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DSE Mathematics (Compulsory Part) Practice Paper: Trigonometry and Coordinate Geometry (12 Questions, 50 Marks, 75 Minutes)
DSE Mathematics: Trigonometry and Coordinate Geometry
DSE Maths: 12 questions, 50 marks, 75 minutes on trigonometric ratios, sine and cosine rules, trigonometric equations, elevation angles, straight lines and circles.
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Paper details Level: S6 | Subject: Mathematics | Time: 75 minutes | Marks: 50 | Questions: 12
Section A (Q1-Q6): Trigonometry
Answer all questions. Show your working.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | In right-angled triangle ABC, angle C = 90°, AB = 17 cm and BC = 8 cm. (a) Find the length of AC. (b) Express sin A, cos A and tan A as fractions. | (a) AC² = 17² – 8² = 289 – 64 = 225, so AC = 15 cm. (2 marks) (b) sin A = BC ÷ AB = 8/17, cos A = AC ÷ AB = 15/17 and tan A = BC ÷ AC = 8/15. (2 marks) |
| 2 | Without using a calculator, evaluate (sin 30° cos 60° + tan 45°) ÷ sin² 45°. | sin 30° = cos 60° = 1/2, tan 45° = 1 and sin² 45° = 1/2. (2 marks) The numerator is 1/4 + 1 = 5/4, so the expression equals (5/4) ÷ (1/2) = 5/2. (2 marks) |
| 3 | The three sides of a triangle are 7 cm, 8 cm and 9 cm. Find the largest angle, correct to 0.1°. | The largest angle is opposite the longest side, 9 cm. By the cosine rule, cos C = (7² + 8² – 9²) ÷ (2 × 7 × 8) = 32 ÷ 112 = 2/7. (2 marks) Hence C = 73.4°, correct to 0.1°. (2 marks) |
| 4 | Two sides of a triangle are 6 cm and 10 cm and the included angle is 30°. (a) Find the area of the triangle. (b) Find the length of the third side, correct to 3 significant figures. | (a) Area = (1/2) × 6 × 10 × sin 30° = 15 cm². (2 marks) (b) Third side² = 6² + 10² – 2 × 6 × 10 × cos 30° = 136 – 60√3 ≈ 32.08, so the third side is about 5.66 cm. (2 marks) |
| 5 | Solve the equation 2 sin² x + sin x − 1 = 0 for 0° ≤ x ≤ 360°. | Let u = sin x. Then 2u² + u – 1 = 0, that is (2u – 1)(u + 1) = 0, so u = 1/2 or u = -1. (2 marks) sin x = 1/2 gives x = 30° or 150°; sin x = -1 gives x = 270°. The solutions are 30°, 150° and 270°. (2 marks) |
| 6 | From point A on horizontal ground, the angle of elevation of the top of a tower is 30°. After walking 40 m towards the tower to point B, the angle of elevation becomes 60°. Find the height of the tower, correct to 0.1 m. | Let the height of the tower be h m and let T be its foot. Then AT = h ÷ tan 30° = h√3 and BT = h ÷ tan 60° = h ÷ √3. (2 marks) Since AT – BT = 40, we get h(√3 – 1/√3) = 40, so h = 20√3 ≈ 34.6 m. (2 marks) |
Section B (Q7-Q9): Coordinate Geometry
Answer all questions using coordinate geometry methods.
| # | Question | Answer and marking notes |
|---|---|---|
| 7 | A(-2, 3) and B(4, -5) are given. (a) Find the length of AB. (b) Find the coordinates of the midpoint of AB. (c) Find the slope of the line AB. | (a) AB = √[(4 + 2)² + (-5 – 3)²] = √(36 + 64) = 10. (2 marks) (b) Midpoint = (( -2 + 4 ) ÷ 2, ( 3 – 5 ) ÷ 2) = (1, -1); the slope of AB is (-5 – 3) ÷ (4 + 2) = -4/3. (2 marks) |
| 8 | The equation of a straight line L is 3x + 4y – 12 = 0. (a) Find the x-intercept and the y-intercept of L. (b) Find the distance from the point (2, 1) to L. | (a) Putting y = 0 gives x = 4 and putting x = 0 gives y = 3, so the intercepts are 4 and 3. (2 marks) (b) Distance = |3(2) + 4(1) – 12| ÷ √(3² + 4²) = 2 ÷ 5 = 0.4. (2 marks) |
| 9 | The equation of a circle C is x² + y² – 6x + 8y – 11 = 0. (a) Find the centre and the radius of C. (b) Determine whether the point (3, 2) lies inside, on or outside C. | (a) x² – 6x + y² + 8y = 11, so completing the square gives (x – 3)² + (y + 4)² = 11 + 9 + 16 = 36; the centre is (3, -4) and the radius is 6. (2 marks) (b) The distance from (3, 2) to the centre is √(0² + 6²) = 6, which equals the radius, so the point lies on the circle. (2 marks) |
Section C (Q10-Q12): Combined Applications
Answer all questions and explain your reasoning.
| # | Question | Answer and marking notes |
|---|---|---|
| 10 | How many points do the line 3x + 4y = 25 and the circle x² + y² = 25 have in common? If there is only one point, find its coordinates. | The distance from the centre (0, 0) to the line 3x + 4y – 25 = 0 is 25 ÷ √(9 + 16) = 25 ÷ 5 = 5, which equals the radius, so the line is a tangent and there is exactly one point in common. (3 marks) The point is (3, 4), since 3(3) + 4(4) = 25 and 3² + 4² = 25. (2 marks) |
| 11 | (a) Prove that (1 – cos²θ) ÷ (sin θ cos θ) = tan θ. (b) Given that tan θ = 3/4 and θ is acute, find sin θ and cos θ. | (a) LHS = (1 – cos²θ) ÷ (sin θ cos θ) = sin²θ ÷ (sin θ cos θ) = sin θ ÷ cos θ = tan θ. (3 marks) (b) With tan θ = 3/4 and θ acute, take opposite 3, adjacent 4 and hypotenuse 5, so sin θ = 3/5 and cos θ = 4/5. (2 marks) |
| 12 | Two boats leave a lighthouse O at the same time, with OA = 12 km, OB = 9 km and angle AOB = 120°. (a) Find the distance AB. (b) Find the area of triangle OAB, correct to 3 significant figures. | (a) AB² = 12² + 9² – 2 × 12 × 9 × cos 120° = 225 + 108 = 333, so AB = √333 ≈ 18.2 km. (2 marks) (b) Area = (1/2) × 12 × 9 × sin 120° = 54 × (√3 ÷ 2) ≈ 46.8 km². (2 marks) |
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