Mathematics Grade 12 P2 Eastern Cape Mock 2023 Memo (Eng)

NATIONAL SENIOR CERTIFICATE NASIONALE SENIORSERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2023 MATHEMATICS P2/WISKUNDE V2 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 18 pages. Hierdie nasienriglyn bestaan uit 18 bladsye.

2 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2023) QUESTION 1/VRAAG 1 Data Set / Datastel: 29 27 24 31 22 19 30 182 ✓✓ mean / gemiddelde 1.1.1(a) Mean / Gemiddelde = 7 = 26 (2) 1.1.1(b) Standard deviation / standaardafwyking = 4,14 ✓ SD / standaardafwyking (1) 1.1.2 ✓✓ calculations {26 – 4,14; 26 + 4,14} = {21, 86; 30, 14} berekeninge 5 players / spelers ✓ answer / antwoord (3) 1.1.3 Rugby team has the same average number of push-ups. ✓✓ for any two valid The rugby team results were clustered around the mean comments using SD because of the smaller standard deviation. and the mean Rugbyspan het dieselfde gemiddelde aantal opstote. Vir enige twee geldige Die rugbyspan se uitslae was rondom die gemiddelde opmerkings in gebruik van gegroepeer agv die kleiner standaardafwyking. standaardafwyking en die gemiddelde (2) 1.2.1 50% ✓ answer / antwoord (1) 1.2.2 Mean / Gemiddelde ✓ answer / antwoord Distribution skewed to the right. (positively skewed) ✓ reason / rede Verspreiding is skeef na regs (positief skeef) (2) [11] QUESTION 2/VRAAG 2 Third Term % 71 80 59 38 41 98 80 88 91 94 64 94 70 42 64 Derde Kw. % Final Term % 74 77 58 41 42 98 78 92 85 92 68 96 73 52 71 Finale Kw. % 2.1 a = 9, 035 b = 0, 895 yˆ = 9,035 + 0,895x ✓ for a / vir a ✓ for b / vir b ✓ for equation / vir vergelyking (3) 2.2 r = 0, 98 ✓ answer / antwoord (1) 2.3.1 𝑦 = 9,035 + 0,895(48) ✓ substitution / vervanging 𝑦 ≈ 52 ✓ answer / antwoord (2) 2.3.2 correlation is very strong OR 48 is within domain of ✓ answer / antwoord regression line. korrelasie is baie sterk OF 48 is binne die gebied van die regressie-lyn. (1) 2.4.1 50% is outside the domain of the line (data set) OR ✓ answer / antwoord (50 ; 80) is an outlier. 50% is buite die gebied van die lyn (datastel) OF (50 ; 80) is ’n uitskieter (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P2/WISKUNDE V2 3 OR/OF Alternative reason: 9,035 + 0,895(50) = 54 and not 80 2.4.2 Increase the gradient / Vermeerder die gradiënt ✓ answer / antwoord (1) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

4 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2023) QUESTION 3 / VRAAG 3 3.1 T(0; -2) ✓ answer / antwoord (1) 3.2.1 2−(−2) 4 ✓ substitution / vervanging mRT = = − = −2 −2−0 2 ✓ answer / antwoord (2) 3.2.2 tan 𝜃 = 1 1 ✓ for / vir tan 𝜃 = 2 2 ∴ 𝜃 = 26,57° ✓ for / vir 𝜃 = 26,57° tan 𝛼 = −2 ✓ for / vir tan 𝛼 = −2 ∴ 𝛼 = 116,57° ✓ for / vir 𝛼 = 116,57° ∴ 𝑅𝑇̂𝑆 = 116, 57° − 26,57° = 90° ✓ 𝑅𝑇̂𝑆 = 90° OR / OF OR / OF 1 1 𝑚 𝑇𝑆 = ✓ 𝑚 𝑇𝑆 = 2 2 ✓ 𝑚𝑅𝑇 = −2 𝑚𝑅𝑇 = −2 ✓✓ product of gradients 𝑚 𝑇𝑆 × 𝑚𝑅𝑇 = −1 produk van gradiënte 𝑅𝑇̂𝑆 = 90° ✓ 𝑅𝑇̂𝑆 = 90° (5) 3.3 1 𝑦 = 2𝑥 + 𝑐 ✓ gradient / gradiënt 1 ✓ substitution of point R or D 2 = 2 (−2) + 𝑐 vervanging van punt R of D 𝑐=3 1 ∴ 𝑦 = 2𝑥 + 3 ✓ answer / antwoord (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P2/WISKUNDE V2 5 3.4 RDST is a parallelogram (opposite sides parallel) ✓ substitution in the MP formula RDST is ’n parallelogram (teenoorst. sye ewewydig) vervanging in die MP formule Midpoint of DT / Middelpunt van DT: ✓ S/R 4+0 5−2 3 ( ; ) = (2; ) ✓ answer / antwoord 2 2 2 Midpoint of RS is the same as midpoint of DT (diagonals bisect each other.) Middelpunt van RS is dieselfde as die middelpunt van DT (hoeklyne halveer mekaar) 3 Midpoint of RS / Middelpunt van RS: (2; 2) OR / OF OR / OF 𝑆(6 ; 1) Midpoint of RS / Middelpunt van RS: ✓✓ 𝑆(6 ; 1) 6−2 1+2 3 ( ; ) = (2; ) ✓ answer / antwoord 2 2 2 (3) 3.5 N(0; 3) ✓ coordinates of N RN = √22 + 12 = √5 koördinate van N RT = √22 + 42 = √20 ✓ for / vir RN ✓ for / vir RT 1 Area /𝑂𝑝𝑝𝑒𝑟𝑣𝑙𝑎𝑘𝑡𝑒 = × √20 × √5 ✓ for the answer / 2 = 5 square units / eenhede2 vir die antwoord OR / OF OR / OF TN = 5 units / eenhede ✓ TN = 5 units / eenhede Height / Hoogte = 2 units / eenhede ✓ Height/hoogte = 2 units / 1 eenhede Area /𝑂𝑝𝑝𝑒𝑟𝑣𝑙𝑎𝑘𝑡𝑒 = × 5 × 2 ✓ sub. into formula 2 Area /𝑂𝑝𝑝𝑒𝑟𝑣𝑙𝑎𝑘𝑡𝑒 = 5 square units / eenhede2 vervanging in formule ✓ answer / antwoord (4) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

6 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2023) QUESTION 4 / VRAAG 4 4.1.1 ON = √(6 − 0)2 + (−8)2 ✓ substitution in correct formula vervanging in korrekte formule = √100 = 10 units / eenhede ✓ answer / antwoord (2) 4.1.2 ON = OS (tangents from the same point) ✓ S and/en R (raaklyne vanaf dieselfde punt) ✓ answer / antwoord ∴ 𝑝 = 10 units / eenhede Answer only = full marks (2) 4.1.3 ON ⊥ NM (tan – radius) / (raaklyn – radius) ✓ S and/en R −8 mON = 6 = 3 −4 ✓ gradient of ON / gradiënt van ON 3 ✓ gradient of NM mNM = 4 gradiënt van NM No penalty for omitting first line. (3) 4.1.4 q+8 mNM = 10 − 6 = 4 3 ✓ for subs and equating vir vervanging en gelyk stel q+8 3 = 4 4 q = −5 ✓ answer / antwoord (2) 4.2 MS = r = 5 units / eenhede ✓ radius / radius ✓ centre sub/ vervang middelpunt (𝑥 − 10)2 + (𝑦 + 5)2 = 25 ✓ answer / antwoord CA marking applies (p and q values) (3) 4.3 𝑘 = 5 OR/OF 𝑘 = 15 ✓ k = 5 OR/OF ✓k = 15 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P2/WISKUNDE V2 7 4.4 Coordinates of the point directly opposite N is C. Koördinate van die punt regoor N is C. ✓ formula and sub / 𝑥+6 𝑦−8 formule en vervanging C ( 2 = 10 ; 2 = −5) ✓ for x-coordinate C(14; −2) vir x-koördinaat Equation of the tangent at C: ✓ for y-coordinate Vergelyking van die raaklyn by C: vir y-koördinaat 4 𝑦 + 2 = − 3 (𝑥 − 14) ✓ substitution / vervanging 4 50 𝑦 = −3𝑥 + 3 ✓ for the answer / 50 vir die antwoord ∴0<𝑡< 3 ✓ for the value of 𝑡. vir die waarde van t (6) 4.5 They will not touch.  answer / antwoord The new circle is the old circle shifted up by 11.  any valid reason / enige geldige antwoord Hulle sal nie raak nie. Die nuwe sirkel is die ou sirkel 11 eenhede opwaarts geskuif. (2) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

8 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2023) QUESTION 5/VRAAG 5 5.1.1 sin 594  = sin 234   − sin 54  = − sin 54  –p = −p (2) 0 5.1.2 cos 36 =𝑝 ✓ answer / antwoord (2) 5.1.3 cos 18  = sin 72   for / vir sin 72 = sin 2(36 )  for / vir sin 2(36°) = 2 sin 36  cos 36   for / vir 2 sin 36 cos 36  1 − p 2  p  = 2    1  1    ( = 2 p 1− p2 ) (  for / vir 2 p 1 − p 2 ) cos 18 cos 36 = cos 2(18) ✓ 𝑝 = 2 𝑐𝑜𝑠 2 18 − 1 ✓ 2 𝑐𝑜𝑠 2 18 = 𝑝 + 1 𝑝+1 ✓ 𝑐𝑜𝑠 2 18 = 2 ✓ 𝑝+1 cos 18 = √ 2 OR/OF ✓ cos 18 = cos(54 − 36) ✓ cos 18 = cos 54 cos 36 + sin 54 sin 36 ✓ cos 18 = √1 − 𝑝2 . 𝑝 + 𝑝. √1 − 𝑝2 ✓ (4) cos 18 = 2𝑝 √1 − 𝑝2 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P2/WISKUNDE V2 9 5.2 cos140 − sin(90 −  ) sin 410 + cos( − ) cos(90 + 50) − cos  = sin 50 + cos  − sin 50 − cos   – sin 50  cos  = sin 50 + cos   sin 50  cos  −(sin 50 + cos  )  for the common factor = (sin 50 + cos  ) vir die gemene faktor = −1  for the answer vir die antwoord Please take note that: sin 50 and − sin 50 can be changed to cos 40 and − cos 40 (6) 5.3 cos( x + 65 ). cos( x + 20 ) − sin( x + 245 ). sin( x + 20 ) = cos( x + 65 ). cos( x + 20 ) + sin( x + 65 ). sin( x + 20 )  reduction / reduksie = cos[( x + 65 ) − ( x + 20 )]  compound angle = cos 45  saamgestelde hoek 1  cos 45  = 2  answer / antwoord (4) 5.4 1 1 𝑐𝑜𝑠 2 𝑥 − 𝑠𝑖𝑛2 𝑥 =  cos 2 x = 2 2 1  for / vir 2𝑥 in both cos 2 x = 2 quadrants / in beide 2𝑥 = 60° + 360°. 𝑘 or/of 2𝑥 = 300° + 360°. 𝑘 kwadrante 𝑥 = 30° + 180. 𝑘 or/of 𝑥 = 150° + 180. 𝑘 𝑘𝜖𝑍  𝑥 = 30° + 180. 𝑘  𝑥 = 150° + 180. 𝑘 OR / OF OR / OF  multiplying by 2 and using 2 𝑐𝑜𝑠 2 𝑥 − 2 𝑠𝑖𝑛2 𝑥 = 1 identity / vermenigvuldig met 2 𝑐𝑜𝑠 2 𝑥 − 2 𝑠𝑖𝑛2 𝑥 = 𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠 2 𝑥 2 en gebruik van identiteit 3 𝑠𝑖𝑛2 𝑥 = 𝑐𝑜𝑠 2 𝑥 1  3 𝑠𝑖𝑛2 𝑥 = 𝑐𝑜𝑠 2 𝑥 𝑡𝑎𝑛2 𝑥 = 3 1 1  𝑡𝑎𝑛 𝑥 = √3 𝑡𝑎𝑛 𝑥 = √3 𝑥 = 30° (reference angle / verwysingshoek)  answer / antwoord 𝑥 = 30° + 180° . 𝑘 , 𝑘 ∈ 𝑍 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

10 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2023) 5.5.1 sin 2𝜃 .tan 𝜃 LHS = cos 2𝜃+1  2 sin  cos  sin θ 2 sin θ cos θ . = cos θ sin  2 cos2 θ−1+1  cos  2sin2 θ =  2 cos2  −1 2cos2 θ 2sin2 𝜃 = tan2 θ  2cos2 𝜃 (4) 5.5.2 cos 2𝜃 + 1 = 0  cos 2𝜃 + 1 = 0 cos 2𝜃 = −1  cos 2𝜃 = −1 2𝜃 = 180°  2𝜃 = 180° 𝜃 = 90°  𝜃 = 90° (4) [30] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P2/WISKUNDE V2 11 QUESTION 6 / VRAAG 6 6.1 ∆ABC ≡ ∆ABD (SAS) / (SHS) ✓ S and/en R (1) 6.2 𝑥 ✓S cos α = AC 𝑥 AC = ✓ answer / antwoord cos α (2) 6.3 𝑥 2 𝑥 2 𝑥 𝑥 ✓ cos rule / cos-reël 𝐶𝐷2 = (cos 𝛼) + (cos 𝛼) − 2 (cos α) (cos α) cos(180° − 2𝛽) 𝑥2 𝑥2 𝑥2 𝐶𝐷2 = 𝑐𝑜𝑠2 𝛼 + 𝑐𝑜𝑠2𝛼 + 2 (𝑐𝑜𝑠2 𝛼) cos 2𝛽 2𝑥 2 ✓ simplification 𝐶𝐷2 = 𝑐𝑜𝑠2 𝛼 (1 + cos 2𝛽) vereenvoudiging 2𝑥 2 𝐶𝐷2 = 𝑐𝑜𝑠2 𝛼 (1 + (1 − 2𝑠𝑖𝑛2 𝛽)) ✓ double angles 2𝑥 2 𝐶𝐷2 = 𝑐𝑜𝑠2 𝛼 × 2𝑐𝑜𝑠 2 𝛽 expansion 4𝑥 2 𝑐𝑜𝑠2 𝛽 dubbelhoeke uitbreiding 𝐶𝐷2 = 𝑐𝑜𝑠2 𝛼 2𝑥 cos 𝛽 ∴ 𝐶𝐷 = ✓ simplification cos 𝛼 vereenvoudiging OR / OF OR / OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

12 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2023) 𝐴𝐷̂𝐵 = 𝛽 ̂𝐵 = 𝛽 ✓ for /vir 𝐴𝐷 𝐶𝐴̂𝐷 = 1800 − 2𝛽 and/en 𝐶𝐷 𝐴𝐶 𝐶𝐴̂𝐷 = 1800 − 2𝛽 0 = sin(180 − 2𝛽) sin 𝛽 𝐴𝐶 sin 2𝛽 ✓ use of sin rule 𝐶𝐷 = gebruik van sinus-reël sin 𝛽 𝑥. 2 sin 𝛽 cos 𝛽 𝐶𝐷 = ✓ substitution of AC cos 𝛼 sin 𝛽 2 𝑥 cos 𝛽 vervanging van AC 𝐶𝐷 = ✓ simplification / cos 𝛼 vereenvoudiging (4) 6.4 2𝑥 cos 𝛽 𝐶𝐷 = cos 𝛼 ✓ substitution / 2(25) cos 65,62° 𝐶𝐷 = vereenvoudiging cos 30° 𝐶𝐷 = 23, 83 cm ✓ answer / antwoord (2) [9] QUESTION 7 / VRAAG 7 7.1 𝑓(180°) = −0, 71 ✓ 𝑓(180°) ∴ Range: 𝑇𝑒𝑟𝑟𝑒𝑖𝑛: 𝑊𝑎𝑎𝑟𝑑𝑒𝑣𝑒𝑟𝑠𝑎𝑚𝑒𝑙𝑖𝑛𝑔: − 0, 71 ≤ 𝑦 ≤ 1 OR / OF [−0,71; 1] ✓ answer / antwoord (2) 7.2 ✓ intercepts / afsnitte ✓shape / vorm ✓ turning points / draaipunte (3) 7.3 Period / Periode = 180° ✓ answer / antwoord (1) 7.4 −45° < 𝑥 < 45° ✓✓ answer / antwoord Accept: −45 ≤ 𝑥 ≤ 45 (2) 7.5 𝑥 = −45° or / of 𝑥 = 135° ✓ 𝑥 = −45° ✓ 𝑥 = 135° (2) 7.6 𝑔(𝑥) = cos(𝑥 + 15°) ✓ answer / antwoord (1) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P2/WISKUNDE V2 13 QUESTION 8 / VRAAG 8 8.1.1 𝑄̂1 = 33° (s in the same segment) ✓S✓R (e in dieselfde segment) 𝑃̂1 = 57° (s of a triangle) / (e van ʼn driehoek) ✓S and/en R (3) 8.1.2 𝑄̂ = 90° ( subtended by the diameter) ✓ S and/en R ( onderspan deur middellyn) 𝑄̂2 = 57° (complementary s / s of a triangle) ✓ S and/en R (komplementêre e / e van ʼn driehoek) (2) 8.2 𝑄𝑇 = 𝑇𝑆 = 8 𝑐𝑚 (line from centre perp to chord) ✓ S and/en R (lyn vanaf middelpunt loodreg op koord) 𝑂𝑄 = 𝑂𝑆 = 10 𝑐𝑚 (radii) / (radiusse) ✓ S and/en R 𝑂𝑄 2 = 𝑇𝑂2 + 𝑄𝑇 2 (Pythagoras) / (Pythagoras) ✓ S and/en R 102 = 𝑂𝑇 2 + 82 𝑇𝑂 = √100 − 64 = 6 cm ✓ answer / antwoord Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

14 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2023) Using Trigonometry 𝑄𝑇 = 𝑇𝑆 = 8 Line from centre perp to chord ✓ 𝑰𝒏 ∆𝑷𝑻𝑺 𝑃𝑇 8 tan 33 = 𝑜𝑟 tan 57 = 8 𝑃𝑇 ✓ 𝑃𝑇 = 5,195 ✓ ∴ 𝑂𝑇 = 4,81 ✓ 𝑰𝒏 ∆𝑻𝑹𝑺 8 𝑅𝑇 tan 33 = 𝑜𝑟 tan 57 = 𝑅𝑇 8 𝑅𝑇 = 12,31 ✓ ∴ 𝑂𝑇 = 2,31 ✓ 𝑰𝒏 ∆ 𝑷𝑺𝑹 ✓ 𝑃𝑆 sin 33 = 20 𝑃𝑆 = 20 sin 33 ✓ 𝑃𝑆 = 10,89 Using Pythagoras in Triangle PTS ✓ 𝑃𝑇 = √10,892 − 82 𝑃𝑇 = 7,39 ✓ 𝑂𝑇 = 10 − 7,39 𝑂𝑇 = 2,61 ✓ OR/OF 𝑆𝑅 ✓ sin 57 = 20 𝑆𝑅 = 20 sin 57 ✓ 𝑆𝑅 = 16,77 Using Pythagoras in triangle TRS 𝑇𝑅 = √16,772 − 82 ✓ 𝑇𝑅 = 14,74 𝑂𝑇 = 14,74 − 10 ✓ 𝑂𝑇 = 4,74 (4) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P2/WISKUNDE V2 15 QUESTION 9 / VRAAG 9 9.1 LK̂N = 90 ✓ answer / antwoord OK̂P = 90 ✓ answer / antwoord (2) 9.2.1 L̂1 = x (tan – chord theorem) / ✓S✓R (raaklyn-koord stelling) (2) 9.2.2 K̂ 1 = x (s opp. = sides) / (e teenoor = sye) ✓S ✓R (2) 9.2.3 ˆ = 90 + x and/en Lˆ = x In  LKP; LKP 1 ✓S✓R  P̂ = 90 − 2 x (s of a triangle) (e van ʼn driehoek) OR / OF OR / OF ̂2 = 900 + 𝑥 ( 𝑒𝑥𝑡  𝑜𝑓 ∆) / (buite  van ) 𝑁 ✓ S/R  P̂ = 90 − 2 x (s of a triangle) ✓ S/R (e van ʼn driehoek) (2) 9.3 OM̂P = 90 ( tan⊥ rad) / (raaklyn ⊥ radius) ✓S&R ✓S OK̂P = 90 (proven) / (bewys) ✓R KOMP is a c.q. (opp. s suppl.) KOMP is ʼn kv (teenoorst. e supplementêr) (3) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

16 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2023) QUESTION 10 / VRAAG 10 10.1 On PQ, mark PG = MN and on PR, mark off PH = MO Join GH ✓ construction In ∆𝑃𝐺𝐻 and ∆𝑀𝑁𝑂 (1) PG = MN (construction) (2) 𝑃̂ = 𝑀̂ (given) (3) PH = MO (construction) ∴ ∆𝑃𝐺𝐻 ≡ ∆𝑀𝑁𝑂 (SAS) ✓ congruency ∴ 𝐺̂1 = 𝑁̂ (Congruency) proof But 𝑄̂ = 𝑁 ̂ (given) ∴ 𝐺̂1 = 𝑄̂ ∴ 𝐺𝐻 ∥ 𝑄𝑅 (corresponding angles formed = ) ✓S ✓R 𝑃𝐺 𝑃𝐻 ∴ = (prop.int; 𝐺𝐻 ∥ 𝑄𝑅) ✓ S and/en R 𝑃𝑄 𝑃𝑅 But PG = MN and PH = MO ✓S 𝑀𝑁 𝑀𝑂 ∴ = 𝑃𝑄 𝑃𝑅 Afrikaans Op PQ, merk af PG = MN en op PR, merk af PH = MO Verbind GH ✓ konstruksie In ∆𝑃𝐺𝐻 en ∆𝑀𝑁𝑂 (1) PG = MN (konstruksie) (2) 𝑃̂ = 𝑀̂ (gegee) (3) PH = MO (konstruksie) ∴ ∆𝑃𝐺𝐻 ≡ ∆𝑀𝑁𝑂 (SHS) ✓ kongruensie ∴ 𝐺̂1 = 𝑁 ̂ (Kongruensie) bewys Maar, 𝑄̂ = 𝑁 ̂ (gegee) ̂ ∴ 𝐺1 = 𝑄 ̂ ✓S ✓R ∴ 𝐺𝐻 ∥ 𝑄𝑅 (ooreenkomstige hoeke gevorm = ) 𝑃𝐺 𝑃𝐻 ✓ S and/en R ∴ = (eweredigheid; 𝐺𝐻 ∥ 𝑄𝑅) ✓S 𝑃𝑄 𝑃𝑅 Maar, PG = MN en PH = MO 𝑀𝑁 𝑀𝑂 ∴ = 𝑃𝑄 𝑃𝑅 (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2023) MATHEMATICS P2/WISKUNDE V2 17 10.2.1 Ŝ1 = V̂1 (alt s; PS∥VT) / (verw. e ; PS || VT) ✓ S and/en R ✓S✓R V̂1 = R̂ 1 (ext. of cq RTVW) / (buite  van kv RTVW) ∴ 𝑆̂1 = 𝑅̂1 (3) 10.2.2 In PWS and/en PSR (1) P̂1 is common / is gemeen ✓S ✓R (2) Ŝ1 = R̂ 1 (proven) / (bewys) (3) 3rd angle of a triangle / (3de hoek van driehoek) ✓ S and/en R PWS ||| PSR (AAA) / (HHH) / () Second mark should be for statement. (3) 10.2.3 In PQW and/en PRQ (1) Q̂1 = R̂ 2 (tan – chord theorem) /(raaklyn – koord stelling) ✓ S and/en R (2) P̂2 is common / is gemeen ✓S (3) 3rd angle of a triangle / 3de hoek van driehoek PQW ||| PRQ (AAA) PQ PW ✓ S and/en R  = (similarity) / (gelykvormig) PR PQ ✓S  PQ2 = PW.PR ✓S (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

18 MATHEMATICS P2/WISKUNDE V2 (EC/SEPTEMBER 2023) PW PS 10.2.4 From / Vanaf 10.2.1 = ( PWS ||| PSR ) PS SR ✓S  PS2 = PW.PR From / Vanaf 10.2.2 PQ2 = PW.PR ✓S  PQ = PS ✓ PW PS First line: = PS PR (3) [20] TOTAL / TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief