Right-angled triangles — 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 P 13 cm NOT TO SCALE D C E 6 cm A 8 cm B The diagram shows a pyramid on a horizontal rectangular base ABCD. The diagonals of ABCD meet at E. P is vertically above E. AB = 8 cm, BC = 6 cm and PC = 13 cm. (a) Calculate PE, the height of the pyramid. [3] (b) Calculate the volume of the pyramid. [The volume of a pyramid is given by 1 × area of base × height.] [2] 3 (c) Calculate angle PCA. [2] (d) M is the mid-point of AD and N is the mid-point of BC. Calculate angle MPN. [3] (e) (i) Calculate angle PBC. [2] (ii) K lies on PB so that BK = 4 cm. Calculate the length of KC. [3]

15 marks

Mark scheme: 6 (a) (AC =) √(82 + 62) M1 1 M1 dep. (PE =) √(132 – [ their AC]2) 2 12 A1 (b) 1 M1 × 6 × 8 × their PE 3 192 (cm3) A1 (c) their PE M1 their CE their PE sin PCA = (0.92307...) Cos = Tan = 13 13 their CE a.r.t. 67.4° (67.380 … ) A1 (d) theirPE M1 4 tan PME = (71.6°) tan MPE = (18.4°) 4 their PE 180 – 2 x anglePME dep M1 2 x angleMPE 36.8° to 36.9° A1 (e) (i) 3 M1 cos PBC = 13 76.7° (76.6576…) A1 (ii) (KC2 = ) 42 + 62 – 2 x 4 x 6 cos(theirPBC) M1 Square root of correct combination M1 dep on first M1. √40.957... or 6.3998 6.40 (cm) A1 15 IGCSE – May/June 2006 0580 and 0581 04

This question in 0580/41 May/June 2006

5 B E 25 cm M O C 14 cm NOT TO SCALE D OBCD is a rhombus with sides of 25 cm. The length of the diagonal OC is 14 cm. (a) Show, by calculation, that the length of the diagonal BD is 48 cm. [3] (b) Calculate, correct to the nearest degree, (i) angle BCD, [2] (ii) angle OBC. [1] (c) DB = 2p and OC = 2q. Find, in terms of p and q, (i) OB, [1] (ii) OD. [1] (d) BE is parallel to OC and DCE is a straight line. Find, in its simplest form, OE in terms of p and q. [2] (e) M is the mid-point of CE. Find, in its simplest form, OM in terms of p and q. [2]  7  (f) O is the origin of a co-ordinate grid. OC lies along the x-axis and q =   .   0  (DB is vertical and DB = 48.) Write down as column vectors (i) p, [1] (ii) BC . [2] (g) Write down the value of │DE │. [1]

16 marks

Mark scheme: 5 (a) Using a right-angled triangle with M1 25 and 7 seen is sufficient (or 50, 14) 25 and 7 252 - 72 oe (or 502 – 142) M1 Must be a correct numerical calculation oe includes trig methods, which can round to 24, then 48 for the E mark (BD) = 48 (or 24 × 2) E1 Dep on M2, correctly established (b) (i) −1  7  M1 If scale drawing seen then M0 cos   × 2 oe  25  147° cao A1 www 2 147.47…. score M1 only (ii) air 32 -34 or ft B1 ft 180 – their 147 (c) (i) q + p oe B1 (ii) q – p oe B1 (d) M1 any correct unsimplified expression OC + CE oe 2q + their (c) (i) e.g. their (q – p) + 2 × their (q + p) p + 3q cao A1 www 2 (e) 1 M1 any correct unsimplified expression OC + OB oe 2 2q + ½ their (c) (i) 0.5p + 2.5q cao A1 www 2 (f) Accept any reasonable notation in both parts (i)  0  B1    24  (ii) B1  7    B1  −24  (g) 50 B1 [16] IGCSE – May/June 2007 0580 and 0581 04

This question in 0580/41 May/June 2007

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

4 (a) For Examiner's 12 cm Use H G NOT TO 6 cm SCALE 14 cm F The diagram shows triangle FGH, with FG = 14 cm, GH = 12 cm and FH = 6 cm. (i) Calculate the size of angle HFG. Answer(a)(i) Angle HFG = [4] (ii) Calculate the area of triangle FGH. Answer(a)(ii) cm2 [2] (b) For Q Examiner's 18 cm Use R NOT TO SCALE 12 cm 117° P The diagram shows triangle PQR, with RP = 12 cm, RQ = 18 cm and angle RPQ = 117°. Calculate the size of angle RQP. Answer(b) Angle RQP = [3]

9 marks

Mark scheme: 6 + 14 − 12 4 (a) (i) (cos (HFG)) = M2 M1 for implicit form 2 × 6 × 14 58.4 (58.41…) A2 A1 for 0.5238… (ii) 0.5 × 6 × 14 × sin (their 58.4) oe M1 ft their (i) 35.8 or 35.77 to 35.78 A1ft Correct or ft their (i) sin(117 ) × 12 (b) (sin (RQP)) = M2 M1 for implicit form 18 36.4 or 36.44... A1  3  k 

This question in 0580/41 May/June 2011

6 For F Examiner's Use NOT TO SCALE C D E 14 cm 36 cm A 19 cm B In the diagram, ABCDEF is a prism of length 36 cm. The cross-section ABC is a right-angled triangle. AB = 19 cm and AC = 14 cm. Calculate (a) the length BC, Answer(a) BC = cm [2] (b) the total surface area of the prism, Answer(b) cm2 [4] (c) the volume of the prism, Answer(c) cm3 [2] (d) the length CE, Answer(d) CE = cm [2] (e) the angle between the line CE and the base ABED. Answer(e) [3]

13 marks

Mark scheme: 6 (a) 23.6 (23.60…) 2 M1 for 142 + 192 (b) 2300 or 2303 to 2304 cao 4 M3 for 2 × ½ × 14 × 19 + 14 × 36 + 19 × 36 + their BC × 36 M2 for 4 of these added M1 for ½ × 14 × 19 (c) 4788 or 4790 cao 2 M1 their triangle area × 36 (d) 43(.0) or 43.04 to 43.05 cao 2 M1 for (their (a))2 + 362 or 362 + 192 + 142  14  (e) 18.9° to 19.02° cao 3 M2 for inv sin   or  their CE   14  inv tan  2 2  or  19 + 36   2 2  19 + 36   inv cos or complete longer  their CE    methods (M1 for clearly identifying angle CEA) IGCSE – May/June 2011 0580 41

This question in 0580/41 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 C Examiner's Use B 26° NOT TO SCALE 79 m 95 m 77° A 120 m D The quadrilateral ABCD represents an area of land. There is a straight road from A to C. AB = 79 m, AD = 120 m and CD = 95 m. Angle BCA = 26° and angle CDA = 77°. (a) Show that the length of the road, AC, is 135 m correct to the nearest metre. Answer(a) [4] (b) Calculate the size of the obtuse angle ABC. Answer(b) Angle ABC = [4] (c) A straight path is to be built from B to the nearest point on the road AC. For Examiner's Calculate the length of this path. Use Answer(c) m [3] (d) Houses are to be built on the land in triangle ACD. Each house needs at least 180 m2 of land. Calculate the maximum number of houses which can be built. Show all of your working. Answer(d) [4]

15 marks

Mark scheme: 6 (a) 1202 + 952 – 2 × 120 × 95 × cos77 M2 M1 for implicit version 135.26 … or 135.3 E2 A1 for 18295 to 18297 their 135 × sin26 sin B sin26 (b) (sinB ) = M2 M1 for = oe 79 their 135 79 48.5 to 48.7 isw A1 131 or 131.3 to 131.5 www4 B1ft ft for 180 – their 48.5 to 48.7 dep on sine rule or sine used (c) (Angle A =) 22.5 to 22.7 B1ft ft 154 – their (b), also accept angle B = 67.3 to 67.5 (ft their (b) – 64) ‘Path’/79 = sin (their A) oe M1 Dep on B1 and their A < 90 eg 79 cos 67.4 30.2 to 30.5 www3 A1 1 (d) × 120 × 95 × sin 77 oe M1 (5554) 2 Their area ÷ 180 M1 Dep on area attempt 30.8 to 30.9 A1 30 B1ft ft their 30.8 to 30.9 truncated dep on at least M1 earned After M2 answer 30 www scores A1B1 Answer 30 ww scores 0 IGCSE – October/November 2011 0580 41

This question in 0580/41 Oct/Nov 2011

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

6 For Q P Examiner's Use NOT TO SCALE 3 cm D C 4 cm A B 12 cm The diagram shows a triangular prism of length 12 cm. The rectangle ABCD is horizontal and the rectangle DCPQ is vertical. The cross-section is triangle PBC in which angle BCP = 90°, BC = 4 cm and CP = 3 cm. (a) (i) Calculate the length of AP. Answer(a)(i) AP = cm [3] (ii) Calculate the angle of elevation of P from A. Answer(a)(ii) [2] (b) (i) Calculate angle PBC. For Examiner's Use Answer(b)(i) Angle PBC = [2] (ii) X is on BP so that angle BXC = 120°. Calculate the length of XC. Answer(b)(ii) XC = cm [3]

10 marks

Mark scheme: 6 (a) (i) 13 or 13.0 www 3 M1 for 32 + 42 oe Equiv if find AC first and M1 for 12 2 + their (3 2 + 4 2 ) 3 3 (ii) 13.32 to 13.35 or 13.3 2 M1 for sin = or tan = oe their AP their AC 3 (b) (i) 36.86 to 36.87 or 36.9 2 M1 for tan (PBC) = oe 4 4 sin their (b )( i ) (ii) 2.770 to 2.774 or 2.77 3 M2 for or M1 for correct sin 120 implicit eqn IGCSE – October/November 2011 0580 43

This question in 0580/43 Oct/Nov 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

4 For Examiner′s D C Use 32° 70 m NOT TO SCALE 40° A B 55 m The diagram shows a school playground ABCD. ABCD is a trapezium. AB = 55 m, BD = 70 m, angle ABD = 40° and angle BCD = 32°. (a) Calculate AD. Answer(a) AD = … m [4] (b) Calculate BC. Answer(b) BC = … m [4] (c) (i) Calculate the area of the playground ABCD. For Examiner′s Use Answer(c)(i) … m2 [3] (ii) An accurate plan of the school playground is to be drawn to a scale of 1: 200 . Calculate the area of the school playground on the plan. Give your answer in cm2. Answer(c)(ii) … cm2 [2] (d) A fence, BD, divides the playground into two areas. Calculate the shortest distance from A to BD. Answer(d) … m [2] _____________________________________________________________________________________

15 marks

Mark scheme: 4 (a) 45.[0] or 45.01 to 45.02 nfww 4 M2 for 552 + 702 – 2.55.70 cos 40 or M1 for correct implicit equation A1 for 2026. …. (b) 84.9 or 84.90 to 84.92 4 B1 for angle BDC = 40 soi 70 sin (their 40 ) M2 for sin 32 or M1 for correct implicit equation (c) (i) 4060 or 4063 to 4064 nfww 3 1 1 M2 for (55 × 70 sin 40 ) + 2 2 (70 × their (b ) sin (180 − their 40 − 32)) oe or M1 for correct method for one of the triangle areas (ii) 1020 or 1015 to 1016 2FT FT their (c) (i) ÷ 4 oe correctly evaluated or M1 their (c) (i) ÷ figs 4 oe distance (d) 35.4 or 35.35… nfww 2 M1 for sin 40 = or better 55 1 or for (55 × 70 sin 40) = (70 × distance) ÷ 2 2 or better IGCSE – October/November 2013 0580 41

This question in 0580/41 Oct/Nov 2013

2 A fi eld, ABCD, is in the shape of a quadrilateral. For Examiner′s A footpath crosses the fi eld from A to C. Use C 26° B NOT TO SCALE 55 m 65° 32° 122° A D 62 m (a) Use the sine rule to calculate the distance AC and show that it rounds to 119.9 m, correct to 1 decimal place. Answer(a) [3] (b) Calculate the length of BC. Answer(b) BC = … m [4] (c) Calculate the area of triangle ACD. For Examiner′s Use Answer(c) … m2 [2] (d) The fi eld is for sale at $4.50 per square metre. Calculate the cost of the fi eld. Answer(d) $ … [3] _____________________________________________________________________________________

12 marks

Mark scheme: 62 × sin 122 2 (a) 119.94[…] nfww 3 M2 for sin 26 AC 62 or M1 for = oe sin 122 sin 26 SC2 for correct answer from alternative methods (b) 109 or 108.7 to 108.8 nfww 4 M2 for 119.9..2 + 552 – 2 × 119.9.. × 55cos65 A1 for 11827[·…] or 11834 to 11835[·…] or M1 for implicit version (c) 1970 or 1969 to 1970.4 2 M1 for ½ × 119.9.. × 62 × sin 32 (d) 22300 or 22310 to 22320 3 M2 for (their (c) + 0.5 × 55 × 119.9.. × sin65) × 4.5 or M1 for their (c) + 0.5 × 55 × 119.9.. × sin65

This question in 0580/43 Oct/Nov 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

7 NOT TO SCALE 75 cm 55 cm 120 cm The diagram shows a water tank in the shape of a cuboid measuring 120 cm by 55 cm by 75 cm. The tank is fi lled completely with water. (a) Show that the capacity of the water tank is 495 litres. Answer(a) [2] (b) (i) The water from the tank fl ows into an empty cylinder at a uniform rate of 750 millilitres per second. Calculate the length of time, in minutes, for the water to be completely emptied from the tank. Answer(b)(i) … min [2] (ii) When the tank is completely empty, the height of the water in the cylinder is 112 cm. NOT TO SCALE 112 cm Calculate the radius of the cylinder. Answer(b)(ii) … cm [3] (c) x cm NOT TO SCALE 75 cm cm 145 55 cm 120 cm A rod of length 145 cm is placed inside the water tank. One end of the rod is in the bottom corner of the tank as shown. The other end of the rod is x cm below the top corner of the tank as shown. Calculate the value of x. Answer(c) x = … [4] (d) Calculate the angle that the rod makes with the base of the tank. Answer(d) … [3] __________________________________________________________________________________________

14 marks

Mark scheme: 7 (a) (i) 120 × 55 × 75 [= 495000] M1 M1 ÷ 1000 [= 495] or 495[l] × 1000 = 495000[ml] (b) (i) 11 2 M1 for 495000 ÷ 750 [÷ 60] oe [660] After 0 scored, SC1 for answer figs 11 figs 495 (ii) 37.5 or 37.50 to 37.51 3 M2 for oe 112π 2 figs 495 or M1 for [112r = ] or π 2 figs 495 [ πr = ] or better 112 (c) 15 4 B3 for answer 60 or M3 for 75 – 145 2 − (55 2 + 120 2 ) oe M2 for 145 2 − ( 55 2 + 120 2 ) oe or M1 for 55 2 + 120 2 2 + 120 2 /145) oe, e.g. (d) 24.4[4..] to 24.45 3 M2 for cos–1 ( 55 or sin −(751 – their (c))/145 or tan −((751 – their (c))/ 55 2 + 120 2 ) or M1 for cos = 55 2 + 120 2 /145 oe or sin = (75 – their (c))/145 or tan = (75 – their (c))/ 55 2 + 120 2

This question in 0580/42 Oct/Nov 2014

5 (a) Andrei stands on level horizontal ground, 294 m from the foot of a vertical tower which is 55 m high. (i) Calculate the angle of elevation of the top of the tower. Answer(a)(i) … [2] (ii) Andrei walks a distance x metres directly towards the tower. The angle of elevation of the top of the tower is now 24.8°. Calculate the value of x. Answer(a)(ii) x = … [4] (b) The diagram shows a pyramid with a horizontal rectangular base. NOT TO y SCALE 4 m 3 m 4.8 m The rectangular base has length 4.8 m and width 3 m and the height of the pyramid is 4 m. Calculate (i) y, the length of a sloping edge of the pyramid, Answer(b)(i) y = … m [4] (ii) the angle between a sloping edge and the rectangular base of the pyramid. Answer(b)(ii) … [2]

12 marks

Mark scheme: 55 5 (a) (i) 10.6 or 10.59... 2 M1 for tan = oe 294 55 (ii) 175 or 174.9[…] to 175.[1…] 4 M2 for [adj =] oe tan 24 8. or M1 for implicit version and M1 dep on at least M1 for 294 − their adj 2 2  1 2 2  2 8.4 + 3  (b) (i) 4.9 or 4.89 to 4.9 4 M3 for 4 +  or M2 for 12 8.4 2 + 32 or M1 for 8.4 2 + 3 2 or 4.2 2 + 5.1 2 4 (ii) 54.7 or 54.71 to 54.722 2 M1 for sin = their 9.4

This question in 0580/41 May/June 2015

7 (a) P NOT TO SCALE 8.4 cm Q 7.6 cm 62° R In the triangle PQR, QR = 7.6 cm and PR = 8.4 cm. Angle QRP = 62°. Calculate (i) PQ, Answer(a)(i) PQ = … cm [4] (ii) the area of triangle PQR. Answer(a)(ii) … cm2 [2] (b) North H NOT TO North SCALE 63 km G J The diagram shows the positions of three small islands G, H and J. The bearing of H from G is 045°. The bearing of J from G is 126°. The bearing of J from H is 164°. The distance HJ is 63 km. Calculate the distance GJ. Answer(b) GJ = … km [5]

11 marks

Mark scheme: 7 (a) (i) 8.27 or 8.269... nfww 4 M2 for 6.7 2 + 4.8 2 − 2 × 6.7 × 4.8 × cos ( 62) oe or M1 for implicit form A1 for [PQ2 =] 68.3 to 68.5 (ii) 28.2 or 28.18.. 2 M1 for 0.5 × 7.6 × 8.4 × sin 62 oe (b) 55.8 or 55.78 to 55.79 nfww 5 B1 for [HGJ] = 81 B1 for [GHJ] = 61 63 M2 for [GJ = ] × sin (their 61) sin (their 81) or M1 for implicit form After M0, SC1 for final answer of 68.1…

This question in 0580/41 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

8 (a) A cylindrical tank contains 180 000 cm3 of water. 45 cm The radius of the tank is 45 cm. NOT TO SCALE Calculate the height of water in the tank. Answer(a) … cm [2] (b) D C NOT TO 70 cm SCALE 40 cm 150 cm A 50 cm B The diagram shows an empty tank in the shape of a horizontal prism of length 150 cm. The cross section of the prism is an isosceles trapezium ABCD. AB = 50 cm, CD = 70 cm and the vertical height of the trapezium is 40 cm. (i) Calculate the volume of the tank. Answer(b)(i) … cm3 [3] (ii) Write your answer to part (b)(i) in litres. Answer(b)(ii) … litres [1] (c) The 180 000 cm3 of water flows from the tank in part (a) into the tank in part (b) at a rate of 15 cm3/s. Calculate the time this takes. Give your answer in hours and minutes. Answer(c) … h … min [3] (d) D 70 cm C x cm NOT TO F E SCALE 40 cm h cm A 50 cm B The 180 000 cm3 of water reaches the level EF as shown above. EF = x cm and the height of the water is h cm. (i) Using the properties of similar triangles, show that h = 2(x – 50). Answer(d)(i) [2] (ii) Using h = 2(x – 50), show that the shaded area, in cm2, is x2 – 2500. Answer(d)(ii) [1] (iii) Find the value of x. Answer(d)(iii) x = … [2] (iv) Find the value of h. Answer(d)(iv) h = … [1]

15 marks

Mark scheme: 28 (a) 28.3 or 28.29… 2 M1 for 180 000 ÷ (π × 45 ) 1 (b) (i) 360 000 3 M2 for (70 + 50 ) × 40 × 150 oe 2 1 or M1 for (70 + 50 ) × 40 oe 2 or their area of ABCD × 150 dependent on their area being two dimensional (ii) 360 1FT FT their (b)(i) ÷ 1000 (c) 3 h 20 min 3 M2 for 180 000 ÷ 15 ÷ 60 (implied by 200) or M1 for 180 000 ÷ 15 (implied by 12000) or correct conversion of their seconds into h and min h 12 ( x − 50 ) (d) (i) = oe M1 i.e. a correct statement from similar figures 40 10 which must contain h, x and numbers h = 2 ( x − 50 ) A1 Answer established with at least one more step and no errors or omissions 1 (ii) ( x + 50 ) 2( x − 50) M1 2 2 − 2500) ×150 = 180 000 or better (iii) 60.8 or 60.82 to 60.83 2 M1 for ( x (iv) 21.7 or 21.65 to 21.66 1FT FT for 2(their (d)(iii) – 50) evaluated only if x > 50 Qu Answers Mark Part Marks  2 13 

This question in 0580/43 May/June 2015

3 (a) T 60 m NOT TO SCALE 50 m B 130° A 70 m C A, B and C are points on horizontal ground. BT is a vertical pole. AT = 60 m, AB = 50 m, BC = 70 m and angle ABC = 130°. (i) Calculate the angle of elevation of T from C. Answer(a)(i) … [5] (ii) Calculate the length AC. Answer(a)(ii) AC = … m [4] (iii) Calculate the area of triangle ABC. Answer(a)(iii) … m2 [2] (b) Y 12 cm NOT TO SCALE 22 cm X 45 cm A cuboid has length 45 cm, width 22 cm and height 12 cm. Calculate the length of the straight line XY. Answer(b) XY = … cm [4] __________________________________________________________________________________________

15 marks

Mark scheme: 3 (a) (i) 25.4 or 25.35… nfww 5 M2 for 60 2 − 50 2 oe soi by 33.1 to 33.2 or M1 for TB2 + 502 = 602 oe and theirTB M2 for tan = oe 70 or B1 for recognising angle TCB as required angle (ii) 109 or 109.0 to 109.1 4 M2 for 502 + 702 – 2 × 50 × 70 × cos130 M1 for implicit cos rule A1 for 11 899 to 11 900 (iii) 1 340 or 1 340.0 to 1 341 2 1 M1 for × 50 × 70 × sin130 oe 2 (b) 51.5 or 51.50 to 51.51 4 M3 for [XY] = 45 2 + 22 2 + 12 2 or M2 for [XY 2 =] 452 + 222 + 122 soi by 2 653 or M1 for 452 + 222 oe or 452 + 122 oe or 122 + 222 oe

This question in 0580/41 Oct/Nov 2015

4 A NOT TO SCALE B D C The diagram shows a tent ABCD. The front of the tent is an isosceles triangle ABC, with AB = AC. The sides of the tent are congruent triangles ABD and ACD. (a) BC = 1.2 m and angle ABC = 68°. Find AC. Answer(a) AC = … m [3] (b) CD = 2.3 m and AD = 1.9 m. Find angle ADC. Answer(b) Angle ADC = … [4] (c) The floor of the tent, triangle BCD, is also an isosceles triangle with BD = CD. Calculate the area of the floor of the tent. Answer(c) … m2 [4] (d) When the tent is on horizontal ground, A is a vertical distance 1.25 m above the ground. Calculate the angle between AD and the ground. Answer(d) … [3]

14 marks

Mark scheme: 6.0 4 (a) 1.6[0] or 1.601 to 1.602 3 M2 for oe cos 68 6.0 or M1 for cos 68 = AC 9.1 2 + 3.2 2 − their6.1 2 (b) 43.5 or 43.6 or 43.49 to 43.56 4 M2 for 2 × 9.1 × 3.2 or M1 for implicit statement A1 for [cos = ] 0.724 to 0.726 1 (c) 1.33 or 1.332...nfww 4 M2 for 3.22 − ( × 2.1 )2 2 or M1 for 3.22 = h 2 + (5.0 × 2.1 )2 and M1 for 12 × 2.1 × their .222 (their 2.22 must come from attempt at Pythag or from trig in triangle BCD) .125 (d) oe 41.1 or 41.13 to 41.14 3 M2 for sin = 9.1 or M1 for correct angle identified

This question in 0580/42 Oct/Nov 2015

7 NOT TO SCALE 210° 88cmcm O A B 8 cm 72° C The diagram shows a design for a logo made from a sector and two triangles. The sector, centre O, has radius 8 cm and sector angle 210°. AC = 8 cm and angle ACB = 72°. (a) Show that angle OAB = 15°. [2] (b) Calculate the length of the straight line AB. AB = … cm [4] (c) Calculate angle ABC. Angle ABC = … [3] (d) Calculate the total area of the logo design. … cm2 [6] (e) The logo design is an enlargement with scale factor 4 of the actual logo. Calculate the area of the actual logo. … cm2 [2]

17 marks

Mark scheme: 7 (a) 360 – 210 [= 150] M1 (180 – 150) ÷ 2 [= 15] M1 or 150 ÷ 2 [=75] and 180 – 75 – 90 [ =15] (b) 15.5 or 15.45 to 15.46 nfww 4 M3 for 2 × 8 cos 15 oe or M2 for 8 cos 15 oe x or M1 for = cos 15 oe 8 8 × sin72 (c) 29.5 or 29.4 or 29.39 to 29.50.. 3 M2 for [sin ABC = ] their (b ) sin ABC sin72 or M1 for = oe 8 their (b ) 210 2 (d) 194 or 193.7 to 194.1 nfww 6 M2 for × π × 8 360 or M1 for [k] π × 8 2 seen and M1 for ½ × 82 × sin 150 oe and M2 for ½ × 8 × their (b) × sin(108 – their (c)) oe or B1 for [angle CAB=] 108 – their (c) (e) 12.1 or 12.11 to 12.13 2FT FT their (d) ÷ 42 oe 2  1  M1 for 42 or   soi  4 

This question in 0580/42 May/June 2016

6 D North 170 m NOT TO SCALE C 33° 180 m A 220 m B The diagram shows five straight footpaths in a park. AB = 220 m, AC = 180 m and AD = 170 m. Angle ACB = 90° and angle DAC = 33°. (a) Calculate BC. BC = … m [3] (b) Calculate CD. CD = … m [4] (c) Calculate the shortest distance from D to AC. … m [2] (d) The bearing of D from A is 047°. Calculate the bearing of B from A. … [3] (e) Calculate the area of the quadrilateral ABCD. … m2 [3]

15 marks

Mark scheme: 6 (a) 126 or 126.4 to 126.5 3 M2 for 220 2 − 180 2 oe or M1 for BC2 + 1802 = 2202 oe (b) 99.9 or 99.86 to 99.87 4 M2 for 1802 + 1702 – 2 × 180 × 170 cos33 180 2 + 170 2 − CD 2 or M1 for cos33 = 2 × 180 × 170 A1 for 9970 or 9973 to 9974 dist (c) 92.6 or 92.58 to 92.59 2 M1 for = sin33 oe 170 180 (d) 115.1 or 115.0 to 115.1 3 M1 for cos = oe 220 M1dep for 47 + 33 + their angle BAC (e) 19700 or 19708 to 19720 3 M1 for 0.5 × 180 × 170 × sin33 oe or 0.5 × 180 × their (c) oe M1 for 0.5 × 180 × their (a) oe or 0.5 × 180 × 220 × sin(their BAC) oe

This question in 0580/41 Oct/Nov 2016

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

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

7 In this question, all measurements are in metres. 6 NOT TO x SCALE 2x – 3 The diagram shows a right-angled triangle. (a) Show that 5x2 - 12x - 27 = 0. [3] (b) Solve 5x2 - 12x - 27 = 0. Show all your working and give your answers correct to 2 decimal places. x = … or x = … [4] (c) Calculate the perimeter of the triangle. … m [2] (d) Calculate the smallest angle of the triangle. … [2]

11 marks

Mark scheme: 7(a) x2 + (2x – 3)2 = 62 oe M1 or x2 + 4x2 – 6x – 6x + 9 = 36 4x2 – 6x – 6x + 9 or better B1 5x2 – 12x – 27 = 0 A1 Dep on M1B1 with no errors or omissions 7(b) 2 B2 2 −−( 12) ± ( − 12) − 4(5)( − 27) B1 for ( −12) − 4(5)( −27) or for 2 × 5 2  12  or better  x −  oe  10  −−( 12) + q −−( 12) − q 2 12  12  27 or oe or oe or ±   + 2 × 5 2 × 5 10  10  5 or both – 1.42, 3.82 final answers B2 B1 for each If B0, SC1 for answers – 1.4 or –1.415… to – 1.415 and 3.8 or 3.815 to 3.815… or answers –1.41 and 3.81 or – 1.42 and 3.82 seen in working or for –3.82 and 1.42 as final ans 7(c) 14.4 or 14.5 or 14.44 to 14.46 2 2FT for 3 × their positive root + 3 evaluated to 3sf or better M1 for 3 × their positive root + 3 oe 7(d) 39.5 or 39.46 to 39.54… 2 M1 for trig statement seen to find either angle their x their (2 x − 3) sin = oe or sin = oe 6 6

This question in 0580/42 May/June 2018

6 (a) E 6 cm D NOT TO SCALE A 12 cm C 6 cm B In the pentagon ABCDE, angle ACB = angle AED = 90°. Triangle ACD is equilateral with side length 12 cm. DE = BC = 6 cm. (i) Calculate angle BAE. Angle BAE = … [4] (ii) Calculate AB. AB = … cm [2] (iii) Calculate AE. AE = … cm [3] (iv) Calculate the area of the pentagon. … cm2 [4] (b) S R P Q 5 cm NOT TO SCALE D C 4 cm A 8 cm B The diagram shows a cuboid. AB = 8 cm, BC = 4 cm and CR = 5 cm. (i) Write down the number of planes of symmetry of this cuboid. … [1] (ii) Calculate the angle between the diagonal AR and the plane BCRQ. … [4]

18 marks

Mark scheme: 6(a)(i) 116.6 or 116.56 to 116.57 4 6 M1 for sin[ EAD ] = oe 12 6 M1 for tan[ BAC ] = oe 12 B1 for [angle DAC] = 60 6(a)(ii) 13.4 or 13.41 to 13.42 2 M1 for 12 2 + 6 2 6(a)(iii) 10.4 or 10.39… 3 2 2 M2 for 12 − 6 or M1 for AE 2 + 6 2 = 12 2 6(a)(iv) 130 or 129.5… to 129.6 4 M1 for 0.5 × 6 ×theirAE oe M1 for 0.5 × 12 × 12 × sin60 oe M1 for 0.5 × 6 × 12 oe 6(b)(i) 3 1 6(b)(ii) 51.3 or 51.30 to 51.34… 4 8 8 M3 for tan = or sin = oe 4 2 + 5 2 4 2 + 5 2 + 8 2 or M2 for 4 2 + 5 2 or 4 2 + 5 2 + 8 2 or M1 for angle ARB clearly indicated

This question in 0580/43 May/June 2018

5 NOT TO SCALE 18 cm h cm x° 6 cm The diagram shows a prism with length 18 cm and volume 253.8 cm3. The cross-section of the prism is a right-angled triangle with base 6 cm and height h cm. (a) (i) Show that the value of h is 4.7 . [3] (ii) Calculate the value of x. x = … [2] (b) Calculate the total surface area of the prism. … cm2 [6]

11 marks

Mark scheme: 5(a)(i)  6  3 For M3 no errors at any stage [h =] 253 8. ÷ 18 ÷   or 1  2  M2 for 253.8 = × 6 × h × 18 oe (no 2 253.8 × 2 [h =] or previous errors) 6 × 18 1 253.8 or M1 for triangle area = ×6 × h soi [h =] 2 6 18 × 2 5(a)(ii) 38.1 or 38.06 to 38.08 2 7.4 M1 for tan = oe 6 5(b) 358 or 357.9 to 358 6 M1 for 6 2 + 7.4 2 M1 for 6 2 + 4.7 2 × 18 [× 2] M1 for 6 × 18 [× 2] M1 for 4.7 × 18 1 M1 for 2 × × 6 × 4.7 oe 2

This question in 0580/41 Oct/Nov 2018

8 (a) D 5 cm 6 cm NOT TO B C X SCALE 8 cm 10 cm A In the diagram, AB and CD are parallel. AD and BC intersect at right angles at the point X. AB = 10 cm, CD = 5 cm, AX = 8 cm and BX = 6 cm. (i) Use similar triangles to calculate DX. DX = … cm [2] (ii) Calculate angle XAB. Angle XAB = … [2] (b) T S 85° 75° w° NOT TO v° O SCALE y° R x° P Q P, Q, R, S and T lie on the circle, centre O. Angle PST = 75° and angle QTS = 85°. Find the values of v, w, x and y. v = … w = … x = … y = … [6] (c) Two containers are mathematically similar. The surface area of the larger container is 226 cm2 and the surface area of the smaller container is 94 cm2. The volume of the larger container is 680 cm3. Find the volume of the smaller container. … cm3 [3]

13 marks

Mark scheme: 8(a)(i) 4 2 M1 for correct method using similar triangles 10 8 e.g. = oe 5 DX 8(a)(ii) 36.9 or 36.86 to 36.87 2 6 6 8 M1 for tan = or sin = or cos = oe 8 10 10 8(b) [v = ] 150 B1 [w = ] 15 B2 FT (180 – their v) ÷ 2 M1 for 180 – 2w = their v oe or angle POQ = 180 – their v oe [x = ] 15 B1 FT their w [y = ] 10 B2 M1 for angle TPS = 5° or angle TXS = 20° or OXP = 20° or TXP = 160° (where X is where OT and PS intersect) 8(c) 182 or 182.4… 3 32  94   V  M2 for = oe      226   680  226 94 or M1 for ratio of lengths = or or 94 226 V 2 94 3 better or for = oe 680 2 226 3

This question in 0580/43 Oct/Nov 2018

3 North C D 170 m 120 m NOT TO 150 m SCALE E 50 m A 100 m B The diagram shows a field ABCDE. (a) Calculate the perimeter of the field ABCDE. … m [4] (b) Calculate angle ABD. Angle ABD = … [4] (c) (i) Calculate angle CBD. Angle CBD = … [2] (ii) The point C is due north of the point B. Find the bearing of D from B. … [2] (d) Calculate the area of the field ABCDE. Give your answer in hectares. [1 hectare = 10 000 m2] … hectares [4]

16 marks

Mark scheme: 3(a) 530 4 B3 for [DE] = 130 m and [DC] = 80 m or B2 for [DE] = 130 m or [DC] = 80 m or M1 for 502 + 1202 or 1702 – 1502 3(b) 52.9 or 52.89… 4 100 2 + 150 2 − 120 2 M2 for 2 × 100 × 150 or M1 for 1202 = 1002 + 1502 – 2 × 100 × 150cos(…) 181 A1 for 0.603 or 0.6033…or 300 3(c)(i) 28.1 or 28.07… 2 15 M1 for cos = oe 17 3(c)(ii) 331.9 or 331.9… 2 FT 360 – their (c)(i) M1 for 360 – their (c)(i) oe 3(d) 1.5[0] or 1.498… nfww 4 1 M1 for × 50 × 120 oe 2 1 M1 for × 100 × 150sin(their (b)) oe 2 1 M1 for × 150 ×theirCD oe 2 1 or × 150 × 170 × sin their (c)(i) 2 If 0 scored, SC1 for dividing their area by 10 000

This question in 0580/41 May/June 2019

8 (a) D 64° 53° 16.5 cm NOT TO SCALE 12.4 cm A 95° B C The diagram shows two triangles ABD and BCD. AD = 16.5 cm and BD = 12.4 cm. Angle ADB = 64°, angle BDC = 53° and angle DBC = 95°. (i) Find AB. AB = … cm [4] (ii) Find BC. BC = … cm [4] (b) y° 3.8 cm NOT TO SCALE 7.7 cm The diagram shows a sector of a circle of radius 3.8 cm. The arc length is 7.7 cm. (i) Calculate the value of y. y = … [2] (ii) Calculate the area of the sector. … cm2 [2]

12 marks

Mark scheme: 8(a)(i) 15.7 or 15.70... 4 2 2 M2 for 16 5. + 12 4. − 2 × 16 5. × 12 4. × cos 64 or M1 for implicit form A1 for 246 to 247 8(a)(ii) 18.7 or 18.68 to 18.69 4 B1 for 32 or angle DBM = 37 or angle CBM = 58 12 4. × sin 53 M2 for oe sin 32 or M1 for implicit form oe 8(b)(i) 116.1 or 116.08 to 116.09... 2 y M1 for × 2 × π × 3.8 = 7.7 oe 360 8(b)(ii) 14.6 or 14.61 to 14.63… 2 their (b)(i) 2 M1 for × π × 3.8 oe 360

This question in 0580/42 May/June 2019

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

4 (a) (i) Calculate the external curved surface area of a cylinder with radius 8 m and height 19 m. … m2 [2] (ii) This surface is painted at a cost of $0.85 per square metre. Calculate the cost of painting this surface. $ … [2] (b) A solid metal sphere with radius 6 cm is melted down and all of the metal is used to make a solid cone with radius 8 cm and height h cm. (i) Show that h = 13.5 . 4 3 [The volume, V, of a sphere with radius r is V = r r .] 3 1 2 [The volume, V, of a cone with radius r and height h is V = r r h .] 3 [2] (ii) Calculate the slant height of the cone. … cm [2] (iii) Calculate the curved surface area of the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = r rl .] … cm2 [1] (c) Two cones are mathematically similar. The total surface area of the smaller cone is 80 cm2. The total surface area of the larger cone is 180 cm2. The volume of the smaller cone is 168 cm3. Calculate the volume of the larger cone. … cm3 [3] (d) The diagram shows a pyramid with a P square base ABCD. DB = 8 cm. NOT TO P is vertically above the centre, X, of SCALE the base and PX = 5 cm. D C X A B Calculate the angle between PB and the base ABCD. … [3]

15 marks

Mark scheme: 4(a)(i) 955 or 955.0 to 955.2 2 M1 for 2 × π × 8 × 19 oe 4(a)(ii) 812 or 811.7 to 811.9... 2 FT their (i) × 0.85 M1 for their (i) × 0.85 or their (i) × 85 4(b)(i) 4 3 M2 4 3 1 2 × π × 6 M1 for × π × 6 = × π × 8 × h 3 3 3 or cancelling clearly 1 2 × π × 8 3 seen to reach 13.5 4(b)(ii) 15.7 or 15.69... 2 M1 for 82 + 13.52 or better 4(b)(iii) 394 or 395 or 394.3 to 394.6... 1 FT π × 8 × their (b)(ii) 4(c) 567 3 3 168  80  2 M2 for =   oe or better V  180  1 1  180  2  80  2 or M1 for   or   oe seen or  80   180  better 4(d) 51.3 or 51.34... 3 5 M2 for tan = oe 4 or M1 for recognition of angle PBX

This question in 0580/41 Oct/Nov 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

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

4 A solid metal cone has radius 1.65 cm and slant height 4.70 cm. (a) Calculate the total surface area of the cone. [The curved surface area, A, of a cone with radius r and slant height l is A = rrl .] … cm2 [2] (b) Find the angle the slant height makes with the base of the cone. … [2] (c) (i) Calculate the volume of the cone. 1 2 [The volume, V, of a cone with radius r and height h is V = rr h .] 3 … cm3 [4] (ii) A metal sphere with radius 5 cm is melted down to make cones identical to this one. Calculate the number of complete identical cones that are made. 4 3 [The volume, V, of a sphere with radius r is V = rr .] 3

12 marks

Mark scheme: 4(a) 32.9 or 32.91 to 32.92… 2 M1 for π × 1.65 × 4.7 + π × 1.652 4(b) 69.4 or 69.44 to 69.45 2 M1 for cos = 1.65 ÷ 4.7 oe 4(c)(i) 12.5 or 12.54 to 12.55 4 1 2 2 2 M3 for × π × 1.65 × 4.7 − 1.65 oe 3 or M2 for 4.7 2 − 1.65 2 oe or for 4.7 × sin(their (b)) oe or M1 for 1.65 2 + h 2 = 4.7 2 oe h or for = sin(their (b)) oe 4.7 4(c)(ii) 41 nfww 4 B3 for 41.7… to 41.9 4 3 or M2 for × π × 5 ÷their 12.5 3 4 3 or M1 for × π × 5 3 After M2 scored, M1 for truncating their decimal number of cones seen to an integer answer

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

5 All the lengths in this question are in centimetres. x + 1 A F D NOT TO 2x E SCALE x + 3 B C 4x – 5 The diagram shows a shape ABCDEF made from two rectangles. The total area of the shape is 342 cm2. (a) Show that x 2 + x - 72 = 0 . [5] (b) Solve by factorisation. x 2 + x - 72 = 0 x = … or x = … [3] (c) Work out the perimeter of the shape ABCDEF. … cm [2] (d) Calculate angle DBC. Angle DBC = … [2]

12 marks

Mark scheme: 5(a) ( 4 x − 5 )( x + 3 ) + ( x + 1)( x − 3 ) = 342 M2 M1 for ( 4 x − 5 )( x + 3 ) or ( x + 1)( x − 3 ) or or for 2 x ( 4 x − 5 ) or ( 3 x − 6 )( x − 3 ) 2 x ( 4 x − 5 ) − ( 3 x − 6 )( x − 3 ) = 342 4 x 2 + 12 x − 5 x − 15 oe and M2 M1 for each x 2 + x − 3 x − 3 oe seen OR 8 x 2 − 10 x and 3 x 2 − 15 x + 18 seen 5 x 2 + 5 x − 18 = 342 leading to A1 no errors or omission x 2 + x − 72 = 0 5(b) ( x + 9 )( x − 8 ) M2 B1 for (x + a)(x + b) where ab = – 72 or a + b = 1 and a, b are integers 8, −9 B1 5(c) 86 2 FT for 12 × their x − 10 (x positive) B1 for any one of 27, 11, 16 seen or for 2 x + 2 x + 4 x − 5 + 4 x − 5 oe or better soi 5(d) 22.2 or 22.16 to 22.17 2 11 their x + 3 M1 for tan = or 27 4 × their x − 5

This question in 0580/43 May/June 2020

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

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

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

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

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

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

10 H G NOT TO E F SCALE 6 cm 24 cm D C A B The diagram shows a cuboid ABCDEFGH. CG = 6 cm, AG = 24 cm and AB = 2BC. (a) Calculate AB. AB = … cm [4] (b) Calculate the angle between AG and the base ABCD. … [3]

7 marks

Mark scheme: 10(a) 20.8 or 20.76 to 20.79 4 B3 for [BC =] 10.4 or 10.38 to 10.39… or 6 3 oe or M2 for (2x)2 + x2 + 62 = 242 oe or M1 for 242 – 62 oe or x2 + 62 oe or (2x)2 + 62 oe, or x2 + (2x)2 oe or SC2 for final answer of 12 5 or 26.8 or 26.83… OR  x  2 M3 for x2 + + 62 = 242 oe    2   x  2 or M2 for x2 +    2   x  2 or M1 for x2 + 62 oe or + 62 oe or    2  242 – 62 oe 10(b) 14.5 or 14.47 to 14.48 3 6 M2 for sin […] = oe 24 or M1 for recognising the correct angle GAC

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

10 D 16.5 cm NOT TO SCALE A 31° 12.3 cm C B The diagram shows a quadrilateral ABCD. AC = 12.3 cm and AD = 16. 5 cm . Angle BAC = 31° , angle ABC = 90° and angle ACD = 90° . (a) Show that AB = 10.54 cm, correct to 2 decimal places. [2] (b) Show that angle DAC = 41.80° correct to 2 decimal places. [2] (c) Calculate BD. BD = … cm [3] (d) Calculate angle CBD. Angle CBD = … [4] (e) Calculate the shortest distance from C to BD. … cm [4]

15 marks

Mark scheme: 10(a) AB M1 cos31 = oe 12.3 10.543... A1 10(b) 12.3 M1 cos = oe 16.5 41.801 to 41.802 A1 10(c) 16.7 or 16.8 or 16.74 to 16.75… 3 2 2 M2 for 10.54 + 16.5 −2 10.54  16.5  cos(31 + 41.8) or for 6.332 + 112 −2 6.33  11  cos(180 − 31) OR M1 for 10.54 2 + 16.5 2 −2 10.54  16.5  cos(31 + 41.8) or for 6.332 + 112 −2 6.33  11  cos(90 + 90 − 31) oe A1 for 280 or 281 or 280.4 to 280.6 10(d) 18.9 to 20.7… nfww 4 BC M1 for sin31 = oe or better and 12.3 CD sin 41.8[0] = oe 16.5 M2dep on M1 for their ( c ) 2 + 6.34 2 − 10.998 2 cos [DBC] = 2  their ( c )  6.34 or M1dep on M1 for 10.9982 = their (c)2 +6.342 – 2 × their (c) ×6.34 × cos DBC 10(e) 2.05 to 2.24… nfww 4 BC M1 for sin31 = oe or better 12.3 CD or sin 41.8[0] = oe 16.5 dist M2dep on M1 for = sin(their angle CBD ) theirBC dist or = sin(their angle CDB ) theirCD or M1 for recognition of shortest distance

This question in 0580/42 Feb/March 2023

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

4 (a) P NOT TO SCALE 8 cm Q R 24 cm (i) Calculate the area of triangle PQR. … cm2 [2] (ii) Calculate angle PRQ. Angle PRQ = … [2] (b) NOT TO SCALE 11 cm 6 cm The diagram shows a half-cylinder of radius 6 cm and length 11 cm. Calculate the volume of the half-cylinder. … cm3 [2] (c) T T D C C 44 cmcm S S O NOT TO 15 cm X X SCALE A B A B 20 cm (i) ABCD is a rectangle with AB = 20 cm and BC = 15 cm. S, X and T are points on a circle centre O, such that DSA and DTC are tangents to the circle. The radius of the circle is 4 cm and TX is a diameter of the circle. The shape DSXT is removed from the corner of the rectangle, leaving the shaded shape shown in the second diagram. Calculate the area of the shaded shape. … cm2 [5] (ii) Calculate the perimeter of the shaded shape. … cm [3]

14 marks

Mark scheme: 4(a)(i) 96 2 1 M1 for  24  8 2 4(a)(ii) 18.4 or 18.43... 2 8 M1 for tan x  oe 24 4(b) 622 or 622.0 to 622.1 … 2 1 2 1 2 M1 for [ 2] 6  11 or 2  6 [ 11] 4(c)(i) 246 or 246.2 to 246.3... 5 270 2 M4 for 15  20 4 4   4 oe 360 OR 270 2 M2 for  4 oe 360 or M1 for k   4 2 , where k  1 M1 for15  20 or 4  4 oe 4(c)(ii) 80.8 or 80.9 or 80.84 to 80.85... 3 M1 for 15  20  11  16 oe 3 M1 for 2 4 oe 4

This question in 0580/41 May/June 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

8 (a) NOT TO SCALE O 6 cm 135° B A 2 cm C D The diagram shows a shape made from a major sector AOB and triangles OBC and AOD. OB = 6cm , BC = 2cm , obtuse angle AOC = 135° and angle BCO = 90° . (i) Show that angle BOC = 19.5° , correct to 1 decimal place. [2] (ii) Calculate the area of the major sector AOB. … cm2 [3] (iii) C is the midpoint of OD. Calculate AD. … cm [5] (iv) Calculate the total area of the shape. … cm2 [4] (b) A sector of a circle has radius 8 cm and area 160cm2. A mathematically similar sector has radius 20 cm. Calculate the area of the larger sector. … cm2 [3]

17 marks

Mark scheme: 8(a)(i) 2 M1 sin[BOC] = or better oe 6 19.47… A1 8(a)(ii) 64.6 or 64.55 to 64.58 3 360 − 135 − 19.5 2 M2 for π 6 oe 360 k 2 or M1 for π 6 oe 360 8(a)(iii) 16.1 or 16 10 to 16.13 5 2 2 M2 for 2  6 − 2 oe or 2  6 cos 19.5 oe or M1 for OC2 + 22 = 62 oe or 6 cos 19.5 or better AND M2 for 62 + their OD2 – 2  6  their OD  cos 135 OR M1 for 62 + their OD2 – 2  6  their OD  cos 135 A1 for 259 to 260 8(a)(iv) 94.2 or 94.3 or 94.15 to 94.27… 4 M1 for ½  6  their OD  sin 135 oe nfww M1 for ½  6  2  sin(90 – 19.5) oe or for ½  their OC  2 M1dep for their (a)(ii) + their two triangle areas 8(b) 1000 cao 3 2 2  20   8  M2 for 160  or 160 ÷ oe      8   20   20  2  8  2 or M1 for   or   oe  8   20  OR sector angle 2 M2 for  π20 360 160 or M1 for  360 oe or better π82 OR percentage 2 M2 for  π20 oe or better 100 160 or M1 for [ 100] oe or better π8 2

This question in 0580/43 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 North B C 11 km 25° NOT TO A SCALE 32 km 14 km 55° H The diagram shows the positions of two lighthouses A and B, a boat C and a harbour H . C is due east of B. (a) Find the bearing of the harbour from boat C. … [1] (b) (i) Show that angle CBH = 100c . [1] (ii) Show that BH = 13.7 km, correct to 1 decimal place. [3] (c) Calculate the bearing of A from B. … [5] (d) At 1 pm boat C sails 32 km directly to the harbour at a speed of 10 knots. (i) Calculate the time when boat C arrives at the harbour. Give this time correct to the nearest minute. [1 knot = 1.852 km/h] … [4] (ii) Calculate the distance of boat C to the harbour when boat C is at the shortest distance from lighthouse B. … km [3]

17 marks

Mark scheme: 6(a) 245 1 6(b)(i) 180 – (55 + 25) [=100] M1 6(b)(ii) 32  sin 25 M2 sin 25 sin100 oe M1 for  oe sin100 BH 32 13.73… A1 6(c) 258 or 257.9 to 258.0… 5 B4 for 67.9 to 68.0… OR  112  13.7 2  14 2  M2 for [cos =]   2  13.7  11   A1 for 0.3738 to 0.376 or M1 for 142 = 112 + 13.72 – 2× 11 × 13.7 × cos B M1dep on at least M1 for 190 + their angle B 6(d)(i) 2 44 pm or 14 44 cao 4 B3 for 1 hour 44 or 1 hour 43.6 to 1 hour 43.8 or 104 or 103.6 to 103.8 or B2 for 1.727 to 1.73 32 or M2 for  60 10  1.852 or M1 for 32 ÷ (10 × 1.852) 6(d)(ii) 7.857 to 7.88 3 x M2 for  cos55 oe 13.7 or M1 for dist to H occurs when perpendicular from B meets CH soi

This question in 0580/41 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

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

13 T NOT TO SCALE 12.6 m E 1.5 m A P 33 m The diagram shows two vertical poles, AE and PT, standing on horizontal ground, AP. Calculate the angle of elevation of the point T from the point E. … [3]

3 marks

Mark scheme: 13 18.6 or 18.59… 3 12.6 − 1.5 M2 for tan = oe 33 or M1 identifying the correct angle of elevation or for 18.6 or 18.59… seen and spoiled After 0 scored SC1 for [angle ETP =] 71.4 or 71.4[0…] or 71.41

This question in 0580/41 Oct/Nov 2025

24 A B NOT TO 6.4 cm D C SCALE 13 cm F E 5.1 cm H G The diagram shows a cuboid ABCDEFGH. AE = 6.4 cm, EH = 5.1 cm and AG = 13 cm. (a) Calculate EF. EF = … cm [3] (b) Calculate the angle between the line AG and the base EFGH of the cuboid. … [3]

6 marks

Mark scheme: 24(a) 10.1 or 10.10… 3 M2 for EF2 + 6.42 + 5.12 = 132 or better or M1 for 6.42 + 5.12 or 132 – 6.42 or 132 – 5.12 24(b) 29.5 or 29.49… to 29.54… 3 6.4 M2 for sin [… =] oe 13 or M1 for identifying angle AGE

This question in 0580/43 Oct/Nov 2025