Non-right-angled triangles — Paper 4 · IGCSE Mathematics - International 0607

9 8 cm B A NOT TO SCALE 6 cm 12 cm C The diagram shows triangle ABC. (a) Use the cosine rule to find angle ABC. Angle ABC = … [3] (b) Use the sine rule to find angle BAC. Angle BAC = … [3]

6 marks

Mark scheme: 9(a) 8 2 + 6 2 − 2 2 M2 M1 for 122 = 82 + 62 – 2 × 8 × 6 cos[...] [cosx =] oe 2 × 6 × 8 117.3 or 117.2 to 117.3 B1 9(b) 6 × sin(their (a)) M2 6 12 [sin = ] oe M1 for = oe 12 sin A sin(their (a)) 26.4 or 26.5 or 26.37 to 26.46 B1

This question in 0607/41 May/June 2017

7 A ship sails 65 km on a bearing of 310° from A to B. It then changes course and sails 40 km on a bearing of 250° from B to C. The ship then returns to A. (a) On the diagram, sketch the path of the ship from A. On your diagram show the bearings and distances. North A [3] (b) Find angle ABC. … [1] (c) Calculate AC and show that it rounds to 91.8 km, correct to the nearest tenth of a kilometre. [3] (d) Find the bearing of C from A. … [4]

11 marks

Mark scheme: 7(a) Correct skettch showing bearingsb 3 B1B for 310° bearingb apprrox correct (2270 to 360) aand and distancees markedm B1B for 250° bearingb apprrox correct (1180 to 270) aand markedm B1B for distannces correctlyy marked 7(b) 120 1 7(c) 402 + 652 – 22×40×65×coos their 120 M1 theirt 120 muust be betweeen 0 and 180 40 2 + 655 2 − [ ]2 AllowA cos1220 = 2 × 400 × 65 91.78 to 91.779 A2 A1A for 8425 or 5 337 7(d) 288 or 287.88... 4 40sin ( their120 ) M2M for oe 91.8 sinθ sin(thheir 120) oro M1 for = oe 404 991.8 IfI cosine rulee used, M2 fofor explicit exxpression or M1M for impliicit. A1A for 22.2 oro 22.16 to 222.17… IfI 0 scored SC2 for answwer 108 or 1007.8…

This question in 0607/42 May/June 2017

7 A NOT TO SCALE 35° 8 cm B D 6 cm 9 cm C (a) Calculate AB. AB = … cm [3] (b) Calculate angle BCD. Angle BCD = … [3]

6 marks

Mark scheme: 7(a) 9.77 or 9.766… 3 8 M2 for oe cos35 8 or M1 for cos35 = oe AB 7(b) 60.6 or 60.61… 3 6 2 + 9 2 − 82 M2 for 2 × 6 × 9 or M1 for 82 = 6 2 + 9 2 −×2 6 × 9cos C

This question in 0607/43 May/June 2017

10 North 90 m D A 35° NOT TO SCALE 120 m C 115 m 65° B The diagram shows a school playing field, ABCD, which is on horizontal ground, with D due East of A. (a) Find the bearing of (i) C from A, … [1] (ii) A from C. … [2] (b) Calculate the length of CD. CD = … m [3] (c) Calculate angle BAC. Angle BAC = … [3] (d) (i) Calculate the area of the school playing field. … m2 [4] (ii) In the school office there is a plan of the school playing field. It is drawn to a scale of 1 : 500. Calculate the area of the school playing field on the plan. Give your answer in cm2. … cm2 [3] Question 11 is printed on the next page.

16 marks

Mark scheme: 10(a)(i) 125 1 10(a)(ii) 305 2 FT their (i), Dep on (i) < 180 M1 for 180 + their(i), Dep on (i) < 180 10(b) 69.3 or 69.32 to 69.33 3 M1 for 902 + 1202 – 2 × 90 × 120 cos 35 A1 for 4806… 10(c) 60.3 or 60.28 to 60.29 3 115sin 65 M2 for oe 120 115 120 or M1 for = oe sin BAC sin65 10(d)(i) 8730 or 8728 to 8730… 4 M1 for 0.5 × 90 × 120 × sin35 M2 for 0.5 × 120 × 115 × sin(180 – 65 – their (c)) oe or M1 for angle ACB = 180 – 65 – their (c) 10(d)(ii) 349 or 349.1 to 349.2… 3 FT their (d)(i) ÷ 25 M2 for their (i) ÷ 25 oe or M1 for squaring scale oe or for figs 349 or 3491 to 3492

This question in 0607/42 Oct/Nov 2017

9 E 50° NOT TO SCALE 13 cm D 70° A B C 15 cm In the diagram, ABC is a straight line, AE = BE = 13 cm and BC = 15 cm. Angle EAB = 70°, angle EBD = 90° and angle BED = 50°. Calculate (a) the length of the perpendicular line from E to AB, … cm [2] (b) the length BD, BD = … cm [2] (c) the length CD, CD = … cm [4] (d) the area of the quadrilateral ACDE. … cm2 [3]

11 marks

Mark scheme: 9(a) 12.2 or 12.21 to 12.22 2 [ ] M1 for sin70 = oe 13 9(b) 15.5 or 15.49… 2 BD M1 for tan50 = oe 13 9(c) 5.32 or 5.316 to 5.319… 4 B1 for [angle DBC = ] 20 M1 for (theirBD)2 + 152 – 2 × their BD × 15cos(their DBC) A1 for 28.26 to 28.30… 9(d) art 195 3 M2 two of 5.0 × 13 × 13 × sin 40 oe 0.5 × 13 × their BD oe 0.5 × 15 × their BD × sin(their 20) or M1 for one of above

This question in 0607/41 May/June 2018

9 B 9.1 cm NOT TO SCALE 8.2 cm A 11 cm C (a) Show that angle BAC = 47.0° , correct to 1 decimal place. [3] (b) Use the sine rule to find angle ABC. Angle ABC = … [3] (c) Find the area of triangle ABC. … cm2 [2] (d) Find the length of the perpendicular from B to AC. … cm [2]

10 marks

Mark scheme: 9(a) 112 + 9.12 − 8.2 2 M2 M1 for 8.2 2 = 112 + 9.12 − 2 × 11 × 9.1 × cos[ ] [cos A] = 2 × 11 × 9.1 46.98 to 46.99 A1 9(b) 11 M2 8.2 11 [sin B = ] × sin 47.0 M1 for = 8.2 sin 47 sin B 78.8 or 78.74 to 78.84 A1 If 0 scored then SC1 for correct answer from cosine rule or other method 9(c) 36.6 or 36.54 to 36.60… 2 M1 for 0.5 × 9.1 × 11 × sin47.0 or M1 for 0.5 × 9.1 × 8.2 × sin( their (b)) or M1 for 0.5 × 8.2 × 11 × sin (180 – 47 – their (b)) 9(d) 6.65 or 6.66 or 6.647 to 6.656… 2 M1 for 9.1 × sin47.0 oe or their (c) ÷ (0.5 × 11)

This question in 0607/42 May/June 2018

10 North North NOT TO SCALE A 120 m B In the diagram, point B is due east of point A. (a) Point C is on a bearing of 060° from A and a bearing of 325° from B. Calculate the distance BC. BC = … m [4] (b) Point D is South of AB. D is 80 m from A and 90 m from B. Calculate the bearing of D from B. … [4]

8 marks

Mark scheme: 10(a) 60.2 or 60.22 to 60.23. 4 B1 for angle ACB = 95 120sin(theirCAB ) M2 for oe sin(their ACB ) BC 120 or M1 for = sin(theirCAB ) sin(their ACB ) 10(b) 228 or 228.1 to 228.2 nfww 4 120 2 + 90 2 − 80 2 M2 for cos[ABD] = 2 × 120 × 90 or M1 for 802 = 1202 + 902 – 2 × 120 × 90 cosABD A1 for 41.8 or 41.80 to 41.81

This question in 0607/42 Oct/Nov 2018

11 F NOT TO SCALE B 41° 5.5 m 6.2 m A C D The diagram shows four points A, B, C and D on horizontal ground. There is a vertical flagpole, FB, held in place by straight wires AF, CF and DF. BCD is a straight line, AB = 5.5 m, BC = 6.2 m and angle FAB = 41°. (a) Show that FB = 4.781 m, correct to 3 decimal places. [2] (b) Calculate angle FCB. Angle FCB = … [2] (c) Angle CDF = 18°. Show that CD = 8.514, correct to 3 decimal places. [3] (d) Angle ABC = 78°. Find AD. AD = … m [3] (e) Find the area of triangle ABD. … m2 [2]

12 marks

Mark scheme: 11(a) FB M1 sin49 sin41 tan 41 = oe e.g. = 5.5 5.5 FB = 4.7810.. [= 4.781] A1 11(b) 37.6 or 37.63 to 37.64 2 4.781 M1 for tan[ FCB ] = oe 6.2 11(c) 4.781 M2 4.781 [CD = ] − 6.2 oe M1 for tan18 = oe tan18 BD 8.5144... A1 11(d) 14.6 or 14.58 to 14.60 3 B2 for 212.7... or M1 for [ AD2 =] 5.52 + (8.514 + 6.2)2 − 2 × 5.5 × (8.514 + 6.2) × cos78 11(e) 39.5 or 39.6 or 39.54 to 39.6[0] 2 M1 for 0.5 × 5.5 × (8.514 + 6.2) × sin 78

This question in 0607/43 May/June 2019

7 North North 10 km B 150° NOT TO A SCALE 12 km 18 km C 21 km D The diagram shows four villages A, B, C and D and five straight roads connecting them. B is 10 km due east of A. C is 12 km from B on a bearing of 150°. D is 21 km from C and 18 km from A. (a) Calculate the distance AC and show that your answer rounds to 19.08 km, correct to 2 decimal places. [4] (b) Using the sine rule, calculate angle ACB and show that your answer rounds to 27.0°, correct to 1 decimal place. [3] (c) Calculate the bearing of D from C. … [4] (d) A straight path, BP, connects B to the closest point, P, on AC. Calculate the length of this path. … km [2] (e) The area within triangle ABC is grassland. Calculate the area of this grassland. … km2 [2]

15 marks

Mark scheme: 7(a) [Angle ABC = ] 120 B1 10 2 + 12 2 − 2 × 10 × 12cos(their ABC ) M1 19.078 to 19.079 A2 A1 for 364 7(b) 10sin120 M2 19.08 10 M1 for = oe 19.08 sin120 sin ACB 26.99... A1 M1 only and A1 imply M2 A1 7(c) 249.8 to 250[.0] 4 212 + 19.082 − 182 M2 for [cos ACD = ] oe 2 × 21 × 19.08 or M1 for 18 2 = 212 + 19.08 2 −×2 21 × 19.08 × cos( ACD ) M1 for 360 – (30 + 27 + their ACD) oe 7(d) 5.45 or 5.446 to 5.448 2 M1 for 12sin27.0 oe 7(e) 52[.0] or 51.94 to 51.99... 2 1 M1 for × 19.08 × their (d) 2 1 or for × 10 × 12 × sin120 oe 2 1 or for × 19.08 × 12 × sin 27 2

This question in 0607/41 Oct/Nov 2019

9 A 58° NOT TO SCALE 14 cm 12 cm O B N C A, B and C are points on the circle, centre O. ON is perpendicular to BC. AB = 14 cm, AC = 12 cm and angle BAC = 58°. (a) Show that BC = 12.73 cm, correct to 2 decimal places. [3] (b) Explain why angle BON = 58°. … … [1] (c) Calculate OB, the radius of the circle. OB = … cm [3] (d) Calculate the area of the shaded segment. … cm2 [3]

10 marks

Mark scheme: 9(a) 142 + 122 – 2 × 14 × 12 × cos58 M1 12.725 to 12.726 A2 or A1 for 161.9... 9(b) Angle at centre = 2 × angle at 1 circumference oe 9(c) 7.49 or 7.5[0] or 7.51 or 7.487 to 7.506 3 6.365 M2 for oe sin58 6.365 or M1 for sin 58 = oe OB 9(d) 31.3 to 31.9 nfww 3 116 M2 for × π × (their (c))2 360 1 − × (their (c))2 × sin116 oe 2 116 or M1 for × π × (their (c))2 oe 360 1 or × (their (c))2 × sin116 oe 2

This question in 0607/43 Oct/Nov 2019

11 (a) C NOT TO SCALE 11 cm 60° A B 12 cm Calculate the shortest distance from B to AC. … cm [7] (b) V NOT TO SCALE h cm S R M 6 cm P 8 cm Q The diagram shows a pyramid on a rectangular base PQRS. The diagonals of the base meet at M and V is vertically above M. PQ = 8 cm, QR = 6 cm and VM = h cm. The volume of the pyramid is 112 cm3. (i) Show that h = 7. [2] (ii) Calculate the length of VR. VR = … cm [3] (iii) K is the mid-point of PS and L is the mid-point of QR. Calculate angle KVL. Angle KVL = … [3]

15 marks

This question in 0607/41 May/June 2020

8 North B 5.37 km NOT TO SCALE North A C 48° 6.13 km 6.42 km D The diagram shows four points A, B, C and D on horizontal ground. B is due North of C and C is due East of A. (a) Find the bearing of (i) D from A, … [1] (ii) A from D. … [1] (b) Calculate angle ABC. Angle ABC = … [2] (c) Calculate the area of quadrilateral ABCD. … km 2 [3] (d) Calculate CD. CD = … km [3] (e) Angle ACD is acute. Find the bearing of D from C. … [4]

14 marks

Mark scheme: 8(a)(i) 138 1 8(a)(ii) 318 1 FT their (i) + 180 8(b) 48.8 or 48.78… 2 6.13 M1 for tan[ x = ] 5.37 8(c) 31.1 or 31.08… 3 M2 for 6. 13 × 5. 37 1 + × 6. 13 × 6. 42 × sin48 2 2 6.13 × 5. 37 or M1 for or 2 1 × 6. 13 × 6.42 × sin48 2 8(d) 5.11 or 5.111… 3 B2 for 26.1… or M1 for 6.132 + 6.42 2 − 2 × 6.13 × 6.42 × cos48 8(e) 201 or 200.9 to 201.1… 4 B3 for 69[.0] or 68.89 to 68.90 or M2 for sin 48 sin C = × 6.42, [C = 69.0] their (d) sin C sin 48 or M1 for = 6.42 their (d)

This question in 0607/42 May/June 2020

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 B 46° NOT TO SCALE D 78° 15° 8.1 cm 9.6 cm A C ABC and ADC are triangles. AD = 8.1 cm and CD = 9.6 cm. Angle ABC = 46° , angle ADC = 78° and angle BAD = 15° . (a) Find AC. AC = … cm [3] (b) Show that angle DAC = 57° , correct to the nearest degree. [3] (c) Find BC. BC = … cm [3] (d) Find the area of quadrilateral ABCD. … cm2 [4]

13 marks

Mark scheme: 10(a) 11.2 or 11.19 to 11.20 3 M2 for 8.12 + 9.6 2 − 2 × 8.1 × 9.6 × cos78 OR M1 for 8.12 + 9.6 2 − 2 × 8.1 × 9.6 × cos78 A1 for 125 or 125.4... 10(b) 9.6 × sin78 M2 9.6 their (a) sin DAC = oe M1 for = oe their (a) sin DAC sin78 56.97 to 57.05... A1 10(c) 14.8 or 14.79 to 14.81 3 their (a) × sin(57 + 15) M2 for BC = sin46 BC their (a) or M1 for = oe sin ( 57 + 15 ) sin 46 10(d) 35.0 to 35.3 4 B1 for angle ACB = 62 soi M1 for area ABC = 0.5 × their (a) × their (c) × sin their (62) or 0.5 × their (c) × (13.7or13.74to 13.75) × sin 46 or 0.5 × their (a) × (13.7or13.74to13.75) × sin(57 + 15) M1 for area ADC = 0.5 × 8.1 × 9.6 × sin78 oe

This question in 0607/41 Oct/Nov 2020

8 North B NOT TO SCALE 17 km 142° C North 4 km A Rani sails in a boat race around a triangular course. She sails from A to B to C and then directly back to A. B is due north of C. (a) Find the bearing Rani sails on from C to A. … [1] (b) Show that AB = 20.3 km, correct to 1 decimal place. [3] (c) Calculate the bearing of B from A. … [3] (d) Rani starts the race at 08 57 and returns to A at 12 33. Calculate the average speed of her boat in km/h. … km/h [3]

10 marks

Mark scheme: 8(a) 218 1 8(b) 42 + 172 – 2 × 4 × 17 × cos142 M2 M1 for implicit cosine rule 20.30… A1 8(c) 007 or 006.92 to 006.98 3 4sin142 M2 for sin B = oe 20.3 4 20.3 or M1 for = oe sin B sin142 OR 17sin142 M2 for sin A= oe 20.3 17 20.3 or M1 for = oe sin A sin142 8(d) 11.5 or 11.47… 3 B1 for 3 h 36 min or 3.6 h seen 4 + 17 + 20.3 M1 for their 3.6

This question in 0607/42 Feb/March 2021

5 P NOT TO SCALE 10 m 20 m C B 35° A A, B and C are points on horizontal ground. BP is a vertical pole. BC = 20 m and BP = 10 m. Angle PAB = 35°. (a) Show that PC = 22.36 m correct to 2 decimal places. [2] (b) Show that AB = 14.28 m correct to 2 decimal places. [2] (c) Calculate AP. AP = … m [2] (d) Angle ABC = 125°. Calculate AC. AC = … m [3] (e) Calculate angle APC. Angle APC = … [3]

12 marks

Mark scheme: 5(a) 202 + 102 M1 22.360 to 22.361 A1 5(b) 10 M1 sin35 sin55 tan35 = oe = , i.e correct implicit AB 10 AB 14.281... A1 5(c) 17.4 or 17.43... 2 10 M1 for sin35 = oe AP or 14.282 + 102 5(d) 30.5 or 30.52... 3 M1 for 20 2 + 14.282 −×2 20 × 14.28 × cos125 A1 for 931.5 to 931.6... 5(e) 99.2 to 99.5 3 M2 for [cos = ] their 30.5 22.36 2 + ( their 17.4 ) 2 − ( 2 ) 2 × 22.36 × ( their 17.4 ) or M1 for ( their 30.5 ) 2 = 22.36 2 + ( their17.4 ) 2 −×2 22.36 × ( their17.4 ) × cos APB

This question in 0607/42 May/June 2021

11 B 48° 120 m 45 m NOT TO SCALE C A 28° 54 m D Angles ACB and ACD are obtuse. (a) Show that AC = 95.9 m correct to the nearest 0.1 metre. [3] (b) Find angle ACD. Angle ACD = … [4] (c) The area of triangle ABD is 5137m2. Calculate the area of triangle BCD. … m2 [4]

11 marks

Mark scheme: 11(a) 2 2 M2 M1 for [AC2] = 1202 + 452 – 2 × 45 × 120 [ AC = ] 120 + 45 −×2 45 × 120 × cos48 × cos 48 95.90 to 95.91 A1 11(b) 95.5 or 95.50 to 95.51… 4 95.9 × sin 28 M2 for sin ADC = 54 95.9 54 or M1 for = oe sin ADC sin28 M1 for 180 − 28 −their ADC 11(c) 552 or 553 or 554 or 551.5 to 553.9 4 M3 for 5137 – 0.5 × 120 × 45 × sin 48 – 0.5 × 95.9 × 54 × sin(their ACD) OR M1 for area ABC = 0.5 × 45 × 120 × sin 48 or M1 for area ABD = 0.5 × 95.9 × 54 × sin ACD OR 120 × sin48 M2 for sin ACB = 95.9 120 95.9 or M1 for = sin ACB sin48 M1 for area BCD = 0.5 × 45 × 54 × sin(360 – their (b) – their ACB)

This question in 0607/43 Oct/Nov 2021

9 (a) x cm 4 cm NOT TO SCALE 40° Calculate the value of x. x = … [3] (b) C 8 cm 9 cm NOT TO SCALE A B 10 cm (i) Calculate angle ABC. Angle ABC = … [3] (ii) T is the point on AB that is the shortest distance from C. Calculate BT. BT = … cm [3] (c) Another triangle PQR has QR = 12 cm, PR = 7 cm and angle PQR = 35°. Calculate the difference between the two possible values of angle QPR. … [5]

14 marks

Mark scheme: 9(a) 6.22 or 6.222 to 6.223 3 4 M2 for oe sin40 4 or M1 for sin 40 = oe x 9(b)(i) 49.5 or 49.45 to 49.46 3 9 2 + 10 2 − 8 2 M2 for [cos=] oe 2.9.10 or M1 for 82 = 92 + 102 – 2 × 9 × 10 cos(...) 9(b)(ii) 5.85 or 5.845 to 5.851... 3 BT M2 for = cos(their(b)(i)) oe or better 9 or M1 for CT drawn and right angle at T 9(c) 21[.0] or 20.98 to 21.00 5 12sin35 M2 for 7 7 12 or M1 for = oe sin35 sin P A1 for 79.5 or 79.50 to 79.51 M1 for 180 – their 79.5 If 0 scored, SC1 for diagram showing the two angles

This question in 0607/42 Feb/March 2022

12 B NOT TO A SCALE 80° 9 m 16 m D 115° 22 m C (a) Calculate the area of triangle BCD. … m2 [2] (b) Calculate angle ADB. Angle ADB = … [6]

8 marks

Mark scheme: 12(a) 160 or 159.5… 2 1 M1 for  16  22  sin115 oe 2 12(b) 84[.0] or 84.02 to 84.03 6 B3 for 32.2 or 32.21.. soi OR M2 for 16 2  22 2 2 16  22  cos115 or M1 for 16 2  22 2  2  16  22  cos115 AND 9sin80 M2 for dependent on their (32.21) cosine rule used sin ABD sin80 or M1 for  9 their (32.21)

This question in 0607/41 May/June 2022

7 North B NOT TO SCALE 535 m 420 m C A 28° 750 m D The diagram shows four points A, B, C and D. B is due north of C and C is due east of A. AC = 420 m, AD = 750 m, BC = 535 m and angle CAD = 28°. (a) Find the bearing of (i) D from A, … [1] (ii) A from D. … [1] (b) Calculate AB. AB = … m [2] (c) Calculate CD. CD = … m [3] (d) Calculate the area of quadrilateral ABCD. … m2 [3] (e) Angle ACD is obtuse. Find the bearing of D from C. … [4]

14 marks

Mark scheme: 7(a)(i) 118 1 7(a)(ii) 298 cao 1 7(b) 680 or 680.1 to 680.2 2 M1 for 4202 + 5352 7(c) 427 or 427.3 to 427.4 3 M2 for  CD   420 2  750 2  2  420  750  cos28 or M1 for  CD 2  420 2  750 2 2 420  750  cos28   7(d) 186000 or 186200 to 186300 3 M1 for area ABC = 0.5  420  535 M1 for area ACD = 0.5  420  750 x sin28 7(e) 145 or 145 to 146 nfww 4 750  sin 28 M2 for sin ACD  theirCD 750 theirCD or M1 for  sin ACD sin28 And M1 for 360 – 90 – (180 – their acute C) OR 420 2  427.367 2  750 2 M2 for cos ACD = 2  420  427.367 or M1 for 7502 = 4202 + 427.372 – 2  420  427.37 cos C And M1 for 360 – 90 – their obtuse C

This question in 0607/42 May/June 2022

8 North NOT TO B 120° SCALE North 65 km C 55° A The diagram shows the route of a ship between three ports, A, B and C. The bearing of B from A is 055° and the bearing of C from B is 120°. BC = 65 km . The ship takes 7 hours to sail from A to B. It sails at a speed of 20 km/h. (a) Find the distance AB. … km [1] (b) Show that angle ABC = 115° . [1] (c) (i) Calculate the distance CA. … km [3] (ii) Calculate the bearing of A from C. … [4] (d) The ship takes 3.6 hours to sail from B to C. It then sails from C to A at a speed of 21.5 km/h. Find the average speed for the complete journey from A to B to C and back to A. … km/h [3]

12 marks

Mark scheme: 8(a) 140 1 8(b) 360 – (120 + 125) or 60 + 55 1 or 180 + 55 – 120 8(c)(i) 178 or 177.5... 3 M2 for ((their140) 2  65 2  2  ( their140)  65  cos115) OR M1 for (their 140)2 + 652 – 2 × (their 140) × 65 × cos115 8(c)(ii) 254 or 255 or 254.3 to 254.5... 4 their140sin115 M2 for sin[C ]  oe their178 sin[ C ] sin115 or M1 for  oe their140 their178 A1 for 45.18 to 45.63 M1 for 360 – 60 – their C oe calculated as answer 8(d) 20.3 or 20.26 to 20.31... 3 their140  65  their178 M2 for their178 7  3.6  21.5 their178 or M1 for 21.5 totaldistance or for clear indication of totaltime

This question in 0607/43 May/June 2022

11 F X E NOT TO SCALE B A C D The diagram shows a triangular prism ABCDEF. X is a point on FE. AB = 8 m , AD = 15 m , AF = 10 m , EC = 6 m and FX = 5 m . Angle ABF = 90° and angle DCE = 90° . (a) Calculate angle CDE. Angle CDE = … [2] (b) Calculate AC. AC = … m [2] (c) Calculate angle CXA. Angle CXA = … [5] (d) Calculate the area of triangle CXA. … m2 [2] Question 12 is printed on the next page.

11 marks

Mark scheme: 11(a) 36.9 or 36.86 to 36.87 2 6 6 M1 for tan[CDE ] = or sin[CDE ] = 8 10 8 or cos[CDE ] = 10 11(b) 17 2 M1 for [ AC 2 ] = 15 2 + 8 2 11(c) 96.2 or 96.16… 5 M1 for AX 2 = 52 + 10 2 M1 for CX 2 = 10 2 + 6 2 M2 dep for their AX 2 + theirCX 2 − their AC 2 [cos CDX =] 2  their AX  theirCX their AC 2 = their AX 2 + theirCX 2 or M1dep for −2 their AX  theirCX  cos CDX both dependent on Pythagoras or trigonometry used for AX and CX 11(d) 64.8 or 64.81 to 64.82 2 M1 dep for area CXA = 0.5  theirCX  their AX  sin(theirCXA) dependent on Pythagoras or trigonometry used for AX and CX

This question in 0607/41 Oct/Nov 2022

10 15 cm NOT TO SCALE A 5 cm 8 cm B C 11 cm Triangle ABC is the cross-section of a prism of length 15 cm. AB = 5 cm , AC = 8 cm and BC = 11 cm . (a) Show that the area of triangle ABC = 18.33 cm 2 correct to 2 decimal places. [4] (b) Find the volume of the prism. … cm3 [1] (c) Find the total surface area of the prism. … cm2 [2] (d) A mathematically similar prism has a volume of 500 cm 3. Calculate the total surface area of this similar prism. Give your answer correct to 2 significant figures. … cm2 [3]

10 marks

Mark scheme: 10(a) 5 2 + 8 2 − 112 M2 M1 for 112 = 52 + 82 – 2 × 5 × 8 × cosA [cos A =] oe 2 5 8 5 2 + 112 − 8 2 or [cos B =] oe or 82 = 52 + 112 – 2 × 5 × 11 × cosB 2 5 11 112 + 8 2 − 5 2 or [cos C =] oe or 52 = 112 + 82 – 2 × 11 × 8 × cosC 2  11  8 0.5 × 5 × 8 × sin(their A) oe M1 or 0.5 × 5 × 11 × sin(their B) oe or 0.5 × 11 × 8 × sin(their C) oe 18.330... A1 Dep on no errors seen and on M2 and M1 awarded 10(b) 275 or 274.9... 1 10(c) 397 or 396.6 to 396.7 2 M1 for 8 × 15 + 11 × 15 + 5 × 15 + 2 × 18.33 10(d) 590 cao 3 2  500  3 M2 for  ( their (c))   oe  their (b)  1  500  3 or M1 for   oe soi  their(b)   their (c) 3  their (b)  2 or = oe      A   500 

This question in 0607/42 Oct/Nov 2022

8 A E 8 cm NOT TO 16 cm 60° SCALE B 17 cm 32° 18 cm 55° D C The diagram shows a pentagon ABCDE and diagonals BD and BE. (a) (i) Calculate angle BCD. Angle BCD = … [1] (ii) Calculate BC. BC = … cm [3] (b) Calculate angle EBD. Angle EBD = … [3] (c) Calculate the area of the pentagon ABCDE. … cm2 [4] (d) Calculate the shortest distance from C to AE. … cm [4]

15 marks

Mark scheme: 8(a)(i) 93 1 8(a)(ii) 14.8 or 14.76… 3 18sin55 M2 for sin ( their ( i ) ) sin ( their ( i ) ) sin55 or M1 for = oe 18 BC 8(b) 59.7 or 59.65… 3 16 2 + 182 − 17 2 M2 for 2  16  18 or M1 for 172 = 162 + 182 – 2  16  18cos(…) 8(c) 250 or 249.7 to 250.3… 4 1 M1 for 8 16  sin60 oe 2 1 M1 for  16  18  sin theirEBD oe 2 1 M1 for  18  theirBC  sin32 oe 2 8(d) 21[.0] or 21.1 or 20.95 to 21.07 4 Triangle BXC where X is on AB extended and angle BXC = 90° M3 for 8 + their BCcos(180 – 60 – 32 – their(a)) or M2 for their BCcos(180 – 60 – 32 – their(a)) or M1 for angle XBC = 180 – 60 – 32 – their(a) If 0 scored, SC1 for recognition of correct shortest distance, e.g. AX OR M1 for [AC2 =] 82 + (theirBC)2 – 2  8  (theirBC) cos(60 + theirEBD + 32) oe M1 for theirAC theirBC = sin ( 60 + theirEBD + 32 ) sin BAC oe M1 for perp = sin ( 90 − theirBAC ) oe theirAC If 0 scored, SC1 for recognition of correct shortest distance

This question in 0607/43 Oct/Nov 2022

8 North B NOT TO 55 km SCALE A C 37 km 64 km 28° D The diagram shows four points A, B, C and D on level ground. B is due north of A and C is due east of A. (a) Calculate AB. AB = … km [3] (b) Calculate the obtuse angle ACD. Angle ACD = … [3] (c) Find the bearing of (i) D from A … [2] (ii) A from D. … [1] (d) Calculate the area of the quadrilateral ABCD. … km2 [3]

12 marks

Mark scheme: 8(a) 40.7 or 40.69… 3 M2 for 552 – 372 oe soi by 1656 or M1 for 552 = AB2 + 372 oe 8(b) 125.7 or 126 or 125.7… 3 64  sin28 M2 for [sin ACD =] 37 64 37 or M1 for = oe sin ACD sin28 8(c)(i) 116 or 116.2 to 116.3 2 M1 for 180 – 28 – their(b) soi by 26.29 or 26.3 8(c)(ii) 296 or 296.2 to 296.3 1 FT 180 + their (c)(i) 8(d) 1280 or 1277 to 1278 3 M1 for 0.5 × their(a) × 37 M1 for 0.5 × 64 × 37 × sin(180 – 28 – their(b))

This question in 0607/42 Feb/March 2023

9 C D NOT TO 118° SCALE 5 m 12 m 35° A B 16 m (a) B is due east of A. Find the bearing of A from C. … [2] (b) Calculate the area of triangle ABC. … m2 [2] (c) Calculate angle CAD. Angle CAD = … [4] (d) Calculate the length of the straight line BD. … m [3]

11 marks

Mark scheme: 9(a) 235 2 M1 for 180  55 or for 360 – 125 or for 270 – 35 or for 35 or 55 or 125 or 145 correctly indicated at C. 9(b) 55.1 or 55.06... 2 1 M1 for  12  16 sin 35 oe 2 9(c) 40.4 or 40.41... 4 5sin118 M2 for [sinC = ] 12 12 5 or M1 for  oe sin118 sinC M1 dep for 180 – 118 – their C dependent on sine rule used to find angle. 9(d) 15.5 or 15.51... nfww 3 M2 for 5 2  16 2  2 5 16cos(35 theirA) or M1 for 52 + 162 – 2 516cos(35 + theirA) A1 for 241 or 240.6 to 240.7...

This question in 0607/41 May/June 2023

8 A ship sails from port A at a constant speed of 18 km/h on a bearing of 040°. A motorboat sails in a straight line at a constant speed from port B to intercept the ship. Port B is 30 km due south of port A. The ship leaves port A at 08 20 and the motorboat leaves port B at 08 30. The motorboat intercepts the ship at point C at 09 50. North C NOT TO SCALE 40° A 30 km B (a) Find the speed of the motorboat. … km/h [5] (b) Find the bearing on which the motorboat sails. … [3]

8 marks

Mark scheme: 8(a) 40.2 or 40.17 to 40.19 5 B1 for 27 M2 for 30 2  (their 27) 2  2  30  ( their 27)  cos140 oe or M1 for 302 + (their27)2 – 2×30×(their27)×cos 140 M1 for their 53.57 ÷ their time from B to C 8(b) [0]18.9 or [0]18.89 to [0]18.91 3 their 27sin140 M2 for oe their 53.57 sin sin140 or M1 for  their 27 their 53.57

This question in 0607/42 May/June 2023

11 C NOT TO SCALE 8 cm 47° A B 10 cm (a) Calculate the area of triangle ABC. … cm2 [2] (b) Calculate the shortest distance from C to AB. … cm [3] (c) Show that BC = 7.41 cm correct to 2 decimal places. [3] (d) C NOT TO SCALE 8 cm O 47° A B 10 cm In triangle ABC, O is the centre of the circle that passes through A, B and C. Calculate the radius of this circle. … cm [4]

12 marks

Mark scheme: 11(a) 29.3 or 29.25... 2 1 M1 for 8 10sin47 2 11(b) 5.85 to 5.86 3 distance M2 for sin 47 = oe or for 8 2 their(a) oe 10 or M1 for recognition of shortest distance 11(c) 2 2 M2 M1 for 82 + 102 – 2  8  10cos47 oe 8 + 10 −2 8 10cos47 A1 for 54.9 or 54.88... 7.408... A1 11(d) 5.06 or 5.07 or 5.064 to 5.066 4 1 2  7.41 M3 for oe sin 47 or 1 2  7.41 M2 for sin47 = oe radius or M1 for angle BOC = 94 soi

This question in 0607/41 Oct/Nov 2023

7 A NOT TO SCALE 5 cm 7.63 cm B 7 cm 12° D C In triangle ACD, AB = 5 cm , AD = 7. 63 cm and BD = 7 cm . Angle BDC = 12° . (a) Show that angle ABD = 77.0° correct to 1 decimal place. [3] (b) Calculate the area of triangle ABD. … cm2 [2] (c) Calculate BC. … cm [4]

9 marks

Mark scheme: 7(a) 7 2 + 5 2 − 7.632 M2 M1 for 7.632 = 7 2 + 52 −2 7 5 cos ABD cos ABD = 2 7 5 ABD = 76.96... A1 no errors or omissions 7(b) 17.1 or 17.[0] or 17.04 to 17.05… 2 M1 for 0.5 5 7 sin77 7(c) 1.61 or 1.605 to 1.606… 4 7sin12 M3 for oe sin (77 − 12) BC 7 or M2 for = sin12 sin(77 − 12) or B1 for [ ACD =]65

This question in 0607/42 Oct/Nov 2023

5 B North NOT TO SCALE 123 m 154 m North A 27° 183 m C 106° D The diagram shows a field ABCD, with a straight path AC. The bearing of C from A is 122° . (a) Calculate the bearing of D from C. … [3] (b) Show that angle ABC = 81.9° correct to one decimal place. [3] (c) Find the total area of the field ABCD. … m2 [5]

11 marks

Mark scheme: 5(a) 255 cao 3 B2 for 75 correctly referenced at C or D OR B1 for angle ACD = 47 B1 for angle ACN(orth) = 58 or ACS(outh) = 122 5(b) 1232 + 154 2 − 1832 M2 M1 for 1832 = 1232 + 1542 – 2 × 123 × [cos] = 154 × cos[…] 2  123  154 81.87... A1 5(c) 15 200 or 15 150 to 15 161 5 183sin27 M2 for CD = sin106 CD 183 or M1 for = sin27 sin106 OR 183sin their 47 M2 for AD = sin106 AD 183 or M1 for = sin their 47 sin106 AND M1 for 1 × 154 × 123 × sin 81.9 2 M1 for 1 × 183 × their CD × sin their47 2 or 1 × 183 × their AD × sin 27 2 or 12 × their CD × their AD × sin 106

This question in 0607/42 Feb/March 2024

8 8 m A B 73.4° NOT TO 36° SCALE 7 m C D The diagram shows a shape ABDC formed from triangle ABC and a sector of a circle BCD, centre B. (a) Show that BC = 9.0 m , correct to 1 decimal place. [3] (b) Use the sine rule to find angle BCA. Angle BCA = … [3] (c) Find the area of triangle ABC. … m2 [2] (d) Find the area of the shaded region. … m2 [3] (e) Find the perimeter of the shape ABDC. … m [2]

13 marks

Mark scheme: 8(a) M2 or M1 for [ BC 2 ]  82  7 2 2 8 7 cos73.4 [ BC  ] 8 2  7 2  2 8 7 cos73.4 M1 for 81.[00…] 9.00[0…] A1 8(b) 8  sin73.4 M2 8 9 [sin...]  M1 for  oe 9 sin BCA sin73.4 58.4 or 58.41… B1 8(c) 26.8 or 26.83… 2 1 M1 for 7 8 sin73.4 oe 2 8(d) 1.64 or 1.641 to 1.642 3 36 M1 for areasector BCD  π 9 9 360 M1 for area triangle BCD  0.5 9 9 sin36 8(e) 29.7 or 29.65 to 29.66 NFWW 2 36 M1 for  2π  9 oe 360

This question in 0607/41 May/June 2024

7 NOT TO SCALE A B 12 cm 12 cm C The diagram shows a logo made from an isosceles triangle and two semicircles. The perimeter of the logo is 37 cm. (a) Show that the diameter of each semicircle is 4.14 cm, correct to 3 significant figures. [2] (b) Calculate angle ACB. Angle ACB = … [3] (c) Calculate the area of the logo. … cm2 [3] (d) A mathematically similar logo has an area of 35 cm2. Calculate the perimeter of this logo. … cm [3]

11 marks

Mark scheme: 7(a) πd = 37 – 12 – 12 oe M1 4.137 to 4.138... A1 7(b) 40.3 or 40.4 3 4.14 M2 for 2 × sin-1 oe or 40.33 to 40.36… 12 12 2  12 2   2  4.14  2 or cos ACB  2  12  12 4.14 or M1 for sin(...) = 12 2  12 2  2  12  12  cos ACB or  2  4.14  2  12 7(c) 60[.0] or 60.1 3 1 M1 for  12  12  sin( their b ) oe or 60.01 to 60.13 2 1  4.14  2 M1 for 2    oe 2  2  7(d) 28.2 or 28.3 3 35 or 28.23 to 28.26 M2 for 37  oe their c 35 their c or M1 for or their c 35 35  p  2 or   their c  37 

This question in 0607/43 May/June 2024

11 North North B NOT TO SCALE 72 km North 85 km A 102 km C A, B, and C are three ports. The bearing of B from A is 040c. (a) Show that angle ABC = 80.6c , correct to 1 decimal place. [3] (b) Find the bearing of B from C. … [2] (c) A ship leaves port A at 13 00. It sails directly towards C at a speed of 32 km/h. At point P the ship is at its shortest distance from B. Find the time when the ship reaches point P. Give your answer correct to the nearest minute. … [6]

11 marks

Mark scheme: 11(a) 72 2  85 2  102 2 M2 M1 for 1022 = 722 + 852 – 2 × 72 × 85 × cos[B] [cosB] = 2  72  85 80.57... A1 11(b) 319.4 2 B1 for 40.6 or 139.4 11(c) 14 17 6 85sin80.6 M2 for sin [A] = 102 72 2  102 2  85 2 or cos[A] = 2  72  102 85 102 or M1 for  sin A sin80.6 or 85 2  72 2  102 2  2  72  102cos A M1 for [AP =] 72 cos their A oe M1 for [time =] their AP ÷ 32 M1 adding their time to 13 00 If 0 scored, SC1 for showing BP on diagram with right angle correctly placed on AC

This question in 0607/43 May/June 2024

5 B NOT TO 26.3 cm SCALE 115° A C The area of triangle ABC is 262 cm 2. (a) Show that AC = 22.0 cm , correct to 1 decimal place. [2] (b) Find BC. BC = … cm [3] (c) Use the sine rule to find angle ABC. Angle ABC = … [3] (d) Find the length of the perpendicular line from A to the line BC. … cm [2]

10 marks

Mark scheme: 5(a) 1 M1  26.3  AC  sin115 = 262 2  262  2  A1 AC = = 21.98  = 22.0     26.3  sin115  5(b) 40.8 or 40.78 to 40.80… 3 M2 for  BC = 26.32 + 222 −2 26.3  22  cos115 or M1 for  BC 2 =  26.32 + 22 2 −2 26.3  22  cos115   5(c) 22  sin115 M2 22 their 40.8 sin ABC = oe M1 for = oe their 40.8 sin ABC sin115 29.2 or 29.3 or 29.23 to 29.26 B1 5(d) 12.8 to 12.9 2 M1 for 262 = 0.5 x their 40.8 Or for x = 26.3  sin(their 29.2)

This question in 0607/41 Oct/Nov 2024

9 B 78.2° NOT TO A SCALE 43.2° 110.9° D C 9.9 cm Triangle ABC is isosceles with AB = BC . (a) Show that AC = 13.5 cm correct to 3 significant figures. [3] (b) Calculate the length AB. … cm [3] (c) Find the area of ABCD. … cm2 [3]

9 marks

Mark scheme: 9(a) 9.9  sin110.9 M2 9.9 AC [ AC ] = M1 for = sin43.2 sin43.2 sin110.9 13.51… seen A1 9(b) 10.7 or 10.70 to 10.71… 3 0.5  13.5 M2 for [ AB ] = sin(0.5  78.2) 0.5  13.5 or M1 for sin(0.5  78.2) = AB OR 13.5  sin 12 (180 − 78.2) M2 for [ AB ] = sin78.2 13.5 AB or M1 for = sin78.2 sin 12 (180 − 78.2) OR 2 13.52 M2 for x = 2  (1 − cos78.2) or M1 for 13.52 = x 2 + x 2 −2 x x cos78.2 9(c) 85.2 to 85.4 3 M1 for 1 2  (their AB) 2  sin78.2 M1 for 12  13.5  9.9  sin(180 − (110.9 + 43.2))

This question in 0607/43 Oct/Nov 2024

14 C B 75° NOT TO SCALE 10 cm 65° A D 14 cm Triangle BCD is isosceles. (a) Find the area of triangle ABD. … cm2 [2] (b) Find the shortest distance from D to AB. … cm [3] (c) Find the perimeter of ABCD. … cm [6]

11 marks

Mark scheme: 14(a) 63.4 or 63.44… 2 1 M1 for  10  14  sin 65 2 14(b) 12.7 or 12.68 to 12.69 nfww 3 x M2 for sin65 = oe 14 1 or their area =  10  x 2 or M1 for recognising shortest distance 14(c) 44.2 or 44.18… to 44.23 6 2 2 M2 for  BD =  10 + 14 −2 10  14cos65 or M1 for 102 + 142 – 2 × 10 × 14 × cos65 their BD sin(theirBDC ) M2 for  BC =  oe sin75 BC their BD or M1 for = oe sin(theirBDC ) sin 75 OR M2 for (theirBD ) 2 + (theirBD ) 2 − 2(theirBD )(theirBD )cos30 or M1 for (theirBD)2 + (theirBD)2 – 2(theirBD)(theirBD)cos30 OR M2 for 2 × theirBD × cos 75 1 2 BC or M1 for cos75 = theirBD M1 for 10 + 14 + their BC + their CD dependent on at least trigonometry used for BC and CD

This question in 0607/42 Feb/March 2025

12 B 68° NOT TO 140 m SCALE 260 m A C The diagram shows a triangular field. AB = 140 m and BC = 260 m. Angle ABC = 68°. (a) Show that AC = 244.8 m correct to 1 decimal place. [3] (b) Giselle walks directly from A to C. Calculate the distance Giselle is from A when she is closest to B. … m [5]

8 marks

Mark scheme: 12(a) 2 2 M2 M1 for 140 2 + 260 2 −2 140  260  cos68 140 + 260 −2 140  260  cos68 244.80… A1 12(b) 24.3 to 24.4 5 M2 for complete explicit method for angle A or angle C or BN 260sin68 sin A = oe 244.8 140 2 + 244.82 − 260 2 or cos A = oe 2  140  244.8 140sin68 or sin C = oe 244.8 260 2 + 244.82 − 140 2 or cos C = oe 2  260  244.8 1  140  260  sin68 2 or BN = oe 1  244.8 2 or M1 for implicit method for angle A or angle C or BN sin A sin68 e.g. = oe 260 244.8 or 2602 = 1402 + 244.82 −2 140  244.8  cos A oe sin C sin68 or = oe 140 244.8 or 1402 = 2602 + 244.82 −2 260  244.8  cosC oe 1 1 or  BN  244.8 = 140  260  sin68 oe 2 2 AND M2 dep for 140 cos their A oe or 244.8 – 260 cos their C oe or 140 2 – theirBN 2 oe dependent on at least M1 above or M1 for clear indication of perpendicular from B to AC

This question in 0607/41 May/June 2025

15 B 30° NOT TO SCALE 12.4 cm A C The area of triangle ABC is 74.4 cm2. AB = 12.4 cm and angle ABC = 30°. (a) Show that BC = 24 cm. [2] (b) Find AC. AC = … cm [3] (c) Find obtuse angle CAB. Angle CAB = … [3] (d) NOT TO Y SCALE X Z Triangle XYZ is similar to triangle ABC. The area of triangle XYZ is 62 cm2. Find YZ. YZ = … cm [3]

11 marks

Mark scheme: 15(a) 74.4 = 0.5  12.4  BC  sin30 oe M1 Use of 24 scores M0 74.4 A1 74.4 [=24] oe oe e.g. 0.5  12.4  sin30 3.1 15(b) 14.6 or 14.63 to 14.64 3 2 2 M2 for 12.4 + 24 −2 12.4  24  cos30 or M1 for [ AC 2 ] = 12.4 2 + 24 2 −2 12.4  24  cos30 15(c) 124.7 to 125.3 3 their 14.6 must come from trig 12.4 2 + (their14.6) 2 − 24 2 M2 for [cos A = ] 2  12.4  their14.6 or M1 for 24 2 = (their14.6) 2 + 12.4 2 −2 their14.6  12.4cos A OR 24  sin30 M2 for [sin A] = oe their 14.6 sin A sin30 or M1 for = oe 24 their 14.6 OR 74.4 M2 for [sin A] = 1  12.4  their 14.6 2 1 or M1 for  12.4  their 14.6sin A = 74.4 2 15(d) 21.9 or 21.90 to 21.92 or 4 30 oe 3 62 74.4 M2 for 24  or 24  74.4 62 62 74.4  YZ 2 62 oe or M1 for or or for  = 74.4 62  24  74.4

This question in 0607/42 May/June 2025

15 NOT TO 75° SCALE y cm 40° 12 cm Calculate the value of y. y = … [4]

4 marks

Mark scheme: 15 11.3 or 11.25 to 11.26 4 B1 for 65 seen 12sin their 65 M2 for sin75 sin75 sin their 65 or M1 for = oe 12 y

This question in 0607/43 May/June 2025

9 North North 178 m A B NOT TO SCALE C ABC is a field on level ground. B is 178 m due east of A. The bearing of C from A is 158˚. The bearing of C from B is 237˚. (a) Calculate the length of CB. … m [4] (b) Calculate the area of the field ABC. … m2 [2]

6 marks

Mark scheme: 9(a) 168 or 168.1… 4 B1 for 33 or 68 or 79 correctly identified 178sin(their 68) M2 for CB = sin(their 79) CB 178 or M1 for = sin(their 68) sin(their 79) 9(b) 8140 or 8150 or 8143 to 8149.6… 2 1 M1 for  178  their168  sin(their 33) oe 2

This question in 0607/41 Oct/Nov 2025

14 B 47° NOT TO SCALE D 82° 12° 12.3 cm 9.8 cm A C ABC and ADC are triangles. AD = 12.3 cm and CD = 9.8 cm. Angle ABC = 47°, angle ADC = 82° and angle BAD = 12°. (a) Find AC. AC = … cm [3] (b) Show that angle DAC = 41.6°, correct to 1 decimal place. [3] (c) BC = 16.1 cm. Find the area of the shaded quadrilateral ABCD. … cm2 [4]

10 marks

Mark scheme: 14(a) 14.6 or 14.62…nfww 3 M2 for AC = 12.32 + 9.82 −2 12.3  9.8  cos82 or M1 for AC 2 = 12.32 + 9.82 −2 12.3  9.8  cos82 Use of angle DAC = 41.6o is 0 marks unless fully correct method to derive it is shown in this part 14(b) 9.8  sin82 M2 sin DAC sin82 sin DAC = oe M1 for = oe their 14.6 9.8 their14.6 41.58 to 41.59 A1 No errors seen 14(c) 55.8 to 56.1 4 B1 for 79.4… M1 for 0.5  their 14.6 16.1 their sin79.4 oe M1 for 0.5 12.3  9.8  sin82 oe

This question in 0607/42 Oct/Nov 2025