# Mathematics Grade 11 P2 Gauteng June 2017 Memo (Afrikaans) > Grade 11 Mathematics Paper 2 memorandum from the Gauteng June 2017 (Afrikaans), 10 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr11-math-p2-afrikaans-june-2017-possible-answers/a4cb0560-cbb2-496a-be03-7442924cb4ae - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Language: Afrikaans - Document type: Memorandum - Year: 2017 - Exam period: Gauteng June - Paper: 2 - Curriculum: CAPS - Pages: 10 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Mathematics](https://www.hlayiso.com/library/grade-11/mathematics) › [Mathematics Grade 11 P2 Gauteng June 2017 Memo (Afrikaans)](https://www.hlayiso.com/library/gr11-math-p2-afrikaans-june-2017-possible-answers/a4cb0560-cbb2-496a-be03-7442924cb4ae) ## About this document Mathematics · Grade 11 · Gauteng June · 2017 · Afrikaans. Memorandum, 10 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 GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINSIALE EKSAMEN JUNIE 2017 GRAAD 11 WISKUNDE VRAESTEL 2 MEMORANDUM 10 bladsye 1 ### Page 2 Downloaded from hlayiso.com Wiskunde Vraestel 2 MEMORANDUM GRAAD 11 GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINSIALE EKSAMEN WISKUNDE Vraestel 2 MEMORANDUM INSTRUKSIES EN INLIGTING A – Akkuraatheid CA – Volgehoue Akkuraatheid S – Stelling R – Rede SR- Stelling/Rede NOTA: • Indien ʼn kandidaat ʼn vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na. • Indien ʼn kandidaat ʼn antwoord DOODGETREK het, maar dit nie oorgedoen het nie, merk die doodgetrekte antwoord. • Volgehoue akkuraatheid word in ALLE aspekte van die memorandum toegepas. Hou op nasien by die tweede berekeningsfout. • Om antwoorde/waardes te veronderstel om ʼn probleem op te los, word NIE toegelaat NIE. 2 ### Page 3 Downloaded from hlayiso.com Wiskunde Vraestel 2 MEMORANDUM GRAAD 11 VRAAG 1 PUNTE: 25 1.1 𝑀𝑀𝑅 = 6−0 ✓ vervang in korrekte formule (3) 4−0 3 6 3 ✓ 𝑚𝑀𝑅 = 2 = = 4 2 3 ✓ vergelyking van MR Vergelyking van lyn MR is: 𝑦 = 𝑥 2 1.2 𝑦 − 5𝑥 + 14 = 0 (4) y = 5x -14 ✓ 𝑚𝑀𝑆 = 5 MS || PR  mPR = 5 ✓𝑚𝑃𝑅 = 5 Vergelyking van lyn PR: 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) 𝑦 − 4 = 5(𝑥 + 2) ✓verv. (2 ; 4) 𝑦 = 5𝑥 + 14 ✓antwoord 1.3 mPR = 5 (5) ∴ tan 𝛼 = 5 ✓tan 𝛼 = 5 𝛼 = 78,69° ✓78,69° 3 mMR = 2 3 tan 𝛽 = 2 𝛽 = 56,31° ✓56,31° ∴ 𝜃 =𝛼−𝛽 ✓∴ 𝜃 = (𝛼 − 𝛽)..(som van hoeke v.∆) 𝜃 = 22,38° ✓22,38° 1.4 3 (4) 𝑦 = 𝑥 en 𝑦 = 5𝑥 + 14 2 3 3 5𝑥 + 14 = 𝑥 ✓stel 5𝑥 + 14 = 𝑥 2 2 10𝑥 + 28 = 3𝑥 7𝑥 = −28 ✓7𝑥 = −28 𝑥 = −4 ✓ 𝑥 = −4 𝑦 = −6 ✓ 𝑦 = −6 R(4 ; 6) 3 ### Page 4 Downloaded from hlayiso.com Wiskunde Vraestel 2 MEMORANDUM GRAAD 11 1.5 𝑑MR = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 (2) ✓verv. in korrekte formule = √(4 + 4)2 + (6 + 6)2 = √64 + 144 ✓antwoord = 4√13 1.6 𝑑PR = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 (5) ✓ verv. in korrekte formule = √(−4 + 2)2 + (−6 − 4)2 ✓√104 OF 2√26 = √104 or 2√26 Oppvl van PMR 1 1 ✓ 2√26. 4√13. sin 22,38° OF = PR. MR. sin 𝜃 2 1 2 1 √104. 4√13. sin 22,38° = 2√26. 4√13. sin 22,38° OF 2 1 2 (✓ afronding) antwoord ✓ 2 √104. 4√13. sin 22,38° = 28 𝑒𝑒𝑛ℎ𝑒𝑑𝑒² 1.7 S(2 ; 4) ✓x=2 (2) ✓ y = 4 VRAAG 2 PUNTE: 26 2.1 𝑦2 = 𝑟2 − 𝑥2 ✓diagram in korrekte kwadrant (4) ∴ 𝑦 2 = (25)2 − (−7)2 y ∴ 𝑦 2 = 576 ∴ 𝑦 = −24 ✓ 𝑦 = −24 −24 −24 ∴ 14 tan 𝜃 = 14 ( ) 0 x ✓tan 𝜃 = −7 −7 25 = 48 ✓antwoord  (-7: y) 2.2 cos(90°+𝑥).sin(180°+𝑥) (6) tan 225°−𝑐𝑜𝑠 2 (−𝑥) (− sin 𝑥)(− sin 𝑥) ✓ (-sinx) (-sinx) = tan 45°−𝑐𝑜𝑠 2 𝑥 ✓ tan 45° sin2 𝑥 = ✓ cos 2 𝑥 1−cos2 𝑥 ✓ tan 45° = 1 sin2 𝑥 = ✓ 1 − cos2 𝑥 = sin2 𝑥 sin2 𝑥 ✓ antwoord =1 4 ### Page 5 Downloaded from hlayiso.com Wiskunde Vraestel 2 MEMORANDUM GRAAD 11 2.3 2 cos 2𝜃 = −0,44 ✓ cos 2𝜃 = −0,22 (6) ∴ cos 2𝜃 = −0,22 ✓ 77,2909° ∴ Verw. hoek = 77,2909 … ° ✓ 2𝜃 = 180° − 77,2909° + 𝑘. 360° 2𝜃 = 180° − 77,2909 … ° + 𝑘. 360° ∴ 𝜃 = 51,35° + 𝑘. 180, 𝑘 ∈ 𝑍 or ✓ 𝜃 = 51,35° + 𝑘. 180, 𝑘 ∈ 𝑍 2𝜃 = 180° + 77,2909 … ° + 𝑘. 360° ✓ 2θ = 180° + 77,2909° + k.360° ∴ 𝜃 = 128,65° + 𝑘. 180°, 𝑘 ∈ 𝑍 ✓ ∴ 𝜃 = 128,65° + 𝑘. 180°, 𝑘 ∈ 𝑍 2.4 tan 𝜃 − sin 𝜃 (5) ∴ LHS = 1 − cos 𝜃 sin 𝜃 sin 𝜃 ✓ tan 𝜃 = sin 𝜃 − 1 = cos 𝜃 cos 𝜃 sin 𝜃−sin 𝜃.cos 𝜃 1 − cos 𝜃 ✓ cos 𝜃 sin 𝜃 − sin 𝜃. cos 𝜃 = cos 𝜃 1 − cos 𝜃 sin 𝜃(1 − cos 𝜃) 1 ✓ sin 𝜃(1 − cos 𝜃) = × ✓× 1 cos 𝜃 1 − cos 𝜃 1−cos 𝜃 sin 𝜃 sin 𝜃 = ✓ = tan 𝜃 cos 𝜃 cos 𝜃 = tan 𝜃 2.5 𝛼 = 90° − 𝛽 ✓cos 20° (5) cos 20° sin(90° − 𝛽) ✓sin 𝛼 = sin(90° − 𝛽) ∴ − sin 70° sin(90° − 𝛽) sin 70° ✓cos 20° = sin 70° = −1 sin 70° =1−1 ✓1 − 1 =0 ✓antwoord 5 ### Page 6 Downloaded from hlayiso.com Wiskunde Vraestel 2 MEMORANDUM GRAAD 11 VRAAG 3 PUNTE:14 3.1 ✓ y-afsnit van f (6) ✓ y-afsnit van g ✓✓ beide draaipunte van g ✓ beide x-afsnitte van f ✓ beide draaipunte van f 3.2 2 ✓✓ antwoord (2) 3.3 2 cos x +1 = 1 – sin x ✓ tan x =-2 (6)  sin x = -2 cos x ✓ 63, 4349°  tan x = -2 ✓ x = 116, 57°  Verw. hoek = 63, 4349....° +k.180° met k 𝜖 Z  x = 180° - 63, 4349....° + k. 180° ✓ -63,43°  x = 116, 57° + k.180° met k є Z ✓ 116, 57°  x = -63,43° OF 116,57° OF 296,57° ✓ 296,57° 6 ### Page 7 Downloaded from hlayiso.com Wiskunde Vraestel 2 MEMORANDUM GRAAD 11 VRAAG 4 PUNTE: 13 4.1 O 1 2 P Q R Konstruksie: Verbind PO en OR ✓ konstruksie(radii, OP en OR) In ∆POQ and ∆ ROQ 1) OQ = OQ..... gemeenskaplik 2) PO = OR ......radii ✓SR ia (PO = OR ..... radii) ̂1 = Q 3) Q ̂ 2 = 90°... gegee ✓S ia (Q̂1 = Q̂ 2 = 90°) ∴ ∆POQ ≡ ∆ROQ … 90°HS ✓SR ∴ PQ = RQ ✓ PQ =RQ (5) 4.2.1 ̂ 1 = 90°... lyn deur middel punt van ⨀ na middel van A ✓SR koord ̂1 + A ̂ = 180° − (O B ̂1 ) .... som van hoeke van ∆ ✓ antwoord = 50° (2) 7 ### Page 8 Downloaded from hlayiso.com Wiskunde Vraestel 2 MEMORANDUM GRAAD 11 4.2.2 Ĉ = 90°..... hoek in semi ⨀ ✓SR ̂ = 180° − (Ĉ + B D ̂ )..... som van hoeke van ∆ = 40° ✓ antwoord (2) 4.2.3 BD = √CD2 + BC 2 ......Pyth. = √402 + 302 = 50 ✓ BD = 50 1 BO = BD .....radii 2 = 25 ✓ BO = 25 1 OA = CD...Midpt. St. 2 = 20 ✓ SR ie (OA = 20 .. Midpt.St) AE = OE – OA = 25 – 20 = 5 eenhede ✓ AE = 5 (4) 8 ### Page 9 Downloaded from hlayiso.com Wiskunde Vraestel 2 MEMORANDUM GRAAD 11 VRAAG 5 PUNTE : 10 5.1 ̂L = ̂J1 = 𝑥........ hoeke teenoor gelyke sye ✓SR ̂ 1 = L̂ = x.......= koorde onderspan gelyke hoeke N ✓S ✓R (3) 5.2 ̂ = Ĵ3 = 𝑦........... hoeke teenoor = sye M ✓SR ̂1 = M K ̂ = y........= koorde onderspan gelyke hoeke ✓SR ̂2 = K Q ̂ 1 + Ĵ1 ... buite hoek van ∆ ✓SR = x +y ̂2 = N P ̂ 1 + ̂J3 ... buite hoek van ∆ ✓SR =x+y ∴ ̂2 = P Q ̂2 = x + y (4) 5.3 ̂4 = P P ̂2 .... regoorstde. hoeke ̂4 = P ✓ S R vir P ̂2 OF ̂4 = Q Q ̂ 2 ....regoorstde. hoeke ̂4 = Q Q ̂2 but ̂2 = P Q ̂2 ........bewys in Vr 5.2 ∴ ̂4 ̂4 = Q P ̂4 ̂4 = Q ✓P ∴ JQ = JP...... sye teenoor = hoeke OF gelykb. ∆ ✓R (3) 9 ### Page 10 Downloaded from hlayiso.com Wiskunde Vraestel 2 MEMORANDUM GRAAD 11 VRAAG 6 PUNTE: 12 6.1 ̂K 2 = ̂T1 ...... hoeke teenoor = sye OF gelyke radii ✓SR ̂T1 = Ĉ......... hoeke in selfde segment ̂ = Ĉ ✓T ∴ K̂ 2 = Ĉ ✓R (3) 6.2 ̂ ...... hoeke teenoor = sye OF gelyke radii ̂2 = A T ✓SR ̂2 = T P ̂... buite hoek van ∆ ̂2 + A ✓SR ̂2 = 2T ̂1 = P K ̂2 ........... hoeke in selfde segment ̂1 = P ✓K ̂2 ̂ ̂2 K1 = 2T ✓R (4) 6.3 ̂4 = 2T P ̂ ........ midpts. hoek = 2 x omtreks hoek ̂4 = 2T ✓P ̂ ✓R ̂= T T ̂1 + T ̂2 ̂1 = Ĉ ....... hoeke in selfde segment T ✓SR 1 ̂2 = T ̂ 1 ..... bewys in Vr.6.2 K 2 1 ̂2 = ✓T ̂1 K 2 ̂4 = 2(Ĉ + 1 K P ̂ ) 2 1 ̂4 = 2(Ĉ + 1 K ̂ ) ✓ P = 2Ĉ + K ̂1 2 1 (5) TOTAAL : 100 10 ## Related documents - [Gr11 Math P1 (Afrikaans) June 2017 Possible Answers](https://www.hlayiso.com/library/gr11-math-p1-afrikaans-june-2017-possible-answers/b8504619-0cb6-476d-944d-f53d2ea22f72) — Mathematics · Memo · 2017 · Afr - [Gr11 Math P2 (English and Afrikaans) November 2017 Answer Book](https://www.hlayiso.com/library/gr11-math-p2-english-and-afrikaans-november-2017-answer-book/b76d8c4b-9b1a-4580-a03a-7d0050e7ab92) — Mathematics · Addendum · 2017 · Afr - [Gr11 Math P2 (Afrikaans) June 2018 Question Paper Errata](https://www.hlayiso.com/library/gr11-math-p2-afrikaans-june-2018-question-paper-errata/f8b25236-b390-468d-a38e-7bca697a99b0) — Mathematics · Errata · 2018 · Afr - [Gr11 Math P2 June 2017 Question Paper](https://www.hlayiso.com/library/gr11-math-p2-june-2017-question-paper/64687c17-ca24-4dad-9367-5ed08ba6ca70) — Mathematics · QP · 2017 - [Gr11 Math P2 (Afrikaans) November 2019 Possible Answers](https://www.hlayiso.com/library/gr11-math-p2-afrikaans-november-2019-possible-answers/d5138457-8292-415c-8ae5-52ff1920ee97) — Mathematics · Memo · 2019 · Afr - [Gr11 Math P1 June 2017 Possible Answers](https://www.hlayiso.com/library/gr11-math-p1-june-2017-possible-answers/19ca7dfb-d358-4c42-bbfc-c68d580cfe21) — Mathematics · Memo · 2017 ## More from Grade 11 Mathematics - [FS Maths Grade 11 June 2025 P1 only](https://www.hlayiso.com/library/fs-maths-grade-11-june-2025-p1-only/82015d8f-bcc0-478e-ad47-d3b295dabb01) — QP · 2025 - [FS Maths Grade 11 June 2025 P2 only](https://www.hlayiso.com/library/fs-maths-grade-11-june-2025-p2-only/5d0e7dd8-cb37-419c-8fe2-0aee27fbdc84) — QP · 2025 - [KZN Maths Grade 11 November 2025 P2 and Memo](https://www.hlayiso.com/library/kzn-maths-grade-11-november-2025-p2-and-memo/a2aac931-a969-41c8-bde2-85d23035d042) — QP + Memo · 2025 - [MATHS P2 ANSWER BOOK GR11 2024 ENG AFR D watermark](https://www.hlayiso.com/library/maths-p2-answer-book-gr11-2024-eng-afr-d-watermark/91fd4f6b-a252-4d30-b6b1-f8f546015336) — Addendum · 2024 · Afr - [Gr 11 Mathematics P1 (Afrikaans) June 2024 Question Paper](https://www.hlayiso.com/library/gr-11-mathematics-p1-afrikaans-june-2024-question-paper/1a0f4a9d-f18a-4aca-af77-5a81acfb8e0c) — QP · 2024 · Afr - [Gr 11 Mathematics P2 (Afrikaans) June 2024 Question Paper](https://www.hlayiso.com/library/gr-11-mathematics-p2-afrikaans-june-2024-question-paper/4934c5d3-e2a1-4d6b-acf6-d00e0623d46d) — QP · 2024 · Afr ## Collection - [Grade 11 past papers](https://www.hlayiso.com/grade-11/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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