Angles — Paper 4 · IGCSE Mathematics 0580

3 S North 7 km 30o NOT TO P R SCALE 55o 15 km 14 km Q The quadrilateral PQRS shows the boundary of a forest. A straight 15 kilometre road goes due East from P to R. (a) The bearing of S from P is 030° and PS = 7 km. (i) Write down the size of angle SPR. [1] (ii) Calculate the length of RS. [4] (b) Angle RPQ = 55º and QR = 14 km. (i) Write down the bearing of Q from P. [1] (ii) Calculate the acute angle PQR. [3] (iii) Calculate the length of PQ. [3] (c) Calculate the area of the forest, correct to the nearest square kilometre. [4]

16 marks

Mark scheme: 3 (a) (i) 60 B1 2 2 + 15 2 − 2 × 7 × 15 cos 60 M2 M1 – if one error in formula (ii) (RS ) = 7 2 13 A2 A1 for (RS ) = 169 www4 (b) (i) 145 B1 (ii) sin Q sin 55 M1 = o.e. 15 14 15 sin 55 M1 Implies previous method sin Q = 14 61.4 A1 www3 (iii) (R =) 63.6 B1 14 sin'63 . '6 M1 their sin(180 – 55 – b(ii)). Could be explicit (PQ ) = sin 55 equivalent cosine rule 15.3 A1 www3 (c) 1 7 . 15 sin'60' + 1 . 15 .'15 . 3'. sin 55 M2 M1 for one correct triangle area in working 2 2 (45.466 + 93.998) 139 or 140 www A2 A1 for 139.4 to 139.5 www4 [16]

This question in 0580/41 Oct/Nov 2005

6 18 m D C NOT TO SCALE 30 m 80º A B 26 m The diagram shows the plan of a garden. The garden is a trapezium with AB = 26 metres, DC = 18 metres and angle DAB = 80°. A straight path from B to D has a length of 30 metres. (a) (i) Using a scale of 1 : 200, draw an accurate plan of the garden. [3] (ii) Measure and write down the size of angle ADB and the size of angle DCB. [2] (iii) A second path is such that all points on it are equidistant from AB and from AD. Using a straight edge and compasses only, construct this path on your plan. [2] (iv) A third path is such that all points on it are equidistant from A and from D. Using a straight edge and compasses only, construct this path on your plan. [2] (v) In the garden, vegetables are grown in the region which is nearer to AB than to AD and nearer to A than to D. Shade this region on your plan. [1] (b) Use trigonometry, showing all your working, to calculate (i) angle ADB, [3] (ii) the length of BC, [4] (iii) the area of the garden. [3]

20 marks

Mark scheme: 6 (a) (i) AB=13 cm and BD=15 cm (± 2 mm) B1 Angle A = 80° (± 2°) B1 A,B,C,D correct within 4 mm B1 Dep. on B2 (ii) Angle ADB correct (57-61°) (± 2°) B1ft Either in working or written on diagram B1ft Angle DCB correct (101-105°) (± 2°) (iii) Acc. bisector of angle A with arcs B2ft B1 for accurate without/wrong arcs (at least 5 cm long) (± 2°)(± 2 mm) (iv) Acc. perp. bisector of AD with at least 1 B2ft B1 for accurate without/wrong arcs pair of arcs (± 2°)(± 2 mm) (at least 5 cm B1 for each if accurate with arcs but short long) (v) ‘Correct’ area shaded below their perp. B1 Dep. on at least B1 in (iii) and B1 in (iv) bisector and below their angle bisector (b) (i) sin D sin 80 M1 No M marks in (b) for measuring + using = lengths from diagram e.g. AD = 20 m 26 30 but allow 13, 15, 9 used for 26, 30, 18 in b 26 sin 80 M1dep dep on 1st M (sin D = ) 30 58.57 to 58.6° A1 www3 (ii) Angle BDC = 41.4 B1 ft Ft 100 – their 58.6 (BC2 =)182 + 302 – 2 × 18× 30 cos'41 '4. M1 Allow 41 or 42 for angle BDC square root of correct collection M1dep Dep on 1st M (413.88… ) 20.3 to 20.35 (m) cao A1 www4 (iii) 5.0 × 26 × 30 sin'41 '4.+ M2 M1 for correct area of one triangle oe (257.9 or 178.6). Must see calc for 5.0 × 18 × 30 sin'41 '4. trapezium height if used (30sin ‘41.4’) Allow 41 or 42 for angle BDC 436 to 437 (m2) cao A1 www3 [20] IGCSE – October/November 2007 0580/0581 04

This question in 0580/41 Oct/Nov 2007

5 E F Q NOT TO SCALE D G O C H P B 12 cm A A circle, centre O, touches all the sides of the regular octagon ABCDEFGH shaded in the diagram. The sides of the octagon are of length 12 cm. BA and GH are extended to meet at P. HG and EF are extended to meet at Q. (a) (i) Show that angle BAH is 135°. [2] (ii) Show that angle APH is 90°. [1] (b) Calculate (i) the length of PH, [2] (ii) the length of PQ, [2] (iii) the area of triangle APH, [2] (iv) the area of the octagon. [3] (c) Calculate (i) the radius of the circle, [2] (ii) the area of the circle as a percentage of the area of the octagon. [3]

17 marks

Mark scheme: 5 (a) (i) 360 ÷ 8 or (8 – 2)× 180 M1 allow 6× 180 IGCSE – May/June 2008 0580, 0581 04 180 – their (360 ÷ 8) o.e. ÷ 8 M1 dependent (ii) 45° used or use implied o.e. E1 Accept sketch with values (b) (i) l M1 For o.e. allow implicit expression = cos45 o.e. 12 A1 (PH = ) 8.49 (8.485….) www2 Accept 72 , 2 18 , 3 8 , 6 2 (ii) (PQ =) 2×their PH + 12 o.e. M1 (PQ =) 29.(0) (28.96–29.00) ft www2 A1 ft ft their PH accept surd form (iii) their PH×their PH ÷2 o.e. M1 (Area APH =) 36 (35.95–36.1) ft www2 A1 ft ft their PH (iv) (their PQ)2 – 4 × their area of triangle o.e. M2 If M0, M1 for a clear collection of areas (Area octagon = ) 695 (694.0–697.1) cao leading to the octagon possibly without www3 A1 any calculation shown (c) (i) 0.5 of their PQ o.e. M1 e.g. 6 + PH, 6tan67.5° 14.5 (14.47–14.53) cao www2 A1 accept surd form (ii) π × ( their r) 2 M1 (660.5…) their circle area × 100 M1 Dependent on first M1 and circle smaller their octagon area than the octagon 94.8 (94.35 to 95.60) cao www3 A1 [17]

This question in 0580/41 May/June 2008

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

7 (a) Calculate the volume of a cylinder of radius 31 centimetres and length 15 metres. Examiner's Give your answer in cubic metres. Use Answer(a) m3 [3] (b) A tree trunk has a circular cross-section of radius 31 cm and length 15 m. One cubic metre of the wood has a mass of 800 kg. Calculate the mass of the tree trunk, giving your answer in tonnes. Answer(b) tonnes [2] (c) NOT TO plastic SCALE sheet D C E The diagram shows a pile of 10 tree trunks. Each tree trunk has a circular cross-section of radius 31 cm and length 15 m. A plastic sheet is wrapped around the pile. C is the centre of one of the circles. CE and CD are perpendicular to the straight edges, as shown. For (i) Show that angle ECD = 120°. Examiner's Use Answer(c)(i) [2] (ii) Calculate the length of the arc DE, giving your answer in metres. Answer(c)(ii) m [2] (iii) The edge of the plastic sheet forms the perimeter of the cross-section of the pile. The perimeter consists of three straight lines and three arcs. Calculate this perimeter, giving your answer in metres. Answer(c)(iii) m [3] (iv) The plastic sheet does not cover the two ends of the pile. Calculate the area of the plastic sheet. Answer(c)(iv) m2 [1]

13 marks

Mark scheme: 7 (a) 4.53 or 4.526 – 4.530…. 3 SC2 for figs 453 or 4526 – 4530 If SC0, M1 for π × (figs 31)2 × 15 (b) 3.62 to 3.624 ft 2ft M1 for their (a) × figs 8 oe (c) (i) 360 – 2 × 90 – 60 oe 2 E2 The 90’s and the 60 must be clearly justified. Accept in diagram. SC1 for 60 or two 90’s soi in correct positions oe e.g 360 ÷ 3 scores 0 (ii) 0.649 (0.6492 to 0.6493) 2 M1 for π × figs 62 ÷ 3 (iii) 7.53 (7.527 or 7.528….) 3 M1 for their (ii) × 3 M1 (indep) for 18 × figs 31 This M is spoiled by extra lengths. (iv) 112.9 to 113 ft 1ft ft their (iii) × 15

This question in 0580/41 May/June 2010

3 (a) For D Examiner's NOT TO Use SCALE C 40° 30° A B E ABCD is a quadrilateral with angle BAD = 40°. AB is extended to E and angle EBC = 30°. AB = AD and BD = BC. (i) Calculate angle BCD. Answer(a)(i) Angle BCD = [3] (ii) Give a reason why DC is not parallel to AE. Answer(a)(ii) [1] (b) A regular polygon has n sides. 5n Each exterior angle is degrees. 2 Find the value of n. Answer(b) n = [3] (c) For Examiner's Use NOT TO SCALE O C 25° A B The diagram shows a circle centre O. A, B and C are points on the circumference. OC is parallel to AB. Angle OCA = 25°. Calculate angle OBC. Answer(c)Angle OBC = [3]

10 marks

Mark scheme: 3 (a) (i) 50 www3 3 B1 for angle ADB or ABD = 70 B1 for angle DBC = 80 (ii) Angle DCB ≠ angle CBE oe 1 Accept angle CDB ≠ angle ABD 5n 360 (b) 12 B3 M2 for = oe 2 n or M1 for 360 soi (c) 65 www 3 OAC = 25, CAB = 25, OBA = 50, BOC = 50, AOB = 80, AOC = 130 B1 each, max 2 IGCSE – October/November 2011 0580 43

This question in 0580/43 Oct/Nov 2011

4 (a) For B Examiner's Use A C NOT TO 42° SCALE D O E F A, B, C, D, E and F are points on the circumference of a circle centre O . AE is a diameter of the circle. BC is parallel to AE and angle CAE = 42°. Giving a reason for each answer, find (i) angle BCA, Answer(a)(i) Angle BCA = Reason [2] (ii) angle ACE, Answer(a)(ii) Angle ACE = Reason [2] (iii) angle CFE, Answer(a)(iii) Angle CFE = Reason [2] (iv) angle CDE. Answer(a)(iv) Angle CDE = Reason [2] (b) For Examiner's P Use NOT TO 5 cm SCALE O 12 cm Q In the diagram, O is the centre of the circle and PQ is a tangent to the circle at P. OP = 5 cm and OQ = 12 cm. Calculate PQ. Answer(b) PQ = cm [3] (c) C B D E NOT TO SCALE A G F In the diagram, ABCD and DEFG are squares. (i) In the triangles CDG and ADE, explain with a reason which sides and/or angles are equal. Answer (c)(i) [3] (ii) Complete the following statement. Triangle CDG is to triangle ADE. [1]

15 marks

Mark scheme: 4 In all parts of (a) candidates may refer to angles marked in diagram. Allow if clear even if reason is more complicated as long as it is full. Reasons dependent on correct answers (a) (i) 42 1 Not alternate segment Alternate oe 1 (ii) 90 1 Allow diameter semicircle oe 1 (iii) 42 1 same arc same segment oe 1 (iv) 138 1 key words must not be spoiled cyclic quad oe 1 (b) 10.9 (10.90 to 10.91) www 3 3 M2 for 12 2 − 5 2 oe i.e explicit or M1 for 12 2 = 5 2 + PQ 2 oe i.e implicit Allow full marks for 119 as final answer Use of trig method must be complete to explicit expression for possible M2 (c) (i) AD = CD and DE = DG 1 Extra pair of sides loses this mark. (Angle) CDG = (angle)ADE 1 Extra pair of angles loses this mark (Sides of) square or 90° + angle ADG R1 As in (a), for all 3 marks allow references to oe diagram if completely clear. R mark dep on at least one pair of sides stated or pair of angles stated (ii) Congruent 1 IGCSE – May/June 2012 0580 42

This question in 0580/42 May/June 2012

6 (a) For Examiner's 114° Use 2x° NOT TO SCALE x° (x – 10)° Find the value of x. Answer(a) x = [3] (b) (i) Write the four missing terms in the table for sequences A, B, C and D. Term 1 2 3 4 5 n Sequence A – 4 2 5 8 3n – 7 Sequence B 1 4 9 16 25 Sequence C 5 10 15 20 25 Sequence D 6 14 24 36 50 [4] (ii) Which term in sequence D is equal to 500? Answer(b)(ii) [2] x 2 − 16 (c) Simplify . 2 x 2 + 7 x − 4 Answer(c) [4]

13 marks

Mark scheme: 6 (a) (x =) 64 www 3 3 B2 for x + 2 x + x = 360 − 114 + 10 or better or M1 for x + 2 x + 114 + x − 10 = 360 (b) (i) –1 1 n 2 oe 1 5n oe 1 2 1 n + 5n oe (ii) 20 2 M1 for their n 2 + 5n = 500 or 20 and 25 seen x − 4 (c) Final answer cao www 4 4 B1 for (x – 4)(x + 4) 2 x − 1 B2 for (2x – 1)(x + 4) or SC1 for (2x + a)(x + b) where either a + 2b = 7 or ab = – 4

This question in 0580/42 May/June 2012

7 For W A X Examiner's Use NOT TO SCALE E B O 7 cm Z D C Y The vertices A, B, C, D and E of a regular pentagon lie on the circumference of a circle, centre O, radius 7 cm. They also lie on the sides of a rectangle WXYZ. (a) Show that (i) angle DOC = 72°, Answer(a)(i) [1] (ii) angle DCB = 108°, Answer(a)(ii) [2] (iii) angle CBY = 18°. Answer(a)(iii) [1] (b) Show that the length CD of one side of the pentagon is 8.23 cm correct to three significant For figures. Examiner's Use Answer(b) [3] (c) Calculate (i) the area of the triangle DOC, Answer(c)(i) cm2 [2] (ii) the area of the pentagon ABCDE, Answer(c)(ii) cm2 [1] (iii) the area of the sector ODC, Answer(c)(iii) cm2 [2] (iv) the length XY. Answer(c)(iv) cm [2] (d) Calculate the ratio area of the pentagon ABCDE : area of the rectangle WXYZ. Give your answer in the form 1 : n. Answer(d) 1 : [5]

19 marks

Mark scheme: 7 (a) (i) 360 Q 5 1 Accept longer correct methods (ii) (180 – 72) 2 M1 Accept [(5 – 2) × 180] or 360 / 5 M1 54 × 2 E1 Then Q 5 180 – 72 E1 (iii) 180 – 90 – 72 1 Accept other methods provided they are fully explained (b) 2 × 7 × sin(72/2) oe M2 M1 for 7 × sin (72/2) oe Alt methods M2 for [DC² =] 7² + 7² – 2.7.7 cos 72 or M1 for implicit version or M2 for (7 sin 72)/sin 54 or M1 for DC/sin 72 = 7/sin 54 oe 8.228 to 8.229 E1 Dep on M2 and with no errors seen IGCSE – October/November 2012 0580 41 (c) (i) 23.3[0..] 2 M1 for ½ × 7 × 7 × sin 72 oe (ii) 116.5 to 116.52 or 117 1 ft ft their (c)(i) × 5 (iii) 30.78 to 30.8 2 M1 for 72/360 × π 7 2 (iv) 12.66 to 12.67 or 12.7 2 M1 for 7 + 7 cos 36 oe [7 + 5.66…] e.g 8.23 cos54 + 8.23 sin72 oe [4.84 + 7.83] (d) 1.43 or 1.432 to 1.453 cao 5 B4 for area of rectangle = 168.3 to 169.2 www or area of triangular corners = 51.6 to 52.5 www or B3 for 13.3 to 13.32 seen or M2 for [ZY =] 8.23 + 2(8.23sin18) oe or 2 (8.23 sin 54) or 2 × 7 sin 72 oe or B1 for [CY =] 2.54[3] or 5.08 to 5.09 seen or [AX =] 6.65 to 6.66 seen

This question in 0580/41 Oct/Nov 2012

8 (a) For Examiner′s D Use 84° NOT TO SCALE E C x° x° 110° 110° A B In the pentagon ABCDE, angle EAB = angle ABC = 110° and angle CDE = 84°. Angle BCD = angle DEA = x°. (i) Calculate the value of x. Answer(a)(i) x = … [2] (ii) BC = CD. Calculate angle CBD. Answer(a)(ii) Angle CBD = … [1] (iii) This pentagon also has one line of symmetry. Calculate angle ADB. Answer(a)(iii) Angle ADB = … [1] (b) A, B and C lie on a circle centre O. Angle AOC = 3y° and angle ABC = (4y + 4)°. NOT TO Find the value of y. SCALE O 3y° C A (4y + 4)° B Answer(b) y = … [4] (c) For Examiner′s S Use NOT TO SCALE R Q 78° P In the cyclic quadrilateral PQRS, angle SPQ = 78°. (i) Write down the geometrical reason why angle QRS = 102°. Answer(c)(i) … [1] (ii) Angle PRQ : Angle PRS = 1 : 2. Calculate angle PQS. Answer(c)(ii) Angle PQS = … [3] (d) NOT TO 7.2 cm2 SCALE 5 cm2 l cm 6.9 cm The diagram shows two similar fi gures. The areas of the fi gures are 5 cm2 and 7.2 cm2. The lengths of the bases are l cm and 6.9 cm. Calculate the value of l. Answer(d) l = … [3] _____________________________________________________________________________________

15 marks

Mark scheme: 8 (a) (i) 118 2 M1 for (3 × 180 – 2 × 110 – 84) [÷ 2] or better (ii) 31 1FT FT (180 – their (i)) ÷ 2 (iii) 22 1FT FT 84 – 2 × their (ii) or 2 × their (ii) – 40, only if positive answer and less than 84 (b) 32 4 B2 for 360 – 3y = 2(4y + 4) oe and B1 for 11y = 352 oe or M1 for angle at centre = 2 × angle at circumference soi (c) (i) Opposite angles [cyclic quad] add to 1 180 (ii) 68 3 M1 for [angle PRS =] 102 ÷ 3 × 2 and M1 for angle PQS = angle PRS or angle PRQ = angle PSQ 5 (d) 5.75 3 M2 for 6.9 × oe 2.7 or M1 for evidence of ratio of areas = (ratio of sides)2 or sf = 1.2 2 2

This question in 0580/41 May/June 2013

2 (a) In this question show all your construction arcs and use only a ruler and compasses to draw For Examiner′s the boundaries of your region. Use This scale drawing shows the positions of four towns, P, Q, R and S, on a map where 1 cm represents 10 km. North P Q Scale: 1 cm to 10 km S R A nature reserve lies in the quadrilateral PQRS. The boundaries of the nature reserve are: ● equidistant from Q and from R ● equidistant from PS and from PQ ● 60 km from R ● along QR. (i) Shade the region which represents the nature reserve. [7] (ii) Measure the bearing of S from P. Answer(a)(ii) … [1] (b) A circular lake in the nature reserve has a radius of 45 m. For Examiner′s Use (i) Calculate the area of the lake. Answer(b)(i) … m2 [2] (ii) NOT TO SCALE A fence is placed along part of the circumference of the lake. This arc subtends an angle of 210° at the centre of the circle. Calculate the length of the fence. Answer(b)(ii) … m [2] _____________________________________________________________________________________

12 marks

Mark scheme: 2 (a) (i) Perpendicular bisector of QR 2 B1 for correct bisector ruled ruled with 2 correct sets of arcs centred Q and R Bisector of angle SPQ ruled with 2 B1 for correct angle bisector ruled correct arcs. (Marks on PS and PQ and correct pair of arcs) Compass drawn arc centre R B2 B1 for any compass drawn arc centre R not used with radius 6 cm (±2 mm) in any construction with no feathering Correct region shaded cao 1dep Dependent on all B4 marks for the correct loci (ii) 217 to 221 1 (b) (i) 6360 or 6361 to 6363 2 M1 for π × 452 210 (ii) 165 or 164.9 to 165 2 M1 for × 2π × 45 360

This question in 0580/43 May/June 2013

4 (a) One angle of an isosceles triangle is 48°. For Examiner′s Use Write down the possible pairs of values for the remaining two angles. Answer(a) … and … … and … [2] (b) Calculate the sum of the interior angles of a pentagon. Answer(b) … [2] (c) Calculate the sum of the angles a, b, c, d, e, f and g shown in this diagram. a g b NOT TO SCALE f c e d Answer(c) … [2] (d) The trapezium, ABCD, has four angles as shown. For Examiner′s All the angles are in degrees. Use B C 3y – 20 4x – 5 NOT TO SCALE 2x + 5 x + y – 10 A D (i) Show that 7x + 4y = 390 . Answer(d)(i) [1] (ii) Show that 2x + 3y = 195 . Answer(d)(ii) [1] (iii) Solve these simultaneous equations. Answer(d)(iii) x = … y = … [4] (iv) Use your answer to part (d)(iii) to fi nd the sizes of all four angles of the trapezium. Answer(d)(iv) … , … , … , … [1] _____________________________________________________________________________________

13 marks

Mark scheme: 4 (a) 48 and 84 2 B1 for each pair 66 and 66 (b) 540 2 M1 for 3 × 180 or (2 × 5 – 4) × 90 or 5 × (180 – 360 ÷ 5) oe (c) 1620 2 M1 for 7 × 360 – their 540 – 360 (d) (i) 2x + 5 + 3y – 20 + 4x – 5 + x + y – 1 Allow partial simplification but not 7x + 4y – 30 = 360 10 = 360 oe (ii) 2x + 5 + 3y – 20 = 180 1 (iii) [x =] 30, [y =] 45 nfww 4 M1 for correct multiplication M1 for correct elimination A1 x = 30 or y = 45 If 0 scored SC1 for correct substitution to find the other variable (iv) 65, 115, 115, 65 1 Accept in any order IGCSE – October/November 2013 0580 43

This question in 0580/43 Oct/Nov 2013

5 S North Scale: 2 cm to 3 km P L In the scale drawing, P is a port, L is a lighthouse and S is a ship. The scale is 2 centimetres represents 3 kilometres. (a) Measure the bearing of S from P. Answer(a) … [1] (b) Find the actual distance of S from L. Answer(b) … km [2] (c) The bearing of L from S is 160°. Calculate the bearing of S from L. Answer(c) … [1] (d) Work out the scale of the map in the form 1 : n. Answer(d) 1 : … [2] (e) A boat B is ● equidistant from S and L and ● equidistant from the lines PS and SL. On the diagram, using a straight edge and compasses only, construct the position of B. [5] (f) The lighthouse stands on an island of area 1.5 cm2 on the scale drawing. Work out the actual area of the island. Answer(f) … km2 [2] __________________________________________________________________________________________

13 marks

Mark scheme: 5 (a) [0]44 to [0]48 1 (b) 12.6 to 13.2 2 B1 for 8.4 to 8.8 seen (c) 340 1 (d) 1 : 150 000 2 M1 for × 100 000 soi (e) Arcs for perp bisector of SL 1 Two pairs of correct arcs Ruled perp bisector of SL 1 Within tolerance of overlay Arcs for bisector of angle PSL 1 Marks on PS and SL plus one pair of correct arcs Ruled bisector of angle PSL 1 Within tolerance of overlay B marked within accuracy 1 Within tolerance of overlay Dep on two correct bisectors drawn (f) 3.375 2 M1 for 1.5 × 1.52 or (2/3)2 seen

This question in 0580/41 May/June 2014

7 (a) E D t ° x° NOT TO SCALE x° C F q° p° 32° Y A B X ABCDEF is a hexagon. AB is parallel to ED and BC is parallel to FE. YFE and YABX are straight lines. Angle CBX = 32° and angle EFA = 90°. Calculate the value of (i) p, Answer(a)(i) p = … [1] (ii) q, Answer(a)(ii) q = … [2] (iii) t, Answer(a)(iii) t = … [1] (iv) x. Answer(a)(iv) x = … [3] (b) R x° Q S y° NOT TO SCALE 63° T U P P, Q, R and S are points on a circle and PS = SQ. PR is a diameter and TPU is the tangent to the circle at P. Angle SPT = 63°. Find the value of (i) x, Answer(b)(i) x = … [2] (ii) y. Answer(b)(ii) y = … [2] __________________________________________________________________________________________

11 marks

Mark scheme: 7 (a) (i) 148 1 (ii) 122 2 B1 for 58 seen at A or 32 seen at Y (iii) 148 1 (iv) 106 nfww 3 B1 for [sum of interior angles =] 720 1 and M1 for {(their 720) − (p+q+t+90)} 2 (b) (i) 63 2 B1 for angle RPS = 27 or 90 at P or at S seen or stated (ii) 54 2 B1 for their x or 63 or letter x at Q seen or state IGCSE – May/June 2014 0580 43 Qu Answers Mark Part Marks

This question in 0580/43 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

9 (a) The points A, B, C and D lie on a circle. AC is a diameter of the circle. ST is the tangent to the circle at A. S A 48° T 27° NOT TO B D y° x° SCALE C Find the value of (i) x, Answer(a)(i) x = … [2] (ii) y. Answer(a)(ii) y = … [2] (b) The diagram shows a shaded shape formed by three semi-circular arcs. The radius of each semi-circle is shown in the diagram. NOT TO SCALE 6 cm 4 cm 2 cm (i) Calculate the perimeter of the shaded shape. Answer(b)(i) … cm [2] (ii) The shaded shape is made from metal 1.6 mm thick. Calculate the volume of metal used to make this shape. Give your answer in cubic millimetres. Answer(b)(ii) … mm3 [5]

11 marks

Mark scheme: 9 (a) (i) 42 2 B1 for BAC = 90 − 48 (ii) 111 2 B1 for 111 or 69 or ACD = 27 correctly placed on diagram or indicated (b) (i) 37.7 or 37.69 to 37.704 nfww 2 M1 for 6π + 4π ± 2π oe (ii) 12100, 12060, 12070, 5 SC4 for answer with figs 121 12062.4 to 12065.6 nfww or 1206 to 1207 OR M2 for total area = 12 π 6 2 + 12 π 4 2 − 12 π 2 2 or 12 π 60 2 + 12 π 40 2 − 12 π 20 2 or M1 for 12 π 6 2 or 12 π 4 2 or 12 π 2 2 or 12 π 60 2 or 12 π 40 2 or 12 π 20 2 A1 for area = 75.39 to 75.41 or 7539 to 7541 and M1 dep for volume = their area × thickness

This question in 0580/41 May/June 2015

2 D NOT TO O SCALE C E 48° A B In the diagram, B, C, D and E lie on the circle, centre O. AB and AD are tangents to the circle. Angle BAD = 48°. (a) Find (i) angle ABD, Answer(a)(i) Angle ABD = … [1] (ii) angle OBD, Answer(a)(ii) Angle OBD = … [1] (iii) angle BCD, Answer(a)(iii) Angle BCD = … [2] (iv) angle BED. Answer(a)(iv) Angle BED = … [1] (b) The radius of the circle is 15 cm. Calculate the area of triangle BOD. Answer(b) … cm2 [2] (c) Give a reason why ABOD is a cyclic quadrilateral. Answer(c) … … [1] __________________________________________________________________________________________

8 marks

Mark scheme: 2 (a) (i) 66 1 (ii) 24 1FT FT 90 – their (a)(i) (iii) 66 2FT FT 90 – their (a)(ii) M1 for [BOD =] 180 – 48 or 180 – 2 × their (a)(ii) (iv) 114 1FT FT 180 – their (a)(iii) (b) 83.6 or 83.60[…] 2 M1 for 1 × 15 × 15 × sin(180 – 48) oe 2 or 1 × 15 × 15 × sin(180 – 2 × their (a)(ii)) oe 2 (c) Opposite angles add up to 180 1 OR Angle in a semicircle [ =90]

This question in 0580/42 May/June 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

6 (a) E D 120° 140° NOT TO SCALE F C A B In the hexagon ABCDEF, AB is parallel to ED and AF is parallel to CD. Angle ABC = 90°, angle CDE = 140° and angle DEF = 120°. Calculate angle EFA. Answer(a) Angle EFA = … [4] (b) D C NOT TO 30° SCALE X 100° B A In the cyclic quadrilateral ABCD, angle ABC = 100° and angle BDC = 30°. The diagonals intersect at X. (i) Calculate angle ACB. Answer(b)(i) Angle ACB = … [2] (ii) Angle BXC = 89°. Calculate angle CAD. Answer(b)(ii) Angle CAD = … [2] (iii) Complete the statement. Triangles AXD and BXC are … . [1] (c) R NOT TO S SCALE Y Q P P, Q, R and S lie on a circle. PR and QS intersect at Y. PS = 11 cm, QR = 10 cm and the area of triangle QRY = 23 cm2. Calculate the area of triangle PYS. Answer(c) … cm2 [2] (d) A regular polygon has n sides. n Each exterior angle is equal to 10 degrees. (i) Find the value of n. Answer(d)(i) n = … [3] (ii) Find the size of an interior angle of this polygon. Answer(d)(ii) … [2]

16 marks

Mark scheme: 6 (a) 100 nfww 4 M3 for a correct calculation that would lead to the answer or B2 two correct relevant different size angles in their diagram or one relevant angle and total in their polygon or angle EDA + angle FAD = 140 or B1 for one relevant angle or total in their polygon (b) (i) 50 2 B1 for angle ADC = 80 or angle BAC = 30 or angle ADB = 50 soi (ii) 41 2FT FT 91 – their (b)(i) B1 for angle XBC = 41 (iii) Similar 1 2    10  2 (c) 27.8 or 27.83 2 M1 for evidence of  11 or 1.21 or    10   11  or 0.826(4…) (d) (i) 60 3 n 360 M2 for = oe 10 n 180( n − 2) n e.g. = 180 − n 10 or B1 for exterior sum = 360 or 180(n – 2) seen (ii) 174 2 their n 360 M1 for or for their n < 1800 10 their n Qu Answers Mark Part Marks

This question in 0580/43 May/June 2015

7 The scale drawing shows the positions of three towns A, B and C on a map. The scale of the map is 1 centimetre represents 10 kilometres. C North A North Scale: 1 cm to 10 km B (a) Find the actual distance AB. Answer(a) … km [1] (b) Measure the bearing of A from B. Answer(b) … [1] (c) Write the scale 1 cm to 10 km in the form 1 : n. Answer(c) 1 : … [1] (d) A national park lies inside the triangle ABC. The four boundaries of the national park are • equidistant from C and B • equidistant from AC and CB • 15 km from CB • along AB. On the scale drawing, shade the region which represents the national park. Leave in your construction arcs. [7] (e) On the scale drawing, a lake inside the national park has area 0.4 cm2. Calculate the actual area of the lake. Answer(e) … km2 [2] __________________________________________________________________________________________

12 marks

Mark scheme: 7 (a) 123 to 127 1 (b) 288 to 292 1 (c) [1:] 1 000 000 1

This question in 0580/41 Oct/Nov 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

5 NOT TO A SCALE North 510 km B 720 km 40° C A plane flies from A to C and then from C to B. AC = 510 km and CB = 720 km. The bearing of C from A is 135° and angle ACB = 40°. (a) Find the bearing of (i) B from C, … [2] (ii) C from B. … [2] (b) Calculate AB and show that it rounds to 464.7 km, correct to 1 decimal place. [4] (c) Calculate angle ABC. Angle ABC = … [3]

11 marks

Mark scheme: 5 (a) (i) 275 2 M1 for 360 – 40 – 45 oe (ii) 095 2FT FT their (a) – 180 M1 for their (a) – 180 oe or 180 – 40 – 45 (b) 464.66 to 464.67 [= 464.7] 4 M2 for 510 2 + 720 2 −×2 510 × 720cos40 or M1 for correct implicit equation A1 for 215 900 to 215 920 510sin(40) (c) 44.9 or 44.86 to 44.87… 3 M2 for 464.7 or M1 for correct implicit equation

This question in 0580/43 May/June 2016

8 (a) North 110° A North NOT TO 38 km 50 km SCALE C B 280° A, B and C are three towns. The bearing of B from A is 110°. The bearing of C from B is 280°. AC = 38 km and AB = 50 km . (i) Find the bearing of A from B. … [2] (ii) Calculate angle BAC. Angle BAC = … [5] (iii) A road is built from A to join the straight road BC. Calculate the shortest possible length of this new road. … km [3] (b) Town A has a rectangular park. The length of the park is x m. The width of the park is 25 m shorter than the length. The area of the park is 2200 m2. (i) Show that x 2 - 25x - 2200 = 0 . [1] (ii) Solve x 2 - 25x - 2200 = 0 . Show all your working and give your answers correct to 2 decimal places. x = … or x = … [4] Question 9 is printed on the next page.

15 marks

Mark scheme: 8(a)(i) 290 2 M1 for 180 + 110 oe 8(a)(ii) 156.8 or 156.7[9..] 5 B1FT for CBA = 10° (their (a) – 280) and B3 for [angle ACB = ]13.2° 50sin(their10) or M2 for [sin C] = 38 50 38 or M1 for = oe sin C sin ( their10) 8(a)(iii) 8.68 or 8.677 to 8.684 3 M2 for [ x = ] 50sin(their10) oe x or M1 for sin ( their10 ) = oe 50 or M1 for a correct right-angled triangle drawn with 50 as hypotenuse 8(b)(i) x (x – 25) = 2200 1 and no errors seen 8(b)(ii) 2 B2 2 −−( 25) ± ( −25) − 4(1)( − 2200) B1 for ( −25) − 4(1)( −2200) or better or 2(1) 2 § 25 · or for ¨ x − ¸ oe better © 2 ¹ −−( 25) + q −−( 25) − q or B1 for or 2(1) 2(1) or both 25 § 25 · 2 or for + or − ¨ ¸ + 2200 2 © 2 ¹ –36.04 and 61.04 final answer B1,B1 If B0B0, SC1 for values in ranges –36.042 to –36.041 and 61.041 to 61.042 seen or for answers –36[.0] or –36.042 to –36.041 and 61[.0] or 61.041 to 61.042 or –36.04 and 61.04 seen in working or for –61.04 and 36.04 as final ans

This question in 0580/41 May/June 2017

2 (a) A P NOT TO 24° SCALE z° B D R y° Q C x° E 38° S PQ is parallel to RS. ABC and ADE are straight lines. Find the values of x, y and z. x = … y = … z = … [3] (b) C NOT TO SCALE D B 42° A The points A, B, C and D lie on the circumference of the circle. AB = AD, AC = BC and angle ABD = 42°. Find angle CAB. Angle CAB = … [3] (c) P NOT TO O SCALE 146° Q R S The points P, Q, R and S lie on the circumference of the circle, centre O. Angle QOS =146°. Find angle QRS. Angle QRS = . … [2]

8 marks

Mark scheme: 2(a) 38 1 118 1 62 1FT FT 180 – their y 2(b) 69 3 B2 for ACB = 42 or B1 for ADB = 42 If zero scored, SC1 for ACB = their ADB 2(c) 107 2 B1 for QPS = 73 or [reflex] QOS = 214

This question in 0580/43 May/June 2017

1 (a) The angles of a triangle are in the ratio 2 : 3 : 5. (i) Show that the triangle is right-angled. [1] (ii) The length of the hypotenuse of the triangle is 12 cm. Use trigonometry to calculate the length of the shortest side of this triangle. … cm [3] (b) The sides of a different right-angled triangle are in the ratio 3 : 4 : 5. (i) The length of the shortest side is 7.8 cm. Calculate the length of the longest side. … cm [2] (ii) Calculate the smallest angle in this triangle. … [3]

9 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 180 ÷ (2 + 3 + 5) × 5 [= 90] 1 with no errors seen 1(a)(ii) 7.05 or 7.053…. 3 x M2 for = sin36 oe or better 12 or B1 for 36 or 54 seen 1(b)(i) 13 2 M1 for 7.8 ÷ 3 soi 1(b)(ii) 36.9 or 36.86 to 36.87 3 B1 for smallest angle identified 3 M1 for sin[ ] = oe 5 7.8 or sin[ ] = oe their ( b )(i) If zero scored, SC1 for calculation of 53.1

This question in 0580/43 Oct/Nov 2017

8 The diagram shows the positions of three cities, Geneva (G), Budapest (B) and Hamburg (H). North H NOT TO 67° SCALE 928 km 864 km B G (a) A plane flies from Geneva to Hamburg. The flight takes 2 hours 20 minutes. Calculate the average speed in kilometres per hour. … km/h [2] (b) Use the cosine rule to calculate the distance from Geneva to Budapest. … km [4] (c) The bearing of Budapest from Hamburg is 133°. (i) Find the bearing of Hamburg from Budapest. … [2] (ii) Calculate the bearing of Budapest from Geneva. … [4]

12 marks

Mark scheme: 8(a) 370 or 370.2 to 370.3 2 M1 for 864 ÷ their time 8(b) 991 or 990.5 … 4 M2 for 8642 + 9282 – 2 × 864 × 928cos 67 or M1 for correct implicit version A1 for 981100 to 981110 8(c)(i) 313 2 M1 for 180 + 133 or 360 – 47 8(c)(ii) [0]79.5 to [0]79.6 … 4 928 × sin67 864 × sin67 M2 for or oe their 991 their 991 or M1 for implicit form of either A1 for [angle HGB =] 59.5 to 59.6 … or [angle HBG =] 53.4 or 53.37 to 53.42 M1 dep for their angle HGB + 20 leading to answer or for 133 – their angle HBG leading to answer

This question in 0580/42 Oct/Nov 2018

2 (a) A C 26° NOT TO SCALE F B x° D E AC is parallel to FBD, ABC is an isosceles triangle and CBE is a straight line. Find the value of x. x = … [3] (b) S P 58° 17° T NOT TO SCALE y° Q The diagram shows a circle with diameter PQ. SPT is a tangent to the circle at P. Find the value of y. y = … [5]

8 marks

Mark scheme: 2(a) 103 3 M1 for angle ABC or angle ACB = 1 (180 − 26 ) 2 oe M1 for angle ABF = 26 or angle CBD or angle FBE = 77 or exterior angle ACB = 103 correctly identified or in correct position 2(b) 75 5 B4 for 105 at a or b or 73 at c and 32 at d or B3 for 58 at m or 58 at e and 17 at k or B2 for 32 at d and 90 soi at (c+k) or 32 at d and 17 at k or 73 at c or B1 for 90 soi at (c + k) or between tangent and radius or 32 at d or 17 at k S P d 58° 17° T c m a y° b k e Q

This question in 0580/42 May/June 2019

7 (a) Show that each interior angle of a regular pentagon is 108°. [2] (b) D E C NOT TO O SCALE M X A B The diagram shows a regular pentagon ABCDE. The vertices of the pentagon lie on a circle, centre O, radius 12 cm. M is the midpoint of BC. (i) Find BM. BM = … cm [3] (ii) OMX and ABX are straight lines. (a) Find BX. BX = … cm [3] (b) Calculate the area of triangle AOX. … cm2 [3]

11 marks

Mark scheme: 7(a) 360 M2 360 180 − or or M1 for or (5 − 2 ) × 180 5 5 (5 − 2 ) × 180 ( 2 × 5 − 4 ) × 90 or 90(2 × 5 – 4) or or or 3 × 180 ÷ 5 5 5 or 6 × 90 ÷ 5 5 × 180 − 360 or 5 × 180 – 360 5 5 − 2 × 180 If 0 scored, SC1 for 5 7(b)(i) 7.05 or 7.053… 3 M2 for 12 × cos54 oe or M1 for implicit form or B1 for length of edge of pentagon = 14.1 to 14.11 If 0 scored, SC1 for right angle at M 7(b)(ii)(a) 22.8 or 22.81 to 22.83… nfww 3 their(b)(i) M2 for oe cos72 or M1 for implicit form oe or B1 for AX = 36.9 or 36.93 to 36.94 7(b)(ii)(b) 179 or 179.1 to 179.3… 3 M2 for 1 × 12 × their AX × sin54 oe 2 or 1 × 12 × theirOX × sin108 oe 2 or 1 × their AX × theirOX × sin18 2 or 12 × 12 2 × sin72 + areaOBX oe or 1 2 × 12 2 × sin72 + area OMB + area MBX oe or M1 for a correct method to find area of one relevant triangle AOB, OMB, MBX, OBX or ONX seen

This question in 0580/42 May/June 2019

1 (a) B NOT TO SCALE 55° q° p° A C 48° D In the diagram, AC and BD are straight lines. Find the value of p and the value of q. p = … q = … [3] (b) The angles of a quadrilateral are x°, (x + 5)° , (2x - 25)° and (x + 10)° . Find the value of x. x = … [3] (c) A regular polygon has 72 sides. Find the size of an interior angle. … [3] (d) D NOT TO P y° SCALE v° C x° O 60° u° 20° A w° B Q A, B, C and D lie on the circle, centre O, with diameter AC. PQ is a tangent to the circle at A. Angle PAD = 60° and angle BAC = 20° . Find the values of u, v, w, x and y. u = … , v = … , w = … , x = … , y = … [6] (e) A, B and C lie on the circle, centre O. Angle AOC = (3x + 22) ° and angle ABC = 5x° . Find the value of x. NOT TO SCALE O (3x + 22)° A C 5x° B x = … [4]

19 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) [p = ] 132 3 B1 for 132 [=p] [q = ] 77 B2 for 77 [=q] or M1 for 180 – (55 + 48) oe or for their p – 55 1(b) 74 3 B2 for 5x – 10 = 360 or M1 for x + ( x + 5) + ( 2 x − 25) + ( x + 10) = 360 or for 5x – 10 = k 1(c) 175 3 360 180( 72 − 2 ) M2 for 180 − or for 72 72 360 or M1 for or for 180 (72 – 2) 72 1(d) [u = ] 30 6 B1 for 30 [v = ] 60 B1 for 60 [w = ] 60 B1 for 60 FT their v [x = ] 120 B1 for 120 FT 2 × their w [y = ] 40 B2 for 40 or B1 for angle BDC = 20 or angle ADO = 30 or angle ADB = 70 1(e) 26 4 B3 for 360 – 22 = 10x + 3x oe or better or for 5x + 1.5x = 180 – 11 oe or better or M2 for 360 – (3x + 22) = 2 × 5x oe 1 or for 5x + (3 x + 22) = 180 oe 2 or SC2 for 360 + 22 = 10x + 3x oe or better or M1 for 180 – 5x, 10x or 360 – (3x + 22) correctly placed on the diagram or identified or for angle A + angle C = 5x

This question in 0580/41 Oct/Nov 2019

2 (a) The diagram shows a triangle and a quadrilateral. All angles are in degrees. NOT TO 3a + 2b SCALE 3b + 10 8a a + 2b 2a b + 50 4b - 2a (i) For the triangle, show that 3a + 5b = 170 . [1] (ii) For the quadrilateral, show that 9a + 7b = 310 . [1] (iii) Solve these simultaneous equations. Show all your working. a = … b = … [3] (iv) Find the size of the smallest angle in the triangle. … [1] (b) Solve the equation 6x - 3 =- 12 . x = … [2] (c) Rearrange 2 (4x - y) = 5x - 3 to make y the subject. y = … [3] (d) Simplify. 2 (27x 9 ) 3 … [2] (e) Simplify. x 2 + 5x x 2 - 25 … [3]

16 marks

Mark scheme: 2(a)(i) 2a + a + 2b + 3b + 10 = 180 1 leading to 3a + 5b = 170 without error or omission 2(a)(ii) 8a + 3a + 2b + b + 50 + 4b – 2a = 360 1 leading to 9a + 7b = 310 without error or omission 2(a)(iii) Correct method to eliminate one variable M1 [a =]15 A2 A1 for each correct value [ b=]25 If 0 scored, SC1 for two values that satisfy one of the equations or for two correct answers with no/incorrect working 2(a)(iv) 30 1 2(b) 1 3 2 M1 for 6x = –12 + 3 or better –1.5 or − 12 or − 2 2(c) 3 x + 3 3 M1 for 8x – 2y = 5x – 3 oe final answer 2 1 or 4 x − y = ( 5 x − 3 ) 2 M1FT for isolating the y term correctly 2(d) 9x6 2 1 M1 for (3x3)2 or 729x18 3 seen ( ) or for 9xk or kx6 as final answer 2(e) x 3 M1 for x(x + 5) final answer nfww M1 for (x – 5)(x + 5) x − 5

This question in 0580/43 Oct/Nov 2019

4 (a) A rectangle measures 8.5 cm by 10.7 cm, both correct to 1 decimal place. Calculate the upper bound of the perimeter of the rectangle. … cm [3] (b) B C D E 80° NOT TO SCALE 9 cm h 40° A 12 cm F ABDF is a parallelogram and BCDE is a straight line. AF = 12 cm, AB = 9 cm, angle CFD = 40° and angle FDE = 80°. (i) Calculate the height, h, of the parallelogram. h = … cm [2] (ii) Explain why triangle CDF is isosceles. … … [2] (iii) Calculate the area of the trapezium ABCF. … cm2 [3] (c) C B 12 cm NOT TO SCALE O 21° D A A, B, C and D are points on the circle, centre O. Angle ABD = 21° and CD = 12 cm. Calculate the area of the circle. … cm2 [5] (d) x° NOT TO 8 cm 9.5 cm SCALE The diagram shows a square with side length 8 cm and a sector of a circle with radius 9.5 cm and sector angle x°. The perimeter of the square is equal to the perimeter of the sector. Calculate the value of x. x = … [3]

18 marks

Mark scheme: 4(a) 38.6 3 M2 for [2 ×] (8.5 + 0.05 + 10.7 + 0.05) or M1 for 8.5 + 0.05 or 10.7 + 0.05 4(b)(i) 8.86 or 8.863… 2 h M1 for = sin 80 or better oe 9 4(b)(ii) ∠CDF = 100 leading to ∠DCF = 40 M1 Implied by 180-(100 + 40) = 40 Or or ∠EDF = 80 leading to ∠DCF = 40 80 – 40 ‘two equal angles’ A1 With no incorrect work seen 4(b)(iii) 66.5 or 66.45 to 66.47… 3 M2 for 0.5(3 + 12) × their (b)(i) or 12 × their (b)(i) – 0.5 × 9 × 9 × sin 100 oe or B1 for DC = 9 or BC = 3 4(c) 130 nfww or 129.6 to 129.8 5 B1 for ∠ACD = 21º or ∠CAD = 69º Method 1 12 M2 for cos 21 = oe AC or M1 for ∠ADC = 90 soi M1 for π(their AC/2)2 OR Method 2 12 r M2 for = oe sin138 sin 21 or M1 for ∠COD = 138 soi M1 for π (their r ) 2 OR Method 3 6 M2 for cos 21 = oe OC or M1 for ∠CXO = 90 soi where X is the point where the perpendicular from O meets the chord CD M1 for π ( their OC) 2 4(d) 78.4 or 78.37 to 78.41 3 M2 for x × 2 × π × 9.5 + 2 × 9.5 = 4 × 8 oe 360 x or M1 for × 2 × π × 9.5 360 After M0, SC1 for 9.5x + 19 = 32 oe

This question in 0580/41 Oct/Nov 2020

3 (a) a° NOT TO 126° SCALE c° b° 63° The diagram shows two straight lines intersecting two parallel lines. Find the values of a, b and c. a = … b = … c = … [3] (b) Q NOT TO SCALE S R 58° x° P Points R and S lie on a circle with diameter PQ. RQ is parallel to PS. Angle RPQ = 58° . Find the value of x, giving a geometrical reason for each stage of your working. … … … x = … [3] (c) NOT TO O SCALE 142° C A y° B Points A, B and C lie on a circle, centre O. Angle AOC = 142° . Find the value of y. y = … [2]

8 marks

Mark scheme: 3(a) 126 3 B1 for each 54 117 3(b) angle [in a] semicircle is 90 B1 Do not accept triangle for angle Allied, co-interior [add to 180] B1 or Angles in triangle [ = 180] and alternate oe 32 B1 3(c) 109 2 B1 for 218 or 71 in correct places or correctly labelled

This question in 0580/42 Feb/March 2021

5 (a) ABCDEFGH is a regular octagon with sides of length 6 cm. The diagram shows part of the octagon. O is the centre of the octagon and M is the midpoint of AB. A M B NOT TO SCALE O (i) (a) Show that angle OAM is 67.5°. [2] (b) Calculate the area of the octagon. … cm2 [4] (ii) Find the area of the circle that passes through the vertices of the octagon. … cm2 [3] (b) NOT TO SCALE 4 m 0.45 m The diagram shows a horizontal container for water with a uniform cross-section. The cross-section is a semicircle. The radius of the semicircle is 0.45 m and the length of the container is 4 m. (i) Calculate the volume of the container. … m3 [2] (ii) NOT TO SCALE 0.3 m The greatest depth of the water in the container is 0.3 m. The diagram shows the cross-section. Calculate the number of litres of water in the container. Give your answer correct to the nearest integer.

17 marks

Mark scheme: 5(a)(i)(a) 8 2 180 8 2    oe M2 8 2 180 8   or 360 2 8 4 90 8   5(a)(i)(b) 174 or 173.8 … 4 M3 for 1 6 2 OM oe or   2 1 2   OA sin45 oe or 1 6 67.5 2   OA sin oe where OA and OM are as in the M2 or M2 for 3 tan67.5  OM oe or for 3 67.5       OA cos or 6 67.5 45 sin sin oe or M1 for tan67.5 3  OM oe or for 3 67.5 cos OA oe or for 45 67.5 6  sin sin OA oe 5(a)(ii) 193 or 193.0 to 193.1 3 M2 for 2 3 67.5        cos oe or M1 for 3 67.5 cos r or 45 67.5 6  sin sin r Question Answer Marks Partial Marks 5(b)(i) 1.27 or 1.272 to 1.273 2 M1 for 2 1 0.45 4 2           or   2 1 0.45 4 2     5(b)(ii) 742 or 743 6 M5 for a method leading to the volume of water e.g. 2 0.15 cos 0.45 4 {2 0.45 360            inv 2 1 0.15 0.45 sin 2 cos 2 0.45                inv } oe OR M2   2 0.15 cos 0.45 2 0.45 360           inv oe or   2 0.15 90 cos 0.45 2 0.45 360            inv oe or M1 for use of 2 0.45 360     oe M2 for 2 1 0.15 0.45 sin 2 cos 2 0.45               inv oe or   1 0.15 0.15 0.45 sin cos 2 2 0.45                 inv oe Question Answer Marks Partial Marks 5(b)(ii) or M1 for use of 2 1 0.45 2  × sinθ oe or  1 2 0.15 0.45 sinβ 2     oe If 0 scored, SC1 for invcos 0.15 0.45       or invsin 0.15 0.45       or 2 2 0.45 0.15  soi

This question in 0580/41 May/June 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

1 (a) 42° NOT TO SCALE x° The diagram shows an isosceles triangle with the base extended. Find the value of x. x = … [3] (b) The diagram shows three lines meeting at a point. The ratio a : b : c = 3 : 4 : 5. Find the value of c. a° NOT TO c° b° SCALE c = … [3] (c) A regular pentagon has an exterior angle, d. A regular hexagon has an interior angle, h. d Find the fraction . h Give your answer in its simplest form. … [4] (d) S R x° ( x + 20)° NOT TO SCALE ( 3x – 40)° Q ( 2x – 5)° P Show that PQRS is a cyclic quadrilateral. [5] (e) B A 50° 9 cm NOT TO O SCALE The diagram shows a circle of radius 9 cm, centre O. The minor sector AOB, with sector angle 50°, is removed from the circle. Calculate the length of the major arc AB. … cm [3]

18 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) 111 3  42 M2 for 180 –180 oe or 42 + 2 180  42 oe 2 180  42 or M1 for oe 2 1(b) 150 3 M1 for k ÷ (3 + 4 + 5) [×p] where p = 1, 3, 4 or 5 5 or oe 12 B1 for 360 used 1(c) 3 4 72 cao nfww B3 for 5 120 or B2 for [d = ] 72 or [h = ] 120 or M1 for 360 ÷ 5 oe isw or 180 – (360 ÷ 6) isw or for (6 – 2) × 180 [÷ 6] 1(d) x + 2x – 5 + x + 20 + 3x – 40 = 360 M1 Accept equivalent equation e.g. 7x – 25 = 360 7x = 360 + 5 – 20 + 40 or better M1 FT their equation, accept e.g. 7x = 385 x = 55 B1 55 and 125 B1dep Dep on M1M1B1 or 105 and 75 Accept 55 + 3 × 55 – 40 = 180 or 2 × 55 – 5 + 55 + 20 = 180 If B0 scored, SC1 for 55, 75, 105 and 125 Opposite angles sum to 180 oe A1 Dep on M1M1B1B1 [so PQRS is a cyclic quadrilateral ] 1(e) 48.7 or 48.69 to 48.70… 3 360  50 M2 for 2 π oe9 360 50 or M1 for  2  π oe9 360

This question in 0580/42 May/June 2023

2 (a) 38° NOT TO SCALE a° b° The diagram shows a straight line intersecting two parallel lines. Find the value of a and the value of b. a = … b = … [2] (b) Calculate the interior angle of a regular 12-sided polygon. … [2] (c) N NOT TO SCALE f ° P O g° 56° A B M The diagram shows a circle, centre O. The points M, N and P lie on the circumference of the circle. AMB is a tangent to the circle at M. Find the value of f and the value of g. f = … g = … [3] (d) NOT TO SCALE 24° k ° 27° The diagram shows a cyclic quadrilateral. Find the value of k. k = … [2]

9 marks

Mark scheme: 2(a) 142 2 B1 for each 142 FT angle b = their angle a 2(b) 150 2 360 M1 for oe isw 12 or 180  12  2  oe isw 2(c) 56 B1 34 B2 M1 for angle at centre = 2 × their 56 oe soi or for angle OMB = 90 oe soi 2(d) 51 2 B1 for opp angle = 129 soi

This question in 0580/42 May/June 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

5 A B (a) In triangle ABC, AB = 8 cm, AC = 7 cm and BC = 5 cm. Using a ruler and compasses only, construct triangle ABC. The side AB has been drawn for you. [2] (b) Measure angle ACB. … [1] (c) Triangle ABC is a scale drawing of a field. (i) The scale is 1 : 10 000. Find the actual distance from A to B. Give your answer in kilometres. … km [1] (ii) B is due east of A. Find the bearing of A from B. … [1]

5 marks

Mark scheme: 5(a) Triangle accurately completed 2 B1 for accurate triangle with no/incorrect arcs or with arcs at C. for accurate triangle with arcs but with AC and BC reversed 5(b) 80 to 85 1 FT their triangle if 0 scored in (a) 5(c)(i) 0.8 1 5(c)(ii) 270 1

This question in 0580/41 Oct/Nov 2025