Non-right-angled triangles — Paper 4 · IGCSE Mathematics 0580

4 For Examiner's Use North NOT TO North SCALE A 126° B North 250 m 23° P The diagram shows three straight horizontal roads in a town, connecting points P, A and B. PB =250 m, angle APB = 23° and angle BAP = 126°. (a) Calculate the length of the road AB. Answer(a) AB = m [3] (b) The bearing of A from P is 303°. Find the bearing of (i) B from P, Answer(b)(i) [1] (ii) A from B. Answer(b)(ii) [2]

6 marks

Mark scheme: 250 AB 250 4 (a) ( AB = ) × sin 23 (s.o.i by 120…) M2 M1 for = o.e. (implicit) sin 126 sin 23 sin 126 121 (120.7 to 121) (m) c.a.o. www3 A1 (b) (i) 280 B1 (ii) (0)69 c.a.o. B2 SC1 for answer 249 [6]

This question in 0580/41 May/June 2009

5 Examiner's D Use 30° C NOT TO SCALE 24 cm 40° 40° A 26 cm B ABCD is a quadrilateral and BD is a diagonal. AB = 26 cm, BD = 24 cm, angle ABD = 40°, angle CBD = 40° and angle CDB = 30°. (a) Calculate the area of triangle ABD. Answer(a) cm2 [2] (b) Calculate the length of AD. Answer(b) cm [4] (c) Calculate the length of BC. Answer(c) cm [4] (d) Calculate the shortest distance from the point C to the line BD. Answer(d) cm [2]

12 marks

Mark scheme: 5 (a) 200.5… to 201 www 2 2 M1 for 0.5 × 24 × 26 sin 40 oe A1 (b) 17.2 (0….) www 4 4 M2 for 262 + 242 – 2 × 26 × 24 cos 40 26 2 + 24 2 − BD 2 or M1 for cos 40 = 2 × 24 × 26 A2 or A1 for 295.976.. (c) 12.8 (12.77…) www 4 4 B1 for Angle C = 110 soi accept on diagram 24 sin 30 M2 for ( BC ) = oe or sin 110 sin 110 sin 30 M1 = oe i.e. a correct implicit 24 BC statement soi A1 (d) 8.208 to 8.230 www 2 2 M1 for their (c) × sin40 oe IGCSE – May/June 2010 0580 41

This question in 0580/41 May/June 2010

5 Examiner's North Use A NOT TO SCALE 180 km 115 km 90 km T H 30° 70° R The diagram shows some straight line distances between Auckland (A), Hamilton (H), Tauranga (T) and Rotorua (R). AT = 180 km, AH = 115 km and HT = 90 km. (a) Calculate angle HAT. Show that this rounds to 25.0°, correct to 3 significant figures. Answer(a) [4] (b) The bearing of H from A is 150°. Find the bearing of (i) T from A, Answer(b)(i) [1] (ii) A from T. Answer(b)(ii) [1] For (c) Calculate how far T is east of A. Examiner's Use Answer(c) km [3] (d) Angle THR = 30° and angle HRT = 70°. Calculate the distance TR. Answer(d) km [3] (e) On a map the distance representing HT is 4.5cm. The scale of the map is 1 : n. Calculate the value of n. Answer(e) n = [2]

14 marks

Mark scheme: 180 + 115 90 5 (a) (cos) M2 M1 for correct implicit expression 902 = …… 2 × 180 × 115 24.98 – 24.99 A2 A1 for (cos) = 0.9064… (b) (i) 125(.0….) ft 1ft ft 150 – their (a) (ii) 305(.0….) ft 1ft ft 180 + their (b)(i) (c) 180sin (54.98 to 55) M2 B1 for 54.98 to 55 or 35 to 35.02 soi in correct or 180cos (35 to 35.02) oe position. or 180sin (360 – their (b)(ii)) Provided either angle is acute or 180cos(their (b)(i) – 90) oe 147(.4….) cao www 3 A1 90 sin 30 TR 90 (d) M2 M1 for = or other correct sin 70 sin 30 sin 70 implicit equation 47.9 (47.88 – 47.89) cao www 3 A1 (e) 2 000 000 oe 2 Allow 1 : 2 000 000 as answer. SC1 figs 2 in answer which could be a ratio. 4 3

This question in 0580/42 May/June 2010

6 For L 5480 km Examiner's Use D NOT TO 165° 3300 km SCALE C The diagram shows the positions of London (L), Dubai (D) and Colombo (C). (a) (i) Show that LC is 8710 km correct to the nearest kilometre. Answer(a)(i) [4] (ii) Calculate the angle CLD. Answer(a)(ii) Angle CLD = [3] (b) A plane flies from London to Dubai and then to Colombo. For It leaves London at 01 50 and the total journey takes 13 hours and 45 minutes. Examiner's The local time in Colombo is 7 hours ahead of London. Use Find the arrival time in Colombo. Answer(b) [2] (c) Another plane flies the 8710 km directly from London to Colombo at an average speed of 800 km/h. How much longer did the plane in part (b) take to travel from London to Colombo? Give your answer in hours and minutes, correct to the nearest minute. Answer(c) h min [4]

13 marks

Mark scheme: 6 (a) (i) 54802 + 33002 – 2 × 5480 × 3300 M2 (75 856 005) M1 for implicit version × cos165 8709.5.. E2 If E0, A1 for 75800000 to 75900000 sin 165 sin L sin 165 (ii) (sinL =) × 3300 M2 M1 for = oe (allow 8709.5.) 8710 3300 8710 (0.09806…) Could use cosine rule using 8710 or better – M2 for explicit form or M1 for implicit form (allow 5.6 to 5.63 for A mark) 5.6 (5.62 to 5.63) A1 www3 (b) 22 35 or 10 35 pm 2 Accept 22 35 pm B1 for 15 35 or 3 35 pm seen or answers 22h 35 mins or (0)8 35(am) or 10 35(am) (c) 8710 ÷ 800 M1 10.88 to 10.9 with no conversion to A1 Implied by correct final ans 2hrs 52 mins if not h/min shown or 10 (hrs) 52 (mins) to 10 (hrs) 54 (mins) oe 13 hrs 45 mins – their time in hrs and M1 Dep on first M1 mins oe e.g. 13 hrs 45mins – 11 hrs 29 mins or 13.75 – their decimal time and a or 13.75 – 10.9 then 2hrs 51 mins correct conversion to hrs and mins or minutes 2 hr 52 mins cao A1 www4 (2 hrs 51.75 mins) IGCSE – October/November 2010 0580 42

This question in 0580/42 Oct/Nov 2010

2 For R Examiner's 4 km Q Use NOT TO SCALE 7 km 4.5 km 85° S 40° P The diagram shows five straight roads. PQ = 4.5 km, QR = 4 km and PR = 7 km. Angle RPS = 40° and angle PSR = 85°. (a) Calculate angle PQR and show that it rounds to 110.7°. Answer(a) [4] (b) Calculate the length of the road RS and show that it rounds to 4.52 km. Answer(b) [3] (c) Calculate the area of the quadrilateral PQRS. [Use the value of 110.7° for angle PQR and the value of 4.52 km for RS.] Answer(c) km2 [5]

12 marks

Mark scheme: 4 + 5.4 − 7 2 (a) (cosQ =) o.e. M2 M1 for 72 = 42 + 4.52 – 2 × 4 × 4.5 × cos(Q) 2 × 4 × 5.4 110.74…. E2 If E0 then A1 for – 0.354(1….) 7 sin 40 RS 7 (b) ( RS = ) M2 M1 for = o.e. sin 85 sin 40 sin 85 4.516 … E1 Can be implied by second M (c) Angle R = 55° B1 (May be seen on diagram) 0.5 × 7 × 4.52 × sin(their 55) o.e. M1 (12.95 – 13.0) their 55 is (180 – 40 – 85) 0.5 × 4 × 4.5 × sin110.7 o.e. M1 (8.418 – 8.42) (s = 7.75) Triangle PRS + Triangle PQR M1 Dependent on M1, M1 21.4 (21.36 – 21.42) A1 www 5 IGCSE – October/November 2010 0580 43 3 (a) 5x2 – x or x(5x – 1) 2 M1 for x2 + 3x or 4x2 – 4x correct (b) 27x9 2 B1 for 27 or for x9 (c) (i) 7x7(1 + 2x7) 2 M1 for any correct partially factorised expression or 7x7(1 + ...) (ii) (y + w)(x + 2a) 2 M1 for x(y + w) + 2a(y + w) or y(x + 2a) + w(x + 2a) (iii) (2x + 7)(2x – 7) 1 2 ( )

This question in 0580/43 Oct/Nov 2010

3 For Examiner's NOT TO Use SCALE x cm 2x cm (x + 5) cm The diagram shows a square of side (x + 5) cm and a rectangle which measures 2x cm by x cm. The area of the square is 1 cm2 more than the area of the rectangle. (a) Show that x2 – 10x – 24 = 0 . Answer(a) [3] (b) Find the value of x. For Examiner's Use Answer(b) x = [3] (c) Calculate the acute angle between the diagonals of the rectangle. Answer(c) [3]

9 marks

Mark scheme: 3 (a) (x + 5)2 – 2x2 = 1 oe M1 Equiv means equation in the three parts, allowing (x + 5)2 expanded (x + 5)2 = x2 + 10x + 25 or B1 x2 + 5x + 5x + 25 x2 + 10x + 25 – 2x2 = 1 E1 For final line reached without any errors or 0 = x2 – 10x – 24 omissions after any previous line with (x + 5)2 expanded (b) 12 3 M2 for (x – 12)(x + 2) or full correct expression from formula. Allow SC1 for ( x + a )( x + b ) and ab = – 24 or a + b = – 10 then SC1 ft (dependent on quadratic factors or two roots from formula) for correct selection of +ve root, if only one +ve. Answer of 12 and –2 scores M2 only (c) 53.1 to 53.2 www 3 3 M2 for 2 × tan −1 ( 12 ) o.e. i.e. any complete method or M1 for tan = 12 o.e. i.e. any correct method leading to any angle in diagram (expressions can be implicit and bod which angle is being worked out) (Implied by 26.56 to 26.57 or 26.6, 63.43 to 63.44 or 63.4, 126.8 to 126.9) 53 or 127 without working score 0 6 2 + 8 2 9 2

This question in 0580/43 May/June 2011

4 For A Examiner's Use NOT TO 8 cm 6 cm SCALE O B C 9 cm The circle, centre O, passes through the points A, B and C. In the triangle ABC, AB = 8 cm, BC = 9 cm and CA = 6 cm. (a) Calculate angle BAC and show that it rounds to 78.6°, correct to 1 decimal place. Answer(a) [4] (b) M is the midpoint of BC. (i) Find angle BOM. Answer(b)(i) Angle BOM = [1] (ii) Calculate the radius of the circle and show that it rounds to 4.59 cm, correct to 3 significant For figures. Examiner's Use Answer(b)(ii) [3] (c) Calculate the area of the triangle ABC as a percentage of the area of the circle. Answer(c) % [4]

12 marks

Mark scheme: 6 + 8 − 9 4 (a) ( cos( A)) = M2 M1 for correct implicit equation with cosA 8.6.2 78.58… www 4 A2 A1 for 0.1979 to 0.198 (this implies M2) (b) (i) 78.6 1 Allow 78.58… 5.4 5.4 (ii) r = oe M2 (M1 for sin(78.6) = ) sin( 786.) r Allow 78.58… or their angle BOM for M2 or M1 4.590 to 4.591 cao www 3 A1 (c) 35.5 (35.48 to 35.57…) cao www 4 4 M1 Area triangle = 0.5 × 6 × 8 × sin (78.6) oe Allow 78.58.. (23.52..) M1 Circle = π × .4 59 2 Allow 4.590 to 4.591 (66.15 to 66.22…) M1 (dependent) % = triangle / circle × 100 Dependent on first 2 M’s IGCSE – May/June 2011 0580 43

This question in 0580/43 May/June 2011

6 For Examiner′s A Use B 30° 52° E 15.7 cm NOT TO SCALE 16.5 cm C 23.4 cm D In the diagram, BCD is a straight line and ABDE is a quadrilateral. Angle BAC = 90°, angle ABC = 30° and angle CAE = 52°. AC = 15.7 cm, CE = 16.5 cm and CD = 23.4 cm. (a) Calculate BC. Answer(a) BC = … cm [3] (b) Use the sine rule to calculate angle AEC. Show that it rounds to 48.57°, correct to 2 decimal places. Answer(b) [3] (c) (i) Show that angle ECD = 40.6°, correct to 1 decimal place. For Examiner′s Use Answer(c)(i) [2] (ii) Calculate DE. Answer(c)(ii) DE = … cm [4] (d) Calculate the area of the quadrilateral ABDE. Answer(d) … cm2 [4] _____________________________________________________________________________________

16 marks

Mark scheme: 15 7. 6 (a) 31.4 3 M2 for sin 30 or M1 for correct implicit statement 15 7. × sin 52 (b) [sinE =] M2 M1 for correct implicit statement 16 5. 48.573… A1 (c) (i) [∠ACE = ] 180 – 52 – 48.57 M1 [= 79.43] [∠ECD = ] 40.57… A1 (ii) 15.3 or 15.27 to 15.281 www 4 M2 for [(DE)2 =] 16.52 + 23.42 – 2 × 16.5 × 23.4cos(40.6 or 40.57) or M1 for full correct implicit statement A1 for 233 to 234 (d) 466 or 466.34 to 466.5 4 M1 for 0.5 × 15.7 × their 31.4 sin(90 – 30) oe M1 for 0.5 × 15.7 × 16.5 sin(128 – their 48.6 or 48.57) oe M1 for 0.5 × 16.5 × 23.4 sin (40.6 or 40.57) oe IGCSE – May/June 2013 0580 41 Qu. Answer Mark Part marks x + 2 9

This question in 0580/41 May/June 2013

3 C 90 m D NOT TO 80 m SCALE 95 m 49° A 55° B The diagram shows a quadrilateral ABCD. Angle BAD = 49° and angle ABD = 55°. BD = 80 m, BC = 95 m and CD = 90 m. (a) Use the sine rule to calculate the length of AD. Answer(a) AD = … m [3] (b) Use the cosine rule to calculate angle BCD. Answer(b) Angle BCD = … [4] (c) Calculate the area of the quadrilateral ABCD. Answer(c) … m2 [3] (d) The quadrilateral represents a fi eld. Corn seeds are sown across the whole fi eld at a cost of $3250 per hectare. Calculate the cost of the corn seeds used. 1 hectare = 10 000 m2 Answer(d) $ … [3] __________________________________________________________________________________________

13 marks

Mark scheme: 80 sin 55 80 x 3 (a) 86.8 or 86.83…. 3 M2 for or M1 for = sin 49 sin 49 sin 55 oe 95 2 + 90 2 − 80 2 (b) 51.2 or 51.15 to 51.16 4 M2 for [cos =] oe .2 95. 90 or M1 for 80 2 = 95 2 + 90 2 − .290.95. cos BCD 10 725 143 A1 for or etc. or 0.627….. 17 100 228 (c) 6700 or 6698 to 6703 3 M2 for 0.5 × 80 × their(a) × sin(180-55-49) oe [3368 – 3370…] [If AB used then AB= 102.8 to 103] + 0.5 × 90 × 95 × sin(their(b)) oe [3329 – 3332] or M1 for one of these triangle area methods oe (d) 2180 or 2176 to 2179 3FT FT their (c) × 0.325 correctly evaluated to 3 3250 sf or better M2 for their (c) × 10 000 or SC1 FT for figs 218 or figs 2176 to 2179 IGCSE – May/June 2014 0580 42 Qu Answers Mark Part Marks

This question in 0580/42 May/June 2014

8 North NOT TO SCALE P 58 km L North 74 km Q A ship sails from port P to port Q. Q is 74 km from P on a bearing of 142°. A lighthouse, L, is 58 km from P on a bearing of 110°. (a) Show that the distance LQ is 39.5 km correct to 1 decimal place. Answer(a) [5] (b) Use the sine rule to calculate angle PQL. Answer(b) Angle PQL = … [3] (c) Find the bearing of (i) P from Q, Answer(c)(i) … [2] (ii) L from Q. Answer(c)(ii) … [1] (d) The ship takes 2 hours and 15 minutes to sail the 74 km from P to Q. Calculate the average speed in knots. [1 knot = 1.85 km/h] Answer(d) … knots [3] (e) Calculate the shortest distance from the lighthouse to the path of the ship. Answer(e) … km [3] __________________________________________________________________________________________

17 marks

Mark scheme: 8 (a) Angle LPQ = 32 soi B1 582 + 742 – 2 × 58 × M2 M1 for correct implicit cos rule 74 cos their P A2 A1 for 1560.3 to 1560.4 or 1560 39.50[1...] 58 sin their P sin PQL sin( their P ) (b) sin PQL = oe M2 M1 for = oe 395. 58 395. 51.1 or 51.08 to 51.09 B1 (c) (i) 322 2 M1 for 180 + 142 oe (ii) [0]13[.1] or 13.08 to 13.09 1FT FT their (b) – 38 (d) 17.8 or 17.77 to 17.78 3 M1 for 74 ÷ 2.25 oe soi by 32.888… to 3 sf or better M1 for dist or speed ÷ 1.85 (e) 30.7 or 30.73 to 30.74… 3 M2 for 58 sin their P oe or 39.5 sin their (b) x or M1 for = sin their P oe 58 x or = sin their (b) 395.

This question in 0580/42 Oct/Nov 2014

5 (a) X NOT TO 5.4 cm SCALE 62° Y 16 cm Z Show that the area of triangle XYZ is 38.1 cm2, correct to 1 decimal place. Answer(a) [2] (b) NOT TO 48° SCALE 6.7 cm x° 8.4 cm Calculate the value of x. Answer(b) x = … [4] (c) North A NOT TO SCALE P B Ship A is 180 kilometres from port P on a bearing of 063°. Ship B is 245 kilometres from P on a bearing of 146°. Calculate AB, the distance between the two ships. Answer(c) … km [5] __________________________________________________________________________________________

11 marks

Mark scheme: 5 (a) 1 × 16 × 4.5 × sin 62 oe M1 2 A1 38.14... 7.6 × sin 48 (b) 95.6 or 95.64 to 95.65 4 M2 for 4.8 or M1 for implicit form and M1dep for 180 − 48 − their 36.4 (c) 286 or 285.7 to 285.8 5 B1 for [Angle APB=] 83° M2 for 180 2 + 245 2 − 2 × 180 × 245 × cos their 83 or M1 for implicit form and A1 for [AB2 =] 81676[.1...] After 0 scored, SC2 for ans 406.87 to 406.88 or 406.9 or 407 if 146° used in cos rule Or SC1 for 180 2 + 245 2 − 2 × 180 × 245 × cos 146 4

This question in 0580/42 Feb/March 2015

6 The diagram shows the positions of two ships, A and B, and a coastguard station, C. North A B NOT TO 95.5 km SCALE 83.1 km 101° C (a) Calculate the distance, AB, between the two ships. Show that it rounds to 138 km, correct to the nearest kilometre. Answer(a) [4] (b) The bearing of the coastguard station C from ship A is 146°. Calculate the bearing of ship B from ship A. Answer(b) … [4] (c) L North 46.2 km NOT TO SCALE 21° 45° B At noon, a lighthouse, L, is 46.2 km from ship B on the bearing 021°. Ship B sails north west. Calculate the distance ship B must sail from its position at noon to be at its closest distance to the lighthouse. Answer(c) … km [2] __________________________________________________________________________________________

10 marks

Mark scheme: 955. + 831. AB 6 (a) 95.52 + 83.12 – 2 × 95.5 × 83.1 × M2 M1 for cos 101 = 2 × 955. × 831. cos 101 138.0… A2 A1 for 19054.[…] also implies M2 (b) 110 or 109.7 to 109.8 4 B3 for 36.2 or 36.20 to 36.24[1..] 831. × sin 101 or M2 for [sin =] oe 138[.0..] or M1 for correct implicit version After M0, SC1 for angle ABC = 42.76 to 42.8 (c) 18.8 or 18.79[…] 2 M1 for 46.2 × cos(45 + 21) oe After M0, SC1 for answer 42.2 or 42.20 to 42.21

This question in 0580/42 May/June 2015

5 K 680 km 65° 40° D North NOT TO SCALE 2380 km M 1560 km C The diagram shows some distances between Mumbai (M), Kathmandu (K), Dhaka (D) and Colombo (C). (a) Angle CKD = 65°. Use the cosine rule to calculate the distance CD. Answer(a) CD = … km [4] (b) Angle MKC = 40°. Use the sine rule to calculate the acute angle KMC. Answer(b) Angle KMC = … [3] (c) The bearing of K from M is 050°. Find the bearing of M from C. Answer(c) … [2] (d) A plane from Colombo to Mumbai leaves at 21 15 and the journey takes 2 hours 24 minutes. (i) Find the time the plane arrives at Mumbai. Answer(d)(i) … [1] (ii) Calculate the average speed of the plane. Answer(d)(ii) … km/h [2] __________________________________________________________________________________________

12 marks

Mark scheme: 5 (a) 2180 or 2181…. nfww 4 M2 for 680 2 + 2380 2 − 2 × 680 × 2380 cos 65 oe or M1 for correct implicit cosine formula A1 for 4 760 000 or 4 758 000 to 4 759 000 (b) 78.7 or 78.71… 3 2380 sin 40 M2 for 1560 or 1560 2380 M1 for = oe sin 40 sin M (c) 309 or 308.7… 2FT FT 230 + their (b) B1FT 50 + their (b) for 129 or 128.7… [i.e. for C from M] (d) (i) 23 39 oe 1 (ii) 650 2 M1 for 1560 ÷ journey time

This question in 0580/43 Oct/Nov 2015

10 (a) r cm NOT TO SCALE w° r cm The area of this sector is r2 square centimetres. Find the value of w. w = … [3] (b) NOT TO r cm SCALE x° r cm 7r r The perimeter of this sector is 2r + centimetres. 10 Find the value of x. x = … [3] (c) y° q cm q cm NOT TO SCALE cm The perimeter of the isosceles triangle is 2q + q 3 centimetres. Find the value of y. y = … [4]

10 marks

Mark scheme: r 10 (a) 115 or 114.5 to 114.6 3 M2 for 2 or better π r 360 w 2 2 or M1 for × π × r = r 360 x 7π r (b) 126 3 M2 for × 2π r [ + 2 r ] = [2 r + ] or better 360 10 x or M1 for × 2π r 360 (c) 120 4 B3 for 2y = 60 or x (base angle) = 30 OR = M3 for cos x or sin y 3 1 oe or cos y = − 2 2 ( 2 ) oe y q 3 or M2 for cos x or sin = ( 2 ) 2 q 2 q 2 + q 2 − q 3 ( ) or [cos y] = oe 2 × q × q or M1 for  2  2 2 q 3 = q + q − 2 × q × q cos y oe  ( )    2 2  1  2 After M0, SC1 for [ h = ]q − q 3  or for  2  q replaced by 1, 2, 4, etc.

This question in 0580/41 Oct/Nov 2016

3 D 180 m North C NOT TO 85° SCALE 240 m A 50° B The diagram shows a field, ABCD. AD = 180 m and AC = 240 m. Angle ABC = 50° and angle ACB = 85°. (a) Use the sine rule to calculate AB. AB = … m [3] (b) The area of triangle ACD = 12 000 m2. Show that angle CAD = 33.75°, correct to 2 decimal places. [3] (c) Calculate BD. BD = … m [5] (d) The bearing of D from A is 030°. Find the bearing of (i) B from A, … [1] (ii) A from B. … [2]

14 marks

Mark scheme: 240sin85 sin50 sin85 3 (a) M2 or M1 for = oe sin50 240 AB 312 or 312.1 …. B1 1 (b) × 180 × 240 × sin A = 12000 M1 2 24000 33.748 to 33.749 A2 A1 for sin = or better or 0.555 or 0.556 43200 or 0.5 or 0.5555 to 0.5556 (c) 328 or 328.3 to 328.5 5 B1 for [angle A =] 78.75 seen M2 for 180 2 + (their AB ) 2 −×2 180 × their AB × cos78.75 180 2 + (theirAB ) 2 − x 2 or M1 for cos78.75 = 2 × 180 × (theirAB ) A1 for 107 800 to 107 900 (d) (i) 108.75 or 108.7 or 108.8 1 (ii) 288.75 or 288.7 or 288.8 2FT FT 180 + their (d)(i) M1 for 180 + their (d)(i) or 360 – (180 – their(d)(i))

This question in 0580/42 Oct/Nov 2016

6 D C NOT TO SCALE 40 km B L 61.1 km North 92.1 km A The diagram shows the position of a port, A, and a lighthouse, L. The circle, centre L and radius 40 km, shows the region where the light from the lighthouse can be seen. The straight line, ABCD, represents the course taken by a ship after leaving the port. When the ship reaches position B it is due west of the lighthouse. AL = 92.1 km, AB = 61.1 km and BL = 40 km. (a) Use the cosine rule to show that angle ABL = 130.1°, correct to 1 decimal place. [4] (b) Calculate the bearing of the lighthouse, L, from the port, A. … [4] (c) The ship sails at a speed of 28 km/h. Calculate the length of time for which the light from the lighthouse can be seen from the ship. Give your answer correct to the nearest minute. … h … min [5]

13 marks

Mark scheme: 6 (a) 40 2 + 61.12 − 92.12 [cosABL =] 2 × 40 × 61.1 M2 M1 for correct implicit version 130.11… 7873 A2 A1 for [cosABL =] –0.644… or – or 12220 3149.2 – 4888 40sin130.1 61.1sin130.1 (b) [0]59.5 or 59.50 to 59.511 4 M2 for or 92.1 92.1 or sin A sin130.1 sin L sin130.1 M1 for = or = 40 92.1 61.1 92.1 and A1 for 19.39 to 19.4… or 30.48 to 30.49…

This question in 0580/43 Oct/Nov 2016

8 (a) In triangle TXZ, TX = 12.5 cm and angle TZX = 37°. Y is a point on the line XZ such that TY = 9.9 cm, angle XTY = 23° and angle TYZ = 72°. T 23° NOT TO SCALE 12.5 cm 9.9 cm 72° 37° X Y Z (i) Calculate XY. XY = … cm [4] (ii) Calculate TZ. TZ = … cm [3] (b) The diagram shows a shape made up of three identical sectors of a circle, each with sector angle 65°. The perimeter of the shape is 20.5 cm. 65° 65° NOT TO SCALE 65° Calculate the radius of the circle. … cm [4]

11 marks

Mark scheme: 8 (a) (i) 5.14 or 5.135 to 5.142 nfww 4 M2 for [ XY 2 = ] 12.52 + 9.92 − 2 × 12.5 × 9.9 × cos 23 or M1 for implicit version A1 for 26.4 to 26.5 OR B1 for [XYT = ] 108 or [TXY = ] 49 12.5sin 23 M2 for oe sin(180 − 72) sin(180 − 72) sin23 or M1 for = oe 12.5 XY

This question in 0580/42 Feb/March 2017

8 P NOT TO SCALE 9 cm D C N 6 cm M A 8 cm B The diagram shows a pyramid on a rectangular base ABCD. AC and BD intersect at M and P is vertically above M. AB = 8 cm, BC = 6 cm and PM = 9 cm. (a) N is the midpoint of BC. Calculate angle PNM. Angle PNM = … [2] (b) Show that BM = 5 cm. [1] (c) Calculate the angle between the edge PB and the base ABCD. … [2] (d) A point X is on PC so that PX = 7.5 cm. Calculate BX. BX = … cm [6]

11 marks

Mark scheme: 8(a) 66[.0] or 66.03 to 66.04 2 9 M1 for tan = oe 4 8(b) 2 2 1 2 2 M1 Any alternative method must be full and complete and 3 + 4 or 6 + 8 result in exactly 5 2 8(c) 60.9 or 60.94 to 60.95 2 9 M1 for tan = oe 5 8(d) 5.83 or 5.84 or 5.827 to 5.840 6 2 2 2 2 M1 for [PB or PC = ] 9 + 5 or [XC =] 9 + 5 – 7.5 3 M1 for angle BPX = 2 × invsin oe their PB B1 for [ PB or PC =] 106 = 10.29 to 10.30 or XC = 2.79 to 2.8[0] or angle BPX = 33.9 or 33.86 to 33.90… M2 for ( their PB ) 2 + 7.5 2 − 2 × their PB × 7.5 × cos ( their BPX ) oe or M1 for correct implicit equation

This question in 0580/42 May/June 2017

10 B 8.5 cm 12.5 cm NOT TO 60° x cm A C SCALE 46° 76° 58° D The diagram shows a quadrilateral ABCD. (a) The length of AC is x cm. Use the cosine rule in triangle ABC to show that 2x2 – 17x – 168 = 0. [4] (b) Solve the equation 2x2 – 17x – 168 = 0. Show all your working and give your answers correct to 2 decimal places. x = … or x = … [4] (c) Use the sine rule to calculate the length of CD. CD = … cm [3] (d) Calculate the area of the quadrilateral ABCD. … cm2 [3]

14 marks

Mark scheme: 10(a) M2 x 2 + 8.5 2 − 12.5 2 12.52 = x2 + 8.52 – 2 × x × 8.5cos60 oe isw M1 for cos60 = 2 × x × 8.5 156.25 = x2 + 72.25 – 8.5x A1 or better 2x2 – 17x – 168 = 0 A1 with no errors or omissions 10(b) 2 2 2 [ −− ]17 ± ([ − ]17) − 4 ( 2 )( −168 ) B1 for ([ − ]17) − 4(2)( −168) or better seen 2 × 2 p + or − q and if in form r B1 for p = [− −] 17 and r = 2 × 2 14.35, –5.85 final answers 1, 1 SC1 for 14.352 to 14.353 and –5.853 to –5.852 seen or 14.3 or 14.4 and –5.8 or –5.9 as final answers or −14.35 and 5.85 as final answers or 14.35 and –5.85 seen in working 10(c) 12.2 or 12.17… nfww 3 their 14.35 × sin46 M2 for sin58 sin 46 sin58 or M1 for = CD their14.35 10(d) 138 or 137.5 to 137.8 nfww 3 M1 for 0.5 × their 14.35 × 8.5sin60 M1 for 0.5 × their 14.35 × their12.2 × sin76

This question in 0580/41 Oct/Nov 2017

3 B North NOT TO 70 m SCALE 100 m C A 40° 110 m D The diagram shows a field ABCD. (a) Calculate the area of the field ABCD. … m2 [3] (b) Calculate the perimeter of the field ABCD. … m [5] (c) Calculate the shortest distance from A to CD. … m [2] (d) B is due north of A. Find the bearing of C from B. … [3]

13 marks

Mark scheme: 3(a) 7040 or 7035. … 3 1 M1 for × 100 × 70 oe 2 1 M1 for × 100 × 110 × sin 40 oe 2 3(b) 374 or 375 or 374.4 to 374.5…. 5 2 2 M2 for 110 + 100 −×2 110 × 100 × cos40 oe or M1 for implicit form A1 for 5250 or 5247. … (or 72.4 or 72.43 to 72.44) M1 for 70 2 + 100 2 3(c) 64.3 or 64.27 to 64.28 nfww 2 distance M1 for sin40 = oe 100 3(d) 235 3 B2 for [angle ACB = ] 34.99 to 35 or [angle ABC = ] 55[.0…] 70 or M1 for tan[ ACB ] = 100 100 or tan[ ABC ] = or equivalent trig ratio 70

This question in 0580/42 Oct/Nov 2017

6 (a) r h NOT TO SCALE 10 cm The diagrams show a cube, a cylinder and a hemisphere. The volume of each of these solids is 2000 cm3. (i) Work out the height, h, of the cylinder. h = … cm [2] (ii) Work out the radius, r, of the hemisphere. 4 3 [The volume, V, of a sphere with radius r is V = r r .] 3 r = … cm [3] (iii) Work out the surface area of the cube. … cm2 [3] (b) NOT TO 7 cm SCALE 40º 10 cm (i) Calculate the area of the triangle. … cm2 [2] (ii) Calculate the perimeter of the triangle and show that it is 23.5 cm, correct to 1 decimal place. Show all your working. [5] (c) NOT TO SCALE cº 9 cm The perimeter of this sector of a circle is 28.2 cm. Calculate the value of c. c = … [3]

18 marks

Mark scheme: 6(a)(i) 25.5 or 25.46… 2 M1 for π × 52 × h = 2000 oe 6(a)(ii) 9.85 or 9.847… 3  2  M2 for [r3=] 2000 ÷  π  oe  3  2 or M1 for πr3 = 2000 oe 3 6(a)(iii) 952 or 952.4…. 3 3 2 M2 for [6 ×] 2000 or M1 for 3 2000 or 6 times their area of one face 6(b)(i) 22.5 or 22.49… 2 1 M1 for × 7 × 10 × sin40 2 6(b)(ii) √(102 + 72 – 2 × 10 × 7 cos40) + 7 M3 M2 for 102 + 72 – 2 × 10 × 7 cos40 + 10 or M1 for correct implicit cosine rule 23.46… A2 A1 for 6.46… or 41.7 to 41.8 6(c) 64.9 or 64.92 to 64.94 3 c M2 for 28.2 – 2 × 9 = × 2 × π × 9 oe 360 c or M1 for × 2 × π × 9 soi 360

This question in 0580/43 Oct/Nov 2017

5 A O NOT TO SCALE 8 cm 7 cm 78° C B The diagram shows a design made from a triangle AOC joined to a sector OCB. AC = 8 cm, OB = OC = 7 cm and angle ACO = 78°. (a) Use the cosine rule to show that OA = 9.47 cm, correct to 2 decimal places. [4] (b) Calculate angle OAC. Angle OAC = … [3] (c) The perimeter of the design is 29.5 cm. Show that angle COB = 41.2°, correct to 1 decimal place. [5] (d) Calculate the total area of the design. … cm2 [4]

16 marks

Mark scheme: 5(a) 8² + 7² − 2 × 7 × 8 × cos78 oe M2 M1 for correct implicit version 9.471.. to 9.472 A2 A1 for 89.7… 5(b) 46.3 or 46.29 to 46.30… 3 7sin78 M2 for [sin OAC = ] 9.47 sin OAC sin78 or M1 for = 7 9.47 5(c) 29.5 – (7 + 8 + 9.47) M1 360 × (29.5 − (7 + 8 + 9.47)) M3 x M2 for × 2 × π × 7 = their arc length 2 × π × 7 360 oe x or M1 for × 2 × π × 7 oe 360 41.15 to 41.171.. B1 5(d) 45[.0] or 44.98 to 45.01 nfww 4 M3 for 41.2 2 ½ × 8 × 7 × sin 78 oe + × π × 7 oe 360 OR M1 for ½ × 8 × 7 × sin 78 oe or ½ × 8 × 9.47 × sin their (b) oe 41.2 2 M1 for × π × 7 oe 360

This question in 0580/42 May/June 2018

7 (a) R 130.6° NOT TO 8.9 cm 12.5 cm SCALE P Q Calculate the area of triangle PQR. … cm2 [2] (b) 18 cm B A 11.6 cm 21.3 cm 123.5° D NOT TO SCALE C In the diagram, AB = 18 cm, BC = 21.3 cm and BD = 11.6 cm. Angle BDC = 123.5° and angle ABC is a right angle. (i) Calculate angle BCD. Angle BCD = … [3] (ii) Calculate AD. AD = … cm [5]

10 marks

Mark scheme: 7(a) 42.2 or 42.23 … 2 1 M1 for × 9.8 × 125. × sin 1306. oe 2 7(b)(i) 27[.0] or 27.00 to 27.01 3 116. × sin 1235. M2 for 213. 11 6. 21 3. or M1 for = oe sin BCD sin 123 5. 7(b)(ii) 15.9 or 15.90 to 15.91 5 M1 for angle ABD = their angle BCD + 33.5 and M2 for 11.6 2 + 18 2 −×2 11.6 × 18 × cos ( theirABD ) or M1 for implicit version A1 for 252.9 to 253

This question in 0580/41 Oct/Nov 2018

6 D 80° NOT TO 8 cm SCALE C 13 cm 4 cm A 11 cm B (a) Calculate angle ACB. Angle ACB = … [4] (b) Calculate angle ACD. Angle ACD = … [4] (c) Calculate the area of the quadrilateral ABCD. … cm2 [3]

11 marks

Mark scheme: 6(a) 52[.0] or 52.02… 4 132 + 4 2 − 112 M2 for [cos = ] 2 × 13 × 4 or M1 for 112 = 13 2 + 4 2 − 2 × 13 × 4 cos(...) A1 for 64 [cos–1 =]104 oe or 0.615 or 0.6153 to 0.6154 6(b) 62.7 or 62.69 to 62.70 4 −1 8sin80  M3 for 180 – sin   – 80 oe  13  8 sin 80 or M2 for sin A = 13 13 8 or M1 for = oe sin 80 sin A A1 for 37.3 or 37.30… If 0 scored, M1 for 180 – 80 – their A 6(c) 66.7 or 66.68 to 66.71 3 M1 for 5.0 × 13 × 4 × sin(theirACB) oe M1 for 5.0 × 8 × 13 × sin(their ACD) oe

This question in 0580/43 Oct/Nov 2018

9 (a) C NOT TO SCALE A D B 58 m In the diagram, BC is a vertical wall standing on horizontal ground AB. D is the point on AB where AD = 58 m. The angle of elevation of C from A is 26°. The angle of elevation of C from D is 72°. (i) Show that AC = 76.7 m, correct to 1 decimal place. [5] (ii) Calculate BD. BD = … m [3] (b) Triangle EFG has an area of 70 m2. EF : FG = 1 : 2 and angle EFG = 40°. (i) Calculate EF. EF = … m [4] (ii) A different triangle PQR also has an area of 70 m2. PQ : QR = 1 : 2 and PQ = EF. Find angle PQR. Angle PQR = … [1] Question 10 is printed on the next page.

13 marks

Mark scheme: 9(a)(i) ∠ ACD = 46 soi B2 B1 for angle ADC = 108 or angle DCB = 18 or ∠CDE = 44 soi 58sin108 M2 sin108 sin their 46 M1 for = oe sin their 46 x 58 76.68… nfww A1 9(a)(ii) 10.9 or 10.91 to 10.94 3 B2 for [AB =] 68.9 or 68.91 to 68.94 or M2 for a correct explicit statement for AB or BD AB or M1 for = cos26 oe 76.7 9(b)(i) 10.4 or 10.43 to 10.44 4 70 M3 for oe sin 40 or M2 for x2 × sin 40 = 70 oe or M1 for 1 x × 2x × sin 40 = 70 2 9(b)(ii) 140 1

This question in 0580/43 May/June 2019

5 North NOT TO SCALE A 120 m 150 m B 180 m C The diagram shows a triangular field, ABC, on horizontal ground. (a) Olav runs from A to B at a constant speed of 4 m/s and then from B to C at a constant speed of 3 m/s. He then runs at a constant speed from C to A. His average speed for the whole journey is 3.6 m/s. Calculate his speed when he runs from C to A. … m/s [3] (b) Use the cosine rule to find angle BAC. Angle BAC = … [4] (c) The bearing of C from A is 210°. (i) Find the bearing of B from A. … [1] (ii) Find the bearing of A from B. … [2] (d) D is the point on AC that is nearest to B. Calculate the distance from D to A. … m [2]

12 marks

Mark scheme: 5(a) 4.29 or 4.285 to 4.286 3 150 M2 for 450 120 180 − − 6.3 4 3 or M1 for [time =] 120 ÷ 4 or 180 ÷ 3 or 150 + 180 + 120 450 ÷ 3.6 or 3.6 = total time 5(b) 82.8 or 82.81 to 82.82 using cosine 4 150 2 + 120 2 − 180 2 M2 for rule 2 × 150 × 120 or M1 for 180 2 = 120 2 + 150 2 − 2 × 120 × 150 cos(...) 4500 A1 for oe 36000 5(c)(i) 127.2 or 127.1 to 127.2 or 127 1 FT 210 – their (b) 5(c)(ii) 307.2 or 307.1 to 307.2 or 307 2 FT 180 + their(c)(i) M1 for 180 + their (c)(i) 5(d) 15 or 14.99 to 15.04 2 dist M1 for cos ( their ( b ) ) = oe 120

This question in 0580/41 Oct/Nov 2019

6 (a) A NOT TO 79° SCALE 8 m 13 m C B The diagram shows triangle ABC. (i) Use the cosine rule to calculate BC. BC = … m [4] (ii) Use the sine rule to calculate angle ACB. Angle ACB = … [3] (b) NOT TO D SCALE (x + 4) m F 30° (4x - 5) m E The area of triangle DEF is 70 m2. (i) Show that 4x 2 + 11x - 300 = 0 . [4] (ii) Use the quadratic formula to solve 4x 2 + 11x - 300 = 0 . Show all your working and give your answers correct to 2 decimal places. x = … or x = … [4] (iii) Find the length of DE. DE = … m [1]

16 marks

Mark scheme: 6(a)(i) 13.9[0…] from cosine rule 4 M2 for 82 + 132 – 2 × 8 × 13cos79 13 2 + 8 2 − BC 2 or M1 for cos 79 = 2 × 8 × 13 A1 for 193 … 6(a)(ii) 66.6 or 66.60… to 66.65 from sine 3 13 × sin 79 M2 for [sin ACB = ] rule their ( a )(i ) sin ACB sin 79 or M1 for = oe 13 their ( a )(i ) 6(b)(i) 1 M1 ( x + 4)( 4 x − 5) sin 30 = 70 2 4x2 + 16x – 5x – 20 = 280 M2 Dep on M1 B1 for 4x2 + 16x – 5x – 20 or better Leading to 4x2 + 11x – 300 = 0 A1 with no errors or omissions seen 6(b)(ii) 2 B2 − 11 ± 11 − 4 × 4 × −300 B1 for 112 − 4 ( 4 )( − 300 ) or better 2 × 4 − 11 + q − 11 − q or for or 2 × 4 2 × 4 –10.14 and 7.39 B2 B1 for each or SC1 for final answers –10.1 or –10.144 to –10.143 and 7.4 or 7.393 to 7.394 or –10.14 and 7.39 seen in working or for –7.39 and 10.14 as final answer 6(b)(iii) 11.4 or 11.39… 1 FT their positive root + 4

This question in 0580/42 Oct/Nov 2019

4 B 107 m C NOT TO SCALE 158 m 132 m 86 m North 116° D A The diagram shows a field, ABCD, on horizontal ground. (a) There is a vertical post at C. From B, the angle of elevation of the top of the post is 19°. Find the height of the post. … m [2] (b) Use the cosine rule to find angle BAC. Angle BAC = … [4] (c) Use the sine rule to find angle CAD. Angle CAD = … [3] (d) Calculate the area of the field. … m2 [3] (e) The bearing of D from A is 070°. Find the bearing of A from C. … [2]

14 marks

Mark scheme: 4(a) 36.8 or 36.84… 2 h h 107 M1 for = tan19 or = oe 107 sin19 sin71 or better 4(b) 42.1 or 42.12… from cosine rule 4 158 2 + 132 2 − 107 2 M2 for [ cos BAC = ] 2 × 158 × 132 or M1 for implicit version A1 for [ cos BAC = ]30939 or 0.7417… 41712 4(c) 35.8 or 35.84… from sine rule 3 86 × sin116 M2 for [ = 0.58557...] 132 sin CAD sin116 or M1 for = oe 86 132 4(d) 9670 or 9669 to 9676 3 1 M2 for × 158 × 132 × sin ( their ( b ) ) oe 2 1 and × 86 × 132 × sin ( 64 − their ( c ) ) oe 2 or M1 for either area 4(e) 214.2 or 214.1… or 214 2 M1 for [180 +]70–their (c) oe

This question in 0580/43 Oct/Nov 2019

8 (a) S 25° R 72° NOT TO SCALE 6 cm 34° Q 7.4 cm P The diagram shows a quadrilateral PQRS formed from two triangles, PQS and QRS. Calculate (i) QR, QR = … cm [3] (ii) PS, PS = … cm [3] (iii) the area of quadrilateral PQRS. … cm2 [4] (b) X H G E F NOT TO SCALE 16 cm D C 18 cm A B 20 cm The diagram shows an open box ABCDEFGH in the shape of a cuboid. AB = 20 cm, BC = 18 cm and AE = 16 cm . A thin rod AGX rests partly in the box as shown. The rod is 40 cm long. (i) Calculate GX, the length of the rod which is outside the box. GX = … cm [4] (ii) Calculate the angle the rod makes with the base of the box. … [3]

17 marks

Mark scheme: 8(a)(i) 2.67 or 2.666… 3 6 × sin 25 M2 for sin72 or M1 for implicit version 8(a)(ii) 4.14 or 4.140… 3 M1 for 6 2 + 7.4 2 − 2 × 6 × 7.4 × cos34 A1 for 17.1 to 17.2 8(a)(iii) 20.4 or 20.35 to 20.36… 4 B1 for angle SQR = 83 M1 for 1 × 6 ×their (a)(i) × sin their (180–72–25) 2 oe 1 M1 for × 6 × 7.4 × sin 34 oe 2 8(b)(i) 8.7[0] or 8.695… 4 B3 for 980 oe or 31.3 or 31.30… or M3 for 40 – 20 2 + 18 2 + 16 2 oe or M2 for 20 2 + 18 2 + 16 2 oe or M1 for any correct attempt at 2-dimensional Pythagoras’ e.g. 182 + 162 8(b)(ii) 30.7 or 30.73 to 30.74… 3 16 M2 for [sin =] oe 20 2 + 18 2 + 16 2 or B1 for identifying angle GAC

This question in 0580/42 Feb/March 2020

7 North B 80 m NOT TO A SCALE 72° 115 m C The diagram shows the positions of three points A, B and C in a field. (a) Show that BC is 118.1 m, correct to 1 decimal place. [3] (b) Calculate angle ABC. Angle ABC = … [3] (c) The bearing of C from A is 147°. Find the bearing of (i) A from B, … [3] (ii) B from C. … [2] (d) Mitchell takes 35 seconds to run from A to C. Calculate his average running speed in kilometres per hour. … km/h [3] (e) Calculate the shortest distance from point B to AC. … m [3]

17 marks

Mark scheme: 7(a) [BC2 =] 802 + 1152 – 2 × 80 × M1 115 cos 72 oe 118.06… A2 A1 for 13939… 7(b) 67.8 or 67.9 or 67.83 to 67.88 3 115 × sin72 M2 for [sin B =] oe 118.1 115 118.1 or M1 for = oe sin B sin72 7(c)(i) 255 3 B1 for bearing of B from A is 75 soi M1 for 180 + 75 oe 7(c)(ii) [00]7.2 2 M1 for their (c)(i) – their (b) –180 7(d) 11.8 or 11.82 to 11.83 3 M1 for 115 ÷ 35 oe M1 for their speed in m/s × 60 × 60 ÷ 1000 7(e) 76.1 or 76.08 to 76.09 3 distance M2 for = sin72 oe 80 or M1 for distance required is perpendicular to AC soi

This question in 0580/41 May/June 2020

4 S NOT TO SCALE 55° P 150 m 25° 45° R 120 m Q The diagram shows two triangles. (a) Calculate QR. QR = … m [3] (b) Calculate RS. RS = … m [4] (c) Calculate the total area of the two triangles. … m2 [3]

10 marks

Mark scheme: 4(a) 65.4 or 65.36 to 65.37 3 M1 for 1502 + 1202 – 2 × 150 × 120 cos 25 A1 for 4270 or 4272 to 4273 4(b) 125 or 124.7 to 124.8 4 B1 for [angle S =] 80 150sin55 M2 for sin their 80 sin their 80 sin55 or M1 for = oe 150 RS 4(c) 10 400 or 10 410 to 10 440 nfww 3 1 M1 for × 120 × 150sin25 oe 2 1 M1 for × 150 × their (b) sin45 oe 2

This question in 0580/42 May/June 2020

6 (a) C 54° x° NOT TO SCALE 5.3 cm 11 cm G 6.9 cm 42° A B The diagram shows triangle ABC with point G inside. CB = 11 cm, CG = 5.3 cm and BG = 6.9 cm. Angle CAB = 42° and angle ACG = 54°. (i) Calculate the value of x. x = … [4] (ii) Calculate AC. AC = … cm [4] (b) NOT TO 2.5 cm SCALE 15 cm Water flows at a speed of 20 cm/s along a rectangular channel into a lake. The width of the channel is 15 cm. The depth of the water is 2.5 cm. Calculate the amount of water that flows from the channel into the lake in 1 hour. Give your answer in litres. … litres [4]

12 marks

Mark scheme: 6(a)(i) 29.5 or 29.50… 4 112 + 5.32 − 6.9 2 M2 for 2 × 11 × 5.3 or M1 for 6.92 = 112 + 5.32 − 2 × 11 × 5.3 cos x A1 for 0.87[0…] oe 6(a)(ii) 13.4 or 13.38… 4 B1FT 84 − their (a)(i) 11 M2 for × sin their 54.5 sin42 or M1 for implicit form 6(b) 2700 4 M2 for 15 × 2.5 × 20 × 60 × 60 or M1 for 15 × 2.5 × 20 M1 for their volume ÷ 1000 If 0 scored, SC1 for figs 27 with no working

This question in 0580/43 May/June 2020

6 D 287.9 m North NOT TO 205.8 m SCALE C 168 m 38° 192 m A B The diagram shows a field, ABCD, on horizontal ground. BC = 192 m, CD = 287.9 m, BD = 168 m and AD = 205.8 m. (a) (i) Calculate angle CBD and show that it rounds to 106.0°, correct to 1 decimal place. [4] (ii) The bearing of D from B is 038°. Find the bearing of C from B. … [1] (iii) A is due east of B. Calculate the bearing of D from A. … [5] (b) (i) Calculate the area of triangle BCD. … m2 [2] (ii) Tomas buys the triangular part of the field, BCD. The cost is $35 750 per hectare. Calculate the amount he pays. Give your answer correct to the nearest $100. [1 hectare = 10 000 m2] $ … [2]

14 marks

Mark scheme: 6(a)(i) 106.01 to 106.02 4 M2 for 192 2 + 168 2 − 287.9 2 [cos[∠CBD] =] oe 2 × 192 × 168 or M1 for the implicit form A1 for –0.276 to – 0.275 6(a)(ii) 292.0 or 291.98 to 291.99 1 6(a)(iii) 310.0 or 310.03 to 310.04 5 168 × sin(90 − 38) M2 for [sin A =] 205.8 sin A sin(90 − 38) or M1 for = 168 205.8 A1 for [A =] 40.0 or 40.03 to 40.04 M1 dep for 270 + their angle DAB oe 6(b)(i) 15 500 or 15 501 to 15 503. … 2 M1 for 0.5 × 192 × 168 × sin(106) oe 6(b)(ii) 55 400 2 FT 3.575 × their (b)(i) oe rounded to nearest 100 M1 for figs 35 75 × figs their (b)(i) or figs 554 or figs 5541 to figs 5543

This question in 0580/41 Oct/Nov 2020

5 C B 65° NOT TO SCALE 4.4 cm 9.7 cm A 8.6 cm 42° D (a) Calculate angle ADB. Angle ADB = … [3] (b) Calculate DC. DC = … cm [4] (c) Calculate the shortest distance from C to BD. … cm [3]

10 marks

Mark scheme: 5(a) 27[.0] or 26.97… nfww 3 8.6 2 + 9.7 2 − 4.4 2 M2 for [cos = ] 2 × 8.6 × 9.7 or M1 for implicit form 5(b) 9.19 or 9.192 to 9.193 4 B1 for [angle BCD =] 73 seen 9.7 × sin65 M2 for oe sin (180 − 65 − 42) sin(180 − 65 − 42) sin65 or M1 for = oe 9.7 DC 5(c) 6.15 or 6.149 to 6.151… 3 d M2 for = sin42 oe their 9.19 or M1 for right angle between line from C to BD and BD soi

This question in 0580/42 Feb/March 2021

5 A NOT TO SCALE 10.6 cm 58° 78° B C X 6.4 cm The diagram shows triangle ABC. X is a point on BC. AX = 10.6 cm, XC = 6.4 cm, angle ABC = 58° and angle AXB = 78°. (a) Calculate AC. AC = … cm [4] (b) Calculate BX. BX = … cm [4] (c) Calculate the area of triangle ABC. … cm2 [3]

11 marks

Mark scheme: 5(a) 13.5 or 13.47… 4 B1 for angle 102 seen M2 for 10.6 2 + 6.4 2 − 2 × 10.6 × 6.4 × cos (180 − 78 ) OR M1 for 10.6 2 + 6.4 2 − 2 × 10.6 × 6.4 × cos (180 − 78 ) A1 for 181.5… 5(b) 8.68 or 8.682 to 8.683 nfww 4 B1 for angle = 44 10.6 M2 for sin(180 – 58 – 78) × oe sin 58 sin(180 − 58 − 78) sin58 or M1 for = oe x 10.6 5(c) 78.2 or 78.17 to 78.19… 3 1 M2 for × 10.6 × ( 6.4 + their 8.68 ) × sin ( 78 ) 2 oe OR 1 M1 for × 10.6 × 6.4 × sin(180 – 78) oe 2 1 M1 for × 10.6 × their 8.68 × sin78 oe 2

This question in 0580/41 May/June 2021

6 B 16 m NOT TO A 57° 32 m SCALE 19 m C 75° D The diagram shows a quadrilateral ABCD made from two triangles, ABD and BCD. (a) Show that BD = 16.9 m, correct to 1 decimal place. [3] (b) Calculate angle CBD. Angle CBD = … [4] (c) Find the area of the quadrilateral ABCD. … m2 [3] (d) Find the shortest distance from B to AD. … m [3]

13 marks

Mark scheme: 6(a) 2 2 M2 or M1 for 162 + 192 – 2 × 16 × 19cos57 16 + 19 – 2 × 16 × 19cos57 oe A1 for 285.8 to 285.9 16.90 to 16.91 A1 6(b) 74.3 or 74.30 to 74.33 4 16.9 × sin75 M2 for [sin ... =] oe 32 16.9 32 or M1 for = oe sin C sin75 B1 for [angle BCD =] 30.7 or 30.67 to 30.69… or M1dep for 105 – their angle BCD 6(c) 388 or 387.7 to 387.9… nfww 3 1 M1 for × 16 × 19 × sin 57 oe 2 1 M1 for × 16.9 × 32 × sin their (b) oe 2 6(d) 13.4 or 13.41 to 13.42 nfww 3 x M2 for = sin57 oe 16 or M1 for distance required is perpendicular to AD soi

This question in 0580/42 May/June 2021

3 (a) C 38.6 m 56.5 m D B 94° 78.4 m NOT TO SCALE 46.1 m 64° E A ABCDE is a pentagon. (i) Calculate AD and show that it rounds to 94.5 m, correct to 1 decimal place. [2] (ii) Calculate angle BAC. Angle BAC = … [3] (iii) Calculate the largest angle in triangle CAD. … [4] (b) Q L 34.3 cm P NOT TO 21.5 cm SCALE 111° R N M 27.6 cm Triangle PQR has the same area as triangle LMN. Calculate the shortest distance from R to the line PQ. … cm [3]

12 marks

Mark scheme: 3(a)(i) AD M1 = tan 64 oe or better 46.1 94.51 to 94.52 A1 3(a)(ii) 46[.0] or 45.96… nfww 3 sin94 M2 for 56.5 × oe 78.4 56.5 78.4 or M1 for = oe sin BAC sin94 3(a)(iii) 102.3 or 102.4 or 102.34 to 102.38 4 38.6 2 + 78.4 2 − 94.5 2 M2 for [cosC = ] 2 × 38.6 × 78.4 or M1 for 94.5 2 = 38.6 2 + 78.4 2 − 2 × 38.6 × 78.4 × cos C and A1 for –0.214 or –0.2144 to –0.2137 If 0 scored, SC2 for [CAD =] 23.5 or 23.51 to 23.52 or for [CDA =] 54.1 or 54.14… 3(b) 16.2 or 16.15… 3 1 1 M2 for × 21.5 × 27.6sin111 = × 34.3 × d 2 2 oe 1 or M1 for × 21.5 × 27.6sin111 seen or 2 1 × 34.3 × d oe soi 2

This question in 0580/42 Oct/Nov 2021

6 D 100° NOT TO SCALE 50° C 12 cm A 8 cm 11 cm B (a) Calculate AD. AD = … cm [3] (b) Calculate angle BAC and show that it rounds to 40.42°, correct to 2 decimal places. [4] (c) Calculate the area of the quadrilateral ABCD. … cm2 [3] (d) Calculate the shortest distance from B to AC. … cm [3]

13 marks

Mark scheme: 6(a) 9.33 or 9.334... 3 12sin50 M2 for sin100 sin100 sin50 or M1 for = oe 12 AD 6(b) 112 + 12 2 − 8 2 M2 M1 for [cos =] 2 2 2 2 × 11 × 12 8 = 11 + 12 − 2 × 11 × 12cos( BAC ) 40.415... A2 201 67 A1 for 0.761... or or 264 88 6(c) 70.8 or 70.77 to 70.79... 3 M1 for 1 × 12 × their (a) × sin(180 − 100 − 50) 2 1 M1 for × 12 × 11 × sin(40.42) 2 6(d) 7.13 or 7.131 to 7.132... 3 dist M2 for = sin(40.42) 11 or M1 for recognition that shortest distance is perpendicular to AC

This question in 0580/43 Oct/Nov 2021

4 (a) C NOT TO SCALE 29.5 cm 35.3 cm 51.6° A B 45 cm In triangle ABC, AB = 45 cm, AC = 29.5 cm, BC = 35.3 cm and angle CAB = 51.6°. (i) Calculate angle ABC. Angle ABC = … [3] (ii) Calculate the area of triangle ABC. … cm2 [2] (b) R NOT TO SCALE S 32 cm 47 cm 56° P 60° 32 cm Q The diagram shows a quadrilateral PQRS formed from two triangles, PQS and QRS. Triangle PQS is isosceles, with PQ = PS = 32 cm and angle SPQ = 56°. QR = 47 cm and angle SQR = 60°. (i) Calculate SR. SR = … cm [4] (ii) Calculate the shortest distance from P to SQ. … cm [3]

12 marks

Mark scheme: 4(a)(i) 40.9 or 40.91… 3 29.5sin51.6 M2 for [sin ABC= ] oe 35.3 35.32 + 452 − 29.52 or for [cos ABC = ] 2 × 35.3 × 45 29.5 35.3 or M1 for = oe sin ABC sin51.6 or for correct implicit cosine rule 4(a)(ii) 520 or 520.0 to 520.2… 2 FT their (a)(i) if used provided working shown M1 for 0.5 × 29.5 × 45 × sin51.6 oe or for 0.5 × 35.3 × 45 × sin(their (a)(i)) or for 0.5 × 35.3 × 29.5sin (180–51.6–their(a)(i)) 4(b)(i) 41.2 or 41.21 to 41.23 4  1  M1 for SQ = 2 × 32 × sin  × 56  oe  2  or 32 2 + 32 2 − 2 × 32 × 32 × cos56 oe 32sin56 or oe sin((180 -56) ÷ 2) M2 for − 2 × 47 × their SQ × cos60 SR 2 = 47 2 + ( their SQ 2 ) or M1 for implicit form 4(b)(ii) 28.3 or 28.25 to 28.29… 3 M2 for 32 × sin62 oe or M1 for recognition that line from P is perpendicular to SQ

This question in 0580/42 Feb/March 2022

7 D NOT TO SCALE 12 km 9 km 14 km A C 25° 32° 123° B (a) Calculate angle ACD. Angle ACD = … [4] (b) Show that BC = 7.05 km , correct to 2 decimal places. [3] (c) Calculate the shortest distance from B to AC. … km [3] (d) Calculate the length of the straight line BD. BD = … km [4] (e) C is due east of A. Find the bearing of D from C. … [2]

16 marks

Mark scheme: 7(a) 39.6 or 39.57 … 4 M2 for [cos =] 2 2 2 14 12 9 2 14 12     or M1 for 92 = 142 + 122 – 2 × 14 × 12 × cos ACD A1 for 0.7708... or 0.771 or 37 48 oe 7(b) 14sin25 sin123 M2 M1 for sin123 sin 25 14 BC  oe 7.054… A1 7(c) 3.74 or 3.735 to 3.739 3 M2 for 7.05 × sin 32 or M1 for recognition that the line from B is perpendicular to AC 7(d) 11.8 or 11.83 to 11.85 4 M1 for 32 + their(a) soi M2 for  2 2 12 7.05 2 12 7.05 cos( 32)      their a or M1 for    2 2 2 12 7.05 cos 32 2 12 7.05       BD their a 7(e) 309.6 or 309.57... 2 FT 270 + their(a) M1 for 270 + their(a) oe

This question in 0580/41 May/June 2022

4 6.4 cm D C 38° NOT TO SCALE 10.9 cm 45° A B ABCD is a trapezium with DC parallel to AB. DC = 6.4 cm, DB = 10.9 cm, angle CDB = 38° and angle DAB = 45°. (a) Find CB. CB = … cm [3] (b) (i) Find angle ADB. Angle ADB = … [1] (ii) Find AB. AB = … cm [3] (c) Calculate the area of the trapezium. … cm2 [3]

10 marks

Mark scheme: 4(a) 7.06 or 7.058… or 7.059 3 2 2 M2 for 6.4  10.9 2 6.4  10.9  cos38 oe OR M1 for 6.42 + 10.92 – 2  6.4  10.9  cos 38 oe A1= 49.8... 4(b)(i) 97 1 4(b)(ii) 15.3[0…] 3 10.9  sin their 97 M2 for [AB =] sin45 sin their 97 sin45 or M1 for  oe AB 10.9 4(c) 72.8 to 72.81… 3 M2 for 1 1  6.4   10.9  sin38  their 15.3  10.9  sin38 2 2 oe or M1 for 12  6.4  10.9  sin38 oe or 12  their15.3  10.9  sin38 oe or M1 for height =10.9  sin38 oe

This question in 0580/42 May/June 2022

7 B 82 m NOT TO A 76° SCALE 55 m C The diagram shows a field ABC. (a) Calculate BC. BC = … m [3] (b) Calculate angle ACB. Angle ACB = … [3] (c) A gate, G, lies on AB at the shortest distance from C. Calculate AG. AG = … m [3] (d) A different triangular field PQR has the same area as ABC. PQ = 90 m and QR = 60 m. Work out the two possible values of angle PQR. Angle PQR = … or … [5]

14 marks

Mark scheme: 7(a) 87.[0] or 86.98 to 86.99 3 2 2 M2 for 82  55 2 82  55  cos76 oe OR M1 for 82 2  55 2  2  82  55  cos76 oe A1 for 7570 or 7566 to 7567 7(b) 66.1 or 66.2 or 66.13 to 66.17 3 82  sin76 M2 for oe their (a) or M1 for 82 their (a) oe sin C sin76 7(c) 13.3 or 13.30 to 13.31 3 M2 for AG = 55 cos 76 oe or M1 for recognition that CG is perpendicular to AB 7(d) 54.1 or 54.13… 5 B4 for 54.1 or 54.13… and or 125.9 or 125.86 to 125.87 125.9 or 125.86 to 125.87 0.5  82  55  sin76 M3 for [sin Q =] oe 0.5  90  60 or M2 for 0.5  82  55  sin 76 = 0.5  60  90  sin Q oe or M1 for 0.5  82  55  sin 76 oe or for 0.5  60  90 sin Q = their area of ABC If B4 not scored then SC1 for two angles seen that sum to 180 (from use of sine ratio) but not 0 and 180.

This question in 0580/43 May/June 2022

8 Q NOT TO 4 m SCALE A 15 m C P 8 m 3 m 20 m B The diagram shows triangle ABC on horizontal ground. AC = 15 m , BC = 8 m and AB = 20 m . BP and CQ are vertical poles of different heights. BP = 3 m and CQ = 4 m . AQ and PQ are straight wires. (a) Show that angle ACB = 117.5° , correct to 1 decimal place. [4] (b) Calculate the area of triangle ABC. … m2 [2] (c) Calculate the length of AQ. … m [2] (d) Calculate the angle of elevation of Q from P. … [3] (e) Another straight wire connects A to the midpoint of PQ. Calculate the angle between this wire and the horizontal ground. … [5]

16 marks

Mark scheme: 8(a) 15 2 + 8 2 − 20 2 M2 M1 for 202 = 152 + 82 − 2.15.8cos ( ) [cos = ] 2.15.8 117.54 to 117.55 A2 37 111 A1 for − or − or –[0].4625 80 240 8(b) 53.2 or 53.19 to 53.23 2 M1 for 0.5  8  15  sin(117.5) oe 8(c) 15.5 or 15.52 to 15.53 2 M1 for 152 + 42 oe 8(d) 7.1 or 7.13 or 7.125 to 7.126 3 4 − 3 M2 for tan [P]= oe or for 7.1 or 8 7.13 or 7.125 to 7.126 seen or M1 for vertical line = 4 – 3 soi After 0 scored SC1 for correct angle identified 8(e) 11.5 nfww or 11.48 to 11.49... 5 B1 for height of 3.5 soi M2 for 15 2 + 4 2 − 2.15.4cos(117.5) 15 2 + 4 2 − (...) 2 or M1 for cos117.5 = 2.15.4 3.5 M1 for tan = oe their 17.216... After M0 scored SC1 for correct angle identified

This question in 0580/41 Oct/Nov 2022

7 (a) R 39.4 cm 38.2 cm NOT TO SCALE P Q 46.5 cm (i) Calculate angle QPR. Angle QPR = … [4] (ii) Find the shortest distance from Q to PR. … cm [3] (b) The diagram shows a cuboid. H G 20 cm E F NOT TO SCALE D C 21 cm A B 29 cm (i) Calculate the length AG. AG = … cm [3] (ii) Calculate the angle between AG and the base ABCD. … [3] (c) North K NOT TO SCALE North 112 km 96° M L The diagram shows the positions of a lighthouse, L, and two ships, K and M. The bearing of L from K is 155° and KL = 112 km . The bearing of K from M is 010° and angle KML = 96° . Find the bearing and distance of ship M from the lighthouse, L. Bearing … Distance … km [5]

18 marks

Mark scheme: 7(a)(i) 52.[0] or 52.01… 4 39.4 2 + 46.5 2 − 38.2 2 M2 for [cosP = ] oe 2  39.4  46.5 or M1 for 38.2 2 = 39.4 2 + 46.52 −2 39.4  46.5  cos P oe A1 for 0.616 or 0.6155… 7(a)(ii) 36.6 or 36.64 to 36.65 3 d M2 for = sin(their 52.01) oe 46.5 or M1 for recognition that the line from Q is perpendicular to PR 7(b)(i) 41[.0] or 41.01… nfww 3 M2 for 292 + 212 + 202 oe or better or M1 for 292 + 212 oe or 292 + 202 oe or 212 + 202 oe or better 7(b)(ii) 29.2 or 29.18 to 29.2 3 20 M2 for sin[GAC] = oe their AG or M1 for angle GAC identified 7(c) bearing 286 B2 B1 for angle MLK = 49 or for angle MKL = 35 correctly identified or angle from North to ML = 106 distance 64.6 or 64.59… B3 112  sin(their 35) M2 for oe sin(96) or M1 for the implicit form

This question in 0580/42 Oct/Nov 2022

8 A NOT TO SCALE 9.5 cm O 10 cm B 7.7 cm D C E A, B and C are points on the circle, centre O. DE is a tangent to the circle at C. AC = 10 cm , AB = 9.5 cm and BC = 7.7 cm . (a) Show that angle ABC = 70.2° , correct to 1 decimal place. [4] (b) Find (i) angle AOC Angle AOC = … [1] (ii) angle ACO Angle ACO = … [1] (iii) angle ACD. Angle ACD = … [1] (c) Calculate the radius, OC, of the circle. OC = … cm [3] (d) Calculate the area of triangle ABC as a percentage of the area of the circle. … % [4]

14 marks

Mark scheme: 8(a) 9.52 + 7.7 2 − 10 2 M2 M1 for 102 = 9.52 + 7.72 – 2×9.5×7.7cosB oe or better [cos B = ] oe 2  9.5  7.7 70.206 to 70.207 or 70.21 to 70.22 A2 2477 A1 for oe or 0.339 or 0.3386…. 7315 8(b)(i) 140.4 1 8(b)(ii) 19.8 1 FT (180 – their (b)(i)) ÷ 2 8(b)(iii) 70.2 1 FT 90 – their (b)(ii) 8(c) 5.31 or 5.314 to 5.315 3 5 M2 for oe cos their(b)(ii) 5 or M1 for = cos(their (b)(ii)) oe r 8(d) 38.8 or 38.9 or 38.78 to 38.85 4 0.5  9.5  7.7  sin70.2 M3 for [ 100] their( (c)) 2 OR M1 for 0.5 × 9.5 × 7.7 × sin70.2 M1 for (their (c)2)

This question in 0580/43 Oct/Nov 2022

2 F NOT TO SCALE D E C A B The diagram shows a solid triangular prism ABCDEF of length 15 cm. AB = 6.4 cm, EB = 5.7 cm and the volume of the prism is 145 cm3. (a) Show that angle EBA = 32° , correct to the nearest degree. [3] (b) Find the length of EA. … cm [3] (c) Calculate the shortest distance from E to AB. … cm [3] (d) Calculate the angle BF makes with the base, ABCD, of the prism. … [4] (e) The prism is made of plastic with density 938 kg/m3. Calculate the mass of the prism in grams. [Density = mass ' volume ] … g [3]

16 marks

Mark scheme: 2(a) 145 M2 M1 for 145 = 12  6.4  5.7  sin x  15 oe [sin =] 1 2  6.4  5.7  15 1 or for  6.4 h 15  145 and sin x h 2 5.7 32.0[0] A1 If M0, SC1 for 145 = 0.5  6.4  5.7  sin32  15 oe 2(b) 3.4[0] or 3.402 to 3.403 nfww 3 2 2 M2 for 6.4  5.7  2  6.4  5.7  cos  32  OR M1 for 6.4 2  5.7 2  2  6.4  5.7  cos  32  A1 for 11.6 or 11.57 to 11.58 2(c) 3.02 or 3.020 to 3.021 3 M2 for sin  32  x 5.7 80 2  50 2  2  80  50  cos75 or M1 for recognition that the line from E is perpendicular to AB e.g. right angle seen or 1  6.4 h 2 2(d) 10.8 or 10.9 or 10.84 to 10.85... 4 their(c) M3 for [sin =] 15 2  5.7 2 their (c) or  tan   5.7  cos 32) 2  15 2   2 oe or M2 for 15 2  5.7 2 or  5.7  cos32  2  15 or M1 for recognition of correct angle 2(e) 136 or 136.0... 3 1000 M2 for 938  145  oe 1000000 or M1 for figs 136 or 13601

This question in 0580/41 May/June 2023

7 North C NOT TO SCALE 60 km 87 km 38° B A The diagram shows the straight roads between town A, town B and town C. AC = 60 km , CB = 87 km and B is due east of A. The bearing of C from A is 038°. (a) Show that angle ACB = 95.1° , correct to 1 decimal place. [5] (b) Without stopping, a car travels from town A to town C then to town B, before returning directly to town A. The total time taken for the journey is 3 hours 20 minutes. Calculate the average speed of the car for this journey. Give your answer in kilometres per hour. … km/h [6]

11 marks

Mark scheme: 7(a) Angle CAB = 52 B1 1 60sin their 52  M3 60sin their 52 180 – 52 – sin M2 for [sin[...]  ] oe    87  87 60 87 or M1 for  oe sin B sin their 52 95.08… A1 7(b) 77.1 or 77.08 to 77.11 6 B4 for dist travelled = 256.9 to 257[.0…] or B3 for [AB =] 109.9 to 110[.0…] or M3 for 60 + 87 + 60 2  87 2 – 2  60  87  cos 95.1 oe or M2 for 60 2  87 2 – 2  60  87  cos 95.1 oe or AB2 = 12093. … to 12097. … 87sin95.1 or oe sin their 52 or M1 for AB2 = 602 + 872 – 2 × 60 × 87 × cos 95.1 oe sin95.1 sin their 52 or  oe AB 87 20 M1 for their total distance ÷ 3 oe 60

This question in 0580/42 May/June 2023

5 (a) X Y NOT TO SCALE 4.8 m A 20.4° C 5.6 m B ABC is a scalene triangle on horizontal ground. AYX is a straight vertical post, held in place by two straight wires XB and YC. AC = 4.8 m, BC = 5.6 m and angle ACB = 20.4°. (i) Calculate AB. AB = … m [3] (ii) Angle XBA = 64°. Calculate AX. AX = … m [2] (iii) AY = 2.9 m. Calculate the area of triangle YAC. … m2 [2] (b) R 30° NOT TO SCALE 8 cm 75° P M Q In triangle PQR, M is the midpoint of PQ. RM = 8 cm, angle PRM = 30° and angle RMQ = 75°. Calculate PQ. PQ = … cm [5]

12 marks

Mark scheme: 5(a)(i) 2[.00] or 2.002 to 2.003 nfww 3 M2 for 4.8 2  5.6 2  2  4.8  5.6  cos20.4 OR M1 for 4.82 + 5.62 – 2  4.8  5.6  cos 20.4 A1 for 4.01[17…] or 4.012 5(a)(ii) 4.1[0] or 4.11 or 4.100 to 4.107 2 AX M1 for tan64  cao their (a)(i) AX their (a)(i) or for  oe sin64 sin(90  64) 5(a)(iii) 6.96 2 1 M1 for  4.8  2.9 oe 2 5(b) 11.3 or 11.31.. 5 8 M4 for 2×  sin30 sin(45) or B4 for PM = 5.65[685...] or 5.66 or better OR B1 for angle RPM = 45° 8 M2 for  sin30 sin(their 45) or M1 for implicit form

This question in 0580/43 May/June 2023

5 (a) D NOT TO 83.2 m SCALE 38° C B A 54.5 m ACD is a right-angled triangle. B is on AC and BC = 54.5 m. AD = 83.2 m and angle ABD = 38° . Calculate angle ACD. Angle ACD = … [5] (b) F G E EFG is a right-angled triangle. A circle can be drawn that passes through the three vertices of the triangle. On the diagram, mark the position of the centre of the circle with a cross. Explain how you decide. … … [2] (c) N R NOT TO 5 cm SCALE 4 cm Q 6 cm M P L In triangle LMN, the ratio angle L : angle M : angle N = 4 : 5 : 6. In triangle PQR, PQ = 6 cm , PR = 4 cm and QR = 5 cm . Calculate the difference between the largest angle in triangle PQR and the largest angle in triangle LMN. … [7]

14 marks

Mark scheme: 5(a) 27.3 or 27.32 to 27.33 5 83.2 M4 for tan[ACD] = oe 83.2 + 54.5 tan38 or 83.2 M3 for [AC =] +54.5 oe tan38 or for [CD =] 2  83.2  2  83.2  54.5 + − 2(54.5) cos(180 − 38)      sin38   sin38  oe or 83.2 83.2 M2 for [AB =] oe or for [BD =] oe tan38 sin 38 83.2 83.2 or M1 for tan38 = oe or sin38 = oe AB BD 5(b) Centre marked at midpoint of B2 B1 for marking the centre at mid-point of FG FG. and Angle in a semi-circle is 90 5(c) 10.8 or 10.81 to 10.82 7 B2 for 72 180 or M1 for [ 6] 4 + 5 + 6 and, for triangle PQR B4 for [angle R=]82.8 or 82.81 to 82.83 5 or B3 for [cosR =] oe or better 40 4 2 + 5 2 − 6 2 or M2 for 2 4 5 or M1 for 62 = 42 + 52 – 245cosR After 0 scored for triangle PQR, SC1 for [P =] 55.8 or 55.77 to 55.78 or  Q = 41.4 or 41.40 to 41.41

This question in 0580/41 Oct/Nov 2023

7 (a) X 2.8 m NOT TO R SCALE 7.1 m P Q The diagram shows a right-angled triangle PQR on horizontal ground. X is vertically above R and the angle of elevation of X from P is 21°. XR = 2.8 m and RQ = 7.1 m. (i) Calculate the angle of elevation of X from Q. … [2] (ii) Calculate PQ. … m [3] (b) M 9.1 cm NOT TO SCALE 32° L K 16.7 cm Calculate the acute angle KML. Angle KML = … [3] (c) C 21.5 cm NOT TO SCALE A 12.3 cm B D The area of triangle ABC is 62.89 cm 2. (i) Show that angle BAC = 28.4°, correct to 1 decimal place. [2] (ii) Calculate BC. … cm [3] (iii) AB is extended to a point D such that angle BDC = 90°. Calculate BD. … cm [3]

16 marks

Mark scheme: 7(a)(i) 21.5 or 21.52... 2 2.8 M1 for tan(…) = oe 7.1 7(a)(ii) 10.2 or 10.17 to 10.18 3 2  2.8 2 oe M2 for  + 7.1  tan21  2.8 or M1 for = tan21 oe PR 7(b) 76.5 or 76.52 to 76.53 3 16.7sin32 M2 for [sin =] oe 9.1 9.1 16.7 or M1 for = oe sin32 sin M 7(c)(i) 1 M1  12.3  21.5sin(...) = 62.89 or better 2 28.40 to 28.41… A1 7(c)(ii) 12.2 or 12.17 to 12.18 3 M2 for 12.32 + 21.52 – 2 12.3  21.5  cos28.4 OR M1 for 12.32 + 21.52 – 2 × 12.3 × 21.5 × cos28.4 A1 for 148 or 148.2 to 148.3 7(c)(iii) 6.6[0] to 6.62 3 M2 for 21.5cos28.4 – 12.3 or M1 for 21.5cos28.4

This question in 0580/42 Oct/Nov 2023

6 A NOT TO SCALE 17.2 cm 54° 68° B C M 12.8 cm The diagram shows triangle ABC with AB = 17.2 cm. Angle ABC = 54° and angle ACB = 68°. (a) Calculate AC. AC = … cm [3] (b) M lies on BC and MC = 12.8 cm. Calculate AM. AM = … cm [3] (c) Calculate the shortest distance from A to BC. … cm [3]

9 marks

Mark scheme: 6(a) 15[.0] or 15.00 to 15.01 3 17.2 M2 for  sin54 oe sin68 sin54 sin68 or M1 for = oe AC 17.2 6(b) 15.7 or 15.65 to 15.66 3 M2 for their152 + 12.82 −2 their15 12.8  cos68 OR M1 for their152 + 12.82 −2 their15  12.8  cos68 A1 for 244.9 to 245.2 6(c) 13.9 or 13.90 to 13.92 3 x x M2 for = sin54 oe or = sin 68 17.2 their15 oe or M1 for distance required is the perpendicular from A to BC soi

This question in 0580/42 Feb/March 2024

6 D 6.5 cm 26° NOT TO C 64° SCALE A 42° 10.4 cm B ABCD is a quadrilateral with AB = 10.4 cm and AD = 6.5 cm. Angle DAB = 64° , angle BDC = 26° and angle DBC = 42° . (a) Show that BD = 9.55 cm, correct to 2 decimal places. [3] (b) (i) Show that angle BCD = 112° . [1] (ii) Calculate CD. CD = … [3] (c) Find the shortest distance from D to AB. … cm [3]

10 marks

Mark scheme: 6(a) 2 2 M2 M1 for 10.42 + 6.52 – 2× 10.4 × 6.5 × 10.4  6.5  2  10.4  6.5  cos64 cos64 A1 for 91.1 to 91.2 9.546 to 9.547 A1 6(b)(i) 180   26  42  B1 6(b)(ii) 6.89 or 6.888 to 6.892... 3 9.55 M2 for  sin 42 oe sin112 sin112 sin 42 or M1 for oe 9.55 CD 6(c) 5.84[2…] 3 x M2 for  sin64 oe 6.5 or M1 for identifying shortest distance from D is perpendicular to AB

This question in 0580/42 May/June 2024

6 (a) H NOT TO SCALE 4 m G F 1.5 m The diagram shows a ladder, GH, on horizontal ground, leaning against a vertical wall, HF. GF = 1.5 m and HF = 4 m . Calculate the length of the ladder, GH. … m [2] (b) W NOT TO SCALE 120 m V 50 m W is 120 m north of V and 50 m east of V. Calculate the bearing of V from W. … [3] (c) B NOT TO SCALE D A C In the quadrilateral ABCD, AD = DC = 5 cm and AB = BC . Angle ABD = 25° and angle BAD = 15° . Calculate the perimeter of the quadrilateral ABCD. … cm [5] (d) S 8 cm R 110° 11 cm NOT TO SCALE 14 cm P 10 cm Q PQRS is a quadrilateral. Calculate angle PQR. Angle PQR = … [5]

15 marks

Mark scheme: 6(a) 4.27 or 4.272... 2 M1 for 42 + 1.52 oe 6(b) 203 or 202.6… 3 B2 for [angle at W = ] 22.6... or for [angle at V =] 67.4 or 67.38… 5 12 or M1 for tan = or oe 12 5 6(c) 25.2 or 25.20 to 25.21[0] 5 B4 for [BC or AB = ] 7.6[0] or 7.604 to 7.605 OR M3 for a complete explicit method 5sin140 leading to AB or BC, e.g. sin25 OR M2 for a complete implicit method leading to AB or BC, e.g. sin 25 sin140  oe 5 BC or AB and M1 (dep on AB from trig) for 2  their AB + 10 OR B1 for any relevant angle E.g. BDA or BDC = 140, DAE or DCE = 50 or ADE or CDE = 40 or ADC = 80 6(d) 79.5 or 79.6 or 79.54 to 79.55... 5 B2 for [PR2 =] 245 or 245.1 to 245.2 or [PR =] 15.65 to 15.66 or 15.7 or M1 for [PR2 = ] 112 + 82 – 2  11  8  cos110 M2 for [cosPQR = ] 10 2  14 2  (their PR ) 2 oe 2  10  14 or M1 for (their PR)2 = 102 + 142 – 2  10  14cosPQR oe

This question in 0580/43 May/June 2024

6 The diagram shows a field ABCD. A straight path AC goes across the field. D 830 m 106° C NOT TO 420 m SCALE A 62° 1150 m B (a) Show that AC = 1028 m, correct to the nearest metre. [3] (b) Angle ACB is obtuse. Calculate angle ACB. Angle ACB = … [4] (c) Part of the field, triangle ACD, is sold for $41 500. Calculate the cost of 1 hectare of this part of the field. Give your answer correct to the nearest dollar. [1 hectare = 10 000 m2] $ … [4]

11 marks

Mark scheme: 6(a) 2 2 M2 or M1 for 420 + 830 −2 420  830  cos106 oe 420 2 + 830 2 −2 420  830  cos106 oe A1 for 1 057 474 …. 1028.3... A1 6(b) 99[.0] or 98.98 to 99.1[0…] 4 B3 for 80.89 to 81.02 1150sin62 or M2 for sin[ ACB =] oe 1028 1028 1150 or M1 for = oe sin62 sin ACB 6(c) 2477 cao nfww 4 B3 for answer 2476.9… or M2 for 1 P  420  830  sin106  = 41 500 2 10000 oe 1 or M1 for  420  830  sin106 oe 2

This question in 0580/41 Oct/Nov 2024

7 (a) C 60° North B NOT TO SCALE 85 m 129 m 39° A 72 m D The diagram shows a field, ABCD with B north of A. BD is a path across the field. AB = 85 m, AD = 72 m, BD = 129 m, angle BDC = 39° and angle BCD = 60°. (i) Show that angle CBD = 81°. [1] (ii) Calculate CD. … m [3] (iii) Show that angle ABD = 31.6°, correct to 1 decimal place. [4] (iv) Find the shortest distance from A to BD. … m [3] (v) Find the bearing of B from C. … [2] (vi) Trees are planted in the field. The number of trees planted is 1100 per hectare. Calculate the total number of trees planted in the field. [1 hectare = 10 000 m2] … [4] (b) A rectangle has an area of 9400 cm2, correct to the nearest 100 cm2. The length of the rectangle is 80 cm, correct to the nearest 10 cm. Calculate the upper bound of the width of the rectangle. … cm [3]

20 marks

Mark scheme: 7(a)(i) 180 – 60 – 39 [ = 81] 1 7(a)(ii) 147 or 147.1… 3 129sin(81) M2 for oe sin60 sin(81) sin60 or M1 for = oe CD 129 7(a)(iii) 85 2 + 129 2 − 72 2 M2 M1 for 72 2 = 852 + 129 2 −2 85  129cos ABD [cos = ] 2  85  129 31.58… A2 A1 for 0.851 to 0.852 9341 or or equivalent fraction 10965 7(a)(iv) 44.5 or 44.51 to 44.54 3 M2 for implicit correct method d e.g. = sin31.6 oe 85 or M1 for recognition that the line from A is perpendicular to BD 7(a)(v) 247 or 247.4… 2 M1 for 180 + (180 – 81 – 31.6) oe or for NBC = 180 – 81 – 31.6 oe or for NCB = 81 + 31.6 oe 7(a)(vi) 972 or 973 4 1 M1 for [ABD]  85  129sin31.6 oe 2 1 or  129  their 44.5 oe 2 1 M1 for [BCD ]  129  their147×sin39 oe 2 their total area M1 for  1100 10000 7(b) 126 nfww 3 9400 + 50 9400 to 9500 M2 for or 70 to 80 80 − 5 or M1 for 9350 or 9450 or 75 or 85 seen

This question in 0580/42 Oct/Nov 2024

12 D A NOT TO SCALE 20° E 8.7 cm 119° 10.9 cm B 11.4 cm C ABCD is a quadrilateral and E is a point on CD. AB = 8.7 cm, BC = 11.4 cm and CE = 10.9 cm. Angle ADE = 90°, angle ABC = 119° and angle CAE = 20°. (a) Show that AC = 17.37 cm, correct to 2 decimal places. [3] (b) Angle AEC is obtuse. Calculate angle ACE. Angle ACE = … [4] (c) Calculate the perimeter of quadrilateral ABCD. … cm [3]

10 marks

Mark scheme: 12(a) M2 M1 for 8.7 2 + 11.4 2 −2 8.7  11.4cos119 8.72 + 11.42 −2 8.7 11.4cos119 A1 for 301.8... 17.372 to 17.373 A1 12(b) 13.[0] or 13.02 to 13.03 4 17.37sin20 M2 for sinE = 10.9 10.9 17.37 or M1 for = oe sin20 sin E M1 for ACE = 180 – 20 – their obtuse AEC oe 12(c) 40.9 or 40.91 to 40.94 3 M1 for a correct implicit trig statement for AD AD e.g. sin(their acute ACE ) = oe 17.37 M1 for a correct implicit statement for CD CD e.g. cos(their acute ACE ) = oe 17.37 or CD2 = 17.372 – (theirAD)2 or for a correct statement for ED eg tan(180 – their obtuse AEC) = theirAD ED

This question in 0580/43 Oct/Nov 2024

17 North NOT TO A SCALE 65 m B 95 m 38° C The diagram shows three points A, B and C on horizontal ground. The bearing of C from A is 145° and angle ACB = 38°. AC = 65 m and BC = 95 m. (a) Find the bearing of B from C. … [2] (b) Show that AB = 59.3 m, correct to 1 decimal place. [3] (c) Angle BAC is obtuse. Work out the bearing of B from A. … [4]

9 marks

Mark scheme: 17(a) 287 2 M1 for North line at C with angle 35 or 145 marked on diagram at C isw or for 360 – 35 – 38 oe 17(b) M2 M1 for 652 + 952 −2 65  95cos38 652 + 952 −2 65  95cos38 A1 for 3518[.0] to 3518.1 59.31… A1 17(c) 244.4 to 244.6 4 B3 for [BAC =] 99.5 or 99.6 or 99.49 to 99.58… isw or for answer 225.4 to 225.5… OR 95sin38 M2 for [sin A = ] oe 59.3 59.32 + 65 2 − 95 2 or [cos A = ] 2  59.3  65 59.3 95 or M1 for = oe sin38 sin A or 952 = 59.32 + 652 −2 59.3  65cosA oe M1dep for 145 + their BAC leading to answer

This question in 0580/42 Feb/March 2025

17 B 56° 15.8 cm NOT TO 9.6 cm 12.1 cm C SCALE 47° A D The diagram shows a quadrilateral ABCD. (a) Calculate AD. AD = … cm [3] (b) Calculate the obtuse angle BDC. Angle BDC = … [4] (c) Calculate the area of the quadrilateral. … cm2 [3]

10 marks

Mark scheme: 17(a) 10.4 or 10.42… 3 2 2 M2 for 9.6 + 12.1 −2 9.6 12.1  cos56 OR M1 for 9.62 + 12.12 − 2  9.6 12.1 cos 56 A1 for 108.6 to 108.7 17(b) 107.2 to 107.3 4 B3 for 72.7[4…] OR −1 sin47  M3 for 180 − sin  15.8    12. 1  sin47 M2 for [sinBDC =] 15.8  12.1 sin BDC sin47 = or M1 for 15.8 12.1 17(c) 89.59 to 89.80… 3 1 M1 for  9.6  12.1  sin56 2 1 M1 for  15.8  12.1  sin(133 – their 2 BDC)

This question in 0580/41 May/June 2025

20 C NOT TO SCALE 10 m 56° 12 m D B 34° A The diagram shows a quadrilateral ABCD. CD = 10 m and DB = 12 m. Angle DBA = 90°, angle CDB = 56° and angle ADB = 34°. (a) Calculate the length of AB. AB = … m [2] (b) Calculate the area of the quadrilateral ABCD. … m2 [3] (c) Calculate the perimeter of the quadrilateral ABCD. … m [5] (d) Calculate the shortest distance from B to the line AD. … m [3]

13 marks

Mark scheme: 20(a) 8.09 or 8.094… 2 AB M1 for tan 34 = oe 12 20(b) 98.3 or 98.28 to 98.31 3 1 M1 for  10  12sin56 oe 2 1 M1 for  12  their (a) oe 2 20(c) 43[.0] to 43.1 5 2 2 M2 for [BC =] 10 +12 − 2×10×12cos56 or M1 for [BC [2] =] 102 +122 – 2×10×12cos56 12 M2 for [AD =] oe cos34 12 or M1 for cos 34 = oe AD 20(d) 6.71 or 6.706 to 6.710… 3 dist M2 for sin 34 = oe or 12 1 1  12  their ( a ) =  theirAD  dist oe 2 2 or M1 for recognition of perpendicular distance

This question in 0580/42 May/June 2025

21 In triangle STU, ST = 8 cm, SU = 9 cm and angle TSU = 50°. Calculate the area of triangle STU. … cm2 [2]

2 marks

Mark scheme: 21 27.6 or 27.57 to 27.58 2 1 M1 for 9 8 sin50 oe 2

This question in 0580/41 Oct/Nov 2025

29 B 13 cm 10 cm NOT TO SCALE y° 14 cm A C 38° 97° D (a) Calculate the value of y. y = … [3] (b) Calculate BD. BD = … cm [5]

8 marks

Mark scheme: 29(a) 63.0 or 63.02 to 63.03 3 10 2 + 14 2 − 132 M2 for [cos y =] oe 2  10  14 or M1 for 132 = 102 + 142 – 2 × 10 × 14 × cos y oe 29(b) 15.1 or 15.13 to 15.14 5 14sin38 M2 for [AD] = sin97 AD 14 or M1 for = oe sin38 sin97 M2 for 102 + (their AD)2 – 2 × 10 × their AD × cos(their y + 180 – 97 – 38) or M1 for angle BAD = their y + 180 – 97 – 38 soi

This question in 0580/41 Oct/Nov 2025

15 A NOT TO D SCALE 140° 112 m 180 m 300 m C B The diagram shows a field, ABCD, in the shape of a quadrilateral. BD is a straight path across the field. (a) Calculate BC. BC = … m [3] (b) Calculate angle DBC. Angle DBC = … [3] (c) The total area of the field, ABCD, is 35 900 m2. Work out the length of the shortest distance from D to AB. … m [4]

10 marks

Mark scheme: 15(a) 392 or 392.4 to 392.5 3 2 2 M2 for 300 + 112 −2 300 112  cos140 OR M1 for 3002 + 1122 − 2  300  112  cos140 A1 for 154 022[…] 15(b) 10.6 or 10.7 or 10.55 to 3 112sin140 M2 for oe 10.69 their (a) 300 2 + ( their (a) ) 2 − 112 2 or cos[ DBC ] = 2  300  their (a) 112 their (a) or M1 for = oe sin DBC sin140 or 112 2 = 300 2 + ( their (a) ) 2 − 2  300  their (a)  cos DBC oe 15(c) 279 or 278.9… 4 M3 for 1  1  (35 900 –  112  300  sin140 ) ÷   180  oe 2  2  OR 1 M1 for  112  300  sin140 oe 2 M1 for recognition that the shortest distance from D is perpendicular to AB

This question in 0580/42 Oct/Nov 2025

18 102° 41.3 cm x° y° 64.5 cm NOT TO SCALE 52.1 cm 70.2 cm (a) Calculate the value of x. x = … [3] (b) Calculate the value of y. y = … [3]

6 marks

Mark scheme: 18(a) 38.8 or 38.77 to 38.78 3 41.3sin102 M2 for [sin x =] oe or better 64.5 64.5 41.3 or M1 for = oe sin102 sin x 18(b) 73.2 or 73.16… 3 64.5 2 + 52.12 − 70.2 2 M2 for 2  64.5  52.1 or M1 for 70.22 = 64.52 + 52.12 –2 × 64.5 × 52.1 × cos y

This question in 0580/43 Oct/Nov 2025