Pythagoras’ theorem and trigonometry — Paper 4 · IGCSE Mathematics - International 0607

11 B NOT TO 8.6 cm SCALE A 9.3 cm C The area of triangle ABC = 23.5 cm2. (a) Show that angle BAC = 36.0°, correct to 1 decimal place. [2] (b) Use the cosine rule to find BC. BC = … cm [3] (c) All the angles in triangle ABC are acute. Use the sine rule to find the largest angle in the triangle ABC. … [3]

8 marks

Mark scheme: 11(a) 0.5 × 8.6 × 9.3 × sin A = 23.5 M1 35.99...[ = 36.0] A1 11(b) 2 2 or M1 for [ x = ] 8.6 + 9.3 −×2 8.6 × 9.3 × cos36 M2 [ x 2 = ] 8.6 2 + 9.32 − 2 × 8.6 × 9.3 × cos36 5.57 or 5.569 to 5.571... A1 M1 with correct answer scores full marks 11(c) 9.3 × sin36 M2 9.3 their ( b ) sin B = or M1 for = their ( b ) sin B sin36 78.8 or 78.9 or 79.[0] A1 M1 with correct answer scores full marks or 78.77 to 78.98...

This question in 0607/41 Oct/Nov 2018

6 NOT TO SCALE 10 cm The diagram shows a regular pentagon, of side 10 cm, with its vertices lying on a circle. (a) Show that the radius of the circle is 8.51 cm, correct to 3 significant figures. [4] (b) Calculate (i) the perimeter of the shaded segment, … cm [3] (ii) the area of the shaded segment. … cm2 [3]

10 marks

Mark scheme: 6(a) 36 or 54 or 72 or 108 or 540 seen B1 5 ÷ cos 54 oe M2 5 or M1 for cos 54 = oe r Starting with 8.51 is M0 8.506 to 8.507 A1 6(b)(i) 20.7 or 20.68 to 20.70 3 72 M2 for × 2 × π × 8.51 + 10 oe 360 72 or M1 for oe soi by ÷ 5 360 6(b)(ii) 11.0 or 11.1 or 11.02 to 11.10... 3 72 M1 for × π × 8.512 oe 360 M1 for 0.5 × 8.512 × sin 72 oe

This question in 0607/42 Oct/Nov 2018

2 y 11 10 9 8 C 7 6 5 4 3 2 1 A 0 x 1 2 3 4 5 6 7 8 9 10 11 12 – 1 – 2 B – 3 – 4 – 5 (a) Describe fully the single transformation that maps triangle A onto triangle B. … [2] 6 (b) Translate triangle A by the vector . [2] e- 3o (c) Triangle A can be mapped onto triangle C by a rotation followed by an enlargement. (i) Use trigonometry to calculate the angle of rotation. … [3] (ii) The scale factor of the enlargement is a where a is an integer. Find the value of a. a = … [3]

10 marks

Mark scheme: 2(a) Reflection 2 B1 for each y = –1 2(b) Triangle at (6, –3), (11, –3), 2  k  6 B1 for translation   or  (10, –1)  −3  k 2(c)(i) 63.4 or 63.43 to 63.44 3 4 B2 for tan [θ ] = oe 2 or B1 for correct angle clearly identified and no other angle seen. 2(c)(ii) 5 3 125 10 5 M2 for or or 5 20 5 or M1 for 10 2 + 5 2 or 4 2 + 2 2 or 12 + 2 2 or 125 or 20 or 5

This question in 0607/41 May/June 2019

8 A 13 cm NOT TO B SCALE 9 cm 6 cm D 11 cm C ABCD is a quadrilateral. (a) Show that BD = 9.22 cm, correct to 3 significant figures. [3] (b) Calculate angle ABD. Angle ABD = … [3] (c) Calculate the total area of the quadrilateral ABCD. … cm2 [3] (d) Calculate the length of the diagonal AC. AC = … cm [3]

12 marks

Mark scheme: 8(a) Correct Pythagoras statement M2 or M1 for [BD]2 + 62 = 112 oe leading to 112 – 62 or 121 – 36 or 85 9.219… A1 9.219… implies M1 A1 8(b) 43.8 or 43.80... nfww 3 9.22 2 + 132 − 9 2 M2 for cos[ABD] = 2 × 9.22 × 13 or better or M1 for 92 = 9.222 + 132 – 2 × 9.22 × 13 cos [ABD] oe 8(c) 69.1 or 69.13 to 69.14... nfww 3 M1 for 0.5 × 9.22 × 6 oe M1 for 0.5 × 9.22 × 13 × sin (their 43.8) oe 8(d) 17.7 or 17.69... 3 M1 for 62 + 132 – 2 × 6 × 13 cos (90 + their 43.8) A1 for 313 or 312.9 to 313.0

This question in 0607/41 May/June 2019

11 B NOT TO SCALE 8 cm C 6.5 cm 64 ° A 9 cm D The diagram shows a quadrilateral ABCD. AB = 8 cm, AD = 9 cm, CD = 6.5 cm and angle BAD = 64°. (a) Calculate BD and show that your answer rounds to 9.05 cm, correct to 2 decimal places. [2] (b) The area of the quadrilateral ABCD is 57.3 cm2. (i) Calculate angle BDC and show that your answer rounds to 58°, correct to the nearest degree. [4] (ii) Calculate angle BCD. angle BCD = … [5] Question 12 is printed on the next page.

11 marks

Mark scheme: 11(a) 82 + 92 – 2 × 8 × 9 × cos64 M1 9.048… [= 9.05] A1 11(b)(i) [Area ABD] = 0.5 × 8 × 9 × sin64 M1 Area BDC = 57.3 – their ABD M1 their 24.94 = 0.5 × 6.5 × 9.05 × sinBDC M1 Angle BDC = 57.98 to 58.02 A1 If 58.0… must see 58.0 not 58 alone 11(b)(ii) 77.4 to 77.54… 5 M1 for 6.52 + 9.052 – 2 × 6.5 × 9.05 × cos58 A2 for 61.8 or 7.86 or A1 for 61.8 seen their 7.86 9.05 M1 for = oe sin58 sin BCD 1 or × their7.86 × 6.5 × sinBCD = their 24.94 2 or 9.052 = 6.52 + (their7.86)2 – 2 × 6.5 × (their7.86) × cos BCD or better

This question in 0607/42 Oct/Nov 2019

11 North A NOT TO SCALE 110° 80 km 120 km North B C The diagram shows the positions of three ports, A, B and C. (a) Calculate BC. BC = … km [3] (b) Use the sine rule to calculate angle ABC. Angle ABC = … [3] (c) The bearing of C from A is 130°. Find the bearing of B from C. … [2] (d) A ship leaves B at 13 50 and sails in a straight line towards C. Its constant speed is 37 km/h. Find the time when it is at its closest point to A. Give your answer correct to the nearest minute. … [5] Question 12 is printed on the next page.

13 marks

Mark scheme: 11(a) 165 or 165.4... 3 M1 for 802 + 1202 – 2 × 80 × 120 × cos 110 A1 for 27 366 to 27 367 11(b) 43[.0] or 42.97 to 43.11 3 120sin110 M2 for their (a) sin ABC sin110 or M1 for = 120 their (a) 11(c) 283 2 FT 240 + their (b) B1 for 27 or 50 or 130 correctly identified at C 11(d) 1525 5 x M1 for cos (their (b)) = 80 A1 for 58.4 or 58.5 or 58.40 to 58.54 M1 for their 58.5 ÷ 37 M1 for correctly converting their time to hours and mins.

This question in 0607/43 May/June 2020

10 A NOT TO SCALE 42° O 142° C P D B A, D, B and C lie on a circle, centre O. AP is a tangent to the circle at A and BP is a tangent to the circle at B. Angle AOB = 142° and angle DAP = 42°. (a) Find the value of (i) angle ABD, Angle ABD = … [1] (ii) angle ACB, Angle ACB = … [1] (iii) angle ADB, Angle ADB = … [1] (iv) angle BAD, Angle BAD = … [1] (v) angle APB. Angle APB = … [1] (b) The radius of the circle is 11 cm. Find the area of triangle ABD. … cm2 [5]

10 marks

Mark scheme: 10(a)(i) 42 1 10(a)(ii) 71 1 10(a)(iii) 109 1 FT 180 – their(ii) 10(a)(iv) 29 1 10(a)(v) 38 1 10(b) 74.2 or 74.21 to 74.25... 5 M1 for [AB =] 2 × 11 cos19 oe theirAB × sin ( their 42 ) M2 for [ AD = ] sin ( their109 ) theirAB × sin ( their 29 ) or [ BD = ] sin ( their109 ) AD theirAB or M1 for = sin their 42 sin their109 BD theirAB or = sin their 29 sin their109 and M1 for [Area =] 0.5 × theirAB × theirAD ×sin(their29) or [Area =] 0.5 × theirAB × theirBD × sin(their42) or [Area =] 0.5 × theirAD × theirBD × sin(their109)

This question in 0607/41 May/June 2021

10 A B C 28° 63° NOT TO SCALE x m D A plane is flying in a straight line ABC at a constant height, x metres, above ground level. The point D is on the ground directly below C. The plane is travelling at a constant speed of 480 km/h. The time taken for the plane to travel from A to B is 18 seconds. x (a) Show that, in metres, AC = + 2400 . tan 63 [3] (b) Find the value of x. x = … [5]

8 marks

Mark scheme: 10(a) x x B1 tan63 = or BC = BC tan63 1000 M1 AB = 480   18 oe 3600 AB = 2400 A1 x AC = + 2400 tan63 10(b) 1750 or 1750.[…] 5 2400tan 28 M4 for oe  tan 28   1 −   tan63  or M3 for tan28 x + 2400tan28 = x oe tan63 x or M2 for tan 28 = oe  x   + 2400   tan63  x or M1 for tan28 = AC OR 2400sin28 M2 for BD = sin35 BD 2400 or M1 for = sin28 sin35 M2 for x = theirBDsin63 x or M1 for sin63 = theirBD

This question in 0607/42 Feb/March 2023

11 A NOT TO SCALE O r cm B C The diagram shows an equilateral triangle ABC touching a circle, centre O and radius r cm. 3 3 2 (a) (i) Show that the area of triangle ABC is r cm 2. 4 [4] (ii) Find an expression, in terms of r and r, for the exact value of the shaded area. … cm2 [1] (b) D A NOT TO SCALE O r cm B C E F Another equilateral triangle DEF is touching the same circle. Find an expression, in terms of r and r, for the exact value of this shaded area. … cm2 [3] (c) Find in its simplest form the ratio perimeter of triangle ABC : perimeter of triangle DEF . … : … [2] Question 12 is printed on the next page.

10 marks

Mark scheme: 11(a)(i) Fully correct method for area M3 2 1  3 r  e.g. [Area] =   2   sin60   2  2  1 2 or Area of BOC = r  sin120 2 1 2 3 = r  2 2 1 2 3 Area of ABC = 3× r  2 2 or Side by cosine rule = r 3 Then 0.5 × a2 × sin 60 or 3 2 3 2 ( side ) = ( r 3) 4 4 3 3 2 = r 4 or M2 for correct method length of side 3 or M1 for cos30 = oe 2 Completion to answer with no errors A1 11(a)(ii) 2 3 3 2 1 πr − r oe 4 11(b) [Area = ] 3 3 r 2 − πr 2 oe 3  2r  2 M2 for 0.5   sin60 oe    tan30  2r or M1 for EF = oe tan30 11(c) 1 : 2 2 B1 for unsimplified

This question in 0607/42 Feb/March 2023

2 (a) Calculate the volume of each shape. (i) A cuboid with a square base of side 5 cm and height 3 cm. … cm3 [2] (ii) A sphere with radius 4 cm. … cm3 [2] (b) A cylinder has volume 120 cm3 and height 6 cm. Calculate its radius. … cm [2] (c) A cone has volume 120 cm3 and height 6 cm. Calculate the length of its sloping edge. … cm [3]

9 marks

Mark scheme: 2(a)(i) 75 2 M1 for 52[3] 2(a)(ii) 268 or 268.0 to 268.1... 2 4 M1 for    43 3 2(b) 2.52 or 2.522 to 2.523... 2 M1 for   r2  6  120 or better 2(c) 7.42 or 7.422 to 7.423 3 1 M1 for    r2  6  120 oe or 3 better M1 for 62 (their r)2 or better

This question in 0607/41 May/June 2023

12 A B D 30° NOT TO SCALE 200 cm C The diagram shows two right‑angled triangles ABC and CBD. AB = BC , AC = 200 cm and angle BDC = 30° . Find the perimeter of triangle ACD. … cm [6]

6 marks

Mark scheme: 12 61.5 or 61.46…. 6 B1 for AB or BC = 10 their BC M2 for CD  oe sin 30 their BC or M1 for sin30  oe CD their BC M2 for BD  oe tan 30 their BC or M1 for tan30  oe BD

This question in 0607/43 May/June 2023

8 (a) E NOT TO 20 cm SCALE D C M A 24 cm F 10 cm B ABCDE is a pyramid with a rectangular base. AB = 10 cm and BC = 24 cm. The length of each sloping edge is 20 cm. Vertex E is vertically above the centre of the base, M. (i) Calculate the length AC. … cm [2] (ii) Calculate EM, the height of the pyramid. … cm [3] (iii) F is the mid-point of BC. Find the angle between EF and the base of the pyramid. … [3] (b) B A NOT TO 7 cm SCALE Cone A is mathematically similar to cone B. The height of cone A is 7 cm and its volume is 66 cm 3. The volume of cone B is 222.75 cm 3. (i) Find the height of cone B. … cm [3] (ii) A sphere also has volume 66 cm 3. Calculate the radius of the sphere. … cm [2]

13 marks

Mark scheme: 8(a)(i) 26 2 M1 for 10 2 + 24 2 or better 8(a)(ii) 15.2 or 15.19 to 15.20 3 2 2 their 26  M2 for 20 −   2  2 their 26  2 M1 for 20 −   2  8(a)(iii) 71.8 or 71.78 to 71.80… 3 231 M2 for tan EFM = or using their height 5 oe ALT M1 for EF 2 = 20 2 − 12 2 5 M1 for cos EFM = their EF or B1 for identifying correct angle or stating EFM 8(b)(i) 10.5 3 222.75 M2 for 3  7 oe 66 222.75  7 3 66 3 or M1 for oe or =   66  l  222.75 8(b)(ii) 2.51 or 2.506 to 2.507… 2 66  3 M1 for 3 4

This question in 0607/42 Oct/Nov 2023

10 (a) Q NOT TO SCALE O R 135° P S P, Q, R and S are points on the circle centre O. Find angle PQR. Angle PQR = … [1] (b) C B D NOT TO SCALE O F 62° A E A, B, C, D and E are points on the circle centre O. AC is parallel to ED. Find the obtuse angle BOD. Angle BOD = … [2] (c) B NOT TO A C SCALE O 7.5 cm E D ABCDE is a regular pentagon. A, B, C, D and E are points on the circle centre O. The length of the perpendicular from O to ED is 7.5 cm. (i) Show that the length of one side of the pentagon is 10.9 cm correct to 3 significant figures. [4] (ii) Calculate the shaded area. … cm2 [4]

11 marks

Mark scheme: 10(a) 45 1 10(b) 124 2 B1 for BED = 62 10(c)(i) [EOD=] 72 or [ODE=] 54 B1 seen or implied by 36 2  7.5  tan36 oe M2 halfside 7.5 M1 for tan(36) = or tan54 = 7.5 7.5 halfside or 2  tan54 10.89…. A1 no errors or omissions 10(c)(ii) 13.1 or 13.11 to 13.13… 4 2 2 2  10.9  M1 r = 7.5 + or better or trig method    2  for r 1 1 2 M1  10.9  7.5 or 2  their r  their sin72 oe 2 their 72 2 M1   their r oe 360

This question in 0607/42 Oct/Nov 2023

11 (a) NOT TO SCALE 12 cm p° 18 cm Calculate the value of p. p = … [2] (b) B NOT TO SCALE 230 m 190 m C A 150 m 180 m D (i) Show that angle ABC = 67.0° correct to 1 decimal place. [4] (ii) Calculate the shortest distance from A to the side BC. … m [3] Question 12 is printed on the next page.

9 marks

Mark scheme: 11(a) 33.7 or 33.69... 2 12 M1 for tan[p] = 18 11(b)(i) 1502 + 1802 M1 192 2 + 230 2 − their (180 2 + 150 2 ) M2 M1 for their (1802 + 1502) = 1902 + 2302 [cos=] – 2 × 190 × 230 cosB 2  190  230 67.03 to 67.04 A1 11(b)(ii) 175 or 174.8 to 174.9... 3 x M2 for sin67[.0] = oe 190 or M1 for distance required is perpendicular to BC oe

This question in 0607/43 Oct/Nov 2023

15 A 4 cm B NOT TO 154° O SCALE C A and C are points on the circle, centre O. BA and BC are tangents to the circle. The radius of the circle is 4 cm. Angle AOC = 154°. (a) Write down the mathematical name of the quadrilateral ABCO. … [1] (b) Calculate the area of the shaded segment. … cm2 [3] (c) Calculate the shortest distance from B to the circumference of the circle. … cm [3]

7 marks

Mark scheme: 15(a) kite 1 15(b) 18.[0] or 17.99… 3 154 2 1 M2 for π 4 − 4 4 sin154 oe 360 2 154 2 1 or M1 for π 4 or 4 4 sin154 oe 360 2 15(c) 13.8 or 13.78… 3 4 M2 for − 4 oe cos77 4 M1 for cos77 = oe or better x or M1 for their OB – 4

This question in 0607/42 Feb/March 2025