Technical Mathematics Grade 10 P2 November Mock 2026 Memo

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 √ 1.1 AB = (2 − (−2))2 + (4 − 1)2 ✓ SF A √ √ = 16 + 9 = 25 ✓ answer CA = 5 units (2) 4−1 3 1.2 mAB = = ✓ SF A 2 − (−2) 4 ✓ answer CA (2) 0−4 4 1.3 mBC = =− ✓ SF A 5−2 3( ) ✓ mBC = − 34 CA 3 4 mAB × mBC = × − = −1 ✓ product = −1 CA 4 3 (3) ∴ AB ⊥ BC ( ) −2 + 5 1 + 0 1.4 M ; ✓ x-value A 2 2 ( ) ✓ y-value A 3 1 (2) M ; OR M(1,5 ; 0,5) 2 2 −3 − 4 1.5 mBD = =7 ✓ mBD = 7 A 1−2 ✓ substitution of point CA y − 4 = 7(x − 2) ✓ answer CA y = 7x − 10 (3) Copyright reserved Please turn over

Technical Mathematics/P2 3 Hlayiso/Mock 2026 NSC – Marking Guidelines 1.6 y = 7(1,5) − 10 = 0,5 ✓ substitution of M CA ∴ M(1,5 ; 0,5) lies on BD (1) 1.7 Square. ✓ square A All four sides are equal (5 units) AND AB̂C = 90◦ . ✓ reason CA (A rhombus with a right angle is a square.) (2) [15] QUESTION 2 2.1.1 sin(37,4◦ + 18,6◦ ) = sin 56◦ ✓ 56◦ A = 0,83 ✓ answer CA (2) 1 2.1.2 − tan 18,6◦ ✓ sec A = cos1 A A cos 37,4◦ = 1,2587 . . . − 0,3365 . . . ✓ answer CA = 0,92 (2) 12 2.2.1 tan θ = , θ in the 3rd quadrant ✓ 3rd quadrant A 5 y ✓ x = −5 and y = −12 A −5 θ x −12 ✓ r = 13 CA 13 ✓ answer CA (4) r2 = (−5)2 + (−12)2 = 169, r = 13 12 sin θ = − 13 13 −5 2.2.2 cosec θ + cot θ = − + ✓ cosec θ = − 13 CA 12 −12 12 13 5 ✓ cot θ = 12 5 CA =− + ✓ answer CA 12 12 8 2 (3) =− =− 12 3 Copyright reserved Please turn over

Technical Mathematics/P2 4 Hlayiso/Mock 2026 NSC – Marking Guidelines 2.3 cos x = 0,65 ✓ cos x = 0,65 A x = cos−1 (0,65) ✓ inverse CA x = 49,46◦ ✓ answer CA (3) [14] QUESTION 3 RQ 3.1.1 sin 6◦ = ✓ correct ratio A 4,5 RQ = 4,5 sin 6◦ ✓ answer CA RQ = 0,47 m (2) PQ 3.1.2 cos 6◦ = ✓ correct ratio A 4,5 PQ = 4,5 cos 6◦ ✓ answer CA PQ = 4,48 m (2) (Pythagoras also accepted) 30 3.2.1 tan 65◦ = ✓ correct ratio A BC 30 ✓ BC as subject CA BC = ✓ answer CA tan 65◦ BC = 13,99 m (3) 30 3.2.2 sin 65◦ = ✓ correct ratio A AC 30 ✓ answer CA AC = = 33,10 m (2) sin 65◦ 30 3.2.3 tan 40◦ = ✓ correct ratio A BD 30 ✓ BD = 35,75 CA BD = = 35,75 m tan 40◦ CD = BD − BC = 35,75 − 13,99 ✓ answer CA CD = 21,76 m (3) [12] Copyright reserved Please turn over

Technical Mathematics/P2 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1 y f: (0◦ ; 2) (270◦ ; 2) ✓ x-intercepts A ✓ turning points (90◦ ; −2) & 2 g 1 (270◦ ; 2) A ✓ shape A 90◦ 180◦ 270◦ 360◦ x g: ✓ end points (0◦ ; 2), −1 (360◦ ; 2) A f ◦ ✓ min. point (180 ; 0) A −2 (90◦ ; −2) ✓ shape A (6) 4.2 2 ✓ answer A (1) 4.3 360◦ ✓ answer A (1) 4.4 0 ≤ y ≤ 2 OR y ∈ [0 ; 2] ✓ critical values CA ✓ notation CA (2) 4.5 0◦ < x < 180◦ OR x ∈ (0◦ ; 180◦ ) ✓ critical values CA ✓ notation CA (2) [12] Copyright reserved Please turn over

Technical Mathematics/P2 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 5 5.1 …equal to half of the third side. ✓ answer A (1) 5.2.1 QR = 2 × ST = 14 cm ✓S A (midpoint theorem) ✓R A (2) 5.2.2 PŜT = Q̂ = 68◦ ✓S A (corresponding ∠s; ST ∥ QR) ✓R A (2) 5.2.3 P̂ = 180◦ − 68◦ − 52◦ = 60◦ ✓S A (sum of ∠s of △) ✓R A (2) 5.3.1 In △ADE and △ABC: Â = Â (common) ✓ Â common S AD̂E = AB̂C (corresp. ∠s; DE ∥ BC) ✓ S/R S/R AÊD = AĈB (corresp. ∠s; DE ∥ BC) ✓ conclusion with reason R ∴ △ADE ||| △ABC (∠∠∠) (3) DE AD 5.3.2 = (||| △s) ✓ ratio with AB = 10 A BC AB 5 4 = ✓ answer CA BC 10 BC = 12,5 units (2) 5.4 Any TWO of: SSS (three sides) ✓ first condition A SAS (two sides and included angle) ✓ second condition A AAS (∠∠S) (2) RHS (90◦ , hypotenuse, side) [14] Copyright reserved Please turn over

Technical Mathematics/P2 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 6 6.1.1 Â = Ĉ (opp. ∠s of ∥m) ✓ S/R S/R 3x + 10◦ = 5x − 30◦ ✓ equation A 2x = 40◦ x = 20◦ ✓ answer CA (3) 6.1.2 Â = 3(20◦ ) + 10◦ = 70◦ ✓ Â = 70◦ CA B̂ = 180◦ − 70◦ = 110◦ ✓ answer with reason CA (co-interior ∠s; AD ∥ BC) (2) 6.2.1 PT̂ Q = 90◦ ✓S A (diagonals of a rhombus are perpendicular) ✓R A (2) 6.2.2 PT = 8 cm and QT = 6 cm (diagonals bisect) ✓ PT = 8 and QT = 6 A PQ2 = 82 + 62 (Pythagoras) ✓ Pythagoras CA PQ2 = 100 ✓ answer CA PQ = 10 cm (3) 6.2.3 4 × 10 = 40 cm ✓ answer CA (1) 6.3 …rectangle. ✓ answer A (1) [12] Copyright reserved Please turn over

Technical Mathematics/P2 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 1,4 7.1.1 r= = 0,7 m ✓ r = 0,7 A 2 ✓ SF CA V = π (0,7)2 (1,8) ✓ answer CA V = 2,77 m3 (3) 7.1.2 A = 2π (0,7)2 + 2π (0,7)(1,8) ✓ 2π r2 substituted CA = 3,08 + 7,92 ✓ 2π rh substituted CA A = 11,00 m2 ✓ answer CA (3) 1 7.2.1 A= × 2,4 × 0,6 ✓ SF A 2 ✓ answer CA A = 0,72 m2 (2) 7.2.2 V = 0,72 × 1,5 ✓ SF CA V = 1,08 m3 ✓ answer CA (2) 7.2.3 AC2 = 2,42 + 0,62 = 6,12 (Pythagoras) ✓ Pythagoras A AC = 2,47 m ✓ answer CA (2) [12] QUESTION 8 8.1.1 0,385 × 60′ = 23,1′ ✓ 23′ A 0,1 × 60′′ = 6′′ ✓ answer CA 47◦ 23′ 6′′ (2) 36 45 8.1.2 18◦ + + ✓ method M 60 3 600 = 18,61◦ (18,6125◦ ) ✓ answer CA (2) π π 8.2 150◦ × ✓ × 180 ◦ M 180◦ 5π ✓ answer CA = rad (2) 6 Copyright reserved Please turn over

Technical Mathematics/P2 9 Hlayiso/Mock 2026 NSC – Marking Guidelines π 8.3 θ = 75◦ × = 1,309 rad ✓ conversion to radians A 180◦ s = rθ ✓ SF CA s = 45 × 1,309 ✓ answer CA s = 58,90 cm (3) [9] TOTAL: 100 Copyright reserved