Non-right-angled triangles — Paper 4 · IGCSE Mathematics (9-1) 0980

8 (a) D 64° 53° 16.5 cm NOT TO SCALE 12.4 cm A 95° B C X The diagram shows two triangles ABD and BCD. DBX is a straight line. 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 the shortest distance from C to the line BX. … cm [5] (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]

13 marks

Mark scheme: 8(a)(i) 15.7 or 15.70... 4 M2 for 16.5 2 + 12.4 2 −×2 16.5 × 12.4 × cos64 or M1 for implicit form A1 for 246 to 247 8(a)(ii) 18.6 or 18.61 to 18.62 5 B1 for [angle C = ]32, angle DBM = 37 or angle CBM = 58 12.4 × sin53 M2 for [BC =] oe sin32 × sin 95 or [DC = ]12.4 oe sin 32 or M1 for implicit form for either BC or DC CN M1 for sin85 = oe where N is the foot theirBC of the perpendicular from C to BX CN or for sin 53 = oe where N is the foot theirDC of the perpendicular from C to BX 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 0980/42 May/June 2019

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

12 marks

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

This question in 0980/41 Oct/Nov 2019

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

10 marks

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

This question in 0980/42 May/June 2020

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

14 marks

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

This question in 0980/41 Oct/Nov 2020

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 0980/42 May/June 2021

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

10 marks

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

This question in 0980/42 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 0980/41 Oct/Nov 2022

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

11 marks

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

This question in 0980/42 May/June 2023

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

14 marks

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

This question in 0980/41 Oct/Nov 2023

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

10 marks

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

This question in 0980/42 May/June 2024

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

11 marks

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

This question in 0980/41 Oct/Nov 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 0980/42 May/June 2025

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

2 marks

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

This question in 0980/41 Oct/Nov 2025

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

8 marks

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

This question in 0980/41 Oct/Nov 2025