Technical Mathematics Grade 10 P2 2026 Memo (Eng)

Mock hlayiso mock exams Practice Series: Grade 10 Exam Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 10 TECHNICAL MATHEMATICS P2 NOVEMBER EXAMINATION 2026 (MOCK) MARKING GUIDELINES MARKS: 100 These marking guidelines consist of 9 pages. Copyright reserved Please turn over

Technical Mathematics/P2 2 Hlayiso/Mock 2026 NSC – Marking Guidelines MARKING CODES A Accuracy CA Consistent accuracy M Method R Rounding NPR No penalty for rounding NPU No penalty for units omitted S Simplification SF Substitution in correct formula S/R Statement/Reason AO Answer only NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • Consistent accuracy (CA) applies in ALL aspects of the marking guidelines where indicated. • No penalty for rounding (NPR) for ALL questions, unless stated otherwise. • In QUESTIONS 5 and 6 (Euclidean geometry) a statement without the correct reason earns no mark for the reason. Accept any valid alternative method. QUESTION 1 5−2 3 1.1 mPQ = = ✓ SF A 0 − (−2) 2 ✓ answer CA (2) 1−5 2 1.2 mQR = =− ✓ mQR = − 32 A 6−0 3( ) ✓ product = −1 CA 3 2 mPQ × mQR = × − = −1 ✓ 90◦ with reason CA 2 3 (3) ∴ PQ ⊥ QR, so PQ̂R = 90◦ (Also accept: PQ2 + QR2 = 13 + 52 = 65 = PR2 , converse of Pythagoras) 3 1.3 m= (parallel lines: equal gradients) ✓ m = 23 CA 2 3 ✓ substitution of R CA y − 1 = (x − 6) ✓ answer CA 2 3 (3) y = x−8 2 3 1.4 −2 = k − 8 ✓ substitution of S CA 2 3 ✓ simplification CA k=6 (2) 2 k=4 Copyright reserved Please turn over

Technical Mathematics/P2 3 Hlayiso/Mock 2026 NSC – Marking Guidelines ( ) 0 + 4 5 + (−2) 1.5 Midpoint of QS = ; = (2 ; 1,5) ✓ midpoint of QS A 2 2 ( ) ✓ midpoint of PR (same) A −2 + 6 2 + 1 (2) Midpoint of PR = ; = (2 ; 1,5) 2 2 Same point ∴ the diagonals bisect each other at M(2 ; 1,5) √ √ √ 1.6 PQ = (0 + 2)2 + (5 − 2)2 = 13 ✓ PQ = √ 13 A √ √ √ QR = (6 − 0)2 + (1 − 5)2 = 52 = 2 13 ✓ QR = 2 13 A Rectangle: the diagonals bisect each other (parallelogram, 1.5) AND PQ̂R = 90◦ (1.2); adjacent ✓ rectangle with reason CA sides PQ ̸= QR, so it is not a square. (3) (Also accept: parallelogram (diagonals bisect / RS ∥ PQ and PS ∥ QR) with a right angle, and PQ ̸= QR so not a square.) [15] Copyright reserved Please turn over

Technical Mathematics/P2 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 2 2.1.1 cos(2(52,3◦ ) − 21,8◦ ) = cos 82,8◦ ✓ 82,8◦ A = 0,13 ✓ answer CA (2) 1 2.1.2 ✓ cosec α = sin1α A sin 52,3◦ × tan 21,8◦ 1,2639 . . . = ✓ answer CA 0,3999 . . . (2) = 3,16 2.2.1 x2 + y2 = r2 ✓ Pythagoras substituted A (−8)2 + t 2 = 172 t 2 = 225 ✓ answer CA t = 15 (t > 0, P in the 2nd quadrant) (2) ( ) 15 8 2.2.2 sin θ + cos θ = + − ✓ both ratios CA 17 17 ✓ answer CA 7 = (2) 17 ( ) 17 −8 2.2.3 cosec θ − cot θ = − ✓ both ratios CA 15 15 ✓ answer CA 25 5 = = (2) 15 3 2 sin 38◦ 2.3 cos(2x − 10◦ ) = = 0,4104 . . . ✓ cos(2x − 10◦ ) = 0,41 A 3 ◦ ✓ 65,77 CA 2x − 10◦ = cos−1 (0,4104 . . .) = 65,77◦ ✓ adding 10◦ CA 2x = 75,77◦ ✓ answer CA x = 37,88◦ (4) [14] QUESTION 3 9,6 3.1.1 AD = = 4,8 m ✓ AD = 4,8 and ratio A 2 BD tan 28◦ = ✓ answer CA 4,8 BD = 4,8 tan 28◦ = 2,55 m (2) Copyright reserved Please turn over

Technical Mathematics/P2 5 Hlayiso/Mock 2026 NSC – Marking Guidelines 4,8 3.1.2 cos 28◦ = ✓ correct ratio CA AB 4,8 ✓ answer CA AB = = 5,44 m (2) cos 28◦ (Pythagoras or sine ratio also accepted) 3.1.3 In △ADE: AÊD = 90◦ ✓ right △ADE identified A DE DE ✓ correct ratio A sin 28◦ = = AD 4,8 ✓ answer CA ◦ DE = 4,8 sin 28 = 2,25 m (3) OR: DB̂E = 62◦ , DE = BD sin 62◦ = 2,25 m 3.2.1 PQRX is a rectangle: PX = 40 m, XR = 15 m ✓ correct ratio A TX tan 22◦ = 40 ✓ T X = 16,16 CA T X = 40 tan 22◦ = 16,16 m ✓ answer CA T R = 16,16 + 15 = 31,16 m (3) 31,16 3.2.2 tan T Q̂R = ✓ correct ratio CA 40 T Q̂R = 37,92◦ ✓ answer CA Angle of depression = T Q̂R = 37,92◦ (alt. ∠s; (2) horizontal lines ∥) [12] QUESTION 4 4.1 y f: ✓ x-intercepts 60◦ , 300◦ A 3 g ✓ min. (180◦ ; −3); end 2 (0◦ ; 1) (360◦ ; 1) points (0◦ ; 1), (360◦ ; 1) A 1 ✓ shape A 90◦ 180◦ 270◦ 360◦ x g: 60◦ 300◦ −1 ✓ asymptotes x = 90◦ , −2 x = 270◦ A −3 f ✓ intercepts 0◦ , 180◦ , 360◦ A (180◦ ; −3) ✓ shape A (6) 4.2 180◦ ✓ answer A (1) Copyright reserved Please turn over

Technical Mathematics/P2 6 Hlayiso/Mock 2026 NSC – Marking Guidelines 4.3 −3 ≤ y ≤ 1 OR y ∈ [−3 ; 1] ✓ critical values CA ✓ notation CA (2) 4.4 h(x) = 2 cos x + 2 ✓ answer CA (1) 4.5 60◦ < x < 90◦ OR x ∈ (60◦ ; 90◦ ) ✓ critical values CA ✓ notation CA (2) [12] QUESTION 5 5.1 …in proportion. ✓ answer A (Accept: proportional / in the same ratio.) (1) 5.2.1 R̂1 = P̂ = 58◦ ✓S A (alt. ∠s; RT ∥ QP) ✓R A (2) 5.2.2 R̂2 = Q̂ = 64◦ ✓S A (corresp. ∠s; RT ∥ QP) ✓R A (2) 5.2.3 R̂3 = 180◦ − 58◦ − 64◦ = 58◦ ✓S CA (∠s on a straight line) OR (sum of ∠s of △) ✓R A (2) 5.2.4 Isosceles triangle: P̂ = R̂3 = 58◦ ✓ isosceles CA ∴ QP = QR (sides opp. equal ∠s) ✓ QP = QR with reason CA (2) Copyright reserved Please turn over

Technical Mathematics/P2 7 Hlayiso/Mock 2026 NSC – Marking Guidelines 5.3.1 In △ADB and △ABC: Â = Â (common) ✓ Â common S AD̂B = AB̂C = 90◦ (given) ✓ 90◦ with reason S/R AB̂D = Ĉ (3rd ∠ of △) ✓ conclusion with reason R ∴ △ADB ||| △ABC (∠∠∠) (3) AB AD 5.3.2 = (||| △s) ✓ correct ratio A AC AB AB2 = AD × AC = 3,2 × 5 = 16 ✓ answer CA AB = 4 units (2) [14] QUESTION 6 6.1.1 Â + B̂ = 180◦ (co-int. ∠s; AD ∥ BC) ✓ S/R S/R 3x − 10◦ + 2x + 15◦ = 180◦ ✓ equation A 5x = 175◦ ✓ answer CA x = 35◦ (3) 6.1.2 B̂ = 2(35◦ ) + 15◦ = 85◦ ✓ B̂ = 85◦ CA Ĉ = 4(35◦ ) − 55◦ = 85◦ ✓ Ĉ = 85◦ CA B̂ = Ĉ: the base angles on the parallel side BC are equal ∴ ABCD is an isosceles trapezium. ✓ conclusion CA (3) (Also accept: D̂ = 180◦ − 85◦ = 95◦ = Â.) 6.2.1 In △KLM and △KNM: KL = KN (given) ✓ both given sides S ML = MN (given) ✓ common side S KM = KM (common) ✓ SSS R ∴ △KLM ≡ △KNM (SSS) (3) Copyright reserved Please turn over

Technical Mathematics/P2 8 Hlayiso/Mock 2026 NSC – Marking Guidelines 6.2.2 LO = 8 cm and K ÔL = 90◦ ✓ LO = 8 and 90◦ with (diagonal KM of the kite bisects LN perpendicularly) reason S/R OR from △KLO ≡ △KNO (SAS, using LK̂O = N K̂O from 6.2.1): LO = ON = 8 cm and K ÔL = K ÔN = 90◦ (∠s on a str. line) √ ✓ KL = 10 and LM = 17 CA KL = 62 + 82 = 10 cm (Pythagoras) √ 2 2 LM = 15 + 8 = 17 cm (Pythagoras) ✓ answer CA Perimeter = 2(10 + 17) = 54 cm (3) [12] QUESTION 7 7.1.1 Area = 12 × 8 − 5 × 3 ✓ method M = 96 − 15 = 81 m2 ✓ answer CA (OR 7 × 8 + 5 × 5 = 81 m2 ) (2) 7.1.2 Area of one tile = 0,6 × 0,6 = 0,36 m2 ✓ 0,36 m2 A 81 ✓ 225 tiles CA Number of tiles = = 225 0,36 With 10%: 225 × 1,1 = 247,5 ✓ 248 (rounded UP) CA ∴ 248 tiles must be bought (3) 7.2.1 R = 0,6 m, r = 0,5 m ✓ both radii A V = π (0,6)2 (2,5) − π (0,5)2 (2,5) ✓ outer volume CA = 2,827 . . . − 1,963 . . . ✓ minus inner volume CA ✓ answer CA V = 0,86 m3 (4) 7.2.2 A = 2π rh = 2π (0,6)(2,5) ✓ r = 0,6 CA A = 9,42 m2 ✓ SF CA (Using r = 0,5 (inner surface): max 2 marks (SF, CA ✓ answer CA answer = 7,85 m2 )) (3) [12] Copyright reserved Please turn over

Technical Mathematics/P2 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1.1 0,745 × 60′ = 44,7′ ✓ 44′ A 0,7 × 60′′ = 42′′ ✓ answer CA 63◦ 44′ 42′′ (2) 17 24 8.1.2 112◦ + + ✓ method M 60 3 600 = 112,29◦ (exact value) ✓ answer CA (2) 7(180◦ ) 180◦ 8.2 + 40◦ − ✓ π = 180◦ substituted M 6 4 = 210◦ + 40◦ − 45◦ ✓ answer CA = 205◦ (2) 0,6 8.3 r= = 0,3 m = 30 cm ✓ r = 30 cm (same units) A 2 s = rθ : 42 = 30θ ✓ θ = 1,4 rad CA θ = 1,4 rad ✓ answer in degrees CA 180◦ θ = 1,4 × = 80,21◦ (3) π [9] TOTAL: 100 Copyright reserved