Mathematics Grade 12 Paper 2 NSC November 2017 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE 12/GRAAD 12 MATHEMATICS P2/WISKUNDE V2 NOVEMBER 2017 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 29 pages. Hierdie nasienriglyne bestaan uit 28 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 2 DBE/November 2017 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 guidelines. 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, merk slegs die EERSTE poging. • As 'n kandidaat 'n antwoord van 'n vraag doodtrek en nie oordoen nie, merk die doodgetrekte poging. • Volgehoue akkuraatheid word in ALLE aspekte van die nasienriglyne toegepas. Hou op nasien by die tweede berekeningsfout. • Aanvaar van antwoorde/waardes om 'n probleem op te los, word NIE toegelaat nie. GEOMETRY 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 a 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 the statement AND reason are both correct. S/R Ken 'n punt toe as beide die bewering EN rede korrek is. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 3 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 Time for 100 m sprint (in seconds) 10,1 10,2 10,5 10,9 11 11,1 11,2 11,5 12 12 12,2 12,5 Tyd vir 100 m-naelloop (in sekondes) Distance of best long jump (in metres) 8 7,7 7,6 7,3 7,6 7,2 6,8 7 6,6 6,3 6,8 6,4 Afstand van beste sprong in verspring (in meter) 1.1 a = 14,343… = 14,34  value of a b = –0,642… = – 0,64  value of b (3) 1.2 y = 14,34 − 0,64(11,7)  substitution correctly = 6,85  answer (2) OR/OF answer y = 6,83 (calculator / sakrekenaar) (2) 1.3 The gradient increases / Die gradient neem toe  increases/neem toe The point (12,3 ; 7,6) lies some distance above the current data.  reasoning in words/ /Die punt (12,3 ; 7,6) lê bokant die huidige data. redenasie in woorde (2) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 4 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 12 13 13 14 14 16 17 18 18 18 19 20 21 21 22 22 23 24 25 27 29 30 36 2.1.1 472 472 x=  23 23 x = 20,52 seconds / sekonde  answer (2) 2.1.2 Q 1 = 16  Q1 Q 3 = 24  Q3 IQR/IKO = Q 3 – Q 1 = 24 – 16 = 8  answer (3) 2.2 20,52 + 5,94 = 26,46  26,46 ∴ > 26,46 ∴ 4 girls/dogters  answer (2) 2.3  whiskers ending at 12 & 36  Q 1 = 16 & Q 3 = 24 (box)  Q 2 = 20 12 14 16 18 20 22 24 26 28 30 36 (3) 2.4.1 Girls / Meisies  answer (1) 2.4.2 Five-number summary of boys: (15 ; 21 ; 23,5 ; 26 ; 38) None of the boys / Nie een van die seuns nie  answer 5 girls completed in less than 15 seconds which was the  reason/rede minimum time taken by the boys. 5 meisies voltooi in minder as 15 sekondes, wat die minimumtyd is wat die seuns geneem het. (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 5 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 y A  1 F 3 ; 3   2 G x E O C (8 ; –4) B (– 6 ; – 5) 3.1.1 y 2 − y1 mFC = x 2 − x1 1 3 − (−4) = 2  substitution of (8 ; –4) 3−8  1 3 & 3; 3  = −  2 2  gradient y = mx + c y − y1 = m( x − x1 ) 3 y=− x+c 2 3 3 − 4 = − (8) + c OR/OF ( y − (−4)) = − ( x − 8)  substitution of m and 2 2 3 (8 ; –4) c =8 y + 4 = − x + 12 2 3 3  equation of AC y=− x+8 y=− x+8 2 2 (4) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 6 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne  substitution of (8 ; –4) (− 4) −  3 1   1 y 2 − y1  2 & 3; 3  mFC = = x 2 − x1 8−3  2 3  gradient = − 2 y = mx + c y − y1 = m( x − x1 )  substitution of m and 1 3  1 3  1 3 = − (3) + c  y − 3  = − ( x − 3) 3; 3  2 2  2 2  2  1 3 9 c =8 OR/OF  y − 3  = − x +  2 2 2  equation of AC 3 3 (4) y=− x+8 y=− x+8 2 2 3.1.2 AC: 3x + 2y = 16 and BG: 7x – 10y = 8 15x + 10y = 80  method /metode: 7x – 10y = 8 solving simultaneously / 22 x = 88 los gelyktydig op x=4  x coordinate (x > 0) 3(4) + 2y = 16 y=2  y coordinate ∴G(4 ; 2) (3) OR/OF 7 8 BG: 7x – 10y = 8 ∴y = x− 10 10 7 8 3  method: equating ∴ x− =− x+8 [CA from 3.1.1] metode: stel vgls gelyk 10 10 2 11 44 x= 5 5  x coordinate (x > 0) x=4 3(4) + 2y = 16  y coordinate y=2 ∴G(4 ; 2) (3) 3.2 xA + 4 y +2 1  equation ito x = 3 and A =3  equation ito y 2 2 2 ∴A(2 ; 5) (2) OR/OF by translation/deur translasie:  equation ito x xA = 3 − (4 − 3) = 2  equation ito y yA = 3 12 + (3 12 − 2) = 5 ∴A(2 ; 5) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 7 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 3.3 The coordinates of the midpt of AB / Die koordinaat van midpt van AB is:  2 + ( −6) 5 + ( −5)   subst A & B into midpt  ;  = ( −2 ; 0 ) formula  2 2   y coordinate = 0 But the y-coordinate of E is 0 ∴ E(–2 ; 0) is the midpoint of AB  E = midpt ∴ EF || BG [midpoint theorem/middelpuntst OR/OF  Reason line divides 2 sides of ∆ in prop/lyn (4) verdeel 2 sye van ∆ in dies verh] OR/OF The coordinates of the midpt of AB / Die koordinaat van midpt van AB is:  2 + ( −6) 5 + ( −5)   subst A & B into midpt  ;  = ( −2 ; 0 )  2 2  formula AE = (−2 − 2) 2 + (0 − 5) 2 = 41  lengths of AE & EB EB = (−2 − (−6))2 + (0 − (−5))2 = 41 ∴ In ∆ABG: AE = EB and AF = FG  AE = EB or E = midpt ∴ EF || BG [midpoint theorem/middelpuntst]  Reason (4) OR/OF Equation of AB:  5 − ( −5)  y − ( −5) =  ( x − ( −6))  2 − ( −6 )  10 15 5 5 y +5= x+ ∴y = x+ 8 2 4 2  equation of AB x-intercept of AB: 5 5 0= x+ ∴ x = −2 4 2 ∴ E(–2 ; 0)  coordinates of E 1 BG: 7x – 10y = 8 3 −0 7 8 7 mEF = 2 = ∴ y= x− 3 − (−2) 10 10 10  gradient of EF 7 7 m EF = m BG = ∴ mBG = 10 10  gradient EF = gradient BG ∴ EF || BG (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 8 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 3.4  1  1 Midpoint of AC =  5 ;    5 ;   2  2 xD + (−6) y + (−5) 1 = 5 and D = 2 2 2 ∴ D(16 ; 6)  x value  y value (4) OR/OF  method finding x by translation/dmv translasie:  method finding y D(16 ; 6)  x value  y value (4) OR/OF − 5 − (−4) 1 5 − (−5) 5 mBC = = and mAB = = −6−8 14 2 − (−6) 4 1 1 34 AD: y − 5 = ( x − 2) ⇒ y = x + 14 14 7 5 5 CD: y + 4 = ( x − 8) ⇒ y = x − 14 4 4 5 1 34  equating x − 14 = x + 4 14 7  x value ∴ x = 16  y value y=6 (4) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 9 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 y R P ( –4 ; 5) M θ x S O K(0 ; –3) 4.1.1 5 − (−3) mPK =  substitution P & K into −4−0 gradient formula = −2  gradient of PK PK ⊥ SR [radius ⊥ tangent/raaklyn]  PK ⊥ SR OR r ⊥ tangent ∴ mPK × mRS = −1 1 ∴ mRS =  answer 2 (4) 4.1.2 1 y = x+c 2 1 5 = (−4) + c OR/OF ( y − 5) = 1 ( x − (−4))  substitution of m and P 2 2 c=7 ( y − 5) = 1 x + 2 2 1 1  equation y= x+7 y = x+7 2 2 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 10 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 4.1.3  − 4 + 0 5 + (−3)  M ;   2 2   x value of M ∴ M(–2 ; 1)  y value of M r 2 = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 r 2 = (−2 + 4) 2 + (1 − 5) 2 ∴ r 2 = 20  r 2 = 20 ∴ ( x + 2) 2 + ( y − 1) 2 = 20 or ( 20 ) 2  equation (4) OR/OF  − 4 + 0 5 + ( −3)   M (– 2 ; 1) M ;  ∴ M(–2 ; 1)  2 2  ( x + 2) + ( y − 1) = r 2 2 2 r 2 = 20 (−4 + 2) 2 + (5 − 1) 2 = r 2 ∴ r 2 = 20  equation ∴ ( x + 2) 2 + ( y − 1) 2 = 20 or ( 20 ) 2 (4) OR/OF  M (– 2 ; 1)  − 4 + 0 5 + ( −3)  M ;  ∴ M(–2 ; 1)  2 2  PK = (−4 − 0) 2 + (5 − (−3)) 2 = 80 r 2 = 20 80 r= = 20 2  equation 2 2 2 (4) ∴ ( x + 2) + ( y − 1) = 20 or ( 20 ) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 11 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 4.1.4 tan θ = mPK = −2  tan θ = −2 ∴ θ = 180° – 63,43° = 116,57°  size of θ PK̂R = 116,57° − 90° [ext ∠ of ΔMOK]  answer = 26,57° (3) OR/OF R 20 10 P 80 K In ∆RPK: PK = (0 − ( −4)) 2 + ( −3 − 5) 2 = 80 PR = ( −4 − 0) 2 + (5 − 7) 2 = 20  lengths of PK, PR & RK RK = 10 PK 2 + KR 2 − PR 2 ( 80 ) 2 + (10) 2 − ( 20 ) 2  correct values into cos PK̂R = = 2.PK.KR 2( 80 )(10) cos rule 2 5 = 5 PK̂R = 26,57°  answer (3) OR/OF 20 80 sin PK̂R = cos PK̂R =  lengths of sides 10 OR/OF 10  ratio PK̂R = 26,57° PK̂R = 26,57°  answer (3) OR/OF 20  lengths of sides tanPK̂R =  ratio 80  answer PK̂R = 26,57° (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 12 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 4.1.5 RS || tangent at K(0 ; – 3 ) 1 ∴ m PS = m tang =  gradient 2 1 ∴ y = x −3  equation 2 (2) OR/OF 1− 5 mPK = = −2 −2+4 mPK × mtan g = −1 [radius ⊥ tangent/raaklyn] 1 ∴ mtan g =  gradient 2 1 ∴y= x−3  equation 2 (2) 4.2 t ∈ (−3 ; 7)  – 3 (A)  7 (CA from 4.1.2) OR/OF  correct inequality −3<t <7 (3)  – 3 (A)  7 (CA from 4.1.2)  correct inequality (3) 4.3 1 RS: y = x+7 ∴ S(–14 ; 0)  coordinates of S 2 SP = (−14 − (−4)) 2 + (0 − 5) 2 = 100 + 25 = 125  length of SP 1  correct base & height Area ∆SMK = . MK . SP 2 into Area rule 1 = ( 20 )( 125 )  correct substitution 2  answer = 25 square units (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 13 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF M(–2 ; 1) θ β S(–14 ; 0) K(0 ; –3) Let β = inclination of SM/ inklinasie van SM 1 RS: y = x + 7 ∴ S(–14 ; 0)  coordinates of S 2 SM = ( −14 − ( −2)) 2 + (0 − 1) 2 = 145  length of SM 1− 0 1 tan β = = ∴ β = 4,76° − 2 − (−14) 12 ∴ SM̂K = 116,57° – 4,76° [ext ∠ of ∆] = 111,81°  size of/grootte v SM̂K 1 Area ΔSMK = (SM)(MK).sinSM̂K 2 1 = ( 145 )( 20 ). sin 111,81° 2  correct substitution into = 24,9985 = 25 square units area rule  answer (5) OR/OF M 145 20 S(–14 ; 0) K(0 ; –3) 1 RS: y = x+7 ∴ S(–14 ; 0) 2  coordinates of S SK = (−14 − 0) 2 + (0 + 3) 2 = 205  length of SK ( 145 ) 2 + ( 20 ) 2 − ( 205 ) 2 2 29 cos SM̂K = =− 2( 145 )( 20 ) 29 SM̂K = 111,80°  size of /grootte v SM̂K 1 Area ΔSMK = (SM)(MK).sinSM̂K 2 1  correct substitution into = ( 145 )( 20 ). sin 111,81° 2 area rule = 24,9985 = 25 square units  answer (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 14 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne OR/OF Produce KS to T 1  coordinates of S RS: y = x + 7 ∴ S(–14 ; 0) 2 SK = (−14 − 0) 2 + (0 + 3) 2 = 205  length of SK & SM 2 2 SM = ( −14 − ( −2)) + (0 − 1) = 145 3 mSK = − ⇒ TŜO = 167,91° 14 1 mSM = ⇒ MŜO = 4,76° 12 MŜK = 180° − 167,91° + 4,76° = 16,85° 1  size of /grootte v MŜK Area ΔSMK = (SM)(SK).sinMŜK 2 1 = ( 145 )( 205 ).sin 16,85° 2  correct substitution into = 24,9985 = 25 square units area rule  answer (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 15 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 5.1 sin(A − 360°).cos(90° + A) cos(90° − A).tan(−A)  sin A sinA(−sinA) = –sin A sinA(− tanA)  sin A sin A  –tan A =  sin A  sinA    tan A =  cos A  cosA = cos A  answer (6) 5.2.1 2 2 t = ( 34 ) − (3) 2  substitution  answer ∴t = – 5 (2) 5.2.2 −5 tan β =  correct ratio 3 (1) 5.2.3 2 cos2β = 2 cos β − 1  compound formula 2  3   substitution = 2  −1  34   9   simplification = 2  − 1  34   answer 16 8 =− OR − (4) 34 17 OR/OF cos 2 β = 1 − 2 sin 2 β  compound formula 2  5   substitution = 1 − 2 −   34   25   simplification = 1 − 2   34   answer 16 8 (4) =− OR − 34 17 OR/OF cos 2 β = cos 2 β − sin 2 β  compound formula 2 2  3   5   substitution =  −−   34   34  9 25  simplification = − 34 34 16 8  answer =− OR − (4) 34 17 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 16 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 5.3.1 LHS = sin(A + B) – sin(A – B) = sin A.cos B + cos A.sin B – (sin A. cos B – cos A. sin B)  compound = sin A.cos B + cos A.sin B – sin A. cos B + cos A. sin B formula  compound = 2cos A.sin B formula = RHS (2) 5.3.2 sin 77° − sin 43° = sin(60° + 17°) − sin(60° − 17°)  60° + 17° = 2cos 60°.sin 17°  60° – 17° 1  simplify = 2 × × sin 17° 1 2  = sin17° 2 (4) OR/OF sin 77° − sin 43° = sin(60° + 17°) − sin(60° − 17°)  60° + 17° = (sin 60° cos17° + cos 60° sin 17°) −  60° – 17°  expansion (sin 60° cos17° − cos 60° sin 17°) 3 1 3 1 1 = cos17° + sin 17° − cos17° + sin 17°  2 2 2 2 2 = sin 17° (4) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 17 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1 y  (–90° ; –3) 1  (0 ; –1) f  x – intercepts: x –210° & 30° –270° –210° –180° –90° 0 30° 90°  shape –1 g –2 –3 (–90° ; –3) (4) 6.2 cos 2 x = 2 sin x − 1 1 − 2 sin 2 x = 2 sin x − 1  cos 2 x = 1 − 2 sin 2 x 2 sin 2 x + 2 sin x − 2 = 0  standard form sin 2 x + sin x − 1 = 0 − b ± b 2 − 4ac  using quadratic sin x = 2a formula − 1 ± 12 − 4(1)(−1) =  substitution into 2(1) quadratic formula −1+ 5 −1− 5 sin x = , since sin x = < −1 has no solution (4) 2 2 6.3 −1+ 5 sin x = = 0,618... 2 Reference ∠ = 38,17° ∴ x = 38,17° + k.360° or x = 141,83° + k.360° ; k ∈ Ζ  38,17° ∴ x = 38,17 ° or − 218,17°  141,83° y = 0,24 –218,17° ∴Points of intersection/snypunte:  0,24 (38,17 ° ; 0,24) and (−218,17° ; 0,24) (4) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 18 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 A y E k B x C D 7.1 AB̂C = 90°  answer (1) 7.2 In ∆ ABE: AB = tan y  correct ratio BE AB = k tan y  value AB In ∆ ABC: AB  correct ratio = sin x AC AB  AC as subject and AC = substitution sin x k tan y = sin x (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 19 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 7.3 180° − 2 x AD̂C = AĈD = = 90° − x  90° − x 2 DC AC = subst into sine rule sin2 x sin (90° − x) DC AC = 2 sin x cos x cos x  2 sin x cos x AC(2 sin x cos x)  cos x DC = cos x k tan y 2 sin x cos x = .  substitution sin x cos x (5) = 2k tan y OR/OF DC 2 = AD 2 + AC 2 − 2AD.AC cos 2 x substitution into cos rule = AC 2 + AC 2 − 2AC 2 cos 2 x = 2AC 2 (1 − cos 2 x)  factorisation = 2AC 2 (1 − 1 + sin 2 x)  1− 2 sin 2 x = 4AC 2 sin 2 x  DC ito AC and sin x DC = 2AC.sin x  k . tan y  = 2 .sin x  substitution  sin x  (5) = 2k . tan y OR/OF DC 2 = AD 2 + AC2 − 2AD.AC cos 2 x correct cos rule 2 2  k tan y   k tan y  = 2  − 2  cos 2 x  substitution  sin x   sin x  2k 2 tan 2 y 2k 2 tan 2 y  1− 2 sin 2 x = − (1 − 2 sin 2 x) sin 2 x sin 2 x 2k 2 tan 2 y 2k 2 tan 2 y  squaring and = 2 − 2 + 4k 2 tan 2 y multiplication sin x sin x DC = 4k 2 tan 2 y  4k 2 tan 2 y = 2k tan y (5) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 20 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 C E 1 2 15° 1 50° 2 D 1 F 2 3 A B 8.1.1 Ê = 50° − 15° = 35° [ext ∠ of ∆/buite ∠ van ∆] S S  R Â = 35° [alt ∠s / verwiss ∠e; CE || AB] (3) OR/OF Ê = 180° − (130° + 15°) = 35° [str line; ∠s of ∆/rt lyn;∠e van ∆]  S Â = 35° [alt ∠s / verwiss ∠e; CE || AB] S  R (3) OR/OF B̂ = 50° [∠s in same segment/∠e in dieselfde segment] S Ĉ 2 + 15° = 50° [alt ∠s / verwiss ∠e; CE || AB] ∴ Ĉ 2 = 35° Â = 35° [∠s in same segment/∠e in dieselfde segment] S  R (3) 8.1.2 Ĉ 2 = 35° [∠s in same segment/∠e in dieselfde segment] SR (2) 8.2 Ĉ 2 = Ê [from 8.1.1 and 8.1.2] S ∴ CF is a tangent to the circle [converse tan chord theorem] R ∴ CF is 'n raaklyn aan die sirkel [omgekeerde raakl koordst] (2) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 21 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 A 3x H E 12y K D F 2x B G 8y C 9.1.1 AF 3 x 3 AG 12 y 3 = = & = = BF 2 x 2 CG 8 y 2 AF AG AF AG ∴ =  = BF CG BF CG ∴ FG || BC [conv prop th/omg eweredigh st. OR line divides 2 sides of ∆ in prop/lyn verdeel 2 sye R v ∆ in dies verh] (2) 9.1.2 AG AH = [prop theorem/eweredigh st; GH || CK OR S  R GC HK line || to 1 side of ∆/lyn || 1 sy van ∆] AG AE = [prop theorem/eweredigh st; GE || CD] S GC ED AH AE (3) ∴ = HK ED 9.2 AE 3 AH 3 = and = ED 2 HK 2 AE 3 15 3  use of ratios = and = 12 2 HK 2 AD = 30  AE = 18 ∴ AE = 18 and HK = 10 KD = AD – AH – HK  HK = 10 ∴ HE = AE – AH = 18 – 15 = 30 – 15 – 10 =3 OR/OF =5 EK = ED – KD  HE = 3 or KD = 5 ∴ EK = HK – HE = 12 – 5 = 10 – 3  EK = 7 =7 =7 (5) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 22 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 O 1 M 2 1 T 2 V 1 W 2 N S 10.1 Line from centre to midpoint of chord/ R lyn vanaf midpt na midpt van koord (1) 10.2.1 OŴT = OŴS = 90° [radius ⊥ tangent/raaklyn] R ∴MN || TS [corresp ∠s =/ooreenkomstige ∠e = R OR co-int ∠s 180°/ko-binne∠e 180° (2) OR alternate ∠s/verwiss ∠e] 10.2.2 M̂ 1 = N̂ 1 [∠s opp = sides/ ∠e teenoor = sye] S M̂1 = T̂ [corresp ∠s/ooreenk ∠e; MN || TS] S ∴ N̂1 = T̂ S ∴TMNS is a cyclic quadrilateral [conv: ext∠ cyclic quad]  R TMNS is 'n koordevierhoek [omgek: buite∠ kdvh] (4) OR/OF M̂ 1 = N̂ 1 [∠s opp = sides/∠e teenoor = sye] S N̂1 = Ŝ [corresp ∠s/ooreenk ∠e; MN || TS] S S ∴ Ŝ = M̂1 ∴TMNS is a cyclic quadrilateral [conv: ext∠ cyclic quad]  R TMNS is 'n koordevierhoek [omgek: buite∠ kdvh] (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 23 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 10.2.3 In ∆OVN and ∆OWS Ô 2 = Ô 2 [common/gemeenskaplik]  S; S; S OR S; S; R OV̂N = OŴS = 90° [from 10.1] ON̂V = OŜW [sum ∠s ∆/som ∠e ∆] ∴∆OVN ||| ∆OWS [∠, ∠, ∠] ∆OVN ||| ∆OWS VN ON VN ON ∴ =  = WS OS WS OS 1 1 But VN = MN [given]  VN = MN 2 2 1 MN ON ∴2 =  substitution WS OS ∴OS.MN = 2ON. WS (5) OR/OF In ∆OVM and ∆OWS  S; S; S OR S; S; R OV̂M = OŴS = 90° [from 10.1] OM̂V = OŜW [sum ∠s ∆/som ∠e ∆] ∴∆OVM ||| ∆OWS [∠, ∠, ∠] ∆OVM ||| ∆OWS OM VM OM VM ∴ =  = OS WS OS WS 1 1 But VN = MN [given]  VN = MN 2 2 1 MN OM ∴2 = WS OS  substitution ∴OS.MN = 2ON. WS [VM = VN] (5) OR/OF If any other 2 ∆s are used, first need to prove that TW = WS by proving ∆OWT ≡ ∆OWS In ∆OVM and ∆OWT   similarity Ô1 = Ô1 [common/gemeenskaplik]  congruency OV̂M = OŴT = 90° [from 10.1] OM̂V = OT̂W [sum ∠s ∆/som ∠e ∆] ∴∆OVM ||| ∆OWT [∠, ∠, ∠]  VN = VM = VM OM 1 ∴ = MN WT OT 2 1 But VN = VM = MN [given] 2 and WT = WS and OT = OS [∆OWT ≡ ∆OWS] 1 MN 2 ON (5) ∴ = WS OS ∴OS.MN = 2ON. WS [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 24 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 11 R M K Q O N P 11.1 Construction: Draw diameter NR and draw RM  construction Konstruksie: Trek middellyn NR en verbind RM ON̂M + MN̂Q = 90° [ radius ⊥ tangent/raaklyn]  S /R NM̂R = 90° [∠ in semi circle/semi-sirkel]  S/ R ∴ MR̂N = 180° − (90° + 90° − MN̂Q) [sum ∠s ∆] = MN̂Q S but MR̂N = MK̂N [∠s same segment/∠e dieselfde segment] S/R ∴ MN̂Q = K̂ (5) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 25 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne R M K O Q N P 11.1 Construction: Draw diameter NR and draw RK  construction Konstruksie: Trek middellyn NR en verbind RK MN̂Q + RN̂M = 90° [radius ⊥ tangent/raaklyn]  S /R  S/ R NK̂R = 90° [∠ in semicircle/semi-sirkel] ∴ MK̂N = 90° − RK̂M S = 90° – RN̂M [∠s same segment/∠e dieselfde S/R segment] (5) ∴ MN̂Q = K̂ Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 26 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne M K Q O N P 11.1 Construction: Draw radii ON and OM  construction Konstruksie: Trek radiusse ON en OM MÔN = 2 K̂ [∠ at centre = 2∠ at circumf/midpts∠ = 2 omtreks∠]  S /R ON̂M + OM̂N = 180° − 2 K̂ [ ∠s of ∆/∠e van ∆] S 180° − 2 K̂ ON̂M = OM̂N = = 90° − K̂ [∠s opp = sides/∠e 2  S/ R teenoor = sye] ON̂Q = 90° [radius⊥tangent/radius⊥raaklyn] S/R (5) ∴ MN̂Q = K̂ Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 27 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 11.2 E D 1 2 3 1 x 2 2 1 A B C 11.2.1(a) Angle in a semi circle/ Hoek in halfsirkel R (1) 11.2.1(b) Exterior ∠of quad = opp interior ∠/ Buite ∠ van vierh = R teenoorst binne ∠ OR/OF Opp ∠s of quad supplementary/ Teenoorst ∠e van vierh is supplementêr (1) 11.2.1(c) tangent chord theorem/ raaklyn koord stelling R (1) 11.2.2(a) In ∆AEC Ê = 180° − (90° + x) [sum ∠s ∆] S = 90° − x D̂ 1 = 180° − (90° + x) [∠s on a straight line] = Ê = 90° − x S ∴ AD = AE [sides opp = ∠s/ sye teenoor = ∠e] R (3) 11.2.2(b) In ∆ADB and ∆ACD Â 2 = Â 2 [common] S D̂ 2 = Ĉ [proven] S B̂2 = D̂ 2 + D̂ 3 [sum ∠e ∆] S ∴∆ADB ||| ∆ACD (3) OR/OF In ∆ADB and ∆ACD Â 2 = Â 2 [common] S D̂ 2 = Ĉ [proven] S ∴∆ADB ||| ∆ACD [∠, ∠, ∠] R (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 28 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 11.2.3(a) AD AB  ratio = [||| ∆s] AC AD AD2 = AC . AB = 3r × r  substitution = 3r2 (2) 11.2.3(b) AD = AE = 3r [from 11.2.2(a) &11.2.3(a)] AB = r and BC = 2r ∴AC = 3r  AC ito r In ∆ACE: AC  trig ratio tan Ê = AE 3r = = 3  simplification 3r ∴ Ê = 60° ∴ D̂1 = 60° [from 11.2.2(a)]  all 3 ∠s = 60° ∴ Â1 = 60° [∠s of ∆ = 180°] (4) ∴ ∆ADE is equilateral/is gelyksydig OR/OF AD DB = [||| ∆s] AC CD 3r DB 3r DB  = = 3r CD 3r CD 1 1  = tan x tan x = 3 3  x = 30° ∴In ∆BDC: x = 30° ∴ Ê = 60° ∴ D̂1 = 60° [from 11.2.2(a)]  all 3 ∠s = 60° ∴ Â1 = 60° [∠s of ∆ = 180°] (4) ∴ ∆ADE is equilateral/is gelyksydig OR/OF AD DB = [||| ∆s] AC CD 3r DB CD = ∴ BD = CD 3r CD 3  BD = 3 DC 2 = BC 2 − DB2 CD 2 = 4r 2 − 3 3DC 2 = 12r 2 − CD 2 4CD 2 = 12r 2 DC = 3r  DC = 3r Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 29 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne EC 2 = EA 2 + AC 2 = 3r 2 + 9r 2 EC = 2 3r  EC = 2 3r ∴ ED = EC – DC = 3r  ED = EA = AD ∴ ED = EA = AD ∴ ∆ADE is equilateral/is gelyksydig (4) [20] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou