Mathematics Grade 12 Paper 2 NSC June 2025 Memo (Afrikaans)

SENIOR CERTIFICATE EXAMINATIONS/ NATIONAL SENIOR CERTIFICATE EXAMINATIONS MATHEMATICS P2/WISKUNDE V2 MAY/JUNE 2025/MEI/JUNIE 2025 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 23 pages. Hierdie nasienriglyne bestaan uit 23 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 2 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/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. LET WEL: • 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. • Aanvaar van antwoorde/waardes 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

Mathematics P2/Wiskunde V2 3 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 134 215 325 326 362 429 515 531 598 610 624 728 923 1 034 1 200 8554  8554 1.1 x= 15 = 570,27  answer (2) 1.2 σ = 291, 03  291,03 (1) 1.3 (279,24 ; 861,3)  (x − σ ; x + σ ) ∴10 premiums  answer (2)  118   k + 100  118 1791×  +  6763 ×   1791× 1.4  100   100  100 = 686, 44 15 k + 100  6763 × 100 sum of new premiums  = 686,44 k + 100 15 6763 × = 8183, 22 100 k + 100 = 1, 209... 100 k + 100 = 120,999... k = 21%  answer (4) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 4 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 Number of items/Aantal 10 3 20 14 17 9 12 18 15 19 items (x) Time (in minutes)/Tyd 5 5 9 7 6 6 8 11 10 12 in minute) (y) 2.1 Scatter plot / Spreidiagram 14  3 points correct 12 Time (in minutes) / 10  6 points correct 8 Tyd (in minute) 6  all points 4 correct 2 0 0 5 10 15 20 25 Number of items / Aantal items (3) 2.2 a = 3,079…  a = 3,08 b = 0,351…  b = 0,35 =yˆ 3, 08 + 0,35 x  equation (3) 2.3 r = 0,74  0,74 (1) 2.4 y = 3,08 + 0,35(13)  substitute x=13 y = 7,63  answer (2) OR y = 7,65 (calculator)  7,65 (2) 2.5 It does not make sense to pack 0 items in 3,08 minutes./  answer Dit maak nie sin dat 0 items in 3,08 minute gepak kan word nie. (1) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 5 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 y T V S(m ; 1) x W R O R/ 3.1 2 x − y + 10 = 0 2 x − 0 + 10 = 0  y=0 x = −5  x = –5 R(–5 ; 0) (2) 3.2 R(–5 ; 0) T(0 ; 10)  T(0 ; 10) (RT) 2 = (−5 − 0) 2 + (0 − 10) 2 OR (RT )2 = 5 2 + 10 2 (Pythag)  subst of R & T into distance formula or Pythagoras RT = 125 units OR/OF RT = 5 5 units  answer (3) 3.3 2 2RT = 5SR 2 2(125) = 5[(m − (−5)) 2 + (1 − 0) 2 ]  length of 2RT2 5[(m + 5) 2 + (1) 2 ] = 250  length of 5SR2 (m + 5) 2 + 1 = 50 m 2 + 10m − 24 = 0 OR/OF (m + 5) 2 = 49  standard form or (m − 2)(m + 12) = 0 m + 5 = ±7 isolating square m = −5 ± 7 m = 2 or m = – 12 m = 2 or m = – 12 N/A N/A ∴m = – 12 ∴m = – 12  negative answer (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 6 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne y T V S(m ; 1) x W R O R/ 1 − 10 3.4 mST =  substitution of S & T − 12 − 0 into gradient formula 3  mST mST = 4 4 1 ∴ mVR = −  mVR = − 3 mST 4 4 y = − x+c OR/OF y − y1 = − ( x − x1 ) 3 3 4 4 0 = − (−5) + c y − 0 = − ( x − (−5))  substitution of R 3 3 20 4 c=− y = − ( x + 5) 3 3 4 20 4 20  equation y =− x− y =− x− 3 3 3 3 (5) 3.5 4 20 VR: y = − x − 3 3 3 ST: y = x + 10 4 3 4 20 x + 10 = − x −  equating VR and ST 4 3 3 9 x + 120 = −16 x − 80 25 x = −200  simplification leading to x = −8 x = −8 y=4 ∴ V(−8 ; 4) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 7 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne y T V S(m ; 1) x R O R/ 3.6 R ′ (5 ; 0 ) VR = [−8 − (−5)]2 + (4 − 0) 2 = 5  length of VR  length of VT VT = (−8 − 0) 2 + (4 − 10) 2 =10 1 Area of RVTR ′ =(VR )(VT ) + 1 (RR ′)(OT ) 2 2 1 1  area ∆VRT = (5)(10) + (10 )(10 ) 2 2  area ∆ RTR ′ = 25 + 50 = 75 units 2  answer OR/OF (5) 3 ST: y = x + 10 4 3 0 = x + 10 4 40 1 x= − or −13  x-intercept of ST 3 3  40  W− ; 0  3   40  55 1 WR' = 5 −  −  = = 18  length of WR ′  3  3 3 Area of RVTR' = Area of ∆TWR' − Area of ∆WVR 1 40  1 40   area ∆ TWR ′ =  5 +  (10 ) −  −5 +  ( 4 ) 2 3  2 3   area ∆ WVR = 75 units 2  answer (5) [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 8 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne y QUESTION/VRAAG 4 E(–1 ; 3) D O M x S(–11 ; –2) T C(–7 ; p) 4.1 CÊM = 90°  answer (1) 0−3 4.2 mME = 3 − ( −1) 3 mME = − 3 4  mME = − 4 4 ∴ mED =  mED 3 4 4 3 = ( −1) + c 3 OR/OF 3 y −= 3 ( x − ( −1) )  substitution of E(–1 ; 3) 4 13 4 13 y = x+ y = x+  equation 3 3 3 3 (4) OR/OF 2 15  ( 5) +  2 DM =  [Pythagoras]  4 25 DM = or 6,25 units 4  13   coordinates of D ∴ D  − ;0   4  3−0 mED =  13  −1 −  −   4 4  mED ∴ mED = 3 4 4 3 = ( −1) + c 3 OR/OF 3 y −= 3 ( x − ( −1) )  substitution of E(–1 ; 3) 4 13 4 13  equation y = x+ y = x+ 3 3 3 3 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 9 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 4 13 4.3 y = x+ 3 3 4 13 0= x+ 3 3 13 xD = −  xD 4  13   xM − xD ∴ DM = 3 −  −   4 25 DM = or 6,25 units  answer 4 (3) OR/OF EM = 5 units  EM = 5 units 15 ED = units 4 2 15 ( 5) +    substitution of EM & ED 2 DM = [Pythagoras]  4 25  answer DM = or 6,25 units 4 (3) 4 13 4.4 EC: = y x+ 3 3 4 13 p = ( −7 ) +  substitution of C(–7 ; p) 3 3 into equation of EC p = −5 (1) OR/OF 4 mEC = 3 p −3 4 =  substitution of C(–7 ; p) −7 − ( −1) 3 into gradient of EC 4 p −3= ( −6 ) 3 p = −5 (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 10 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 4.5 M → E: (x ; y) → (x – 4 ; y + 3) [translation]  method: translation C → S: (–7 ; –5) → (–7 – 4 ; –5 + 3) ∴S(–11 ; –2)  xS = −11  yS = −2 (3) OR/OF M → C: (x ; y) →(x – 10 ; y – 5) [translation]  method: translation E → S: (–1 ; 3) → (–1 – 10 ; 3 –5) ∴S(–11 ; –2)  xS = −11  yS = −2 (3) OR/OF E(–1 ; 3) and C(–7 ; –5)  − 1 + (− 7 ) 3 + (− 5)   ;  [Midpoint of EC]  method: midpoint  2 2  = (− 4 ; − 1) S(x ; y) and M(3 ; 0) xS + 3 yS + 0 = −4 = −1 2 2 xS = −11 yS = −2  xS = −11  yS = −2 ∴S(–11 ; –2) (3) 4.6 rNEW = 5 + 7 rNEW = 12  rNEW = 12 MS = (3 − (− 11))2 + (0 − (− 2))2 MS = 200 or 10 2 or 14,14 units  MS 14,14 > 12 ∴ S(− 11 ; − 2 ) lies outside the circle  conclusion (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 11 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne y E(–1 ; 3) D O M x S(–11 ; –2) T C(–7 ; p) 4.7 Inclination of EM: tan M ˆ = m = −3 ME 4 ref. ∠ = 36,87° Inclination of EM = 143,13°  inclination of EM ∴ EM̂D = 36,87°  EM̂D ˆ = m = 5 or 1 Inclination of CM: tan M CM 10 2 ∴ M̂ = 26,57°  inclination of CM ∴ EM̂T = 26,57° + 36,87° = 63,44°  EM̂T But EM = MT [radii] 180° − 63,44° ∴ ET̂M = 2 ∴ ET̂M = 58,28°  answer (5) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 12 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 y 2 = 13 2 − (− 5) 2 5.1.1 [Pythagoras] y = –12  y = –12 θ sin 2 θ 2  12  13 = −   substitution  13  • (–5 ; y) 144 =  answer 169 (3) OR/OF sin 2 θ = 1 − cos 2 θ  square identity 2  5  substitution sin 2 θ = 1 −  −   13  144 sin 2 θ =  answer 169 (3) 5.1.2 tan (360° − θ ) = − tan θ  − tan θ  − 12  = −   −5  12  answer =− 5 (2) 5.1.3 cos(θ −135°) = cos θ cos 135° + sin θ sin 135°  cpd. ∠ expansion = cos θ (− cos 45°) + sin θ (sin 45°)  reduction  5  2   12  2  5  1   12  1  =  −  −  +  −   OR  −  −  +  −    substitution  13  2   13  2   13  2   13  2  7 2 7 =− =−  answer 26 13 2 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 13 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 5.2 2 cos(180° − x )sin (− x ) 1 − 2 cos 2 (90° − x )  cos(180° − x) = − cos x 2(− cos x )(− sin x )  sin( − x) = − sin x = 1 − 2 sin 2 x  cos 2 (90° − x) = sin 2 x 2 sin x cos x = cos 2 x  1 − 2 sin 2 x = cos 2 x sin 2 x  2 sin x cos x = sin 2 x = cos 2 x = tan 2 x  answer (6) 5.3 (tan 92°)(tan 94°)(tan 96°)...(tan 176°)(tan 178°)  sin 92°  sin 94°  sin 96°   sin 176°  sin 178°  =   ...    quotient identity  cos 92°  cos 94°  cos 96°   cos176°  cos178°   cos 2°  cos 4°  cos 6°   sin 4°  sin 2°   co-ratios =   ...    − sin 2°  − sin 4°  − sin 6°   − cos 4°  − cos 2°   reduction =1  answer (4) OR (tan 92°)(tan 94°)(tan 96°)...(tan 176°)(tan 178°) = (tan 92°)(tan 94°)(tan 96°)....(− tan 4°)(− tan 2°)  reduction  sin 92°  sin 94°  sin 96°   sin 4°  sin 2°   quotient identity =   ... −  −   cos 92°  cos 94°  cos 96°   cos 4°  cos 2°   cos 2°  cos 4°  cos 6°   sin 4°  sin 2°  =   ... −  −   co-ratios  − sin 2°  − sin 4°  − sin 6°   cos 4°  cos 2°  =1  answer (4) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 14 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1 LHS = 2 cos 2 (45° + x ) = 2 cos 2 ( 45° + x ) + 1 − 1  +1–1 = cos  2 ( 45° + x )  + 1  double angle = cos ( 90° + 2 x ) + 1  simplification (− = sin 2 x ) + 1  reduction = 1 − sin 2 x = RHS (4) OR/OF LHS = 2 cos 2 (45° + x ) = 2 ( cos ( 45° + x ) ) 2 = 2(cos 45° cos x − sin 45° sin x ) 2  compound ∠ expansion 2  2 2  = 2 cos x − sin x   subst. special ∠ values  2 2  1 1  = 2 cos 2 x − sin x cos x + sin 2 x   simplification 2 2  2 2 = cos x − 2 sin x cos x + sin x  sin 2 x + cos 2 x = 1 = 1 − sin 2 x = RHS (4) 6.2.1 = sin ( A − B ) − sin ( A + B ) LHS = sin A cos B − cos A sin B − ( sin A cos B + cos A sin B )  sin A cos B − cos A sin B = sin A cos B − cos A sin B − sin A cos B − cos A sin B  − sin A cos B − cos A sin B = −2cos A sin B = RHS (2) 6.2.2 sin 4 x − sin10 x = sin ( 7 x − 3 x ) − sin ( 7 x + 3 x )  4x =7x–3x & 10x =7x+3x = −2cos 7 x sin 3 x  answer (2) 6.2.3 sin 4 x − sin10 x sin = 3x −2 cos 7 x sin 3 x sin 3 x =  substitution 2 cos 7 x sin 3 x + sin 3 x 0 = sin 3 x(2 cos 7 x + 1) = 0  factorisation 1 sin 3 x = 0 or cos 7 x = −  both equations 2 3 x = 0° 7 x 120 = ° or 7 x 240 = ° x = 0° x = 17,14° x = 34,29°  answer N/A  answer (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 15 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 y g –180° 0 180° x f 7.1 Range of f: y ∈ [ −1 ; 3]  y ∈ [ −1 ; 3] (1) OR/OF −1 ≤ y ≤ 3  −1 ≤ y ≤ 3 (1) 7.2 Period of g: 180°  180° (1) 7.3 f increasing: x ∈ (− 180° ; 0°)  x ∈ (− 180° ; 0°) (1) OR/OF –180° < x < 0°  –180° < x < 0° (1) 7.4.1 g ( x ). f '( x ) < 0 x ∈ (− 90° ; 0°) ∪ (0° ; 90°)  x ∈ ( −90° ; 0° )  x ∈ ( 0° ; 90° ) (2) OR/OF –90° < x < 0° or 0° < x < 90°  –90° < x < 0°  0° < x < 90° (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 16 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 1 7.4.2 cos x ≤ − 2 2 cos x + 1 ≤ 0  2 cos x + 1 ≤ 0 x ∈ [− 180° ; − 120°] ∪ [120° ; 180°]  x ∈ [ −180° ; − 120°]  x ∈ [120° ; 180°] (3) OR/OF 2 cos x + 1 ≤ 0  2 cos x + 1 ≤ 0 –180° ≤ x ≤ –120° or 120° ≤ x ≤ 180°  –180° ≤ x ≤ –120°  120° ≤ x ≤ 180° (3) 7.5 g (x ) = sin 2 x h(x ) = sin 2(x − 45°) = sin (2 x − 90°)  sin ( 2 x − 90° ) = − sin (90° − 2 x ) = − cos 2 x  equation of h (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 17 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 D p A y x B C 8.1 p tan y =  correct trig ratio AB p  answer AB = tan y (2) 8.2 In ∆BAC: ˆ sin BAC ˆ sin ACB =  correct use of sine-rule BC AB sin (180° − 2 x ) sin x  substitute BC & AB = 2p  p   tan y    sin 2 x  tan y  = sin x ×    sin (180° − 2 x ) = sin 2 x 2p  p  2sin x cos x  tan y   sin 2 x = 2sin x cos x = sin x ×   2p  p   tan y  2cos x =  (2 p)  p  cos x = tan y (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 18 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne D p A y x B C 8.2 OR/OF In ∆BAC: ˆ BC2 = AB2 + AC2 − 2 ( AB )( AC ) cos BAC  correct use of cos-rule 2 2  p   p   p  p  (2 p) =   cos (180° − 2 x )  substitute AB & BC 2  +  − 2   tan y   tan y   tan y   tan y  2 p2 2 p 2 ( − cos 2 x ) 4 p2 = −  cos (180° − 2 x ) = − cos 2 x tan 2 y tan 2 y 4 p 2 tan 2 y = 2 p 2 (1 + cos 2 x ) 2 1 + 2 cos 2 x − 1  cos = 2 x 2 cos 2 x − 1 tan y = 2 2 2 tan y = cos x tan y = cos x (4) 8.3 cos x = tan y tan y = cos 60°  substitution of 60° tan y = 0,5 y = 26,57°  answer (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 19 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 9.1 D G E H F 9.1.1 HF GF 2 = = [line || to one side of ∆/lyn || een sy van ∆]  S  R EH GD 5 OR [prop theorem; GH||DE/eweredigheidst.; GH || DE] (2) 9.1.2 EH DG 5 = = [line || to one side of ∆/lyn || een sy van ∆] EF DF 7 OR [prop theorem; GH||DE/eweredigheidst.; GH || DE] EH 5 =  S 21 7 EH = 15cm  answer (2) OR/OF HF 2 = [line || to one side of ∆/lyn || een sy van ∆] EF 7 OR [prop theorem; GH||DE/eweredigheidst.; GH || DE] HF 2 =  S 21 7 HF = 6cm EH = 21 − 6 EH = 15cm  answer (2) 9.1.3 ∆FGH ||| ∆FDE [∠∠∠]  S (1) 9.1.4 GH FH GH FG = [|||∆'s] OR/OF = [|||∆'s]  S DE FE DE FD GH 2 =  S DE 7 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 20 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne K 9.2 T L 1 2 4 3 2 2 1 O 1 1 M 2 N 1 2 R P 9.2.1 L̂ 2 = 90° [∠ in semi-circle/ OR L̂1 = R̂ 1 [tan chord theorem/  S  R ∠ in halwe sirkel] raaklyn-koordst.] R̂ 2 = 90° [∠ in semi-circle] R̂ 1 = P̂ [tan chord theorem]  S/R ∠ in halwe sirkel] raaklyn-koordst.] ∴ L̂ 2 = R̂ 2 ∴ L̂1 = P̂ ∴ LT || PR [alt ∠s =/verw. ∠e = ] ∴ LT || PR [corresp. ∠s =/  R ooreenk. ∠e =] (4) 9.2.2 L̂1 = R̂ 1 [tan chord theorem/raaklyn-koordst.]  S R L̂1 = Ô1 [corresp. ∠s; LT || OM/ooreenk. ∠e; LT ||OM]  S/R ∴ R̂ 1 = Ô1  S ∴ L, O, R and M are concyclic. ∴ LORM is a cyclic quadrilateral [converse ∠s in the same seg/  R omgekeerde ∠e in dies. sirkel segm] (5) 9.2.3 O L̂R = M̂ 1 [∠s in the same seg/∠e in dieselfde segment]  S/R 2L̂ 3 = M̂ 1 [∠ at centre = 2 ×∠ at circumference/  S R midpt. ∠ = 2× omtreks ∠ ] ∴ O L̂R = 2L̂ 3 ∴L̂ 4 = L̂ 3  S ∴ LN bisects OL̂R (4) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 21 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 10.1 A P M 1 N ♠ ● K L ♠ ● B C 10.1 Construction: Draw line MN, where M and N are points on AB and AC respectively such that AM = PK and AN = PL.  construction In ∆ AMN and ∆PKL Aˆ = ˆP [given] AM = PK [construction] AN = PL [construction] ∆AMN ≡ ∆PKL [s∠s]  S/R ∴M ˆ = ˆ K 1 But Bˆ = Kˆ [given] ∴Mˆ = ˆ B  S 1 ∴MN || BC [corresp ∠s =/ooreenk. ∠e =]  S/R AB AC ∴ = [line || one side of ∆/lyn || een sy v ∆]  SR AM AN But AM = PK and AN = PL AB AC ∴ = PK PL (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 22 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 10.2 A B F C R O D 10.2.1 In ∆ CFO and ΔCBD Cˆ =C ˆ [common]  S CF̂O = CB̂D [corresp ∠s; BD||FO/ooreenk. ∠e; BD||FO]  S/R CÔF = CD̂B [sum of ∠s in ∆/binne ∠e v ∆] OR [corresp ∠s; BD||FO/ ooreenk. ∠e; BD||FO]  S/R OR R ∆ CFO |||ΔCBD [∠∠∠ ] (3) 10.2.2 FO CO = [||| ∆s]  S/R BD CD FO.CD = CO.BD But RD̂O = FĈO [given] ∴ BD = BC [sides opp equal ∠s/sye teenoor gelyke ∠e ]  S/R ∴ FO.CD = CO.BC (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Mathematics P2/Wiskunde V2 23 DBE/May/June/Mei/Junie 2025 SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne 10.2.3 RD = 6 units [DR = RB] RO 3 = RD 4 3 ∴ RO = (6) 4 RO = 4,5 units  S OR ⊥ BD [line from centre to midpt of chord/  S/R midpt. sirkel; midpt koord] ∴ DO = 6 2 + 4,5 2 [Pythagoras] DO = 7,5 units  S BF DO = [line || to one side of ∆/lyn || een sy v ∆]  S/R BC DC OR/OF [prop theorem; BD||FO/eweredigheidst.;BD||FO] BC = BD = 12 units [sides opp equal ∠s/sye teenoor gelyke ∠e]  S BF 7,5  S ∴ = 12 19,2 7,5 × 12 BF = 19,2 75 BF = units 16 (6) RO 4,5 3 10.2.4 tan RD̂O = = =  ratio RD 6 4 RD̂O = 36,87° ˆ  RDO FĈO = RD̂O = 36,87° [given] ∴ AB̂D = 73,74° [ext ∠ of ∆/buite ∠ v. ∆]  answer (3) OR/OF ˆ CD 2 = BC2 + BD 2 − 2 ( BC )( BD ) cos DBC 2 2 2 ˆ = 12 + 12 − 19, 2 cos DBC  substitution into 2 (12 )(12 ) cosine-rule ˆ = − 7 cos DBC 25 ˆ DBC = 106,26° ˆ  DBC ∴ AB̂D = 73,74° [∠s on a straight line/∠e op ’n reguitlyn]  answer (3) [20] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou