Printable · GCSE Higher · ages 14-16
Pythagoras’ theorem and trigonometric ratios worksheet — GCSE Higher
Fifteen questions on "pythagoras’ theorem and trigonometric ratios" — DfE statement G20. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
part Higher
Pythagoras’ theorem and trigonometric ratios worksheet — GCSE Higher
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- 1.In a right-angled triangle, the side opposite angle θ is 6 cm and the hypotenuse is 10 cm. Work out the size of angle θ. Give your answer correct to 1 decimal place.
- 2.In a right-angled triangle one of the other two angles is 45°, and the side opposite that 45° angle is 7 cm. Work out the length of the hypotenuse. Give your answer to 1 decimal place.
- 3.Points P and Q have coordinates P(0, 0, 0) and Q(5, 7, 9) in a three-dimensional coordinate system, with all lengths in centimetres. Work out the distance PQ. Give your answer correct to 1 decimal place.
- 4.In triangle ABC the angle at A is 90° and BC = 10 cm. For the angle at B, sin B = 3/5. Work out the length of AC.
- 5.In triangle ABC the angle at C is 90° and the hypotenuse AB is 12 cm. H is the point on AB for which CH is perpendicular to AB, and AH = 3 cm. Work out the length of AC.
- 6.A support strut for a shed roof is in the shape of a right-angled triangle. The two shorter sides of the strut are 5 m and 9 m. Work out the length of the sloping strut (the hypotenuse). Give your answer correct to 1 decimal place.
- 7.The top of a clock tower is 25 m above level ground. Oliver stands on the ground 25 m from the foot of the tower. Work out the angle of elevation of the top of the tower from the point where Oliver stands.
- 8.A cuboid has length 3 cm, width 4 cm and height 12 cm. Work out the length of the diagonal that runs from one corner of the cuboid to the opposite corner.
- 9.A flagpole is 12 m tall. Freya stands 9 m from the base of the flagpole on level ground. Work out the angle of elevation of the top of the flagpole from where Freya stands. Give your answer correct to 1 decimal place.
- 10.A ladder of length 7 m leans against a vertical wall. The foot of the ladder is 3 m from the base of the wall. Work out how high up the wall the ladder reaches. Give your answer correct to 1 decimal place.
- 11.A cuboid has a rectangular base measuring 6 cm by 8 cm, and a height of 4 cm. Work out the angle between the diagonal from a base corner to the opposite top corner, and the base of the cuboid. Give your answer correct to 1 decimal place.
- 12.A tent has the cross-section of a right-angled triangle. The sloping side of the cross-section is 10 m long, and it makes an angle of 60° with the horizontal ground. Using the exact value of cos 60°, work out the horizontal distance covered by the sloping side.
- 13.A right-angled triangle has a hypotenuse of 8 cm, and one of its shorter sides is 4 cm. Work out the size of the angle between that 4 cm side and the hypotenuse.
- 14.Two vertical masts stand on level ground, 40 m apart. One mast is 30 m tall and the other is 50 m tall. Work out the straight-line distance between the tops of the two masts. Give your answer to 1 decimal place.
- 15.A right-angled triangle has a hypotenuse of 10 cm and one of its other angles is 45°. Work out the length of one of the two shorter sides. Give your answer to 1 decimal place.
Answer key
- (d) 36.9° — sin θ = opposite/hypotenuse = 6/10 = 0.6, so θ = sin⁻¹(0.6) = 36.86...° ≈ 36.9°. "53.1°" finds the OTHER acute angle in the triangle, 90° − 36.9°, instead of θ itself, as if the two acute angles had been swapped. "31.0°" comes from using the tangent ratio instead of sine, working out tan⁻¹(6/10) = 31.0° with the wrong ratio for the two sides given. "36.8°" rounds sin⁻¹(0.6) = 36.86...° down to 36.8° instead of correctly rounding it up to 36.9°.
- (c) 9.9 cm — Method: the 7 cm side is opposite the 45° angle and the hypotenuse is wanted, so use sin θ = opposite ÷ hypotenuse and rearrange it for the hypotenuse. Working: sin 45° = 7 ÷ h, so h = 7 ÷ sin 45° = 9.899…, which is 9.9 to 1 decimal place. Answer: 9.9 cm. The distractors: 5.0 cm comes from multiplying by sin 45° instead of dividing by it; 14.0 cm comes from doubling the 7 cm side, which is the rule for a side opposite 30° and not one opposite 45°; 7.0 cm comes from reading the two equal sides of a 45° right-angled triangle as including the hypotenuse, when the equal pair is the two shorter sides.
- (c) 12.4 cm — In three dimensions, the distance between two points extends Pythagoras' theorem to three squared terms: PQ² = 5² + 7² + 9² = 25 + 49 + 81 = 155. Taking the square root, PQ = √155 = 12.4 cm (1 d.p.). Using only the x- and y-coordinates, and ignoring the third dimension entirely, gives PQ = √(5² + 7²) = √74 = 8.6 cm (1 d.p.), which is not the distance in 3D space. Adding the three coordinates directly, 5 + 7 + 9 = 21 cm, treats the coordinates as if they were lengths along a single straight path rather than the sides of a right-angled arrangement. Squaring and adding all three coordinates but forgetting to take the square root leaves 155 cm, the squared distance rather than the distance itself.
- (b) 6 cm — Method: the right angle is at A, so BC is the hypotenuse and AC is the side opposite the angle at B; sin B = opposite ÷ hypotenuse therefore gives AC ÷ BC = 3/5. Working: AC ÷ 10 = 3/5, so AC = 10 × 3 ÷ 5 = 6. Answer: 6 cm. The distractors: 8 cm is AB, the side next to the angle at B, which is what cos B = 4/5 produces — the right method used on the wrong side; 3 cm comes from reading the 3 in the ratio as a length and never scaling it up to the 10 cm hypotenuse; 30 cm comes from multiplying by 3 and forgetting to divide by 5.
- (d) 6 cm — Method: CH cuts the triangle into two smaller right-angled triangles, so Pythagoras' theorem can be written in each of the three right-angled triangles and the results combined. Working: HB = 12 − 3 = 9 cm. In triangle ACH, CH² = AC² − 3² = AC² − 9; in triangle CHB, CH² = CB² − 9² = CB² − 81. Setting those equal gives AC² − 9 = CB² − 81. In triangle ABC, AC² + CB² = 12² = 144, so CB² = 144 − AC². Substituting gives AC² − 9 = 144 − AC² − 81, so 2 × AC² = 72 and AC² = 36, giving AC = 6. Answer: 6 cm. The distractors: 36 cm comes from stopping at AC² = 36 and never taking the square root; 9 cm is HB, the other part of the hypotenuse, written down in place of AC; 4 cm comes from working out 12 ÷ 3, treating AH as a scale factor between the two triangles rather than as a length.
- (d) 10.3 m — By Pythagoras' theorem, the hypotenuse = √(5² + 9²) = √(25 + 81) = √106 = 10.29...≈ 10.3 m. "106 m" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "14 m" comes from adding the two shorter sides, 5 + 9, instead of using Pythagoras' theorem at all. "10.2 m" comes from rounding 10.29...m down to 10.2 instead of correctly rounding it up to 10.3.
- (c) 45° — Method: the tower, the ground and the line of sight form a right-angled triangle in which the 25 m height is opposite the angle of elevation and the 25 m along the ground is adjacent to it, so use tan θ = opposite ÷ adjacent. Working: tan θ = 25 ÷ 25 = 1, so θ = tan⁻¹(1). Answer: 45°. The distractors: 90° comes from using sin θ = 25 ÷ 25 = 1, which treats the 25 m along the ground as the hypotenuse when it is the side next to the angle; 1° comes from writing down the value of tan θ as though it were the angle itself; 50° comes from adding the two given lengths, 25 + 25, instead of comparing them.
- (d) 13 cm — Use Pythagoras' theorem in three dimensions: for a cuboid with edges a, b and c the space diagonal d satisfies d² = a² + b² + c². Substitute a = 3, b = 4, c = 12: d² = 3² + 4² + 12² = 9 + 16 + 144 = 169. Take the square root: d = √169 = 13 cm. Adding the three edges directly, 3 + 4 + 12 = 19 cm, ignores that Pythagoras' theorem is about squares, not lengths, and gives 19 cm. Stopping after squaring and adding, without taking the square root, leaves 169 cm — the squared length, not the length itself. Using only the 4 cm and 12 cm edges finds the diagonal of one face, √(4² + 12²) = √160 = 12.6 cm (1 d.p.), and leaves out the third dimension entirely.
- (b) 53.1° — The angle of elevation is opposite the height of the flagpole, 12 m, and adjacent to the distance from its base, 9 m, so tan θ = 12/9 = 1.333..., giving θ = tan⁻¹(1.333...) = 53.13...° ≈ 53.1°. "36.9°" finds the OTHER acute angle of the triangle, 90° − 53.1°, the angle at the top of the flagpole rather than the angle of elevation at Freya's position. "48.6°" comes from wrongly treating 9/12 as a sine ratio and finding sin⁻¹(0.75) = 48.6°, when neither side here is the hypotenuse. "41.4°" comes from wrongly treating 9/12 as a cosine ratio and finding cos⁻¹(0.75) = 41.4°, again without a hypotenuse in the ratio at all.
- (c) 6.3 m — By Pythagoras' theorem, the height = √(7² − 3²) = √(49 − 9) = √40 = 6.32...≈ 6.3 m. "6.4 m" rounds 6.32...m up to 6.4 instead of correctly rounding it down to 6.3. "4.0 m" comes from subtracting the two given lengths directly, 7 − 3 = 4, instead of subtracting their squares. "10.0 m" comes from adding the two given lengths, 7 + 3 = 10, instead of using Pythagoras' theorem at all.
- (a) 21.8° — The base diagonal has length √(6² + 8²) = √(36 + 64) = √100 = 10 cm, using Pythagoras' theorem on the rectangular base. The angle between the space diagonal and the base lies in the right-angled triangle formed by the height (4 cm, opposite), the base diagonal (10 cm, adjacent) and the space diagonal (hypotenuse), so tan(angle) = 4 ÷ 10 = 0.4, giving angle = 21.8° (1 d.p.). Using the height and one base edge instead of the full base diagonal, tan(angle) = 4 ÷ 8 = 0.5, gives 26.6° instead. Inverting the ratio, tan(angle) = 10 ÷ 4 = 2.5, gives 68.2°, the complement of the angle rather than the angle itself. Using the height and the other base edge, tan(angle) = 4 ÷ 6, gives 33.7°.
- (c) 5 m — The horizontal distance is the side adjacent to the 60° angle, and the sloping side is the hypotenuse, so horizontal distance = hypotenuse × cos 60°. The exact value of cos 60° is 1/2, so horizontal distance = 10 × 1/2 = 5 m. Using the sloping side itself as the horizontal distance, without using any trigonometry at all, gives 10 m. Using sin 60° = √3/2 instead of cos 60° finds the vertical height of the tent rather than the horizontal distance: 10 × √3/2 = 5√3 = 8.7 m (1 d.p.). Dividing the sloping side by cos 60° instead of multiplying by it, 10 ÷ 0.5 = 20 m, treats the sloping side as though it were the adjacent side rather than the hypotenuse.
- (d) 60° — Method: the 4 cm side is next to the angle wanted and the 8 cm side is the hypotenuse, so the ratio built from them is cos θ = adjacent ÷ hypotenuse, and the angle comes from the inverse cosine. Working: cos θ = 4 ÷ 8 = 0.5, so θ = cos⁻¹(0.5). Answer: 60°. The distractors: 30° comes from using sin⁻¹(0.5), which treats the 4 cm side as the side opposite the angle when it is the side next to it; 45° comes from assuming the two acute angles of the triangle must be equal; 90° comes from writing down the right angle the question already gives instead of the angle it asks for.
- (a) 44.7 m — Method: the line joining the two tops is the hypotenuse of a right-angled triangle whose horizontal side is the gap between the masts and whose vertical side is the difference in their heights, so Pythagoras' theorem applies. Working: the difference in heights is 50 − 30 = 20 m, so d² = 40² + 20² = 1600 + 400 = 2000 and d = √2000 = 44.721…, which is 44.7 m to 1 decimal place. Answer: 44.7 m. The distractors: 34.6 m comes from subtracting the squares, √(40² − 20²), instead of adding them; 60.0 m comes from adding the two sides of the triangle, 40 + 20, rather than using Pythagoras' theorem; 50.0 m is the height of the taller mast, copied from the question in place of the distance asked for.
- (c) 7.1 cm — Method: a shorter side is opposite the 45° angle and the hypotenuse is known, so sin θ = opposite ÷ hypotenuse gives that side directly. Working: sin 45° = x ÷ 10, so x = 10 × sin 45° = 7.071…, which is 7.1 to 1 decimal place. Answer: 7.1 cm. The distractors: 5.0 cm comes from halving the hypotenuse, which is the rule for the side opposite a 30° angle and not a 45° one; 14.1 cm comes from dividing by sin 45° instead of multiplying by it, which makes a shorter side longer than the hypotenuse; 10.0 cm comes from taking tan 45° = 1 and concluding that the shorter side matches the hypotenuse.
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