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
Answer key: Pythagoras’ theorem and trigonometric ratios worksheet — GCSE Higher
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- (c) 15/17 — Method: cos θ = adjacent ÷ hypotenuse, so find the hypotenuse with Pythagoras' theorem first and then decide which short side is next to θ. Working: the hypotenuse is √(8² + 15²) = √(64 + 225) = √289 = 17 cm. The angle θ is opposite the 8 cm side, so the side next to it is the 15 cm side, and cos θ = 15 ÷ 17. Answer: 15/17. The distractors: 8/17 is sin θ, opposite over hypotenuse, used in place of the cosine; 8/15 is tan θ, opposite over adjacent; 17/15 comes from writing the cosine ratio upside down, as hypotenuse over adjacent.
- (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.
- (d) 7.7 m — The horizontal distance is adjacent to the 50° angle and the zip wire is the hypotenuse, so horizontal distance = 12 × cos 50° = 12 × 0.6428... = 7.71...≈ 7.7 m. "9.2 m" uses the sine ratio instead of cosine, 12 × sin 50° = 9.19...≈ 9.2 m, which actually finds the vertical drop of the zip wire, not the horizontal distance. "15.7 m" comes from dividing by the sine ratio instead of multiplying by the cosine ratio, 12 ÷ sin 50° = 15.66...≈ 15.7 m, both the wrong operation and the wrong ratio. "12.0 m" simply uses the length of the zip wire itself as the horizontal distance, ignoring the angle of 50° altogether.
- (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.
- (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.
- (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.
- (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.
- (b) 13.3 cm — The apex is directly above the centre of the square base, so the height, half the base diagonal, and a slant edge form a right-angled triangle with the slant edge as the hypotenuse. Half the base diagonal is 14 ÷ 2 = 7 cm. Using Pythagoras' theorem, height = √(15² − 7²) = √(225 − 49) = √176 = 13.3 cm (1 d.p.). Using the slant edge itself as the height, without applying Pythagoras' theorem at all, gives 15 cm. Using the full base diagonal (14 cm) instead of half of it gives √(15² − 14²) = √(225 − 196) = √29 = 5.4 cm (1 d.p.), far too short for a pyramid this size. Adding the two squares instead of subtracting them, √(15² + 7²) = √(225 + 49) = √274 = 16.6 cm (1 d.p.), gives a length longer than the slant edge itself, which cannot be the height.
- (d) 21.2 m — Method: the cable is the hypotenuse of a right-angled triangle whose vertical side is the drop from the roof to the bracket and whose horizontal side is 15 m, so use Pythagoras' theorem. Working: the drop is 20 − 5 = 15 m, so c² = 15² + 15² = 225 + 225 = 450 and c = √450 = 21.213…, which is 21.2 m to 1 decimal place. Answer: 21.2 m. The distractors: 25.0 m comes from using the whole 20 m height of the roof as the vertical side and forgetting that the bracket is already 5 m up; 30.0 m comes from adding the two sides of the triangle, 15 + 15, instead of using Pythagoras' theorem; 15.0 m is the horizontal distance on its own, which would be the length of the cable only if it ran level.
- (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°.
- (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.
- (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.
- (a) 69.3 m — Method: each observer gives a right-angled triangle with the mast as the opposite side, so tan θ = 60 ÷ distance and the distance from the foot of the mast is 60 ÷ tan θ; because both stand on the same side, the gap between them is the difference of those two distances. Working: from Amelia, 60 ÷ tan 30° = 103.92… m; from Noah, 60 ÷ tan 60° = 34.64… m; the gap is 103.92… − 34.64… = 69.28… m, which is 69.3 m to 1 decimal place. Answer: 69.3 m. The distractors: 103.9 m is Amelia's own distance from the foot of the mast, written down before the second distance has been taken away; 34.6 m is Noah's distance from the foot of the mast; 138.6 m comes from adding the two distances, which would be right only if the two observers stood on opposite sides of the mast.
- (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.
- (b) 21 cm — By Pythagoras' theorem, the other side = √(29² − 20²) = √(841 − 400) = √441 = 21 cm. "9 cm" comes from subtracting the two given lengths directly, 29 − 20 = 9, instead of subtracting their squares. "441 cm" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "35 cm" comes from adding the squares of the two given lengths instead of subtracting them, √(29² + 20²) = √1241 ≈ 35, treating both given lengths as if they were the two shorter sides rather than a shorter side and the hypotenuse.
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