# Mathematics Grade 12 P2 NSC Supplementary 2012 Memo (Afr) > Grade 12 Mathematics Paper 2 memorandum from the NSC Supplementary 2012 (Afrikaans), 18 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/mathematics-p2-feb-march-2012-memo-afr/a1f1adcd-e13e-4070-ad0a-f4d44d911d7f - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Document type: Memorandum - Year: 2012 - Exam period: NSC Supplementary - Paper: 2 - Curriculum: CAPS - Pages: 18 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 P2 NSC Supplementary 2012 Memo (Afr)](https://www.hlayiso.com/library/mathematics-p2-feb-march-2012-memo-afr/a1f1adcd-e13e-4070-ad0a-f4d44d911d7f) ## About this document Mathematics · Grade 12 · NSC Supplementary · 2012. Memorandum, 18 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 NASIONALE SENIOR SERTIFIKAAT GRAAD 12 WISKUNDE V2 FEBRUARIE/MAART 2012 MEMORANDUM PUNTE: 150 Hierdie memorandum bestaan uit 18 bladsye. Kopiereg voorbehou Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Wiskunde/V2 2 DBE/Feb.–Mrt. 2012 NSS – Memorandum VRAAG 1 1.1 Gemiddeld n 3 102100 ∑x 1 1 102100 = 3 antwoord n 9 (2) = R11 344, 44 1.2 Standaardafwyking 33 antwoord n ∑ (x − x) 1 1 2 = R4 460,97 (2) n 1.3 Waarde van een standaardafwyking bo gemiddeld 3 tel gemiddeld = R11 344,44 + R4 460,97 en std. af. by = R15 805,41 Slegs een persoon het kommisie bo R 15 805,41 verdien. 3 afleiding Dus het slegs 1 persoon 'n gradering van 'goed' ontvang. (2) [6] VRAAG 2 2.1 3 ten minste vier punte korrek 3 al die punte korrek (2) 2.2 Eksponensieel, want die toename in groei wys amper 'n verdubbeling elke 3eksponensieel jaar. 3 rede (2) Kopiereg voorbehou Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Wiskunde/V2 3 DBE/Feb.–Mrt. 2012 NSS – Memorandum 2.3 JAAR 1995 1996 1997 1998 1999 2000 2001 3 ten minste 4 N punte (Getal in 8 17 34 67 135 281 552 korrek miljoene) 3alle punte Log N korrek (korrek tot (2) EEN 6,9 7,2 7,5 7,8 8,1 8,4 8,7 desimale plek) OF (indien slegs logwaardes in table in berekening gebring is) JAAR 1995 1996 1997 1998 1999 2000 2001 N (Getal in 8 17 34 67 135 281 552 3 ten minste 4 miljoene) punte Log N korrek (korrek tot 3alle punte EEN 0,9 1,2 1,5 1,8 2,1 2,4 2,7 korrek desimale (2) plek) 2.4 3 ten minste 4 punte korrek gestip 3 alle punte korrek (2) Kopiereg voorbehou Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Wiskunde/V2 4 DBE/Feb.–Mrt. 2012 NSS – Memorandum OF (indien slegs logwaardes in table in berekening gebring is) 3 ten minste 4 punte korrek gestip 3 alle punte korrek (2) 2.5 Lineêr, want die punte lê baie naby aan 'n reguitlyn 3lineêr 3 rede (2) [9] Kopiereg voorbehou Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Wiskunde/V2 5 DBE/Feb.–Mrt. 2012 NSS – Memorandum VRAAG 3 3.1 40 3 40 (1) 3.2 Tyd, t, in minute Frekwensie 3 vir intervalle in 0 ≤ t<5 3 tabel 5 ≤ t <10 5 3 vir eerste drie 10 ≤ t < 15 10 korrekte 15 ≤ t < 20 15 frekwensies 20 ≤ t < 25 7 3 vir laaste twee korrekte frekwensies (3) 3.3 15 3korrekte frekwensies 3middelpunt- 10 waardes 3geen spasies tussen stawe Frekwensie 5 0 5 10 15 20 25 Tydintervalle (3) [7] VRAAG 4 3 elke korrekte a=7 b = 15 c = 17 d = 23 e = 34 f = 37 g = 42 antwoord (7) OF g = 42 ; a = 7 ; d = 23 ; f = 37 ; b = 15 42 + 7 + 23 + 37 + 15 + 3c 3g = 25 3a 7 3d 3c = 51 3f c = 17 3b e = 34 3c 3e (7) [7] Kopiereg voorbehou Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Wiskunde/V2 6 DBE/Feb.–Mrt. 2012 NSS – Memorandum VRAAG 5 5.1 y 2 − y1 mAD = x 2 − x1 −2−4 3 vir substitusie = 5 −1 6 3 3 vir antwoord = − = − 4 2 (2) 5.2 AD = ( x2 − x1 ) + ( y 2 − y1 ) 2 2 = (5 − 1) 2 + (−2 − 4) 2 3 vir substitusie = 16+ 36 3 52 = 52 (2) 5.3 ⎛ x + x 2 y1 + y 2 ⎞ M =⎜ 1 ; ⎟ ⎝ 2 2 ⎠ 3 x-waarde ⎛1+ 5 4 − 2 ⎞ M =⎜ ; ⎟ ⎝ 2 2 ⎠ 3 y-waarde M = (3 ; 1) (2) 5.4 mBC = mAD lyne is parallel 3 = − 3 waarde mBC 2 y – y1 = m (x – x1) 3 3 subst (–3 ; 1) y – 1 = − (x + 3) 2 2 y − 2 = −3 x − 9 3 vergelyking 3x + 2 y + 7 = 0 (3) OF 3 3 waarde mBC y =− x+c 2 3 3 subst (–3 ; 1) 1 = − (−3) + c 2 7 c=− 3 vergelyking 2 3 7 y =− x− 2 2 3x + 2 y + 7 = 0 (3) Kopiereg voorbehou Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Wiskunde/V2 7 DBE/Feb.–Mrt. 2012 NSS – Memorandum 5.5.1 3 m AD = − A(1 ; 4) 2 3 tan β = mAD 3 tan β = − 2 B(–3 ; 1) α β 3 123,69° β = 180° − 56,31° F E β = 123,69 D(5 ; –2) (2) 5.5.2 − 2 −1 − 3 m BD = = −3 5 − (−3) 8 3 m BD = 3 8 tan α = − 8 α = 180° − 20,56° 3159,44° α = 159,44° FEˆ D = 180° − 159,44° = 20,56° 320,56° 3123,69° EFˆD = 123,69° 335,75° FDˆ E = 180° − (20,56° + 123,69°) = 35,75° (5) 5.6 Koördinate van middelpunt M (3 ; 1) Radius van sirkel: 3 waarde van 1 1 1 radius of AD = (2 13 ) = 13 = 52 3 substitusie in 2 2 2 vergelyking vir Vergelyking van die sirkel is: (x − 3) + ( y − 1) = 13 2 2 sirkelmiddelpunt- vorm (2) OF 3 waarde van r2 r = (3 − 1) + (1 − 4) = 13 2 2 2 3 substitusie in Vergelyking van die sirkel is: vergelyking in (x − 3)2 + ( y − 1)2 = 13 sirkelmiddelpunt vorm (2) 5.7 M(3 ; 1) B(– 3 ; 1) MB = (3 + 3) + (1 − 1) ) 2 2 3 substitusie MB = 6 3 buite (2) Punt B lê buite die sirkel want MB > radius OF M(3 ; 1) B(– 3 ; 1) 3substitusie MB = 3 + 3 = 6 Radius van die sirkel = 13 < 6 3 buite (2) Punt B lê buite die sirkel want MB > radius [20] Kopiereg voorbehou Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Wiskunde/V2 8 DBE/Feb.–Mrt. 2012 NSS – Memorandum VRAAG 6 6.1 Koördinate van middelpunt M (−2 ; 1) 33 koördinate van (1 + 2)2 + (− 2 − 1)2 = 18 = r 2 middelpunt 3 berekening Radius = 18 of 3 2 3 waarde (4) 6.2 −3 m MS = = −1 3 gradiënt MS 3 m MS xm RS = −1 OF raaklyn ⊥ radius m RS = 1 3 gradiënt RS y – y1 = m (x – x1) y + 2 = 1(x – 1) 3 subst (1 ; –2) y = x–3 3 vergelyking (4) OF 3 gradiënt MS −3 m MS = = −1 3 3 gradiënt RS m MS xm RS = −1 m RS = 1 3 subst (1 ; –2) y = x+c − 2 = 1+ c 3 vergelyking c = −3 y = x−3 (4) 6.3 MS 1 = 3 MP = 3MS MP 3 ∴ MP = 3MS MP 2 = 9MS 2 (a + 2) 2 + (b − 1) 2 = 9(32 + 32 ) = 162 (1) 3 vergelyking MS ⊥ SR en PS ⊥ SR ∴ m PS = mMS 3 dieselfde b+2 3 gradiënte = = −1 a −1 − 3 3 gradiënt b + 2 = −a + 1 b = −a − 1 (2) 3 b = -a - 1 Vervang (2) in (1) Kopiereg voorbehou Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Wiskunde/V2 9 DBE/Feb.–Mrt. 2012 NSS – Memorandum (a + 2) 2 + (− a − 1 − 1) 2 = 162 3 substitusie (a + 2) 2 + (a + 2) 2 = 162 2(a + 2) 2 = 162 (a + 2) 2 = 81 a + 2 = 9 of − 9 a = 7 of − 11 3a=7 b = − a − 1 = −8 3 b = -8 P(7 ; − 8) (8) OF MS 1 = MP 3 ∴ MP = 3MS 3 MP = 3MS MP 2 = 9MS 2 (a + 2) 2 + (b − 1) 2 = 9(32 + 32 ) = 162 (1) 3 vergelyking MS ⊥ SR en PS ⊥ SR ∴ m PS = m MS b+2 3 3 dieselfde = = −1 gradiënte a −1 − 3 b + 2 = −a + 1 3 gradiënt b = −a − 1 (2) Vervangt (2) in (1) 3 b = -a - 1 a 2 + 4a + 4 + a 2 + 4a + 4 = 162 2a 2 + 8a − 154 = 0 3 substitusie a 2 + 4a − 77 = 0 (a + 11)(a − 7) = 0 a = 7 of − 11 Maar a > 0 ∴a = 7 3a=7 b = − a − 1 = −8 3 b = -8 P(7 ; − 8) (8) OF Kopiereg voorbehou Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Wiskunde/V2 10 DBE/Feb.–Mrt. 2012 NSS – Memorandum P(a ; b) MSP is 'n reguitlyn (MS ⊥ SR) mPM = − 1 3 MSP is 'n b −1 reguitlyn = −1 3 m PM = −1 a+2 b − 1 = −a − 2 b −1 b = −a − 1.......(1) 3 a+2 PS = 2MS = 2 9 + 9 = 2 18 PS 2 = 4(18) = 72 3 vergelyking 1 (a − 1) 2 + (b + 2) 2 = 72.......(2) 3 vergelyking 2 (a − 1) + (− a − 1 + 2) = 72 2 2 2(a − 1) = 72 2 3 substitusie van 2a − 4a − 70 = 0 2 (a − 1) 2 = 36 vergelyking 1 in a 2 − 2a − 35 = 0 a − 1 = 6 of − 6 vergelyking 2 (a − 7)(a + 5) = 0 OF a = 7 of − 5 33 koördinate a = 7 of a /= −5 a=7 b = −7 − 1 = −8 b = −8 P (7 ; − 8) P(7 ; -8) (8) OF M(−2; 1) 3 33 diagram (−2; −2) 3 S(1 ;−2) 33 (−2; −8) 3 (−2; −2) 3 (1; −8) 6 33 P(7 ; −8) 3 6 (8) (−2; −8) P(a ;b) (1; −8) Kopiereg voorbehou Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Wiskunde/V2 11 DBE/Feb.–Mrt. 2012 NSS – Memorandum P(a ; b) xS − xM y − yM 1 = S = 33 verdeling van x P − xM yP − yM 3 'n lynstuk in 'n −3 3 1 gegewe = = b −1 a + 2 3 verhouding − 9 = b −1 33 substitusie b = −8 3 vergelyking 9=a+2 a=7 3 vergelyking P (7 ;−8) 3 koördinate (8) [16] VRAAG 7 7.1 y 5 4 K 3 Vir korrekte koördinate en 2 benoeming van N 1 elke beeld: x 3 K′ -5 -4 -3 -2 -1 N/ 1 2 3 4 5 3 L′ 3M′ -1 M/ L -2 3 N′ -3 K/ M L/ -4 -5 (4) 7.2.1 Transformasie is nie rigied (star) nie, want die area het verander as 3 nie rigied nie gevolg van die vergroting. 3 grootte nie behou nie (2) 7.2.2 N // (−2 ; − 2) 33 koördinate van N // (2) 7.3 (x ; y) → (–y ; x) → (–2y ; 2x) 3 –y 3x 3 –2y 3 2x (4) 7.4 Area van KLMN : area van K // L// M // N // = 1 : 4 33 antwoord (2) 7.5 Indien die verste punt vanaf die oorsprong in die sirkel gedruk/gestuur 3 K – verste word, dan sal die hele vierhoek in die sirkel gedruk wees. K is die 3 KO = 18 verste weg. 3 antwoord KO = 3 2 + 3 2 = 18 (3) Kopiereg voorbehou Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Wiskunde/V2 12 DBE/Feb.–Mrt. 2012 NSS – Memorandum 1 [17] p.KO = 1 , p = 18 VRAAG 8 8. x Q = x cosθ + y sin θ 3 subst −2 en −3 in x Q = −2 cos135° + (−3) sin 135° korrekte formule vir xQ 2 3 −1 − 2 3 gebruik 135° xQ = − = of or - 0,71 2 2 2 2 3 x-koördinate (in y Q = y cos θ − x sin θ enige formaat) y Q = −3 cos135° − (−2) sin 135° 3 subst −2 en −3 in 3 2 5 5 2 korrekte formule yQ = + = = = 3,54 vir yQ 2 2 2 2 3 vir y-koördinate (in ⎛ −1 5 ⎞ enige formaat) Q⎜ ; ⎟ ⎝ 2 2⎠ (5) OF x Q = x cosθ − y sin θ 3 subst −2 en −3 in korrekte formule x Q = −2 cos(−135°) − (−3) sin( −135°) vir xQ 2 3 −1 − 2 3 gebruik −135° xQ = − = of or - 0,71 3 x-koördinate (in 2 2 2 2 enige formaat) y Q = y cos θ + x sin θ y Q = −3 cos(−135°) + (−2) sin(−135°) 3 subst −2 en −3 in korrekte formule 3 2 5 5 2 yQ = + = = = 3,54 vir yQ 2 2 2 2 3 vir y-koördinate (in ⎛ −1 5 ⎞ enige formaat) Q⎜ ; ⎟ (5) ⎝ 2 2⎠ OF 3 subst −2 en 135° in x ′ = x cos θ − y sin θ korrekte formule − 2 = x cos135° − y sin 135° vir x′ −x y −2= − 2 2 3 vereenvoudiging − 2 2 = −x − y (1) y ′ = y cos θ + x sin θ 3 subst −2 en 135° in − 3 = y cos135° + x sin 135° korrekte formule −y x vir y′ −3= + 2 2 −3 2 = x− y (2) 3 y-koördinaat Los (1) en (2) gelyktydig op: 3 x-koördinaat (5) Kopiereg voorbehou Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Wiskunde/V2 13 DBE/Feb.–Mrt. 2012 NSS – Memorandum − 5 2 = −2 y 5 −1 y= en x= 2 2 OF Gebruik eerste beginsels: Q = (− r cos α ; r sin α ) r 135° α θ r 3 θ 2 3 3 tan θ = Q = (−2 ; − 3) / 2 3 r = 13 3 3 θ = 56,31° tan θ = 2 r = 3 2 + 2 2 = 13 θ = 56,31° 3 Q = (− r cos α ; r sin α ) ∴ α = 135° − 56,31° = 78,69° 3 antwoord Q = (− r cos α ; r sin α ) (5) = (−0,71 ; 3,54) [5] Kopiereg voorbehou Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Wiskunde/V2 14 DBE/Feb.–Mrt. 2012 NSS – Memorandum VRAAG 9 9.1.1 r = 13 3 13 12 cos α = 13 12 3 13 (2) 9.1.2 QÔR = 180° − (90° + α ) 3 180° − (90° + α ) = 90° − α 3 90° − α (2) 9.1.3 QR cos QÔR = 3 OQ 7,5 7,5 cos(90° − α ) = cos(90° − α ) = OQ OQ 7,5 7,5 3 OQ = sin α cos(90° − α ) 7,5 OQ = sin α 5 3 7,5 13 OQ = 5 319,5 13 OQ = 19,5 (4) OF 3 7,5 7,5 sin( RQ̂O) = sin(RQ̂O) = OQ OQ 7,5 7,5 3 ∴ OQ = sin α sin α 5 7,5 3 OQ = 13 5 319,5 13 OQ = 19,5 (4) 9.2 cos x. cos x (− tan x ) LK = 3 cos x − cos x sin x 3 – tan x = cos x. sin x cos x 3 cos x = sin x 3 antwoord = RK (4) [12] Kopiereg voorbehou Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Wiskunde/V2 15 DBE/Feb.–Mrt. 2012 NSS – Memorandum VRAAG 10 10.1 Periode = 120º 3 120° (1) 10.2 sin 3x = -1 3 − 30° x = −30° of x = 90° 3 90° (2) 10.3 Maksimum waarde van f (x) is 1 3 maks van f(x) ∴ Maksimum waarde van h(x) is 0 3 antwoord (2) 10.4 3 − 90° ; 90° y 3 (0° ; 3) g 3 (180° ; − 3) 2 1 f x -90 -60 -30 30 60 90 120 150 180 -1 -2 -3 (3) 10.5 sin 3 x − cos x = 0 3 3 sin 3x = 3 cos x sin 3 x − 3 cos x = 0 ∴ sin 3 x = 3 cos x 3 antwoord Daar is 2 oplossings waar die grafieke van f en g gelyk is (2) 10.6 f(x).g(x) < 0 333 vir elke x∈ (−60º ; 0º) of (60º ; 90º) of (120º ; 180º) interval 3 korrekte hakies OF of korrekte simbole –60°< x < 0° of 60°< x < 90° of 120°< x < 180° (4) [14] Kopiereg voorbehou Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Wiskunde/V2 16 DBE/Feb.–Mrt. 2012 NSS – Memorandum VRAAG 11 11.1.1 sin 61° = p sin 241° = sin (180° + 61°) 1 3 − sin 61° = − sin 61° p = − p 61˚ 3antwoord 1− p (2) 11.1.2 cos 61° = 1 − sin 61° 2 3 identiteit 3 antwoord = 1− p (2) 11.1.3 cos122° = cos 2(61°) 3 dubbelhoek 3 uitbreiding = 2 cos 2 61° − 1 ( = 2 1− p ) −1 2 = 2(1 − p ) − 1 3 antwoord = 2 − 2 p −1 = 1− 2p (3) 11.1.4 cos 73°cos15° + sin 73°.sin15° = cos(73° − 15°) 3 cos(73°-15°) = cos 58° = (cos 180° − 122°) = − (cos 122°) 3 − (cos 122°) = −(1 – 2p) 3 antwoord = 2p −1 (3) 11.2.1 (cos x + sin x ) 2 − (cos x − sin x ) 2 LK = 3 (cos x − sin x )(cos x + sin x ) 2 2 (cos x + sin x ) − (cos x − sin x ) cos 2 x + 2 cos x sin x + sin 2 x − (cos 2 x − 2 sin x cos x + sin 2 x ) = (cos x − sin x )(cos x + sin x ) (cos x − sin x )(cos x + sin x ) 3teller 4 cos x sin x 3 4 cos x sin x = cos 2 x − sin 2 x 3 cos 2 x − sin 2 x 2 sin 2x 3 2 sin 2 x = cos 2 x 3 cos 2 x = 2 tan x = RK (6) 11.2.2 x = 45° ;135° 3 45° 3 135° (2) 11.3.1 sin x = cos 2 x − 1 sin x = 1 − 2 sin 2 x − 1 3 1 – 2sin2 x sin x = −2 sin 2 x 2 sin 2 x + sin x = 0 (1) Kopiereg voorbehou Blaai om asseblief ### Page 17 Downloaded from hlayiso.com Wiskunde/V2 17 DBE/Feb.–Mrt. 2012 NSS – Memorandum 11.3.2 sin x = cos 2x – 1 2 sin2 x + sin x = 0 sin x (2 sin x + 1) = 0 3 sin x (2 sin x + 1) = 0 1 3 sin x = 0 of sin x = sin x = 0 of sin x = – 1 2 – ∴ x = 0° + 180° k; k ∈ Z of x = {210° or 330°} + 360°k; k ∈ Z 2 3 0° + 180° k OF 3 210° 3 330° x = n.180° 3 + 360°k; k ∈ Z x = n.360° − 30° x = (2n + 1).180° + 30°, n ∈ Z (6) 11.4 tan 1˚ × tan 2˚ × tan 3˚ × tan 4˚ × …….× tan 87˚ × tan 88˚ × tan 89˚ ⎛ sin 1° ⎞⎛ sin 2° ⎞ ⎛ sin 45° ⎞ ⎛ sin 88° ⎞⎛ sin 89° ⎞ =⎜ ⎟⎜ ⎟......⎜ ⎟......⎜ ⎟⎜ ⎟ 3 identiteit ⎝ cos 1° ⎠⎝ cos 2° ⎠ ⎝ cos 45° ⎠ ⎝ cos 88° ⎠⎝ cos 89° ⎠ ⎛ sin 1° ⎞⎛ sin 2° ⎞ ⎛ sin 45° ⎞ ⎛ sin(90° − 2°) ⎞⎛ sin(90° − 1°) ⎞ 3 ko-verhoudings =⎜ ⎟⎜ ⎟......⎜ ⎟......⎜⎜ ⎟⎟⎜⎜ ⎟⎟ ⎝ cos 1° ⎠⎝ cos 2° ⎠ ⎝ cos 45° ⎠ ⎝ cos(90° − 2° ⎠⎝ cos(90° − 1°) ⎠ ⎛ sin 1° ⎞⎛ sin 2° ⎞ ⎛ sin 45° ⎞ ⎛ cos 2° ⎞⎛ cos 1° ⎞ =⎜ ⎟⎜ ⎟......⎜ ⎟......⎜ ⎟⎜ ⎟ ⎝ cos 1° ⎠⎝ cos 2° ⎠ ⎝ cos 45° ⎠ ⎝ sin 2° ⎠⎝ sin 1° ⎠ 3 vereenvoudiging = tan 45° 3 vir antwoord =1 (4) OF 3 identiteit tan 89° = cot 1° tan 88° = cot 2° ... 3 ko-verhoudings ∴ produk is (tan 1°. cot 1°)(tan 2°. cot 2°)...(tan 44°. cot 44°). tan 45° 3 vereenvoudiging = 1 × 1 × 1 × ... × 1 = 1 3vir antwoord (4) [29] Kopiereg voorbehou Blaai om asseblief ### Page 18 Downloaded from hlayiso.com Wiskunde/V2 18 DBE/Feb.–Mrt. 2012 NSS – Memorandum VRAAG 12 12 By Δ CBG en ΔCDH: CG² = x² + y² Pythagoras 3 CG² CH² = x² + y² Pythagoras 3 CH² By ΔFAE AE² = x² + x² = 2x² 3 AE² = GH² 3 AE² = GH² By Δ CGH GH² = CG² + CH² - 2 CG.CH. cos GCH CG 2 + CH 2 − GH 2 3 gebruik van cos-reël cos GĈH = 2CG.CH 3 manipulasie van x + y 2 + x 2 + y 2 − 2x 2 2 formule cos GĈH = 2 x2 + y2. x2 + y2 3 substitusie 2y2 cos GĈH = 2y2 2( x 2 + y 2 ) 3 cos GĈH = 2( x 2 + y 2 ) y2 cos GĈH = (8) x2 + y2 [8] TOTAAL: 150 Kopiereg voorbehou ## Related documents - [Mathematics P2 Feb-March 2012 Eng](https://www.hlayiso.com/library/mathematics-p2-feb-march-2012-eng/4a8c1244-86a4-426a-a294-006c3983ca41) — Mathematics · QP · 2012 · Eng - [Mathematics P3 Feb-March 2012 Memo Eng](https://www.hlayiso.com/library/mathematics-p3-feb-march-2012-memo-eng/aeefeca8-fee8-40ef-b720-2c36d96f8734) — Mathematics · Memo · 2012 · Eng - [MATHS P2 MEMO SEPT 12 ENG](https://www.hlayiso.com/library/maths-p2-memo-sept-12-eng/6a675437-ca52-451c-ad30-9b9930a6b1fb) — Mathematics · Memo · 2012 · Eng - [Mathematics P2 (SG) (English) June 2012 QP](https://www.hlayiso.com/library/mathematics-p2-sg-english-june-2012-qp/8050c3b4-c3de-40f3-ad7c-829034658a34) — Mathematics · QP · 2012 · Eng - [WISK V2 MEMO SEPT 12 AFR](https://www.hlayiso.com/library/wisk-v2-memo-sept-12-afr/b3772ba3-1bdc-4236-a9d4-0728d9126d78) — Mathematics · Memo · 2012 · Afr - [Mathematics P1 Feb-March 2013 Memo Eng](https://www.hlayiso.com/library/mathematics-p1-feb-march-2013-memo-eng/ee7e02f2-8218-4df4-a922-7d82fccf8fd5) — Mathematics · Memo · 2013 · Eng ## More from Grade 12 Mathematics - [MATHS P1 ANSWER BOOK GR12 SEPT 2025 English Afrikaans watermark](https://www.hlayiso.com/library/maths-p1-answer-book-gr12-sept-2025-english-afrikaans-watermark/d78c3fd3-dcd5-4f3f-911b-e5635b4686cc) — Addendum · 2025 · Afr - [Maths Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-gauteng/1d6bf7fd-d288-4a01-99f7-2c87e247760c) — QP · 2025 - [Maths Grade 12 NSC P2 QP September 2025 Afr Limpopo](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-afr-limpopo/f2f8d481-5d87-4e89-bb21-6077bbcf403d) — QP · 2025 · Afr - [Maths Grade 12 NSC P1 QP September 2025 Afr Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p1-qp-september-2025-afr-northern-cape/f5b3a4ce-fa91-4475-96e0-0654a87e1405) — QP · 2025 · Afr - [Maths Grade 12 NSC P2 QP September 2025 Eng Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-eng-northern-cape/4ede8207-ca82-4419-bae1-fae734498cc7) — QP · 2025 · Eng - [MATHS P2 ANSWER BOOK GR12 JUNE 2024 Afr+English](https://www.hlayiso.com/library/maths-p2-answer-book-gr12-june-2024-afr-english/4a71e975-4ac4-4f55-8e23-0b8155a5b0c6) — Addendum · 2024 · Eng ## Collection - [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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