Technical Mathematics Grade 10 P2 Eastern Cape November 2018 Memo

Downloaded from hlayiso.com NASIONALE SENIOR SERTIFIKAAT GRADE 10 NOVEMBER 2018 TECHNICAL MATHS P2 MARKING GUIDELINE PUNTE: 100 This marking guideline consists of 11 pages.

Downloaded from hlayiso.com 2 TECHNICAL MATHS P2 (EC/NOVEMBER 2018) QUESTION 1 1.1.1  x  x y  y2  M 1 2; 1   2 2   2  0 4  3   Substitution M  ;   2 2   1  M 1;    Answer (2)  2 1.1.2 y2  y1 mMB  x2  x1 1   (1) mMB  2  Substitution 1  (3) 1  1  2 1 3 1 mMB  8  Answer (2) 1.1.3 1 mMB  8 y  mx  c 1 1  ( 3)  c  Substitution 8 5 c   5 8  c 1 5 8 y  x   Equation 8 8 OR y  y1  m  x  x1  1 y  (1)  x  (3)  8  Substitution 1 3 y  x  1 8 8 1 5  Simplification y  x  8 8  Answer (3) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2018) TECHNICAL MATHS P2 3 1.1.4 𝐶𝐷2 = (𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 52 = ( 𝑘 − 2)2 + (0 − (−4))2  Substitution 52 = ( 𝑘 − 2)2 + (0 + 4)2 25= (𝑘 − 2)2 + 16 k 2  4k  4  9 = 0  Standard form k 2  4k  5  0  Factors  (𝑘 − 5)(𝑘 + 1) = 0 k   1 or k  5 ∴𝑘=5  Answer (4) BC   x2  x1    y2  y1  1.1.5 2 2 BC   2  (3)    4  (1)  2 2 BC  34  BC AD   5  0    0  3 2 2  AD  AD  34 BC  AD  34  BC = AD  BC  AD  ABCD is a parallelogram (Opp. sides of quad are equal)  Conclusion OR 3 mBC   5 03 mAD  50 3 mAD    mBC 5 3  mAD  mBC    BC / / AD 5 3 1 4 mAB   03 3 04 4 mDC   52 3 4 mAB  mDC  4  mAB  mDC  3 3  AB // DC  AB / / DC  ABCD is a parallelogram (Opp. sides of quad are parallel )  Conclusion (4) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com 4 TECHNICAL MATHS P2 (EC/NOVEMBER 2018) QUESTION 2 2.1.1 𝑐𝑜𝑠𝑒𝑐𝐴 + cot 𝐵 1 1    Reciprocals sin A tan B 1 1    Substitution sin 57 tan 39  2.43  Answer (3) 2.1.2 3A 2 cos 2 3(57)  2 cos  Substitution 2  Answer  0.16 (2) 2.2.1 5𝑐𝑜𝑠𝜃 = −3 3 𝑐𝑜𝑠𝜃 = − 5  Correct Diagram 2 2 2 𝑟 =𝑥 +𝑦 52 = (−3)2 + 𝑦 2 𝑦 2 = 25 − 9 = 16 ∴ 𝑦 = −4  y  4 𝑐𝑜𝑠𝜃 + 𝑡𝑎𝑛𝜃 3  4      Substitution 5  3  3 4   5 3 11   Answer 15 (4) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2018) TECHNICAL MATHS P2 5 2.2.2 r r sec    sec   x x 5  Substitution sec  3  Answer 5  sec 𝜃 = − 3  Reciprocal OR 1 sec   cos   Substitution 1  3   Answer 5 5 sec    3 (3) 2.3 2 tan(2𝑥 + 12°) − 3 = 1 2 tan(2𝑥 + 12°) = 4  Transposing 3 tan(2𝑥 + 12°) = 2  tan(2𝑥 + 12°) = 2 2𝑥 + 12° = tan−1(2)  tan−1 (2) 2𝑥 + 12° = 63,43° 2𝑥 = 63,43° − 12° 2𝑥 = 51,43  2x = 51,43° 𝑥 = 25,72  Answer (5) [18] Copyright reserved Please turn over

Downloaded from hlayiso.com 6 TECHNICAL MATHS P2 (EC/NOVEMBER 2018) QUESTION 3 3.1 ˆ  90 ABC ( AB  AD)  Statement  Reason (2) 3.2 ˆ  ˆ DAC BAC  90 ( AB  AD) ˆ  49  BAC ˆ  49 BAC ˆ  41  ACB ( Sum of  ABC )  Statement & Reason ˆ  AB tan ACB BC  AB  45 tan 41  Subst. into tan ratio  AB  39.12m  Answer  AB  3912cm (4) 3.3 ED tan ˆ EAD  AD ED tan 73   Substitution 45  ED  45 tan 73  Simplification  ED  147.19m  Answer  ED  14719cm (3) 3.4 EC  CD  ED CD  45 tan 41 CD  39.12m  CD = 3912 cm  CD  3912cm  3912 cm  14719 cm  EC  3912cm  14719 cm  EC  18631 cm  Answer (3) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2018) TECHNICAL MATHS P2 7 QUESTION 4 4.1.1  x-intercepts  y-intercepts  asymptotes  shape (4) 4.1.2 Period is 180°  180° (2) 4.1.3 𝑦 = −3𝑡𝑎𝑛𝑥  Answer (1) 4.2.1 𝑔(𝑥) = 𝑎 𝑠𝑖𝑛𝑥 3 = 𝑎 cos 0°  Substitution 3 = 𝑎(1) Answer Only: full marks 𝑎=3 Answer (2) 4.2.2 Range is −1 ≤ 𝑦 ≤ 5 -1 5 (2) [11 Copyright reserved Please turn over

Downloaded from hlayiso.com 8 TECHNICAL MATHS P2 (EC/NOVEMBER 2018) QUESTION 5 ′ 5.1.1 P𝑅̂ S= 30° (𝐴𝑙𝑡.  𝑠 ; 𝑃𝑄//RS)  Statement (2)  Reason 5.1.2 E𝑇𝑅̂ 𝑆 = 40° (Corr.  𝑠 ; 𝑃𝑄//RS)  Statement (2)  Reason 5.1.3 𝑃̂ + 𝑄̂ + 𝑃𝑅̂ 𝑄 = 180° (  𝑠𝑢𝑚 ∆)  statement and 30°+40° + 𝑃𝑅̂ 𝑄 = 180° reason. 𝑃𝑅̂ 𝑄 = 180° − 30° − 40° (2) = 110° Answer 5.1.4 PRT  PRS  SRT ( Same )  statement and Reason PRT  300  400  700  Answer OR PRT  Qˆ  Pˆ ( Ext.  of   Sum of 2 opp.int . )  PRT  40  30  PRT  70  (2) 5.1.5 𝑃̂ + 𝑄̂ = 𝑃𝑅̂ 𝑇 (ext.  = 𝑠𝑢𝑚 of 2 opp. Int. angles)  Reason OR ̂ ˆ , therefore 𝑃𝑅̂ 𝑇 𝑃𝑅 𝑇 is an interior angle of triangle PQR is the sum of 𝑄̂ and 𝑃̂ (two opposite interior angles)  Reason (1) [9] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2018) TECHNICAL MATHS P2 9 QUESTION 6 6.1 ABDE is an Isosceles trapezium. Given one pair of sides  Trapezium to be equal and another parallel  Equal pair  Parallel pair (3) 6.2.1 BAE  EDB 180 (Opp s of isosceles Trapezium)  Statement and (2) 2 x  EDB 180 reason  Answer EDB 180  2 x 6.2.2 BAE  AED(base  ' sIsoscelestrapezium)  Statement and reason AED  2 x  Answer (2) 6.3 180° − 2𝑥 = 𝑥 (opp. s of parm.) 3𝑥 = 180°  180° − 2𝑥 = 𝑥 ∴ 𝑥 = 60°  𝑥 = 60° (2) 6.4 𝐴𝐶̂ 𝐵 = 𝐶𝐴̂𝐵 (𝐴𝑙𝑡. 𝑎𝑛𝑔𝑙𝑒𝑠 AE//BD)  Statement & reason Therefore, triangle ABC is isosceles  Isosceles Therefore, ABCE is a Rhombus (adjacent sides are equal,  Conclusion (4) AB=BC)  Reason [13] Copyright reserved Please turn over

Downloaded from hlayiso.com 10 TECHNICAL MATHS P2 (EC/NOVEMBER 2018) QUESTION 7 7.1.1 Proof: In ∆𝑃𝑄𝑂 𝑎𝑛𝑑 𝑀𝑁𝑂 𝑂̂ 𝑖𝑠 𝑐𝑜𝑚𝑚𝑜𝑛  𝑂̂ 𝑖𝑠 𝑐𝑜𝑚𝑚𝑜𝑛 O𝑃̂𝑄 = 𝑂𝑀 ̂ 𝑁 ( 𝑐𝑜𝑟𝑟. 𝑎𝑛𝑔𝑙𝑒𝑠 𝑃𝑄//MN)  Statement & Reason O𝑄̂ 𝑃 = 𝑂𝑁 ̂ 𝑀 ( 𝑐𝑜𝑟𝑟. 𝑎𝑛𝑔𝑙𝑒𝑠 𝑃𝑄//MN)  Statement & Reason ∆𝑃𝑄𝑂///∆ MNO (𝐴𝐴𝐴) ∆𝑃𝑄𝑂///∆ MNO (𝐴𝐴𝐴) (3) 7.1.2 PQ QO PO  Answer   (1) MN NO MO 7.2.1 𝑂𝑄 = 𝑃𝑄 (∆𝑃𝑄𝑂///∆ MNO)  Statement and reason 𝑂𝑀 𝑀𝑁 𝑂𝑄 6 =9  Simplification 12 6 𝑂𝑄 = 9 ×12  Answer (3) 𝑂𝑄 = 8 units 7.2.2 𝑃𝑀 = 𝑀𝑁 𝑂𝑃 𝑃𝑄 𝑃𝑀 9 =6  𝑃𝑀 = 9 19 19 6 9 57 PM= 6 ×19 = 2 ≈ 28,50 𝑢𝑛𝑖𝑡𝑠  Simplification  Answer (3) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2018) TECHNICAL MATHS P2 11 QUESTION 8 8.1.1 107.5° = 107° + 0.5×60  Multiply by 60 = 107 + 30  30’ = 107°30′00"  00” (3) 8.1.2 69°64′89" = 69° + 64 + 89   Divide by 60 60 60×60 and 3600 = 69.1°  Answer (3) 8.2 𝜃 = 𝑠/𝑟  𝜃 = 𝑠/𝑟 35 180 35 = 7× 𝜋  7 180 = 286.48°  Multiply by 𝜋  286.48 (4) 𝜋 8.3 2𝜋 − 9 − 120° 17 17 = 9 𝜋 − 120°  9𝜋 180 17 180 = 9 𝜋× 𝜋 − 120°  Multiply by 𝜋 = 340° − 120°  Answer = 220° (3) [13] TOTAL: 100 Copyright reserved Please turn over