# Mathematics Grade 12 Paper 2 NSC November 2018 Memo > Grade 12 Mathematics Paper 2 memorandum from the NSC November Exam 2018, 24 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/mathematics-p2-nov-2018-final-memo-afr-eng/7b9169bc-bd3a-413d-b53f-c9dc4f5da641 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Document type: Memorandum - Year: 2018 - Exam period: NSC November Exam - Paper: 2 - Curriculum: CAPS - Pages: 24 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematics](https://www.hlayiso.com/library/grade-12/mathematics) › [Mathematics Grade 12 Paper 2 NSC November 2018 Memo](https://www.hlayiso.com/library/mathematics-p2-nov-2018-final-memo-afr-eng/7b9169bc-bd3a-413d-b53f-c9dc4f5da641) ## About this document Mathematics · Grade 12 · NSC November Exam · 2018. Memorandum, 24 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 MATHEMATICS P2/WISKUNDE V2 NOVEMBER 2018 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 24pages. Hierdie nasienriglyne bestaan uit 24 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 2 DBE/November 2018 NSC/NSS – 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 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 3 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 1.1.1 140 items  answer (1) 1.1.2 Modal class/modale klas: 20 < x ≤ 30 minutes  answer OR/OF (1) 20 ≤ x < 30 minutes  answer (1) 1.1.3 Number of minutes taken = 20 minutes  answer (1) 1.1.4 140 – 126 [Accept: 124 to 128]  126 14 orders (12 to 16)  answer Answer only: Full marks (2) 1.1.5 75th percentile is at 105 items  105 =37 minutes [accept 36 – 38 minutes]  answer Answer only: Full marks (2) 1.1.6 Lower quartile is at 35 items =21,5 min [accept 21 – 23 min] Answer only:  lower quartile (Q 1 ) IQR = 37 – 21,5 Full marks = 15,5 min [accept 13 – 17 min]  answer (2) 35 70 75 80 80 90 100 100 105 105 110 110 115 120 125 1.2.1(a) 1420  1420 x= Answer only: Full marks 15 = R 94,666.. = R94,67  answer (2) 1.2.1(b) σ = R 22,691... = R 22,69  answer (2) 1.2.2(a) They both collected the same (equal) amount in  answer tips, i.e. R1 420 over the 15-day period. Hulle albei het dieselfde bedrag met fooitjies (1) ontvang, nl. R1 420 oor die 15 dae-tydperk 1.2.2(b) Mary's standard deviation is smaller than Reggie's which suggests that there was greater variation in the amount of tips that Reggie collected each day  explanation compared to the number of tips that Mary collected each day. Marie se standaardafwyking is kleiner as Reggie s'n wat beteken dat daar groter variasie/verspreiding in die fooitjies was wat Reggie elke dag ontvang (1) het in vergelyking met die getal fooitjies wat Marie elke dag ontvang het. [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 4 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 2.1 251 km/h  answer (1) 2.2.1 r = 0,52 OR C  answer (1) 2.2.2 The points are fairly scattered and the least squares regression line is increasing.  reason (1) Die punte is redelik verspreid en die kleinstekwadrate- regressielyn neem toe. 2.3 There is a weak positive relation hence the height could have an influence  answer (1) Daar is 'n swak positiewe verband, tog kan die lengte 'n invloed hê. OR/OF There is no conclusive evidence that the height of a player will influence his/her tennis serve speed.  answer Daar is geen duidelike bewys dat die lengte van die (1) speler sy/haar afslaanspoed kan beïnvloed nie. OR/OF There is no conclusive evidence that a taller person will serve faster than a shorter person.  answer Daar is geen duidelike bewys dat 'n langer speler (1) vinniger sal afslaan as 'n korter een nie. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 5 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne 2.4 For (0 ; 27,07), it means that the player has a height of 0 m but can serve at a speed of 27,07 km/h. It is impossible for a person to have a height of 0 m.  explanation (1) (0 ; 27,07) beteken dat 'n speler 'n lengte van 0 m kan hê en teen 'n spoed van 27,07 km/h kan afslaan. Dit is onmoontlik om 'n lengte van 0 m te hê. OR/OF This means that the player does not exist and  explanation therefore cannot serve and have a serve speed. (1) Dit beteken dat die speler nie bestaan nie en daarom nie kan afslaan en 'n afslaanspoed hê nie. [5] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 6 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 y K(– 1 ; 2) 78,69o θ P x O N(1 ; –1) L(–5 ; –2) o 78,69 T(–6; – 3) 13 13 o 22,62 M(–3 ; –5) 3.1.1 y − y1 m KN = 2 x 2 − x1 2 − (−1)  correct substitution m KN = Answer only: Full marks −1−1 3  answer =− 2 (2) 3.1.2 3 3 tan θ = m KN = −  tan θ = m KN = − 2 2 θ = 180° − 56,31° Answer only: Full marks θ = 123,69°  answer (2) 3.2 Inclination KL = 123,69° − 78,69° = 45° [ext ∠∆] S tan 45° = m KL = 1  tan 45° = m KL = 1 (2) 3.3 y = x+c 2 = −1 + c  substitute (– 1 ; 2) and m c=3 y=x+3  equation (2) OR/OF y − y1 = 1( x − x1 ) y − 2 = 1( x − (−1))  substitute (– 1 ; 2) and m y = x+3  equation (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 7 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne 3.4 KN = (1 + 1) 2 + (−1 − 2) 2  substitute K and N into distance formula KN = 13 or 3,61 Answer only: Full marks  answer (2) 3.5.1 ( x + 3) 2 + ( y + 5) 2 = 13 ...(1)  equation (1) L is a point on KL y=x+3 …(2) (2) in (1): ( x + 3) 2 + ( x + 3 + 5) 2 = 13  substituting eq (2) x 2 + 6 x + 9 + x 2 + 16 x + 64 = 13 2 x 2 + 22 x + 60 = 0 x 2 + 11x + 30 = 0  standard form ( x + 5)( x + 6) = 0 x = –5 or x = –6 y = –2 or y = –3  x-values L(–5 ; –2) or (–6 ; –3)  y-values (5) OR/OF ( x + 3) 2 + ( y + 5) 2 = 13 ...(1)  equation (1) L is a point on KL y=x+3 ∴x=y–3 …(2) (2) in (1): ( y − 3 + 3) 2 + ( y + 5) 2 = 13  substituting eq (2) y 2 + y 2 + 10 y + 25 = 13 2 y 2 + 10 y + 12 = 0 y2 + 5y + 6 = 0  standard form ( y + 2)( y + 3) = 0 y = –2 or y = –3  y-values (both) x = –5 or x = –6  x-values (both) L(–5 ; –2) or (–6 ; –3) (5) 3.5.2 Midpoint of KM: (– 2 ; – 1,5)  midpoint of KM x +1 y −1 3 ∴ L = −2 and L =− 2 2 2 ∴ L(– 5 ; – 2) x value  y value (3) OR/OF m KN = m LM  m LM = m KN y − (− 5) 3 =− x − (− 3) 2 2( x + 3 + 5) = −3( x + 3) 2 x + 16 = −3x − 9 5 x = −25 Answer only: Full marks x = −5 x value ∴ L(– 5 ; – 2)  y value (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 8 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF N→M: N→K: ( x ; y ) → ( x − 4 ; y − 4) ( x ; y ) → ( x − 2 ; y + 3) ∴ L(– 1 – 4 ; 2 – 4) OR/OF ∴ L(– 3 – 2 ; – 5 + 3) ∴ L(– 5 ; – 2) ∴ L(– 5 ; – 2) OR/OF N→M: N→K: ( x ; y ) → ( x − 4 ; y − 4) ( x ; y ) → ( x − 2 ; y + 3) ∴ L(– 1 – 4 ; 2 – 4) OR/OF ∴ L(– 3 – 2 ; – 5 + 3) ∴ L(– 5 ; – 2) ∴ L(– 5 ; – 2) OR/OF N→M: N→K: ( x ; y ) → ( x − 4 ; y − 4) ( x ; y ) → ( x − 2 ; y + 3) ∴ L(– 1 – 4 ; 2 – 4) OR/OF ∴ L(– 3 – 2 ; – 5 + 3) ∴ L(– 5 ; – 2) ∴ L(– 5 ; – 2) OR/OF N→M: N→K: ( x ; y ) → ( x − 4 ; y − 4) ( x ; y ) → ( x − 2 ; y + 3) ∴ L(– 1 – 4 ; 2 – 4) OR/OF ∴ L(– 3 – 2 ; – 5 + 3) ∴ L(– 5 ; – 2) ∴ L(– 5 ; – 2) OR/OF N→M: N→K:  transformation ( x ; y ) → ( x − 4 ; y − 4) ( x ; y ) → ( x − 2 ; y + 3) ∴ L(– 1 – 4 ; 2 – 4) OR/OF ∴ L(– 3 – 2 ; – 5 + 3) ∴ L(– 5 ; – 2) x value  y value ∴ L(– 5 ; – 2) (3) 3.6 T(–6 ; –3) (from Question 3.5.1)  coordinates of T 2 2 KT = (−1 − (−6)) + (2 − (−3))  length of KT = 50 KN = 13 (CA from 3.4) 1 Area of ΔKTN = KT.KN sinLK̂N 2 1  substitution into area rule = 50 . 13 sin78,69° 2 = 12,50 square units  answer (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 9 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF In ∆KLM: TL 13 = sin22,62° sin78,69° TL = 1,414..  length of TL 2 2 KL = (−1 − (−5)) + (2 − (−2)) = 32 ∴ KT = 7,0708… 1  length of KT Area of ΔKTN = KT.KN sinLK̂N 2 1 = (7,0708). 13 sin78,69° 2  substitution into area rule = 12,50 square units  answer (4) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 10 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 y J S(6 ; 5) F x H K G(m ; n) 4.1 F(3 ; 1) x value  y value (2) 4.2 FS = (6 − 3) 2 + (5 − 1) 2  substitution of F FS = 5 &S  answer (2) 4.3 FH(FS) : HG = 1 : 2 ∴ HG = 2 FH = 10  HG = 10 (1) 4.4 Tangents from common/same point /  answer Raaklyne vanaf gemeenskaplike of dieselfde punt (1) 4.5.1 FĤJ = 90° [tan ⊥ radius / rkl ⊥ radius] SR FJ2 = 202 + 52 [Pyth theorem/stelling] S FJ = 425 or 5 17 or 20,62  answer (4) 4.5.2 2 2 ( x − m) + ( y − n) = 100  answer (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 11 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne 4.5.3 K(22; n) [radius ⊥ tangent]  K(22; n) GK = HG = 10 [radii] FH = FS = 5 [radii] m = 22 – 10 m = 12  value of m F, H and G are collinear [HJ is a common tangent] F, H en G is saamlynig [HJ is 'n gemeemskaplike raaklyn] FG 2 = (12 − 3) 2 + (n − 1) 2  subst. of F and G in distance formula 15 2 = 81 + (n − 1) 2  FG = 15 (n − 1) 2 = 144 n 2 − 2n − 143 = 0  simplification/ n − 1 = ±12 OR/OF (n + 11)(n − 13) = 0 standard form n ≠ 13 or n = −11 n = −11 or n ≠ 13  value of n ∴ G(12; –11)  coordinates of G (7) OR/OF K(22; n) [radius ⊥ tangent]  K(22; n) GK = HG = 10 [radii] FH = FS = 5 [radii] m = 22 – 10 m = 12  value of m Let J(22 ; y): FJ 2 = (22 − 3) 2 + ( y − 1) 2  subst. of F and J in distance formula 425 = 361 + y 2 − 2 y + 1  FJ = 425 0 = y 2 − 2 y − 63  standard form 0 = ( y − 9)( y + 7) ∴ y = 9 or/of y ≠ –7 ∴ n = 9 – 20 = –11  value of n ∴ G(12; –11)  coordinates of G (7) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 12 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 y P(k ; 1) 5 θ • x O R 5.1.1 k 2 = ( 5 ) 2 − 12  substitution into theorem of Answer only: full marks Pythagoras =4 k = −2  answer (2) 5.1.2(a) 1 tan θ = − answer 2 (1) 5.1.2(b) cos(180° + θ ) = − cos θ  reduction 2 Answer only: full marks  answer = 5 (2) 5.1.2(c) a+b sin(θ + 60°) = 20 LHS = sin θ cos 60° + cosθ sin 60°  expansion  1  1   2  3  =    +  −    subst of sin θ  5  2   5  2   subst of cos θ 1− 2 3  both special ∠s = 2 5 1− 2 3  1− 2 3 2 5 = 20 (5) 5.1.3 1 tan θ = − 2 ∴θ = 180° − 26,57° ∴θ = 153,43° θ tan(2θ − 40°) = tan[(2 × 153,43°) − 40°]  substitution = tan 266,87° =18,3  answer (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 13 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne cos x + sin x cos x − sin x 5.2 LHS = − RHS = 2 tan 2 x cos x − sin x cos x + sin x (cos x + sin x )2 − (cos x − sin x )2  single fraction = (cos x − sin x )(cos x + sin x ) cos 2 x + 2 sin x cos x + sin 2 x − cos 2 x + 2 sin x cos x − sin 2 x =  expansion cos 2 x − sin 2 x 2(2 sin x cos x)  simplification (both) = cos 2 x − sin 2 x 2 sin 2 x  double ∠ identity =  double ∠ identity cos 2 x = 2 tan 2 x = RHS OR/OF (5) cos x + sin x cos x − sin x LHS = − RHS = 2 tan 2 x cos x − sin x cos x + sin x (cos x + sin x )2 − (cos x − sin x )2  single fraction = (cos x − sin x )(cos x + sin x ) (cos x + sin x + cos x − sin x)(cos x + sin x − cos x + sin x)  difference of two = squares cos 2 x − sin 2 x (2 cos x)(2 sin x) = cos 2 x − sin 2 x  simplification (both) 2(2 sin x cos x) = cos 2 x − sin 2 x 2 sin 2 x  double ∠ identity = cos 2 x  double ∠ identity = 2 tan 2 x (5) = RHS OR/OF RHS = 2 tan 2 x 2 sin 2 x = cos 2 x 2(2 sin x. cos x)  double ∠ identity = 2 2  double ∠ identity cos x − sin x 4 sin x. cos x = cos 2 x − sin 2 x 1 + 2 sin x. cos x − (1 − 2 sin x. cos x )  identity & method = cos 2 x − sin 2 x (cos x + sin x ) 2 − (cos x − sin x ) 2  factorising numerator = and denominator (cos x + sin x)(cos x − sin x) (cos x + sin x) 2 (cos x − sin x) 2  writing as 2 terms = − (cos x + sin x)(cos x − sin x) (cos x + sin x)(cos x − sin x) cos x + sin x cos x − sin x = − = LHS (5) cos x − sin x cos x + sin x Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 14 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne 5.3 52 ∑ cos A 2 A =38 = cos 2 38° + cos 2 39° + cos 2 40° + ... + cos 2 51° + cos 2 52°  expansion = sin 2 52° + sin 2 51° + sin 2 50° + ... + cos 2 51° + cos 2 52°  co ratio = 7(1) + cos 2 45°  cos 2 45° 2 2  2  1   7 × identity = 7 +   or = 7 +     2   2 1 =7  answer 2 OR/OF (5) 52 ∑ cos A 2 A =38 = cos 2 38° + cos 2 39° + cos 2 40° + ... + cos 2 51° + cos 2 52°  expansion = (cos 2 38° + sin 2 52° ) + (cos 2 39° + sin 2 51°)... + cos 2 45°  pairing 2 = 7(1) + cos 45°  cos 2 45° 2 2  2  1   7 × identity = 7 +   or = 7 +   2     2 1 =7 2  answer (5) [23] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 15 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1 Period = 120°  answer (1) 6.2 3 2 = − 2 tan x  equating 2 3  tan  t  = − 1 2  3  general solution of 3 t = − 45° + k .180° t = 135° + k .180° OR/OF 3 2 2 t t = 90° + k .120° ; k ∈ Z t = − 30° + k .120° ; k ∈ Z 2  general solution of t (3) OR/OF 3 2 = − 2 tan x 2  equating 3  tan  t  = − 1 2   general solution of 3 3 t = 135° + k .360° or/of t = 315° + k .360° 3 2 2 t 2 t = 90° + k .240° or.of t = 210° + k .240° ; k ∈ Z  general solution of t (3) 6.3 y asymptotes: x = ± 60°; x = 180° 2 x-intercepts 0°; ±120°  negative shape x –120° –90° – 60° –30 ° 0° 30° 60° 90° 120° 150° 180°  (90° ; 2) or –2 ( − 30° ; 2) or f (30° ; − 2) or ( − 90° ; − 2) (4) 6.4 x ∈ (−60° ; − 30°] or (60° ; 90°]  interval  interval  notation OR/OF (3) –60°< x ≤ –30° or 60° < x ≤ 90°  interval  interval  notation (3) 6.5 3  g ( x) = − 2 tan  ( x + 40°) = f ( x + 40°)  Translation of 40° 2   to the left Translation of 40° to the left / skuif met 40° links (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 16 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 A 30° 2x 3h h D B C 7.1 AB̂D = 30°  AB̂D = 30° h sin 30° = AB h h AB = OR AB = OR AB = 2h  answer sin 30° 1 (2) 2 OR/OF BÂD = 60°  BÂD = 60° h cos 60° = AB h h AB = OR AB = OR AB = 2h cos 60° 1  answer 2 (2) 7.2 2 2 2 BC = AB + AC − 2AB.AC cos BÂC  use of cosine rule in = (2h) 2 + (3h) 2 − 2(2h)(3h) cos 2 x ∆ABC = 13h 2 − 12h 2 (2 cos 2 x − 1)  substitution = 13h 2 − 24h 2 cos 2 x + 12h 2  double angle identity = 25h 2 − 24h 2 cos 2 x  25h 2 − 24h 2 cos 2 x BC = h 25 − 24 cos 2 x (4) [6] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 17 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 T M R 1 66° 2 2 1 N 1 2 O P 8.1.1 P̂ = M̂ 1 = 66° [tan chord theorem/raaklyn koordst] S R (2) 8.1.2 M̂ 2 = 90° [∠ in semi circle/∠ in halfsirkel] S R (2) 8.1.3 N̂ 1 = 180° − (90° + 66°) [sum of ∠s of /som van ∠e ∆ MNP] S = 24° (1) 8.1.4 Ô 2 = P̂ = 66° [corres. ∠s;/ooreenk ∠e, PM || OR] S R (2) 8.1.5 R̂ + N̂ 1 + N̂ 2 = 180° − 66° [sum of ∠s of/som van ∠e ∆ RNO] = 114° S R̂ = N̂ 1 + N̂ 2 = 57° [∠s opposite = radii/∠e teenoor = radii] S/R ∴ N̂ 2 = 33° S (3) OR/OF PÔR = 114° [∠s on straight line/∠e op reguitlyn] S PN̂R = 57° [∠ at centre = twice ∠ at circumference/ S/R midpts∠ = 2 × omtreks∠] ∴ N̂ 2 = 33° S (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 18 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne 8.2 B A M 20 C 21 D F H 15 G 8.2.1 FC || AB || GH [opp sides of rectangle/teenoorst sye v reghoek] R (1) 8.2.2 AC AF = [line || one side of ∆] OR [prop theorem; FC || GH] S R CH FG [lyn || een sy van ∆] OF [eweredighst; FC || GH] AC 20 = 21 15 20 × 21 AC = 15 = 28  AC DB = AC = 28 [diags of rectangle =/hoeklyne v reghoek = ] S 1 DM = DB = 14 [diags of rectangle bisect/hoekl v reghoek halveer] S 2 (5) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 19 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 19 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 9.1 J M 2 O1 K L 9.1 Constr/Konstr.: Draw KO and MO/Trek KO en MO  construction Proof: Ô1 = 2Ĵ [∠ at centre = twice ∠ at circumference]  S/R [midpts∠ = 2 × omtreks∠] Ô 2 = 2L̂ [∠ at centre = twice ∠ at circumference]  S Ô1 + Ô 2 = 360° [∠s around a point /∠e om 'n punt]  S/R S ∴ 2Ĵ + 2L̂ = 360° ( ) ∴ 2 Ĵ + L̂ = 360° ∴ Ĵ + L̂ = 180° (5) OR/OF  construction Constr/Konstr.: Draw KO and MO/Trek KO en MO Proof: Let Ĵ = x SR Ô1 = 2 x [∠ at centre = twice ∠ at circumference] [midpts∠ = 2 × omtreks∠]  S/R Ô 2 = 360° − 2 x [∠s around a point /∠e om 'n punt] S ∴ L̂ = 180° − x [∠ at centre = twice ∠ at circumference] ∴ Ĵ + L̂ = 180° (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 20 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 20 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne 9.2 A B 1 2 x 1 S 2 3 1 2 3 T 4 1 y 1 2 2 D R C 9.2.1(a) B̂1 = x [∠s in same seg/∠e in dieselfde segm] S R (2) 9.2.1(b) B̂ 2 = y [ext ∠ of cyclic quad/buite∠ koordevh] S R (2) 9.2.2 Ĉ = 180° − ( x + y ) [sum of ∠s of/som v ∠e, ∆ ACR] S SB̂D + Ĉ = x + y + 180° − ( x + y ) SB̂D + Ĉ = 180° S SCDB is a cyclic quad [converse opp angles of cyclic quad] R [omgekeerde teenoorst ∠e koordevh] (3) OR/OF Ŝ1 = T̂2 [∠s in same segment/∠e in dies. segment] S T̂2 = D̂1 + D̂ 2 = BD̂R [ext ∠ of cyc quad/buite∠ koordevh] ∴ Ŝ1 = BD̂R S ∴ SCDB is cyc quad [ext ∠ of quad = opp ∠/buite∠ = tos ∠] R (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 21 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 21 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne 9.2.3 T̂4 = y − 30° [ext ∠ of/buite ∠ ∆ TDR] S T̂1 = y − 30° [vert opp ∠s =/regoorst ∠e =] S y − 30° + x + 100° = 180° [sum of ∠s of/som v ∠e, ∆ AST] ∴ x + y = 110° SB̂D = 110° S ∴ SD not diameter [line does not subtend 90° ∠ ] SD nie 'n middellyn [lyn onderspan nie 90°∠] R (4) OR/OF AŜT = Ĉ + D̂ 2 [ext ∠ of/buite ∠ ∆ SCD] S Ĉ = 100° − 30° = 70° S SB̂D = 180° − 70° [opp ∠s cyclic quad/ teenoorst ∠e kdvh] = 110° S ∴ SD not diameter [line does not subtend 90° ∠ ] SD nie 'n middellyn [lyn onderspan nie 90°∠] R (4) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 22 Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 22 DBE/November 2018 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 A 2 1 1 D x 2 B 1 M 2 C 10.1.1 Â 2 = Â 1 = 90° − x [= chords subtend = ∠s  S R = kde onderspan =∠e] D̂ 2 = x [exterior angle of cyclic quad/buite∠ koordevh.]  S/R ∴ Ĉ 2 = 90° − x [sum of ∠s of/som v ∠e, ∆DCM]  Ĉ 2 = 90° − x ∴ Ĉ 2 = Â1 = 90° − x ∴ MC is a tangent to the circle at C [converse: tan chord th] MC is 'n raaklyn by C [omgekeerde raakl koordst] R (5) OR/OF Â 2 = Â 1 = 90° − x [= chords subtend = ∠s/ SR = kde onderspan =∠e] Ĉ1 + Ĉ 2 = x [sum of ∠s of/som v ∠e, ∆ACM]  Ĉ1 + Ĉ 2 = x ∴ Ĉ1 + Ĉ 2 = B̂ = x ∴ MC is a tangent to the circle at C [converse : tan chord th] MC is 'n raaklyn by C [omgekeerde raakl koordst] R (5) OR/OF In ∆AMC and ∆ACB: Â 2 = Â 1 = 90° − x [= chords subtend = ∠s/ SR = kde onderspan =∠e] AM̂C = AĈB = 90° [given] ∴ Ĉ1 + Ĉ 2 = B̂ = x  Ĉ1 + Ĉ 2 = x Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Text truncated here. 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