# Mathematics Grade 12 P2 NSC November 2025 Memo (Afrikaans) > Grade 12 Mathematics Paper 2 memorandum from the NSC November Exam 2025 (Afrikaans), 26 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/grade-12-mathematics-2025-nsc-november-exam-memorandum-paper-2-afrikaans/2967716b-d8b6-4b78-bc21-976f40b16bfa - Updated: 2026-09-28 - Subject: Mathematics - Grade: Grade 12 - Language: Afrikaans - Document type: Memorandum - Year: 2025 - Exam period: NSC November Exam - Paper: 2 - Curriculum: CAPS - Pages: 26 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 November 2025 Memo (Afrikaans)](https://www.hlayiso.com/library/grade-12-mathematics-2025-nsc-november-exam-memorandum-paper-2-afrikaans/2967716b-d8b6-4b78-bc21-976f40b16bfa) ## About this document Mathematics · Grade 12 · NSC November Exam · 2025 · Afrikaans. Memorandum, 26 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 NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE 12/GRAAD 12 MATHEMATICS P2/WISKUNDE V2 NOVEMBER 2025 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 26 pages./ Hierdie nasienriglyne bestaan uit 26 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Mathematics P2/Wiskunde V2 2 DBE/November 2025 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. 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 ### Page 3 Mathematics P2/Wiskunde V2 3 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 AGE OF CAR SELLING PRICE (IN YEARS) OF CAR (IN RANDS) 2 293 000 3 265 000 3 256 000 4 219 000 4 241 000 4 246 000 6 226 000 6 176 000 7 154 000 7 180 000 8 148 000 1.1 a = 331 397,20 ✓ a = 331 397,20 b = – 22 988,32 ✓ b = – 22 988,32 yˆ = 331 397, 20 − 22 988, 32 x ✓ equation (3) 1.2 yˆ = 331 397, 20 − 22 988,32 ( 5) ✓ substitution = 216 455, 60 ✓ answer (2) OR/OF yˆ = 216 455, 61 (calculator) ✓✓ answer (2) 1.3 The strong correlation ( r = − 0, 95 ) suggests that the data points ✓ strong lie close to the regression line. correlation OR r = − 0, 95 Therefore, the prediction will be valid./ ✓ answer (2) ’n Sterk korrelasie (r = –0,95) dui aan dat die punte naby aan die regressielyn lê. Dus, die voorspelling is geldig. 1.4 The average decrease per year is R22 988,32. ✓ answer Die gemiddelde afname per jaar is R 22 988,32. (1) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Mathematics P2/Wiskunde V2 4 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 TIME, t CUMULATIVE (IN MINUTES) FREQUENCY 0  t  20 16 0  t  40 40 0  t  60 59 0  t  80 67 0  t  100 70 2.1.1 70  70 (1) 2.1.2 No. of people = 67 – 40  67 – 40 = 27  27 (2) 2.1.3 Histogram 70 65  two frequencies 60 correct  all frequencies 55 correct 50  no gaps between Number of people 45 bars 40 35 30 25 24 20 19 16 15 10 8 5 3 0 20 40 60 80 100 Time, t (in minutes) (3) 2.1.4 Skewed to the right OR positively skewed  answer Skeef na regs OF positief skeef (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Mathematics P2/Wiskunde V2 5 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 2.2 11 14 19 20 8 10 2 14 11 + 14 + 19 + 20 + 8 + 10 + 2 + 14 + x  equating using = 12 9 mean x + 98 = 108 x = 10  answer The 9th player scored 10 points  standard  = 5, 23 ( 5, 22812 ) deviation ( x −  ; x +  ) = (12 − 5, 23; 12 + 5, 23) = ( 6, 77;17, 23)  interval 3 players' points were outside one standard deviation of the mean.  answer 3 spelers se punte aangeteken lê buite een standaardafwyking van die gemiddeld. (5) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Mathematics P2/Wiskunde V2 6 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 y S(15 ; 16) P(–1 ; 8) R(12 ; 2) O x T(4 ; –2) Q(–4 ;–6) ( −4 − 12 ) + ( −6 − 2 ) ( −4 − 12 ) + ( −6 − 2 ) 2 2 2 2 3.1 QR =  QR = = 320 = 8 5 units  answer (2) −6 − 2 2 − ( −6 ) 3.2 mQR = OR mQR =  correct substitution of Q ( −4 ; − 6) − 4 − 12 12 − ( −4 ) & R (12 ; 2) into gradient formula 1 1 mQR = mQR =  answer 2 2 (2) 1 3.3 mQR = 2 1 tan  =  tan  = mQR 2  = 26, 57°  answer (2) 1 3.4 mQR = 2 1 1 −6 = ( −4 ) + c OR y−2= ( x − 12)  correct substitution of gradient and 2 2 point Q ( −4 ; − 6) or R (12 ; 2) 1 c = −4 y−2 = x−6 2 1 1 y= x−4 y= x − 4  answer 2 2 (2) 3.5 Q → R : (x ; y) → (x + 16 ; y + 8)  S(15 ; 16)  xS =15  yS =16 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Mathematics P2/Wiskunde V2 7 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 1 3.6 mQR = 2 mPT = − 2  mPT Equation of PT: y = −2 x + c y − y1 = −2( x − x1 ) 8 = −2( −1) + c OR y − 8 = −2( x − ( −1)) c =6 y − 8 = −2 x − 2 y = − 2x + 6 y = − 2x + 6  equation of PT 1 − 2x + 6 = x − 4  equation QR = equation PT 2 −4 x + 12 = x − 8  simplification 5 x = 20 x =4 1 y = ( 4) − 4 2 y = −2 T ( 4; − 2 )  T( xT ; yT ) (5) OR ( −4 − ( −1) ) + ( −6 − 8 ) = 205  PQ = 205 2 2 PQ = (12 − ( −1) ) + ( 2 − 8 ) = 205  PR = 205 2 2 PR =  PQR is isosceles / PQR is ’n gelykbenige   PQR is isosceles  ⊥ height bisects the base QR /  ⊥ height bisects the base QR ⊥ hoogte halveer die basis QR  T is midpoint of QR / T is middelpunt van QR  T ( 4; − 2 )  T ( 4; − 2 ) (5) ( 4 − ( −1) ) + ( −2 − 8 ) 2 2 3.7 PT = PT = 125 = 5 5 units = 11,18 units  length of PT Area of PQRS = QR.PT = 8 5 5 5 ( )( )  substitution of QR and PT  answer = 200 units 2 (3) OR Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Mathematics P2/Wiskunde V2 8 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne ( 4 − ( −1) ) + ( −2 − 8 ) 2 2 PT = PT = 125 = 5 5 units = 11,18 units  length of PT Area of PQR = 1 2 ( )( 8 5 5 5 )  substitution of QR and PT = 100 units2 Area of PQRS = 2  Area of PQR = 200 units 2 answer (3) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Mathematics P2/Wiskunde V2 9 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 y M(p ; q) A D B O x E C 4.1 p = −6  p = −6 (1) 4.2 ˆ = 90° MDB [MC || y-axis] AM = ME = q + 1 [radii]  AM = q + 1 MD = q  MD = q AM2 = AD2 + MD2 [Pythagoras] ( q + 1) = ( q − 1) + q 2 2 2  substitution into Pythagoras q 2 + 2q +1= q 2 − 2q +1+ q 2 q 2 − 4q = 0  standard form q ( q − 4) = 0 q  0 or q = 4 (4) 4.3 AM = 5 units ( x + 6 ) + ( y − 4 ) = 25 2 2  LHS  RHS (2) 4.4 3 units  answer (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Mathematics P2/Wiskunde V2 10 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne ( x + 6 ) + ( 0 − 4 ) = 25 2 2 4.5  substituting y = 0 into equation of circle ( x + 6) = 9 2 x+6=3 or x + 6 = −3 x = −3 x = −9 A( −9 ; 0)  coordinates of A B( −3 ; 0)  coordinates of B (3) OR ( x + 6 ) + ( 0 − 4 ) = 25 2 2  substituting y = 0 into x + 12 x + 36 + 16 − 25 = 0 2 equation of circle x 2 + 12 x + 27 = 0 ( x + 3)( x + 9 ) = 0 x = −3 or x = −9 A( −9 ; 0)  coordinates of A B( −3 ; 0)  coordinates of B (3) OR q −1 = 3 DB = AD = 3 [line from centre ⊥ to chord/  DB = 3 lyn vanuit midpt ⊥ op koord] A( −9 ; 0)  coordinates of A B( −3 ; 0)  coordinates of B (3) 4−0 4.6 mMB = − 6 − ( −3 ) 4 =−  mMB 3 3 mBC =  mBC 4 3 3 y= x+c y − y1 = ( x + x1 ) 4 4 3 3 0 = ( −3 ) + c OR y − 0 = [ x − ( −3)]  substitution of gradient BC 4 4 and coordinates of B 9 3 c= y = ( x + 3) 4 4 3 9 3 9 y= x+ y= x+ 4 4 4 4  answer (4)  9 4.7 C  −6 ; −   xC  yC  4 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Mathematics P2/Wiskunde V2 11 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne y M(p ; q)   A D B O x E C 3 4.8 tan  = 4  =    =  3 mAC = − 4 3 tan  = −  tan  = mAC 4  =  − 36,87  = 143,13  value of  ˆ = 106,26  ACB  answer (4) OR ˆ =m =4 tan MAB MA 3 ˆ MAB = 53,13 ˆ  MAB ˆ = 90° − 53,13 AMD ˆ = 36,87° AMD ˆ  AMD ˆ = 90° MAC [tangent ⊥ radius / raaklyn⊥radius] ˆ = 53,13 ACM ˆ  ACM ˆ =106, 26 [property of kite /eienskappe van vlieër]  ACB  answer (4) OR AD tan ˆ ACD =  trig ratio in ΔACD or ΔBCD DC ˆ = 3 =4 tan ACD ˆ  tan ACD 9 3 4 ˆ ACD = 53,13 ˆ  ACD ˆ =106, 26 [property of kite/eienskappe van vlieër]  ACB  answer (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Mathematics P2/Wiskunde V2 12 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne OR AB = 6 units 15  AC = BC AC = BC = units [tangent from same point/ 4 raaklyne vanuit dieselfde punt] ( AB ) = ( AC ) + (BC)2 − 2 ( AC )( BC ) cos Cˆ 2 2 2 2 ( 6 ) =   +   − 2     cos Cˆ 2 15 15 15 15  substitution into cosine-rule  4   4  4  4  ˆ = − 0,28 cos ACB  simplification ˆ = 106,26 ACB  answer (4) OR ˆ =m =4 tan MAB MA 3 ˆ = 53,13 MAB ˆ  MAB AMBC is a cyclic quad/ AMBC is ’n kvh  AMBC is a cyclic quad/kvh ˆ = 53,13 [s in the same seg/e in dies segm]  MCB ˆ  MCB ˆ =106, 26 [property of kite/eienskappe v vlieër]  ACB  answer (4) [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 Mathematics P2/Wiskunde V2 13 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 5.1.1 r 2 = k 2 + 12 [Pythagoras] r = k 2 +1  third side = k 2 +1 cos 40 k = k 2 +1  answer (2) 5.1.2 2sin 25 cos 25 −2 + 4sin 2 25 sin 50  sin 50 = ( −2 1 − 2 sin 2 25 )  −2 (1 − 2 sin 2 25 ) sin 50 = −2 cos 50  double angle  k   −2  1 =    OR = − tan 50  subst OR quotient identity  k 2 + 1   k +1  2 2 1  answer =− k 2 (5) 5.1.3 sin10 = sin ( 50 − 40 )  sin10 = sin ( 50 − 40 ) = sin 50 cos 40 − cos 50 sin 40  correct expansion  k  k   1  1  =   −    first term  second term  k +1  k +1   k +1  k +1  2 2 2 2 k 2 −1 = 2 k +1 (4) OR sin10 = cos 80  sin10 = cos 80 = cos 2 ( 40 ) = 2 cos 2 40 − 1  correct expansion 2  k   substitution = 2  −1  k +1  2 2k 2 = 2 −1 k +1 k 2 −1 = 2 k +1 (4) OR Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 Mathematics P2/Wiskunde V2 14 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne sin10 = sin ( 60 − 50 )  sin10 = sin ( 60 − 50 ) = sin 60 cos 50 − cos 60 sin 50  correct expansion  3  1   1  k  =    −     first term  second term  2     k +1   2  k +1  2 2 3−k = 2 k2 +1 (4) 5.2.1 sin ( 540 + x ) .cos ( 90 + x ) sin ( − x )  sin(540° + x) = –sin x ( − sin x)( − sin x)  cos(90° + x) = –sin x = ( − sin x)  sin(–x) = –sin x = − sin x  answer (4) 5.2.2 x (180 ;360 )  x (180 ;360 ) (2) OR 180  x  360  180  x  360 (2) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 Mathematics P2/Wiskunde V2 15 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1 LHS =  tan (180 − x )  (1 − cos 2 x ) + cos 2 x = ( − tan x ) ( sin 2 x ) + cos 2 x  tan (180 − x ) = − tan x  sin x   ( sin x ) + cos x = − 2 2  1 − cos 2 x = sin 2 x  cos x   quotient identity sin 3 x =− + cos 2 x cos x sin 3 x − cos 3 x  simplification to a = − cos x single fraction ( sin x − cos x ) ( sin 2 x + sin x.cos x + cos 2 x )  factors for a difference = − cos x of cubes ( sin x − cos x )(1 + sin x.cos x )  sin 2 x + cos 2 x = 1 = − cos x = RHS (6) OR RHS = ( sin x − cos x )(1 + sin x.cos x ) − cos x ( sin x − cos x ) ( cos 2 x + sin 2 x + sin x.cos x ) =  1 = sin 2 x + cos 2 x − cos x sin x cos x + sin x + sin 2 x cos x − cos 3 x − sin 2 x cos x − sin x.cos 2 x 2 3 =  expansion − cos x sin 3 x − cos 3 x  simplification = − cos x sin 3 x = + cos 2 x  split fraction − cos x − sin x = cos x ( ) sin 2 x + cos 2 x  quotient identity ( ) = − tan x 1 − cos 2 x + cos 2 x  sin 2 x = 1 − cos 2 x = LHS (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 16 Mathematics P2/Wiskunde V2 16 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 1 6.2 sin 2 x ; cos 2 x ; sin 2 x 2 1 cos 2 x − sin 2 x = sin 2 x − cos 2 x  cos 2 x − sin 2 x = 1 sin 2 x − cos 2 x 2 2 1 cos 2 x − sin 2 x = ( 2sin x cos x ) − cos 2 x  sin 2 x = 2 sin x cos x 2 cos2 x − sin 2 x = sin x cos x − cos2 x 2cos2 x − sin x cos x − sin 2 x = 0  standard form ( 2cos x + sin x )( cos x − sin x ) = 0  factors 2 cos x = − sin x or cos x = sin x  both equations tan x = −2 or tan x = 1 ref  = 63, 43 or ref  = 45 x = 116, 57 + k .180 or x  45 + k .180 ; k  Z  x = 116,57°  116,57 ° + k .180 ; k  Z OR (7) 1 sin 2 x ; cos 2 x ; sin 2 x 2 1  cos 2 x − sin 2 x = 1 sin 2 x − cos 2 x cos 2 x − sin 2 x = sin 2 x − cos 2 x 2 2 1  sin 2 x = 2 sin x cos x cos x − sin x = ( 2sin x cos x ) − cos 2 x 2 2 2 cos x − sin x = sin x cos x − cos2 x 2 2 cos 2 x − sin 2 x − sin x cos x + cos 2 x = 0 ( cos x − sin x )( cos x + sin x ) + cos x ( cos x − sin x ) = 0  factors ( cos x − sin x )( cos x + sin x + cos x ) = 0  factors cos x = sin x or 2 cos x = − sin x  both equations tan x = 1 tan x = −2 ref  = 45 ref  = 63, 43 x  45 + k .180 ; k  Z x = 116, 57 + k .180  x = 116,57°  116,57 ° + k .180 ; k  Z (7) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 17 Mathematics P2/Wiskunde V2 17 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 7.1 180°  answer (1) 7.2 2  asymptotes  shape 1  intercepts with axes –135° –90° –67,5° –45° 0 22,5° 45° 90° 112,5° 135° f –1 g (3) 7.3 f ( x ) = cos 2 x h ( x ) = cos 2 ( x + 45 ) = cos ( 2 x + 90 )  answer = − sin 2x (1) 7.4 y   −1 ; 1  y   −1 ; 1 OR (1) −1  y  1  −1  y  1 (1) 7.5 tan 2 x − 1 = 0 tan 2 x = 1 2 x = 45 x = 22, 5  x = 22,5° (1 − tan 2 x )( cos 2 x )  0 − ( tan 2 x − 1)( cos 2 x )  0 ( tan 2 x − 1)( cos 2 x )  0  ( tan 2 x − 1)( cos 2 x )  0 x   0 ; 22,5  112,5 ;135 )  first interval  second interval OR 0  x  22, 5 or 112, 5  x  135 (4) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 18 Mathematics P2/Wiskunde V2 18 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 A D 2 Area = 648 cm 18 B 36 C 104° 52,6° 43,97 K L 8.1 Area of ABCD = BC × AB 648 = 2AB× AB  BC = 2AB AB2 = 324  substitution into area of AB = 18 cm rectangle (2) 8.2 2 AC =AB + BC 2 2  Pythagoras = 182 + 36 2  AC 2 = 182 + 36 2 AC = 1620 = 18 5 = 40, 25cm  answer (2) KC BC 8.3 = ˆ sin KBC ˆ sin BKC KC 36 =  substitution into sine rule sin104 sin 52, 6 36 sin104 KC = sin 52, 6 KC = 43, 97 cm  answer (2) 8.4 AK 2 =AB2 + BK 2  Pythagoras = 18 2 + 18 2 AK = 648 = 18 2 cm = 25,46cm  length of AK ˆ KC 2 = AK 2 + AC 2 − 2AK.AC cos KAC ˆ (43,97)2 = (18 2)2 + (18 5)2 − 2(18 2)(18 5)(cos KAC)  substitution into cosine rule ˆ = 0,16... cos KAC  simplification ˆ = 80,60 KAC  answer (4) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 19 Mathematics P2/Wiskunde V2 19 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 K 1 2 O 1 2 T P 9.1 Construction: Draw KO produced ✓ construction Oˆ =K ˆ + Tˆ [ext  of /buite  van Δ] 1 1 But Kˆ = Tˆ [s opp equal sides/e teenoor gelyke sye] ✓ S / R 1 O ˆ = 2K ˆ ✓ S 1 1 ˆ =K O ˆ +P [ext  of /buite  van Δ] 2 2 But K̂ 2 = P [s opp equal sides/e teenoor gelyke sye] Oˆ = 2K ˆ ✓ S 2 2 ˆ +O O ˆ = 2K ˆ + 2K ˆ ✓ S 1 2 1 2 ( ˆ +K =2 K 1 ˆ 2 ) ˆ = 2 TKP  TOP ˆ (5) OR K 1 2 O P T 9.1 Construction: Draw KO 0 ✓ construction ˆ T = K1 ˆ [  s opp. equal sides/e teenoor gelyke ✓ S/R sye] ✓ S  KOT ˆ = 180 − 2K ˆ [sum of  s of  /binne e van Δ] 1 Pˆ = Kˆ [  s opp. equal sides/e teenoor gelyke sye] 2 ✓ S ˆ = 180 − 2K  KOP ˆ [sum of  s of  / binne e van Δ] 2 ✓ S ˆ = 360 − KOT TOP ( ˆ + KOP ˆ ) [  s around a point/e om 'n punt] = 360 − (180 − 2K ˆ ) ˆ + 180 − 2K 1 2 ˆ + 2K = 2K ˆ 1 2 ( ˆ +K =2 K 1 ˆ 2 ) (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 20 Mathematics P2/Wiskunde V2 20 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne ˆ = 2 TKP  TOP ˆ 9.2 P 32° 1 2 O L M F 9.2.1 Ô 2 = 64 [ at centre = 2 × at circumference/ ✓ S ✓R Midpts = 2  Omtreks] (2) OR PLO = Pˆ = 32 [s opp equal radii/ ✓ S/R e teenoor gelyke radiusse] Ô 2 = 64 [ext  of ∆ / buite  van ∆] ✓S (2) 9.2.2 ˆ = 90° PFM [tan ⊥ diameter/raaklyn ⊥ middellyn] ✓ S ✓ R M̂ = 26° [sum of  s of ∆/binne e van Δ] ✓ S (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 21 Mathematics P2/Wiskunde V2 21 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 Q P 1 2 2 1 35° 1 2 R 4 3 1 V S 2 2 1 T 10.1 R̂ = 55 [sum of s in ∆/binne e van Δ] ✓S  QTSˆ = 55 [s in the same seg/e in dieselfde segment] ✓S✓R (3) OR Ŝ2 = 35 [s in the same seg/e in dieselfde segment] ✓S✓R  QTS ˆ = 55 [sum of s in ∆/binne e van Δ] ✓S (3) 10.2 ˆ =125 SPQ [opp s of cyclic quad/teenoorst. e van kvh] ✓S ✓R Sˆ = R 1 ˆ = 55 [given/gegee] ˆ +Sˆ =180 SPQ 1  PQ || SR [co-int s supp/ko-binne e suppl ] ✓R (3) OR Sˆ 1 = R ˆ = 55 [given/gegee] PQR ˆ =125 [opp s of cyclic quad/teenoorst. e van kvh] ✓S ✓R  Q̂1 =125 − 35 = 90 Q ˆ +ˆV = 180 1 1  PQ || SR [co-int s supp/ko-binne e suppl ] ✓R (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 22 Mathematics P2/Wiskunde V2 22 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 10.3 Q̂1 = 90 [co-int s; PQ|| SR/ ko-binne e; PQ ||SR] ✓S  PT is a diameter [converse  in semi-circle/ ✓R chord subtends 90° omgekeerde  in halwe sirkel / koord onderspan 90 ] (2) OR Ŝ2 = 35 [ext  of SVT or sum of s in ∆ ✓ S buite  v Δ of binne e van Δ] ˆ = 90 PST  PT is a diameter [converse  in semi-circle/ ✓R chord subtends 90° omgekeerde  in halwe sirkel / koord onderspan 90 ] (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Text truncated here. 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