Trigonometry & Vectors

In triangle ABC, angle A = 40°, angle B = 65° and side a = 8 cm. Use the sine rule to find side b, correct to 1 decimal place.

• A11.3 cm✓
• B1.4 cm
• C5.7 cm
• D7.3 cm

Answer: The sine rule states b = a × sinB ÷ sinA. Substituting the given values gives b = 8 × sin65° ÷ sin40° = 11.3 cm. Swapping the angles in the ratio gives 5.7 cm, and forgetting to divide by sinA gives 7.3 cm; the remaining distractor drops the side length entirely.

In triangle PQR, angle P = 52°, angle Q = 71° and side p = 15 m (opposite P). Find side r, correct to 1 decimal place.

• A31.9 m
• B9.4 m
• C16.0 m✓
• D8.0 m

Answer: Angle R is found first from the angle sum of a triangle: 180° − 52° − 71° = 57°. The sine rule then gives r = 15 × sin57° ÷ sin52°, which evaluates to 16.0 m. Halving or doubling the correct ratio, or substituting the wrong angle, gives the other values.

In triangle ABC, side a = 9 cm is opposite angle A, and side b = 12 cm is opposite angle B = 80°. Find angle A, correct to 1 decimal place.

• A47.6°✓
• B127.6°
• C42.4°
• D132.4°

Answer: The sine rule can be rearranged to sinA = a × sinB ÷ b when a side and its opposite angle are unknown. Substituting the given values gives sinA = 9 × sin80° ÷ 12 = 0.7386, so A = 47.6°. Taking the obtuse supplementary angle of 132.4°, or a related obtuse miscalculation of 127.6°, gives two of the other results; the remaining distractor comes from a similar slip.

In triangle XYZ, side x = 11 cm is opposite angle X, and side y = 14 cm is opposite angle Y = 95°. Find angle X, correct to 1 decimal place.

• A51.5°✓
• B146.5°
• C48.3°
• D128.5°

Answer: Rearranging the sine rule gives sinX = x × sinY ÷ y. Substituting the given values gives sinX = 11 × sin95° ÷ 14 = 0.7827, so X = 51.5°. Taking the obtuse supplementary angle of 128.5°, or a related obtuse miscalculation of 146.5°, gives two of the other results; the remaining distractor mixes up which side belongs to which angle.

In triangle ABC, side b = 7 cm, side c = 10 cm and the angle between them, angle A, is 55°. Find side a, correct to 2 significant figures.

• A8.3 cm✓
• B4.2 cm
• C15.1 cm
• D68.7 cm

Answer: The cosine rule states a² = b² + c² − 2bc·cosA. Substituting gives a² = 49 + 100 − 140cos55° = 68.7, so a = 8.3 cm. Forgetting to take the square root gives 68.7 cm, and adding instead of subtracting the cosine term gives 15.1 cm; the remaining distractor halves the correct answer.

In triangle ABC, side b = 6 m, side c = 9 m and the angle between them, angle A, is 120°. Find side a, correct to 1 decimal place.

• A13.1 m✓
• B9.4 m
• C171.0 m
• D7.9 m

Answer: For an obtuse included angle, cos120° = −0.5, so the subtraction term −2bc·cosA becomes an addition. Substituting gives a² = 36 + 81 + 54 = 171, so a = 13.1 m. Forgetting to take the square root gives 171.0 m, and treating the angle as acute gives 7.9 m; the remaining distractor comes from a similar slip.

A triangle has sides a = 8 cm, b = 6 cm and c = 5 cm. Find angle A, the angle opposite the longest side, correct to 1 decimal place.

• A78.5°
• B92.9°✓
• C87.1°
• D101.3°

Answer: Rearranging the cosine rule gives cosA = (b² + c² − a²) ÷ 2bc, used to find an angle from three known sides. Substituting gives cosA = (36 + 25 − 64) ÷ 60 = −0.05, so A = 92.9°, an obtuse angle because cosA is negative. Treating cosA as positive or mislabelling the sides gives the other results.

A triangle has sides a = 10 cm, b = 7 cm and c = 9 cm. Find angle A, the angle opposite side a, correct to 1 decimal place.

• A103.8°
• B76.2°✓
• C156.2°
• D146.9°

Answer: Rearranging the cosine rule gives cosA = (b² + c² − a²) ÷ 2bc. Substituting gives cosA = (49 + 81 − 100) ÷ 126 = 0.2381, so A = 76.2°. A sign slip in the numerator gives 103.8°, and related combined sign-and-side slips give 146.9° or 156.2°.

A triangle has two sides of length 8 cm and 11 cm with an included angle of 50°. Find the area of the triangle, correct to 1 decimal place.

• A4.2 cm²
• B44.0 cm²
• C28.3 cm²
• D33.7 cm²✓

Answer: The area of a triangle with two known sides and the included angle is Area = ½ab·sinC. Substituting gives Area = 0.5 × 8 × 11 × sin50° = 33.7 cm². Forgetting the sine factor entirely gives 44.0 cm² (just ½ab), and omitting side a from the formula gives 4.2 cm²; using cosine instead of sine gives the remaining total.

A triangle has two sides of length 15 cm and 9 cm with an included angle of 130°. Find the area of the triangle, correct to 1 decimal place.

• A3.4 cm²
• B67.5 cm²
• C103.4 cm²
• D51.7 cm²✓

Answer: Using Area = ½ab·sinC with a = 15, b = 9 and C = 130° gives Area = 0.5 × 15 × 9 × sin130° = 51.7 cm². Omitting the ½ gives 103.4 cm², and omitting the sine factor altogether gives 67.5 cm²; omitting side a from the formula gives the remaining total.

A triangle has two sides of length 10 cm and 12 cm and an area of 45 cm². Find the included angle, correct to 1 decimal place (take the acute answer).

• A131.4°
• B41.4°
• C148.6°
• D48.6°✓

Answer: Rearranging Area = ½ab·sinC gives sinC = 2 × Area ÷ (ab). Substituting gives sinC = 90 ÷ 120 = 0.75, so the acute angle is C = 48.6°. Taking the obtuse solution of 131.4°, or a related obtuse miscalculation of 148.6°, gives two of the other results; the remaining distractor misplaces the ½.

Without using a calculator, find the exact value of sin60° × cos30°.

• A√3/4
• B√3/2
• C3/4✓
• D1/4

Answer: The exact values sin60° = √3/2 and cos30° = √3/2 come from the standard 30-60-90 triangle. Multiplying gives (√3/2) × (√3/2) = 3/4. Using only one factor gives √3/2, forgetting to square the surd gives √3/4, and multiplying the wrong pair of angles (sin30°×cos60°) gives 1/4.

Without using a calculator, find the exact value of tan45° + sin90°.

• A√3
• B2✓
• C0.5
• D√2

Answer: The exact values tan45° = 1 and sin90° = 1 come from the standard angle facts. Adding these gives 1 + 1 = 2. Misremembering tan45° as √2, or confusing it with tan60° = √3, gives two of the other results; the remaining distractor comes from a similar mix-up of standard values.

What is the exact value of cos60°?

• A1
• B√3/2
• C√2/2
• D1/2✓

Answer: The exact value cos60° = 1/2 comes from the standard 30-60-90 triangle ratios. This is distinct from cos30° = √3/2 and cos45° = √2/2, which are easily confused with it. Mixing up the standard angles gives the other results.

A right-angled triangle has a hypotenuse of 12 cm and one angle of 60°. Find the exact length of the side opposite the 60° angle, giving your answer as a surd.

• A6√2 cm
• B3√3 cm
• C12√3 cm
• D6√3 cm✓

Answer: The side opposite a known angle in a right-angled triangle is found from opposite = hypotenuse × sinθ. Substituting gives opposite = 12 × sin60° = 12 × (√3/2) = 6√3 cm. Using cos60° instead of sin60°, or the wrong hypotenuse multiple, gives the other results.

A cuboid has dimensions 6 cm by 8 cm by 10 cm. Find the length of the space diagonal, correct to 1 decimal place.

• A200.0 cm
• B100.0 cm
• C14.1 cm✓
• D576.0 cm

Answer: The space diagonal of a cuboid is found from d = √(l² + w² + h²), extending Pythagoras' theorem into three dimensions. Substituting gives d = √(36 + 64 + 100) = √200 = 14.1 cm. Squaring the dimensions but forgetting the final square root gives 200.0 cm; using only two of the three dimensions before forgetting the root gives 100.0 cm, and squaring the sum of the edges instead gives 576.0 cm.

A cuboid has a rectangular base 5 cm by 12 cm and a height of 9 cm. Find the angle between the space diagonal and the base, correct to 1 decimal place.

• A34.7°✓
• B8.7°
• C40.1°
• D55.3°

Answer: The base diagonal is found first using Pythagoras: √(5² + 12²) = √169 = 13 cm. The angle to the base is then found from tanθ = height ÷ base diagonal = 9 ÷ 13, giving θ = 34.7°. Using the complementary angle of 55.3°, or a much smaller ratio, gives the other results.

A pyramid has a square base of side 10 cm, and its apex is directly above the centre of the base at a height of 12 cm. Find the angle between a sloping edge and the base, correct to 1 decimal place.

• A59.5°✓
• B67.4°
• C50.2°
• D9.8°

Answer: The horizontal distance from the centre of the square base to a corner is half the base diagonal, (10√2) ÷ 2 = 5√2 cm. The angle to the base is then found from tanθ = height ÷ horizontal distance = 12 ÷ 5√2, giving θ = 59.5°. Using the full diagonal instead of half, or a much smaller ratio, gives the other results.

Ship A sails from a port on a bearing of 050° for 12 km. Ship B sails from the same port on a bearing of 130° for 9 km, so the angle between their paths at the port is 80°. Find the distance between the ships, correct to 1 decimal place.

• A19.1 km
• B363.8 km
• C187.5 km
• D13.7 km✓

Answer: The angle between the two bearings at the port is 130° − 50° = 80°, which becomes the included angle for the cosine rule. Substituting a² = 12² + 9² − 2(12)(9)cos80° gives a² = 187.5, so the distance is 13.7 km. Forgetting to take the square root gives 187.5 km; using 130° as the included angle without subtracting 50° gives 19.1 km, and forgetting the square root on that wrong angle gives 363.8 km.

A triangular field has two sides of length 45 m and 60 m with an included angle of 72°. Find the area of the field, correct to the nearest square metre.

• A1350 m²
• B1284 m²✓
• C2567.9 m²
• D2700 m²

Answer: Using Area = ½ab·sinC with a = 45, b = 60 and C = 72° gives Area = 0.5 × 45 × 60 × sin72° = 1283.9 m², which rounds to 1284 m². Forgetting the ½ gives 2567.9 m², and forgetting the sine factor gives 1350 m²; forgetting both gives 2700 m².

In triangle OAB, OA = a and OB = b. P is the point on OA such that OP = ⅓a, and Q is the point on OB such that OQ = ⅓b. Express vector PQ in terms of a and b.

• A⅓b − ⅓a✓
• Ba − b
• C⅓a − ⅓b
• D⅓a + ⅓b

Answer: P divides OA so that OP = ⅓a, and Q divides OB so that OQ = ⅓b. Vector PQ is found from OQ − OP = ⅓b − ⅓a. This shows PQ is parallel to AB (since AB = b − a), confirming P and Q split the triangle proportionally — a vector-geometry argument. Subtracting in the wrong order gives ⅓a − ⅓b, forgetting to scale by ⅓ gives a − b, and adding instead of subtracting gives ⅓a + ⅓b.

In triangle OAB, OA = a and OB = b. M is the midpoint of AB. Find vector OM in terms of a and b, and state what this shows about M.

• Aa − ½b
• B½a − ½b
• C½a + ½b✓
• Da + ½b

Answer: Vector AB = b − a, and M is the midpoint of AB, so AM = ½(b − a). Adding this to OA gives OM = a + ½(b − a) = ½a + ½b. This is a vector-geometry proof that M sits symmetrically between a and b, consistent with M being the midpoint of AB. Forgetting to halve AB gives a + ½b, and sign slips give the other two options.

In triangle OAB, OA = a and OB = b. Point R is on OA extended such that OR = 2a, and point S is on OB extended such that OS = 2b. Express vector RS in terms of a and b, and state what this proves about RS and AB.

• A2a − 2b
• Bb − a
• C2a + 2b
• D2b − 2a✓

Answer: Vector RS is found from OS − OR = 2b − 2a. Since AB = b − a, this means RS = 2(b − a) = 2AB, proving RS is parallel to AB and exactly twice its length — a vector-geometry proof using an enlargement relationship. Subtracting in the wrong order gives 2a − 2b, forgetting the scale factor gives b − a, and adding instead of subtracting gives 2a + 2b.

In triangle OAB, OA = a and OB = b. C is the point on AB such that AC:CB = 1:3. Express vector OC in terms of a and b.

• A¾a + ¼b✓
• Ba + ¼b
• C¾a − ¼b
• D¼a + ¾b

Answer: Vector AB = b − a, and C divides AB in the ratio 1:3 from A, so AC = ¼(b − a). Adding this to OA gives OC = a + ¼(b − a) = ¾a + ¼b. Applying the ratio from the B end instead gives ¼a + ¾b, using b directly instead of b − a gives a + ¼b, and a sign slip gives ¾a − ¼b.

In triangle OAB, OA = a, OB = b, M is the midpoint of OA and N is the midpoint of OB. Express vector MN in terms of a and b.

• Ab − a
• B½a + ½b
• C½a − ½b
• D½b − ½a✓

Answer: The midpoint of OA gives OM = ½a and the midpoint of OB gives ON = ½b, since a midpoint vector is half the original. Vector MN is found from ON − OM = ½b − ½a. Forgetting the halving, or subtracting in the wrong order, gives the other results.

In triangle OAB, OA = a and OB = b. Point C lies on AB such that AC:CB = 2:1. Express vector OC in terms of a and b.

• A½a + ½b
• Ba + ⅔b
• C⅓a + ⅔b✓
• D⅔a + ⅓b

Answer: Vector AB is found from OB − OA = b − a, and C divides AB in the ratio 2:1 from A, so AC = ⅔(b − a). Adding this to OA gives OC = a + ⅔(b − a) = ⅓a + ⅔b. Using the wrong division ratio or applying it from the wrong end gives the other results.