OxfordAQA International AS and A-level Maths: Sine/cosine rules, radians, arcs (9660) – 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 practise content statement PP1.2 (Trigonometry, first part) in Version 5.2 of the OxfordAQA International AS and A-level Mathematics (9660) specification. The specification puts this content in International AS Unit PSM1 (section 3.2, PP1). Start with the study guide first, or use the revision notes to refresh. See the course hub and the printable checklist. Give lengths and angles to 3 significant figures unless a question says otherwise. Mark totals are shown in bold brackets.

Questions

1. (a) Write 160° in radians, in terms of π. [1] (b) Write 11π/12 radians in degrees. [1] (c) Write 0.9 radians in degrees, to 1 decimal place. [1]

2. In triangle PQR, angle P = 54°, angle Q = 67° and the side opposite P has length 12.6 cm. Find the lengths of the other two sides. [4]

3. In triangle ABC, AB = 5.8 cm, BC = 9.1 cm and angle ABC = 112°. (a) Find the length of AC. [3] (b) Find the area of the triangle. [2]

4. A triangle has sides of length 6 cm, 7 cm and 9 cm. Find the size of its largest angle, and the area of the triangle. [5]

5. A sector of a circle has radius 9.5 cm and angle 0.8 radians. Find its arc length, its area and its perimeter. [4]

6. A circular sector has radius 5 cm, and its curved edge is 8.5 cm long. Find the angle of the sector in radians and the area of the sector. [3]

7. A sector of a circle has perimeter 24 cm and area 36 cm². Find the radius and the angle of the sector. [5]

8. A glass panel is cut from a circle of radius 12 cm by a straight cut AB. The cut AB subtends an angle of 2π/3 at the centre O. The panel is the segment between AB and the minor arc. (a) Find the exact area of the panel. [4] (b) Find the exact perimeter of the panel. [2]

9. (a) State the period of y = tan x and the equations of its asymptotes for 0° ≤ x ≤ 360°. [2] (b) For y = 3 sin 2x, state the period, the maximum value, and the smallest positive value of x at which the maximum occurs. [3]

10. It is given that sin 38° = 0.6157, cos 38° = 0.7880 and tan 38° = 0.7813, all to 4 decimal places. Write down the value of each of the following. [5] sin 142°, cos 142°, sin 218°, cos 322°, tan 218°

11. A walker leaves a camp at P and goes 6.5 km on a bearing of 070° to a lake at Q. She then walks 9 km on a bearing of 175° to a hut at R. (a) Show that angle PQR = 75°. [2] (b) Find the straight-line distance PR. [3] (c) Find the bearing of R from P. [3]

12. A sector of a circle has centre O and radius 10 cm. The chord joining the two ends of the arc has length 16 cm. (a) Use the cosine rule to find the angle of the sector, in radians to 3 significant figures. [3] (b) Find the perimeter of the sector. [2] (c) Find the area of the region enclosed by the chord and the arc. [3]

Answers

1. (a) 160 × π/180 = 8π/9 [1] (b) 11π/12 × 180/π = 165° [1] (c) 0.9 × 180/π = 51.6° [1] Examiner insight: Show the multiplication by π/180 or 180/π, because a bare answer to an angle conversion gives little to credit if a digit is wrong, and an exact answer in terms of π is usually the form asked for.

2. Angle R = 180° − 54° − 67° = 59° [1] q / sin 67° = 12.6 / sin 54° [1] q = 12.6 × sin 67° ÷ sin 54° = 14.3 cm [1] r = 12.6 × sin 59° ÷ sin 54° = 13.3 cm [1] Examiner insight: When two angles are given, find the third before using the sine rule; setting up the ratio with the correct matching pair is the step most likely to be credited even if the value is slightly off.

3. (a) AC² = 5.8² + 9.1² − 2 × 5.8 × 9.1 × cos 112° [1] = 33.64 + 82.81 + 39.5… = 156.0… [1] AC = 12.5 cm [1] (b) Area = ½ × 5.8 × 9.1 × sin 112° [1] = 24.5 cm² [1] Examiner insight: Cos 112° is negative, so the two products add. Writing the substitution line before evaluating lets the method be seen even if a sign slips.

4. Largest angle is opposite the 9 cm side. cos θ = (6² + 7² − 9²) / (2 × 6 × 7) [1] = 4/84 = 0.0476… [1] θ = 87.3° [1] Area = ½ × 6 × 7 × sin 87.3° [1] = 21.0 cm² [1] Examiner insight: Use the unrounded angle in the area, since a rounded angle can shift the third significant figure; identify why you are choosing the 9 cm side so the working is easy to follow.

5. Arc = 9.5 × 0.8 = 7.6 cm [1] Area = ½ × 9.5² × 0.8 = 36.1 cm² [1] Perimeter = 2 × 9.5 + 7.6 [1] = 26.6 cm [1] Examiner insight: A common loss is giving the arc length as the perimeter. Writing 2r + arc makes it clear the radii were included.

6. θ = 8.5/5 = 1.7 rad [1] Area = ½ × 5² × 1.7, or ½ × 5 × 8.5 [1] = 21.25 cm², which is 21.3 cm² to 3 s.f. [1] Examiner insight: The shortcut A = ½rl is a valid alternative, but quote the formula you use so the method is clear.

7. Perimeter: 2r + rθ = 24, so rθ = 24 − 2r [1] Area: ½r(rθ) = 36 [1] ½r(24 − 2r) = 36, so 12r − r² = 36 [1] r² − 12r + 36 = 0, so (r − 6)² = 0 and r = 6 cm [1] θ = (24 − 12)/6 = 2 rad [1] Examiner insight: A repeated root means only one sector fits, so do not hunt for a second solution. Substitute back to check that both the perimeter and the area are satisfied.

8. (a) Sector area = ½ × 12² × 2π/3 = 48π [1] Triangle area = ½ × 12² × sin(2π/3) [1] = 72 × √3/2 = 36√3 [1] Panel = 48π − 36√3 cm² [1] (b) Arc = 12 × 2π/3 = 8π [1] Chord² = 12² + 12² − 2 × 12 × 12 × cos(2π/3) = 288 + 144 = 432, so the chord is 12√3. Perimeter = 8π + 12√3 cm [1] Examiner insight: Exact answers are expected, so keep π and √3 in the working. A decimal such as 88.4 can be given as a check but should not replace the exact form.

9. (a) Period is 180° (π radians) [1] Asymptotes at x = 90° and x = 270° [1] (b) Period = 360° ÷ 2 = 180° [1] Maximum value 3 [1] Maximum when 2x = 90°, so x = 45° [1] Examiner insight: Say how you reached each value, for example dividing 360° by 2, so that a slip in one part does not hide a correct idea in the others.

10. sin 142° = sin(180° − 38°) = 0.6157 [1] cos 142° = −cos 38° = −0.7880 [1] sin 218° = sin(180° + 38°) = −0.6157 [1] cos 322° = cos(360° − 38°) = 0.7880 [1] tan 218° = tan(180° + 38°) = 0.7813 [1] Examiner insight: A rough sketch of the graph or the quadrant diagram prevents sign errors. Questions like this ask for values from the given data, not new evaluations.

11. (a) The bearing of P from Q is 070° + 180° = 250° [1] Angle PQR = 250° − 175° = 75° [1] (b) PR² = 6.5² + 9² − 2 × 6.5 × 9 × cos 75° [1] = 42.25 + 81 − 30.28… = 92.97… [1] PR = 9.64 km [1] (c) sin(QPR) / 9 = sin 75° / 9.642… [1] Angle QPR = 64.4° [1] R lies clockwise from Q as seen from P, so the bearing is 70° + 64.4° = 134.4°, which is 134° to 3 s.f. [1] Examiner insight: On a “show that” the working must reach the stated value from first principles, so write the back bearing explicitly. In part (c), a sketch showing which side of PQ the point R lies on supports adding rather than subtracting.

12. (a) cos θ = (10² + 10² − 16²) / (2 × 10 × 10) [1] = −56/200 = −0.28 [1] θ = 1.85 rad [1] (b) Arc = 10 × 1.8546 = 18.55 [1] Perimeter = 20 + 18.55 = 38.5 cm [1] (c) Sector = ½ × 100 × 1.8546 = 92.73 [1] Triangle = ½ × 100 × sin θ = 48 (since sin θ = 0.96) [1] Segment = 92.73 − 48 = 44.7 cm² [1] Examiner insight: A negative cosine gives an angle above π/2, which is correct here. Use the unrounded angle in later parts, since 1.85 alone would give 92.5 for the sector and shift the final digits.

Where marks are usually lost

• Treating an angle such as 0.9 as degrees when it is in radians, which gives a very different value.
• Choosing the wrong pair in the sine rule, so that a side is divided by the sine of a non-matching angle.
• Dropping the minus sign when the cosine of an obtuse angle appears in the cosine rule.
• Giving the arc length alone when the question asks for the perimeter of a sector.
• Treating a segment as sector minus sector instead of sector minus triangle.
• Giving a decimal for an answer requested in terms of π or in surd form.
• Missing the obtuse second answer when an area or sine rule equation has two solutions.
• Mixing up bearings: adding or subtracting 180° wrongly, or measuring the angle between legs from the wrong side.
• Giving the period of y = sin 2x as 720° or 2π, instead of 180° or π.

Next steps

• Review the revision notes for this unit.
• Reread the study guide where a question went wrong.
• Visit the course hub, and record progress on the printable checklist.
• The free 10-minute diagnostics show you what to work on next.
• Book a free trial class with a teacher: see the trial class page.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section 3.2.1, content statement PP1.2: Trigonometry. Published by OxfordAQA.