OCR A Level Mathematics A: Trigonometric ratios, triangle rules, small angles, graphs and basic equations (H240) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 1.05a to 1.05g and 1.05o of OCR A Level Mathematics A (H240), Version 3.1 (August 2026), for first assessment in 2018. Parts marked A Level only use Stage 2 content: radians, sectors and small angle approximations. The specification allows a scientific or graphical calculator on every paper, so use one where it helps, and show your working. Angles are in degrees unless stated, and answers are to 3 significant figures unless exact values are requested.

Learn the content first with the study guide and the revision notes. Course hub: /boards/ocr/a-level/mathematics/. Printable checklist: /checklists/ocr/a-level/mathematics/. Want a quick baseline first? Use the free diagnostics.

Questions

1. Write down the exact value of:

(a) sin 240° [1] (b) cos(11π/6) (A Level only) [1] (c) tan(−45°) [1]

2. (a) Express 105° in radians, as a multiple of π. (A Level only) [1] (b) Express 5π/9 in degrees. (A Level only) [1] (c) A sector has radius 12 cm and angle 0.65 radians. Find its arc length, area and perimeter. (A Level only) [3]

3. (A Level only) A sector of a circle has area 40 cm² and arc length 10 cm.

(a) Find the radius and the sector angle in radians. [4] (b) Find the area between the arc and the chord joining its ends. [3]

4. In triangle PQR, angle P = 58°, angle Q = 39° and QR = 14.2 cm.

(a) Find the length PR. [2] (b) Find the length PQ. [2]

5. In triangle ABC, AB = 8.4 cm, BC = 11.3 cm and angle ABC = 73°.

(a) Find AC. [3] (b) Find angle BAC. [3] (c) Find the area of the triangle. [2]

6. In triangle ABC, angle B = 40°, AC = 8 cm and AB = 11 cm.

(a) Find the two possible sizes of angle C. [3] (b) Find both possible values of BC. [3]

7. A walker goes from A to B, a distance of 5.4 km on a bearing of 040°. She then walks 6.8 km from B to C on a bearing of 115°. Find how far she is from A, and the bearing of her position C from A. [7]

8. It is given that cos 50° = c.

(a) Write each of these in terms of c: cos 130°, cos 310°, sin 140°. [3] (b) State the period of y = tan x in degrees, and the equation of its asymptote between x = 0° and x = 180°. [2] (c) State the number of solutions of sin x = 0.3 for 0° ≤ x ≤ 720°. [1]

9. (A Level only) In this question θ is small and in radians.

(a) Use the approximation for cos θ to estimate cos 0.1. Find the error in your estimate, to 2 significant figures. [2] (b) Show that (1 − cos 4θ) / (θ tan 2θ) is approximately 4. [3] (c) Use the small angle approximations to form a quadratic equation for the small solution of cos θ + 2 sin θ = 1.1, and solve it to find this solution to 3 significant figures. [3]

10. (a) Solve 5 sin θ = −2 for −180° < θ < 180°. [3] (b) Solve tan 2θ = 1.5 for 0° ≤ θ < 180°. [3]

11. (a) Solve 2 cos²θ + cos θ − 1 = 0 for 0° ≤ θ < 360°. [4] (b) (A Level only) Solve tan(2x + π/4) = 1 for −π ≤ x ≤ π, giving exact answers. [4]

Answers

1. (a) 240° is in quadrant 3 with reference angle 60°, where sine is negative: sin 240° = −√3/2 [1] (b) 11π/6 = 330°, reference angle 30°, cosine positive: √3/2 [1] (c) tan(−45°) = −tan 45° = −1 [1] Examiner insight: Exact value questions are about the reference angle and the sign. A decimal from a calculator is a risky substitute when the question says “exact”.

2. (a) 105 × π/180 = 7π/12 [1] (b) 5π/9 × 180/π = 100° [1] (c) Arc length = rθ = 12 × 0.65 = 7.8 cm [1]; area = ½ × 144 × 0.65 = 46.8 cm² [1]; perimeter = 24 + 7.8 = 31.8 cm [1] Examiner insight: The perimeter needs the two radii as well as the arc. Candidates often stop at the arc length, so read the word “perimeter” twice.

3. (a) Area = ½rs, so 40 = ½ × r × 10 [1], giving r = 8 cm [1]. Then θ = s/r [1] = 10/8 = 1.25 radians [1] (b) Segment = sector − triangle = ½r²θ − ½r² sin θ = ½r²(θ − sin θ) [1] = 32(1.25 − sin 1.25) [1] = 9.63 cm² (with the calculator in radian mode) [1] Examiner insight: A calculator left in degree mode gives a wrong value for sin 1.25 with no warning. Estimate first: sin 1.25 should be a little below 1.

4. (a) PR / sin 39° = 14.2 / sin 58° [1], so PR = 10.5 cm [1] (b) Angle R = 180° − 58° − 39° = 83° [1], so PQ = 14.2 × sin 83° / sin 58° = 16.6 cm [1] Examiner insight: Match each side to its opposite angle before substituting. Writing the ratio down first makes the method clear even if a later number slips.

5. (a) AC² = 8.4² + 11.3² − 2 × 8.4 × 11.3 × cos 73° [1] = 198.25 − 55.50 = 142.75 [1], so AC = 11.9 cm [1] (b) sin A / 11.3 = sin 73° / 11.9476 [1], so sin A = 0.9045 [1]. BC is shorter than AC, so A is smaller than 73° and acute: A = 64.8° [1] (c) Area = ½ × 8.4 × 11.3 × sin 73° [1] = 45.4 cm² [1] Examiner insight: Keep the unrounded value of AC for part (b). Using 11.9 can shift the final angle in the third figure.

6. (a) sin C / 11 = sin 40° / 8 [1], so sin C = 0.8838 [1], giving C = 62.1° or 117.9° [1] (b) Angle A = 180° − 40° − C = 77.9° or 22.1° [1]. BC = 8 sin A / sin 40° [1], giving 12.2 cm or 4.68 cm [1] Examiner insight: This is the ambiguous case. The wording “possible values” signals two answers, and a complete answer shows that 117.9° fits in the triangle (40° + 117.9° is less than 180°).

7. The bearing of A from B is 220°, so angle ABC = 220° − 115° = 105° [1]. AC² = 5.4² + 6.8² − 2 × 5.4 × 6.8 × cos 105° [1] = 75.4 + 19.01 = 94.41 [1], so AC = 9.72 km [1]. sin(∠BAC) / 6.8 = sin 105° / 9.716 [1], so ∠BAC = 42.5° [1]. Bearing of C from A = 040° + 42.5° = 082.5° [1] Examiner insight: A sketch with a north line at each point is the safest way to show where 105° comes from. Quote the bearing with three figures.

8. (a) cos 130° = −cos 50° = −c [1]; cos 310° = cos 50° = c [1]; sin 140° = sin 40° = cos 50° = c [1] (b) Period 180° [1]; asymptote x = 90° [1] (c) Two solutions in each 360° cycle, so 4 [1] Examiner insight: “In terms of c” means no other trig values or decimals in the answer. For each part, name the symmetry you used.

9. (a) cos 0.1 ≈ 1 − 0.01/2 = 0.995 [1]. The calculator value is 0.99500417, so the error is 4.2 × 10⁻⁶ [1] (b) 1 − cos 4θ ≈ 1 − (1 − (4θ)²/2) = 8θ² [1]; tan 2θ ≈ 2θ so the denominator is about 2θ² [1]; the ratio is 8θ² / 2θ² = 4 [1] (c) 1 − θ²/2 + 2θ = 1.1, so θ² − 4θ + 0.2 = 0 [1]; θ = 2 ± √3.8 [1]; the small root is θ = 0.0506 (the root near 3.95 is not small) [1] Examiner insight: In a “show that” question write every substitution line. The final value 4 must come from your working, not be stated at the start.

10. (a) sin θ = −0.4 [1]; principal value θ = −23.6° [1]; the second solution is −180° + 23.6° = −156.4°, so θ = −23.6° or −156.4° [1] (b) 0° ≤ 2θ < 360° [1]; 2θ = 56.3° or 236.3° [1]; θ = 28.2° or 118.2° [1] Examiner insight: Convert the interval before you solve. If you solve for θ first and then double, you will miss or invent solutions.

11. (a) (2 cos θ − 1)(cos θ + 1) = 0 [1]; cos θ = 1/2 gives θ = 60° or 300° [1]; cos θ = −1 gives θ = 180° [1]; no other values, so there are three solutions in total [1] (b) The range for 2x + π/4 is −7π/4 ≤ 2x + π/4 ≤ 9π/4 [1]. tan = 1 when 2x + π/4 = π/4 + nπ [1], so 2x = nπ and the values of n from −2 to 2 are in range [1]; x = −π, −π/2, 0, π/2, π [1] Examiner insight: Each factor of a quadratic gives its own set of solutions, so check you have dealt with both. In (b), the endpoints count because the inequalities are not strict.

Where marks are usually lost

• Calculator in degree mode for a radians question, or the reverse.
• Giving only the principal value of an inverse trig function and missing the other solutions in the interval.
• Dividing by k too early, or not extending the interval when the angle is a multiple such as 2θ or 3θ.
• Forgetting the second triangle in the ambiguous case, or keeping a second angle that pushes the total over 180°.
• Using ½ab sin C when the angle lies outside the two sides used.
• Rounding intermediate values, so that a later answer is off in the third significant figure.
• Writing bearings with fewer than three figures, or measuring from the wrong north line.
• Using sector formulae with an angle in degrees.
• Decimals where an exact value is asked for.

Next steps

• Go back over the revision notes for any weak topics.
• Reread the study guide where an answer surprised you.
• Hub: OCR A Level Mathematics, plus the printable checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, A Level in Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR.