Right-angled triangles — Paper 4 · IGCSE Mathematics 0580

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

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 (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