Technical Mathematics Grade 10 P2 National November 2016 Memo

Downloaded from hlayiso.com GRAAD 12 NATIONAL SENIOR CERTIFICATE GRADE 10 TECHNICAL MATHEMATICS EXEMPLAR 2016 MEMORANDUM MARKS: 100 This memorandum consists of 8 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 2 DBE/2016 CAPS – Grade 10 Exemplar – Memorandum QUESTION 1 1.1 −1 − 2 2 +1 = mAC = or answer 2−0 0−2 3 (1) = − 2 1.2 𝑥1 + 𝑥2 𝑦1 + 𝑦2 formula 𝑀� ; � 2 2 −3 + 2 0 − 1 𝑀� ; � 2 2 1 1 answer 𝑀 �− ; − � 2 2 (2) 1.3 1 subst. into grad. −2 − 2 𝑚𝑀𝐷 = =5 formula 1 5 −2 − 0 subst. into str. 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) Answer only full marks line. formula 𝑦 − 2 = 5(𝑥 − 0) y = 5x + 2 equation 𝑦 = 5𝑥 + 2 (4) 1.4 𝐴𝐵 = �(𝑥2 − 𝑥2 )2 + (𝑦2 − 𝑦1 )2 formula = �(−3 − 𝑡)2 + (0 − (−3))2 �(−3 − 𝑡)2 + 9 = √13 substitution 𝑡 2 + 6𝑡 + 9 + 9 = 13 𝑡 2 + 6𝑡 + 5 = 0 factors (𝑡 + 5)(𝑡 + 1) = 0 answer 𝑡 = −1 (4) 𝐴𝐵 = √13 1.5 𝐷𝐶 = �(0 − 2)2 + (2−(−1))2 length of DC = √13 𝐴𝐷 = �(0 − (−3))2 + (2 − 0)2 length of AD = √13 𝐵𝐶 = �(−1 − 2)2 + (−3−(−1))2 length of AB = √13 � 𝐶 = 90° conclusion All sides are equal and 𝐴𝐷 (4) Hence ABCD is a square. [15] Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 3 DBE/2016 CAPS – Grade 10 Exemplar – Memorandum QUESTION 2 2.1.1 sin ( x + y=) sin ( 43 + 32,5) substitution answer only full marks. answer = 0,97 (2) 2.1.2  x− y  43 − 32,5  sec   = sec    2   2  changing to cos 1 = answer only full marks.  21  answer cos    4 (2) = 1, 00 2.2.1 13sin α + 5 = 0 and 900 < α < 2700 y correct diagram −5 α  13 −5 sin 𝛼 = x 13 -5 𝑥 2 + (−5)2 = 132 13  𝑥 = −12 ∴ 𝑥 = −12 −12 12 answer ∴ cot 𝛼 = = −5 5 (4) 2.2.2 −12 5 −12 cos α + tan α = +  13 12 13 −144 + 65 5 =  156 12 −79 answer = (3) 156 2.3 cot = x tan 530 + sin 2330 1 = tan 530 + sin 2330 1 tan x  tan x 1 tan x = tan 53 + sin 2330 0  1   tan −1 x = tan −1  0 0   tan 53 + sin 233   answer = 62,150 Answer 62.120 two marks (3) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 4 DBE/2016 CAPS – Grade 10 Exemplar – Memorandum QUESTION 3 3.1.1 SQ = 5m – 1,5m = 3,5m  answer (1) 3.1.2 SQ sin 63° = SR 3,5  sin 63° = SR 3,5 sin 63° = SR  answer ∴SR = 3,93m (2) 3.1.3 PQ cos150 = cos150 PR PQ PR =  making PR the cos150 subject of the 5m formula PR = cos150  substitution = 5,18m answer Or PQ OR sin 750 = sin750 PR PQ  making PR the PR = sin 750 subject of the 5m formula = sin 750  substitution = 5,18m answer (4) 3.2.1 AB = tan 52° BC AB  = tan 52° BC 45 ∴ = tan 52° BC 45 45 BC =  BC = tan 520 tan 520 ∴ BC = 35,16 m answer (3) 3.2.2 AB 45 = tan 38° ∴ = tan 38° BD BD 45  = tan 38° BD ∴ BD = 57,60m length of BD CD = 35,16m + 57,60m CD = 92,76m answer (3) [13] Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 5 DBE/2016 CAPS – Grade 10 Exemplar – Memorandum QUESTION 4 4.1 y-int of g shape of g x intercepts of f y intercept of f shape of f (5) 4.2 x = 90o and x = 270o  answer (1) 4.3 𝑦 ∈ [0; 2]  answer with OR 0 ≤ 𝑦 ≤ 2 correct notation (2) 4.4 (180o; 0) 1800 0 (2) 4.5 180o < x < 270o  answer with OR 𝑥 ∈ (180°; 270°) correct notation (2) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 6 DBE/2016 CAPS – Grade 10 Exemplar – Memorandum QUESTION 5 5.1 If they are equiangular OR answer Their corresponding sides are proportional (1) 5.2.1 in ∆ABC and ∆EDC statement and ˆ = ECB ˆ = 760 .....vert opp angles reason ACB  statement and 0 0 ˆ= Eˆ= 180 − 76 = 520.....alt angles, AB || DE A reason 2  statement and ˆ D ˆ= 52 ...alt angles, AB || DE 0 reason B =  statement and ∴∆ABC ||| ∆EDC ( A. A. A ) reason (4) 5.2.2 If at least one pair of corresponding sides is equal. reason (2) 5.3.1 x = 38mm....midpoint theorem value of x reason (2) 5.3.2 = α 46 .....alt ∠s; YZ // DE o value of α reason (2) 5.3.3 ˆ + 460 + 40 DFE = 0 1800 ∠ sum in ∆ ˆ = 1800 − 860 = DFE 940  two statements = ˆ YFZ ˆ DEF vert.opp.∠s and reasons ˆ = 940 answer YFZ Or OR ˆ = 460 + 400 Or EFZ ˆ = 460 + 400 ext. ∠ of ∆DFE  two statements DFY and reasons ˆ = 1800 − 860 ∠s on a str line DFZ or EFY YFZ answer ˆ ∴ YFZ 94 = 0 OR OR  two statements ˆ YFZ= 1800 − (α + β ) sum of ∠s of ∆FYZ and reasons answer ( = 1800 − 460 + 400 ) alt. ∠ =s ; DE // YZ (2) = 940 [13] Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 7 DBE/2016 CAPS – Grade 10 Exemplar – Memorandum QUESTION 6 6.1.1 Both opposites sides are parallel answer (1) 6.1.2 2 x + 20 = 5 x − 40....opp ∠s of //m statement 3 x = 60 o reason simplification x = 20o answer (4) 6.2.1 8x + 20 + 4x + 20 + x – 20 + 5x – 20 = 3600 ( sum of ∠s in quad ) statement 18x = 3600 reason x = 200 simplification answer (4) 6.2.2 𝐴̂ = 8𝑥 + 2 = 8(20°) + 2° = 162° value of A 𝐵� = 4𝑥 + 2 = 4(20°) + 2° = 82° 𝐶̂ = 5𝑥 − 2 = 5(20°) − 2° = 98° value of D 𝐷� = 𝑥 − 2 = 20° − 2° = 18° 𝐴̂ + 𝐷 � = 180° reason 𝐴̂ + 𝐷 � = 162° + 18° = 180° Hence AB // DC co-int ∠s supp OR Therefore ABCD is a trapezium. value of B̂ Or B̂ + Cˆ = 820 + 920 = 1800 value of Ĉ Hence AB // DC co-int ∠s supp ˆ =1800  B̂ + C Therefore ABCD is a trapezium. reason (4) [13] QUESTION 7 7.1 𝐶𝑃2 = 𝑃𝐴2 + 𝐴𝐶 2 (Pythagoras) statement 102 = 62 + 𝐴𝐶 2 reason AC = 8 m substitution answer (4) 7.2 Let the new point be R such that AR = 5 m statement 𝐶𝑅 2 = 𝑅𝐴2 + 𝐴𝐶 2 (Pythagoras) substitution 𝐶𝑅 2 = 52 + 82 answer CR = √89 m (3) [7] Copyright reserved Please turn over

Downloaded from hlayiso.com Technical Mathematics/P2 8 DBE/2016 CAPS – Grade 10 Exemplar – Memorandum QUESTION 8 8.1.1 1220 + 0, 46 × 600  × 600 + 27.6′ = 1220 + 27, 6′ answer only full marks = 1220 + 27′ + 0, 6 × 600  answer = 1220 27′36′′ (3) 8.1.2  59 13  59 83059′13′′ =  83 + +    60 60 × 60  answer only full marks 60 13 = 83,990  602  answer (3) 8.2 s 4 formula θ= = r 6 2 2 =  3 3 2 1800 θ= × conversion 3 π = 38,390 answer (4) 8.3 4π 6π − 150 + 22π 3  3 × 6π + 4π 3 = − 150 3 22π = − 150 3  22  conversion = ×1800  − 150  3  0 answer = 1305 (3) [13] TOTAL: 100 Copyright reserved