Edexcel International GCSE Maths Pythagoras and Trigonometry Past Paper Questions and Solutions (4MA1 Higher)

1. The 52 cm side is opposite the right angle, so it is the hypotenuse. To find a shorter side, subtract the squares.
2. h² = 52² − 30² = 2704 − 900 = 1804
3. h = √1804 = 42.47…

Answer: h = 42.5

1. AC is the hypotenuse (opposite the right angle). From the 65° angle at C, AB is the opposite side, so use sin.
2. sin 65° = ^AB⁄_8.4, so AB = 8.4 × sin 65° = 7.612…

Answer: 7.61 cm

1. The top triangle has hypotenuse 16 cm and one side 12.8 cm. Use Pythagoras for its base: √16² − 12.8² = √256 − 163.84 = √92.16 = 9.6 cm
2. So every triangle has sides 9.6, 12.8 and 16, and the square has side 9.6 cm.
3. The bottom-right triangle’s vertical side is the right side of the square (9.6), so its horizontal side is 12.8. The top-right triangle shares that 12.8 side, so its vertical side is 9.6.
4. The top triangle’s vertical side is 12.8, but 9.6 of it is covered by the top-right triangle. The exposed part is 12.8 − 9.6 = 3.2 cm.
5. Go round the outside: left of square 9.6 + bottom of square 9.6 + three hypotenuses 16 × 3 + exposed edge 3.2
6. 9.6 + 9.6 + 48 + 3.2

Answer: 70.4 cm

1. The perimeter of the shaded shape is AE + EB + BC + CD + DA. We know all of these except CD.
2. CD = AB (opposite sides of a rectangle). Find AB with Pythagoras in the right-angled triangle AEB: AB² = 7.2² + 5.4² = 51.84 + 29.16 = 81, so AB = 9 cm
3. DA = BC = 6 cm
4. Perimeter = 7.2 + 5.4 + 6 + 9 + 6

Answer: 33.6 cm

1. From the 63° angle at P: PR is the adjacent side and PQ is the hypotenuse, so use cos.
2. cos 63° = ^24.3⁄_PQ, so PQ = ^24.3⁄_{cos 63°} = 53.52…

Answer: 53.5 cm

1. The line from A down to the middle of BC is the height, and it splits BC into 14 cm and 14 cm.
2. Pythagoras: height² = 17.5² − 14² = 306.25 − 196 = 110.25, so height = 10.5 cm
3. Area = ½ × base × height = ½ × 28 × 10.5

Answer: 147 cm²

1. Area = ½ × base × height: ½ × 7 × h = 42, so h = 12 cm
2. The height from C cuts the base in half (the triangle is isosceles), making a right-angled triangle with sides 3.5 cm and 12 cm.
3. Pythagoras: y² = 12² + 3.5² = 144 + 12.25 = 156.25

Answer: y = 12.5

1. Draw a perpendicular from A down to DC. This makes a right-angled triangle with hypotenuse AD = 12 cm and angle 60° at D.
2. Height of the trapezium = 12 × sin 60° = 10.392… cm
3. Horizontal distance from D to the foot of the perpendicular = 12 × cos 60° = 6 cm
4. The trapezium is symmetrical, so the same 6 cm is cut off at the other end: AB = 47 − 6 − 6 = 35 cm
5. Area = ½(AB + DC) × height = ½ × (35 + 47) × 10.392… = 426.08…

Answer: 426 cm²

1. From the 24° angle, PQ is the opposite side and QR is the adjacent side, so use tan.
2. tan 24° = ^opposite⁄_adjacent = ^6.5⁄_QR
3. QR = ^6.5⁄_{tan 24°} = 14.599…

Answer: 14.6 cm

1. In right-angled triangle ADB: DB = 16 is opposite the 65° angle.
2. AB (hypotenuse): sin 65° = ¹⁶⁄_AB, so AB = ¹⁶⁄_{sin 65°} = 17.653… cm
3. AD (adjacent): tan 65° = ¹⁶⁄_AD, so AD = ¹⁶⁄_{tan 65°} = 7.460… cm, and AC = 2 × 7.460… = 14.921… cm
4. Perimeter = AB + BC + AC = 2 × 17.653… + 14.921… = 50.22…

Answer: 50.2 cm

1. OM joins the centre to the midpoint of a chord, so OM is perpendicular to RT (the perpendicular from the centre to a chord bisects the chord). Triangle ROM has a right angle at M.
2. RM = 12 ÷ 2 = 6 cm, and RM is opposite the 52° angle. OR is the hypotenuse (the radius).
3. sin 52° = ⁶⁄_r, so r = ⁶⁄_{sin 52°} = 7.614… cm
4. Area = π × 7.614…² = 182.13…

Answer: 182 cm²

1. Area = ½(BC + AD) × height: ½(8.4 + 17.6) × h = 179.4, so 13h = 179.4 and h = 13.8 cm
2. The trapezium is symmetrical, so the bottom overhangs by the same amount on each side: (17.6 − 8.4) ÷ 2 = 4.6 cm
3. Drop a perpendicular from B: a right-angled triangle with height 13.8 and base 4.6. tan A = ^13.8⁄_4.6 = 3, so angle A = 71.565…°
4. BC is parallel to AD, so angles A and B are co-interior and add to 180°: angle ABC = 180° − 71.565…°

Answer: 108.4°

1. Find AB with Pythagoras in triangle ABC (AC is the hypotenuse): AB² = 7.5² − 6² = 56.25 − 36 = 20.25, so AB = 4.5 cm
2. Area of triangle ABC = ½ × base × height = ½ × 6 × 4.5 = 13.5 cm²
3. Area of triangle DAC = 31.5 − 13.5 = 18 cm²
4. Triangle DAC has its right angle at A, so AC and AD are the base and height: ½ × 7.5 × AD = 18
5. AD = 36 ÷ 7.5

Answer: 4.8 cm

1. Triangle ABC is isosceles (AC = BC), so the perpendicular from C meets AB at its midpoint. So AM = 12 ÷ 2 = 6 m.
2. Use Pythagoras’ theorem in the right-angled triangle AMC, where AC = 9 m is the hypotenuse: MC² = 9² − 6² = 81 − 36 = 45
3. MC = √45 = 6.708… m
4. Each length must be a whole number of metres, so round each length up (rounding down would make it too short): MC needs 7 m. AB = 12 m, AC = 9 m and BC = 9 m are already whole numbers.
5. Total length to buy = 12 + 9 + 9 + 7 = 37 m
6. Cost = 37 × 21.50

Answer: 795.50 euros

Part (a)

1. In the right-angled triangle, 100 m is opposite the 20° angle and d is adjacent, so use tan.
2. tan 20° = ¹⁰⁰⁄_d, so d = ¹⁰⁰⁄_{tan 20°} = 274.74…

Answer: d = 275

Part (b)

1. Height from sea level to A: tan 25° = ^height⁄_274.74…, so height = 274.74… × tan 25° = 128.11… m
2. Mast height = 128.11… − 100 = 28.11…

Answer: 28.1 m

Part (a)

1. The parallelogram is two equal triangles, so area = ab sin C = 9.3 × 14.7 × sin 106° = 131.41…

Answer: 131 cm²

Part (b)

1. EG is opposite the 106° angle in triangle EFG. Cosine rule: EG² = 9.3² + 14.7² − 2 × 9.3 × 14.7 × cos 106°
2. = 86.49 + 216.09 + 75.36… = 377.95…
3. EG = 19.44…

Answer: 19.4 cm

1. In triangle DEF: DE is the hypotenuse and EF is adjacent to the 40° angle, so use cos: EF = 12 × cos 40° = 9.192… cm
2. In triangle EFG: from the 28° angle at G, EF is opposite and EG is adjacent, so use tan: tan 28° = ^EF⁄_EG
3. EG = ^9.192…⁄_{tan 28°} = 17.28…

Answer: 17.3 cm

1. We know two equal sides and the angle between them, so use area = ½ab sin C with a = b = EF.
2. ½ × EF² × sin 130° = 74
3. EF² = ¹⁴⁸⁄_{sin 130°} = 193.20…
4. EF = √193.20… = 13.899…

Answer: 13.9 cm

1. In triangle ABC we know AC, BC and angle A (opposite BC). Use the sine rule to find angle B: sin B = ^{24 × sin 64°}⁄₃₁ = 0.6958…, so B = 44.09…°
2. Angle ACB = 180° − 64° − 44.09…° = 71.90…°
3. Angle DCE = angle ACB = 71.90…° (vertically opposite angles are equal)
4. Cosine rule in triangle CDE: DE² = 16² + 19² − 2 × 16 × 19 × cos 71.90…° = 256 + 361 − 188.83… = 428.16…
5. DE = 20.69…

Answer: 20.7 cm

Part (a)

1. C is 10 cm vertically above H. The projection of CF onto the base is FH, a diagonal of the square: FH = √12² + 12² = √288 = 16.97… cm
2. In right-angled triangle CHF: tan(angle) = ^CH⁄_FH = ¹⁰⁄_16.97…
3. Angle = tan⁻¹(0.5892…) = 30.50…°

Answer: 30.5°

Part (b)

1. First find AF. In the trapezium ADEF, AD = 15 is the sloping side, FE = 12 across, so A is √15² − 12² = 9 cm higher than D. AF = 10 + 9 = 19 cm.
2. B is directly above G with BG = AF = 19 cm. The base diagonal EG = √288 cm.
3. In right-angled triangle BGE: BE² = 19² + 288 = 361 + 288 = 649
4. BE = √649 = 25.47…

Answer: 25.5 cm

1. The parallelogram is two identical triangles, so area = PQ × PS × sin P: 6.1 × 3.8 × sin P = 18
2. sin P = ¹⁸⁄_23.18 = 0.7765…, so P = 50.94…° (acute, as stated)
3. QS is opposite angle P in triangle PQS. Use the cosine rule: QS² = 6.1² + 3.8² − 2 × 6.1 × 3.8 × cos 50.94…°
4. QS² = 37.21 + 14.44 − 29.21… = 22.43…
5. QS = 4.737…

Answer: 4.74 cm

1. Area = ½ × AB × BC × sin B: ½ × 4.6 × 8.3 × sin B = 12
2. sin B = ²⁴⁄_38.18 = 0.6286…, so B = 38.94…° (acute)
3. Cosine rule: AC² = 4.6² + 8.3² − 2 × 4.6 × 8.3 × cos 38.94…° = 30.66…
4. AC = 5.537… cm
5. Perimeter = 4.6 + 8.3 + 5.537… = 18.437…

Answer: 18.4 cm

1. We know two sides and the angle between them, so first find AC with the cosine rule.
2. AC² = 9.7² + 12.3² − 2 × 9.7 × 12.3 × cos 115° = 94.09 + 151.29 + 100.84… = 346.22…
3. AC = 18.607… m
4. Now use the sine rule. Angle x at C is opposite AB = 9.7: ^{sin x}⁄_9.7 = ^{sin 115°}⁄_18.607…
5. sin x = ^{9.7 × sin 115°}⁄_18.607… = 0.4724…
6. x = sin⁻¹(0.4724…) = 28.19…

Answer: x = 28.2

1. Use coordinates with D at the origin: across along DC, back along CH, up along DA. Then A = (0, 0, 6), B = (6, 0, 6), C = (6, 0, 0) and H = (6, 6, 0).
2. T = midpoint of AB = (3, 0, 6) and V = midpoint of CH = (6, 3, 0)
3. Differences: across 6 − 3 = 3, back 3 − 0 = 3, down 6 − 0 = 6
4. 3D Pythagoras: TV² = 3² + 3² + 6² = 9 + 9 + 36 = 54

Answer: √54 cm

1. A is directly above F, so the projection of AH onto the base is FH, and AF = 6 cm.
2. FH is a diagonal of the base rectangle (9 by 5): FH = √9² + 5² = √106 = 10.295… cm
3. In right-angled triangle AFH: tan(angle) = ^AF⁄_FH = ⁶⁄_10.295… = 0.5827…
4. Angle = tan⁻¹(0.5827…) = 30.23…°

Answer: 30.2°

1. We need angle DAB in triangle ABD. We know AB and AD but not BD, so first find BD from triangle BCD.
2. In triangle BCD: BC = 12.8 m is opposite the 40° angle at D, and BD is opposite the 62° angle at C. A side–angle pair is known, so use the sine rule.
3. ^BD⁄_{sin 62°} = ^12.8⁄_{sin 40°}, so BD = ^{12.8 × sin 62°}⁄_{sin 40°} = 17.58… m
4. Now use the cosine rule in triangle ABD, since all three sides are known: cos x = ^{13.42 + 15.22 − 17.58…2}⁄_{2 × 13.4 × 15.2}
5. cos x = ^{179.56 + 231.04 − 309.13…}⁄_407.36 = 0.2491…
6. x = cos⁻¹(0.2491…) = 75.57…

Answer: x = 75.6

1. To reach BC we first need the shared side BD. In triangle ABD we know two sides and the angle between them, so use the cosine rule.
2. BD² = 9.4² + 12.8² − 2 × 9.4 × 12.8 × cos 72° = 88.36 + 163.84 − 74.36… = 177.83…
3. BD = 13.335… cm
4. In triangle BCD: BD is opposite the 54° angle at C, and BC is opposite the 39° angle at D. Use the sine rule.
5. ^BC⁄_{sin 39°} = ^BD⁄_{sin 54°}, so BC = ^{13.335… × sin 39°}⁄_{sin 54°} = 10.373…

Answer: 10.4 cm

1. FC is a space diagonal of the cuboid. 3D Pythagoras: FC² = AB² + BC² + AF² (length, width and height).
2. 18² = 9² + BC² + 7², so 324 = 81 + BC² + 49
3. BC² = 324 − 130 = 194
4. BC = √194 = 13.928…

Answer: 13.9 cm

1. BE is an edge of the prism, so it is perpendicular to the face ABC and angle CBE = 90°.
2. In right-angled triangle CBE: BC is opposite the 40° angle and CE = 13 is the hypotenuse: BC = 13 × sin 40° = 8.356… cm
3. Area of triangle ABC = ½ × AC × BC × sin C = 22, so ½ × 6 × 8.356… × sin C = 22
4. sin C = 0.8776…, so C = 61.35…° or 180° − 61.35…° = 118.65…°. Angle ACB is obtuse, so C = 118.65…°
5. Cosine rule: AB² = 6² + 8.356…² − 2 × 6 × 8.356… × cos 118.65…° = 36 + 69.83… + 48.07… = 153.90…
6. AB = 12.40…

Answer: 12.4 cm

1. Let AE = s. Then EF = 2s, the square has side s and the triangle has side s.
2. Height of J above the base = side of square + height of the equilateral triangle = s + ^√3⁄₂s = ^{(2 + √3)s}⁄₂
3. M is the midpoint of GF: from A it is ^s⁄₂ across and 2s along. AM = √(^s⁄₂)² + (2s)² = √¹⁷⁄₄s² = ^√17⁄₂s
4. tan y° = ^JM⁄_AM = ^{(2 + √3)s}⁄₂ ÷ ^√17s⁄₂ = ^{2 + √3}⁄_√17
5. Rationalise: ^{(2 + √3)√17}⁄₁₇ = ^{2√17 + √51}⁄₁₇ = ^{√68 + √51}⁄₁₇

Answer: T = ^{√68 + √51}⁄₁₇

1. Triangle ADE (right angle at D): tan 18° = ^2.2⁄_AD, so AD = ^2.2⁄_{tan 18°} = 6.770… cm
2. E is directly above D, so the angle we want is angle EBD, in the right-angled triangle EDB.
3. BD is the diagonal of the rectangle ABCD: BD² = 6.770…² + 6² = 81.84…, so BD = 9.046… cm
4. tan EBD = ^2.2⁄_9.046… = 0.2431…, so angle EBD = 13.66…°

Answer: 13.7°

1. The angle between a line and a plane is the angle between the line and its projection onto the plane. Drop a perpendicular from D to the base: it lands at M, the midpoint of FE.
2. Height DM: triangle FDE is isosceles with FD = 12 and FM = 5, so DM = √12² − 5² = √119 = 10.908… cm
3. AM on the base: from A go 5 across (to the middle of the width) and 15 along. AM = √5² + 15² = √250 = 15.811… cm
4. In right-angled triangle AMD: tan(angle) = ^DM⁄_AM = ^10.908…⁄_15.811… = 0.6899…
5. Angle = tan⁻¹(0.6899…) = 34.60…°

Answer: 34.6°

1. Set up coordinates to find lengths: put A at (0, 0, 0) with AD along the x-axis, DE going back along the y-axis and AB up the z-axis. Then B = (0, 0, 8), C = (12, 0, 8) and F = (12, 20, 8).
2. M is the centre of the base, so M = (6, 10, 0). P is the midpoint of CF, so P = (12, 10, 8).
3. Use 3D Pythagoras for each side of triangle BMP:
4. BM² = 6² + 10² + 8² = 200
5. MP² = 6² + 0² + 8² = 100
6. BP² = 12² + 10² + 0² = 244
7. Cosine rule for the angle at M: cos BMP = ^{BM2 + MP2 − BP2}⁄_{2 × BM × MP} = ^{200 + 100 − 244}⁄_{2 × √200 × 10} = ⁵⁶⁄_282.84… = 0.19798…
8. Angle BMP = cos⁻¹(0.19798…) = 78.58…°

Answer: 78.6°

1. Area = ½ × AB × AC × sin 30° = ½(2x − 1)(2x + 1) × ½ = ¼(4x² − 1) = x² − 0.25
2. Set equal to the given area: x² − 0.25 = x² + x − 3.75, so x = 3.5
3. So AB = 6 cm and AC = 8 cm.
4. Find BC with the cosine rule: BC² = 6² + 8² − 2 × 6 × 8 × cos 30° = 100 − 83.138… = 16.861…, so BC = 4.106… cm
5. The largest angle is opposite the longest side, AC = 8 cm, so it is angle B.
6. Cosine rule: cos B = ^{62 + 4.106…2 − 82}⁄_{2 × 6 × 4.106…} = ^−11.138…⁄_49.275… = −0.2260…
7. B = cos⁻¹(−0.2260…) = 103.06…°

Answer: 103°