded from hlayiso.com Technical Mathematics/P2 4 NW/June 2019 Memorandum – Grade 10 3.1.1 155 tan A   correct ratio 475 Aˆ  18,07232215  Aˆ  18,07232215  Rounding  18 (3) 3.1.2 EF=155 opp sides of a rectangle CE sin CBˆ E   Substitution into the correct ratio BC CE sin 22  800 CE  800 sin 22  CE  299,69  299,69  CF  299,69  155  CF = 299,69 + 155  Answer  454,69 OR CE cos 68  800 CE  800cos 68  299,69  299,69  155  454,69 (4) 3.2 sin 2 45  cos 60  tan 10. cot10  1  2    1  1 2 1   2     tan 10.  2 2 tan 10  0 1  1 1 2 1 OR  OR cot 10 2 tan 10  1  1 1  Simplification     . cot10  2 2 cot10   Answer  0 1 1 (5) 3.3.1 cos x  0,82  x  34,9  x  34,9  (1) 3.3.2 2 sin 2 x  0,631 0,631  dividing by 2 sin 2 x  2  18,39100097 2 x  18,39100097  Answer x  9.2  (3) Demo NW/JUNE/TEC-MATH/ EMIS/6*******

Downloaded from hlayiso.com Technical Mathematics/P2 5 NW/June 2019 Memorandum – Grade 10 3.3.3 tan( x  20 )  sin 49,5  0,760 tan( x  20 )  0,760  37,23483398 x  20  37,23483398  x  17,2  x  17,2  (3) [19] QUESTION 4 4.1 The sum of two interior opposite angles  Answer (2) 4.2 4.2.1 Rˆ 2  Q  Pˆ1 ... An ext. angle of a triangle  Statement and reason 110  30  P1  Pˆ1  80  Pˆ  80 1 (2) 4.2.2 Let Pˆ2  x  Sˆ1 …. PR = RS Rˆ  2 x  180 … angles of a   Rˆ 2  2 x  180 and the reason 2 2 x  180  110  simplification 2 x  70   Pˆ2  35  x  35  Pˆ2 (3) 4.2.3 PRS is isosceles  … PR = RS  Statement  Reason (2) [9] QUESTION 5 5.1.1 Obtuse triangle  Answer (1) 5.1.2 Right angled triangle  Answer (1) 5.2.1 5.2.1 OQ  2cm …diagonals bisect each other  2cm  Reason (2) POˆ Q  90 ...diagonals intersect each other  5.2.2 at  90 right angles  Reason (2) 5.2.3 QPˆ O  20 ...diagonals of a kite bisect angles  QPˆ O  20 with correct reason  QPˆ S  20  20  40  QPˆ S  40 (2) Demo NW/JUNE/TEC-MATH/ EMIS/6*******

Downloaded from hlayiso.com Technical Mathematics/P2 6 NW/June 2019 Memorandum – Grade 10 5.3.1 5.3.1 CDˆ O  36,87 ….Diagonals of a rhombus bisect  CDˆ O  36,87 (a) angles.  Reason (2) 5.3.1 AOˆ D  90 …Diagonals of a rhombus bisect each   AOˆ D  90  (b) other at right angles  Reason (2) 5.3.2 AO  Correct ratio tan 36,87  ....AOD is a right angled   Answer 8 AO  8  tan 36,87   6cm (2) 5.3.3 AD  8  6 ...Pythagoras 2 2 2  AD  8  6 with reason 2 2 2  100  10 cm  AD  10cm AE  EB converse of midpoint theorem  AE = EB OE  5cm  5 cm (4) [18] Demo NW/JUNE/TEC-MATH/ EMIS/6*******

Downloaded from hlayiso.com Technical Mathematics/P2 7 NW/June 2019 Memorandum – Grade 10 QUESTION 6 6.1 In s NPO & YZO Nˆ  Yˆ  90....given  Statement and reason NOˆ P  YOˆ Z ...Vert opp angles  Statement and reason  Statement and reason Pˆ  Zˆ ..... third angles of a  (3) NPO ||| YZO 6.2 NP NO NP NO  ....Proportional sides   YZ YO YZ YO 23 x   Substitution 92 124 92x  2852 92x 2852 92x 2852    92 92 92 92  x  31 cm (3) [6] Demo NW/JUNE/TEC-MATH/ EMIS/6*******