Mathematics Grade 10 P2 Gauteng June 2016 Memo (English)

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION PROVINCIAL EXAMINATION JUNE 2016 GRADE 10 MATHEMATICS (PAPER 2) MEMORANDUM 7 pages 1

Downloaded from hlayiso.com MEMORANDUM Mathematics (P2) Grade 10 GAUTENG DEPARTMENT OF EDUCATION PROVINCIAL EXAMINATION MATHEMATICS (P2) MEMORANDUM QUESTION 1 1.1 sin y x   cos r r  y y r r   x r x  y r  simplification x (3) Answer only 1/3 1.2.1 PQ 5 sin    answer PR 13 (1) 1.2.2 PR 13 sec    answer QR 12 (1) 1.2.3 PQ 5 tan    answer QR 12 (1) [6] 2

Downloaded from hlayiso.com MEMORANDUM Mathematics (P2) Grade 10 QUESTION 2 2.1 correct quadrant 4 tan  3 3 y  tan    4 x r 5 r 2  x2  y r 2  (4) 2  (3) 2 r 5 5 sin   3 cot   3     3  4   5   5   3   5    3  4     3  4  7  3 7 (5) 2.2 25 cos  2 2 4 substitution  25  5  16   25   25   16 answer (2) [7] 3

Downloaded from hlayiso.com MEMORANDUM Mathematics (P2) Grade 10 QUESTION 3 3.1.1 sin x  2 cos 3 y Do not penalise  sin( 42 )  2 cos( 3  68 ) for rounding off  sin( 42 )  2 cos 204   1,16 answer (2) 3.1.2 3 tan ( x  y ) 2 Do not penalise for rounding  3 tan 2 (42   68 ) off  3 tan 2 110   22,65 answer (2) 3.2.1 2 sin  1,432 Do not penalise  sin  0,716 for rounding off   45,725 answer (2) 3.2.2 tan 3  6,345 3  81,044  tan 1 3   27,015 answer (3) *penailse once [9] only for rounding off in 3.2 4

Downloaded from hlayiso.com MEMORANDUM Mathematics (P2) Grade 10 QUESTION 4 4.2 sin2 45  cos 60  tan10. cot 10  cos 60  1  2 2 2 1     1  1  sin2 45   2  2 2 1 1   1 2 2  tan 10. cot 10  1 1 answer = 1 (4) [9] 5

Downloaded from hlayiso.com MEMORANDUM Mathematics (P2) Grade 10 QUESTION 5 5.1 Pˆ1  Qˆ  Rˆ 2 (exterior angle = sum of interior opposite angles)  reason Pˆ  30   110  1 Pˆ1  110   30   80 answer (2) 5.2 Pˆ2  Sˆ1 (  S opposite equal sides are equal) statement and reason Pˆ  Rˆ  Sˆ  180  2 2 1 (Sum of s of a triangle  180 )  Pˆ2  110   Pˆ2  180  (Given : Rˆ 2  110  and Pˆ2  Sˆ1 ) statement and  2 Pˆ  180   110  2 reason  2 Pˆ2  70 simplification  Pˆ  35 2 (3) OR OR statement and Pˆ2  Sˆ1 (  S opposite equal sides are equal) reason Rˆ  Pˆ  Sˆ 1 2 1 (exterior angle = sum of interior opposite angles) statement and  Pˆ2  35 reason simplification (3) [5] QUESTION 6 In ABC and CDA  Cˆ 1  Aˆ 2 Bˆ  Dˆ (given) AC is common Reason Cˆ 1  Aˆ 2 (alternate angles ; AD // BC ) (AD // BC)  ABC  CDA ( ; ; S ) S + R  AD BC ( ABC  CDA ) AD = BC ABCD is a parallelogram ( one side = // ) reason ( one side = // ) OR OR In ABC and CDA Statement Bˆ  Dˆ (given) AC is common Reason Cˆ 1  Aˆ 2 (alternate angles ; AD // BC ) (AD // BC)  ABC  CDA ( ; ; S ) S + R  AD BC ( ABC  CDA ) AD = BC  AB  DC ( ABC  CDA ) reason (opposite sides =) ABCD is a parallelogram ( opposite sides = ) 6

Downloaded from hlayiso.com MEMORANDUM Mathematics (P2) Grade 10 [5] 7

Downloaded from hlayiso.com MEMORANDUM Mathematics (P2) Grade 10 QUESTION 7 7.1 AO + OC = 4 xy ( given – diagonals bisect) OC = 2 xy  OA = 2 xy BO + OD = 2x 2  2 y 2 ( given – diagonals bisect)  BO = x 2  y 2 BO = x 2  y 2 If rhombus – diagonals bisect at 90  LHS LHS = BC2 RHS = BO2 + OC2  = x2  y2  2   2 = x 2  y 2  (2xy) 2 RHS OR = x4  2x2 y2  y4 = x 4  2x 2 y 2  y 4  4x 2 y 2 = x 4  2x 2 y 2  y 4  OR = x 2  y 2  2 conclusion BOC is a right angled triangle OR Prove AOD as a right angled triangle Diagonals bisect each other at 90/right angles ABCD is a Rhombus (5) 7.2 Rˆ1  120  m (opposite angles of a // )  Rˆ1  120  Rˆ  60 2 (angles on a straight line) T̂  60 (angles opposite equal sides)  T̂  60 Ŝ  60 (sum of angles of a triangle)  Ŝ  60  4 x  60  x  15 x  15 (4) [9] TOTAL: 50 8