O Level Elementary Mathematics Practice Paper 1

TuitionGoWhere Practice Paper - Elementary Mathematics O-Level

Answer Key and Marking Scheme (Version 1)

Total Marks: 40

Section A: Basic Trigonometry and Pythagoras' Theorem (10 marks)

1. (a) (AC = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10) cm. [A1]

(b) (\sin \angle BAC = \frac{\text{opp}}{\text{hyp}} = \frac{BC}{AC} = \frac{6}{10} = \frac{3}{5}). [A1]

2. Let height be (h) m. (h^2 + 2^2 = 5^2) [M1] (h^2 = 25 - 4 = 21) (h = \sqrt{21} \approx 4.58) m (3 s.f.) [A1]

3. (\tan \angle PRQ = \frac{\text{opp}}{\text{adj}} = \frac{PQ}{QR} = \frac{12}{9} = \frac{4}{3}). [A1]

4. Hypotenuse is (x + 2). ((x + 2)^2 = x^2 + (x + 1)^2) [M1] (x^2 + 4x + 4 = x^2 + x^2 + 2x + 1) (x^2 + 4x + 4 = 2x^2 + 2x + 1) (0 = x^2 - 2x - 3) [M1] ((x - 3)(x + 1) = 0) (x = 3) or (x = -1) (reject as length cannot be negative). Sides are 3 cm, 4 cm, and 5 cm. [A1]

5. Let horizontal distance be (d) m. (\tan 28^\circ = \frac{80}{d}) [M1] (d = \frac{80}{\tan 28^\circ} \approx \frac{80}{0.5317} \approx 150) m (3 s.f.) [A1]

Section B: Sine Rule, Cosine Rule, and Area of Triangle (12 marks)

6. (a) Using cosine rule: (BC^2 = 7^2 + 9^2 - 2(7)(9)\cos 65^\circ) [M1] (BC^2 = 49 + 81 - 126 \cos 65^\circ) (BC^2 = 130 - 126(0.4226) = 130 - 53.25 = 76.75) (BC = \sqrt{76.75} \approx 8.76) cm (3 s.f.) [A1]

(b) Area (= \frac{1}{2} \times 7 \times 9 \times \sin 65^\circ) [M1] (= 31.5 \times 0.9063 \approx 28.5) cm² (3 s.f.) [A1]

7. (a) Using sine rule: (\frac{\sin \angle XZY}{10} = \frac{\sin 40^\circ}{8}) [M1] (\sin \angle XZY = \frac{10 \sin 40^\circ}{8} = 1.25 \times 0.6428 = 0.8035) [M1] (\angle XZY = \sin^{-1}(0.8035) \approx 53.5^\circ) or (180^\circ - 53.5^\circ = 126.5^\circ). [A1]

(b) In triangle (XYZ), (XY = 10) cm and (XZ = 8) cm. Since (XY > XZ), the angle opposite (XY) ((\angle XZY)) must be larger than the angle opposite (XZ) ((\angle XYZ = 40^\circ)). Both (53.5^\circ) and (126.5^\circ) are greater than (40^\circ), but the sum of angles in a triangle is (180^\circ). If (\angle XZY = 126.5^\circ), then (\angle YXZ = 180^\circ - 40^\circ - 126.5^\circ = 13.5^\circ), which is valid. However, the ambiguous case of the sine rule requires checking: both values give a possible triangle. The question implies one valid value; typically, the acute angle is taken unless context dictates otherwise. Accept reasoned argument for either value with justification. [A1]

Marking note: Award full credit for identifying both values and explaining the ambiguity, or for selecting one value with valid geometric reasoning.

8. Largest angle is opposite the longest side (70 m). Let this angle be (C). Using cosine rule: (70^2 = 50^2 + 60^2 - 2(50)(60)\cos C) [M1] (4900 = 2500 + 3600 - 6000 \cos C) (4900 = 6100 - 6000 \cos C) (6000 \cos C = 1200) (\cos C = 0.2) (C = \cos^{-1}(0.2) \approx 78.5^\circ) (1 d.p.) [A1]

9. Area (= \frac{1}{2} \times 15 \times 12 \times \sin 30^\circ) [M1] (= 90 \times 0.5 = 45) cm². [A1]

Section C: Bearings and 3D Applications (10 marks)

10. (a) Diagram showing:

• North line at (A), bearing (055^\circ) to (B) (8 km).
• North line at (B), bearing (145^\circ) to (C) (6 km).
• Triangle (ABC) with angle (ABC) labelled.
• Angle between north at (B) and (BA): (180^\circ + 55^\circ = 235^\circ) (back bearing). Angle (ABC = 235^\circ - 145^\circ = 90^\circ). [A2] (1 for correct positions, 1 for correct angle labelling)

(b) Since (\angle ABC = 90^\circ), triangle (ABC) is right-angled at (B). (AC = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10) km. [A2]

(c) (\tan \angle BAC = \frac{6}{8} = 0.75) [M1] (\angle BAC = \tan^{-1}(0.75) \approx 36.9^\circ). Bearing of (C) from (A = 055^\circ + 36.9^\circ = 091.9^\circ). [A1]

11. (a) (AC = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10) cm. [A1]

(b) (EC) is the space diagonal. (EC^2 = AC^2 + AE^2) (since (AE) is perpendicular to base). [M1] (EC^2 = 10^2 + 5^2 = 100 + 25 = 125) (EC = \sqrt{125} = 5\sqrt{5} \approx 11.2) cm (3 s.f.) [A1]

(c) Let (\theta) be the angle between (EC) and the base. In right-angled triangle (EAC): (\tan \theta = \frac{AE}{AC} = \frac{5}{10} = 0.5) [M1] (\theta = \tan^{-1}(0.5) \approx 26.6^\circ) (1 d.p.) [A1]

Section D: Angles of Elevation, Depression, and Multi-step Problems (8 marks)

12. Let height of taller building be (H) m. Difference in height (= H - 25). (\tan 18^\circ = \frac{H - 25}{40}) [M1] (H - 25 = 40 \tan 18^\circ) [M1] (H - 25 = 40 \times 0.3249 = 12.996) (H = 25 + 13.0 = 38.0) m (3 s.f.) [A1]

13. (a) In right-angled triangle (PQR): (\tan 38^\circ = \frac{PQ}{PR} = \frac{PQ}{12}) [M1] (PQ = 12 \tan 38^\circ \approx 12 \times 0.7813 = 9.38) m (3 s.f.) [A1]

(b) In right-angled triangle (PQS): (\tan 52^\circ = \frac{PQ}{PS} = \frac{PQ}{9}) (PQ = 9 \tan 52^\circ \approx 9 \times 1.2799 = 11.52) m.

Note: The two values of (PQ) are inconsistent, indicating the points (R) and (S) are on opposite sides of (Q).

Using (PQ = 12 \tan 38^\circ \approx 9.375) m: In (\triangle PQR): (QR = \frac{PQ}{\tan 38^\circ} = 12) m (given). In (\triangle PQS): (QS = \frac{PQ}{\tan 52^\circ} = \frac{9.375}{1.2799} \approx 7.32) m. [M1]

Since (R), (Q), and (S) are collinear, and (R) and (S) are on opposite sides of (Q): (RS = RQ + QS = 12 + 7.32 = 19.3) m (3 s.f.) [A2]

Alternative: Use (PQ = 9 \tan 52^\circ \approx 11.52) m, then (QR = 11.52 / \tan 38^\circ \approx 14.74) m, (QS = 9) m, (RS = 14.74 + 9 = 23.7) m. Award marks for consistent working with either value, noting the ambiguity in the problem statement.

END OF ANSWER KEY