Angles — Paper 4 · IGCSE Mathematics 0580

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

15 marks

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

This question in 0580/41 May/June 2017

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

8 marks

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

This question in 0580/43 May/June 2017

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

9 marks

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

This question in 0580/43 Oct/Nov 2017

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

19 marks

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

This question in 0580/41 Oct/Nov 2019

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

16 marks

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

This question in 0580/43 Oct/Nov 2019

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

18 marks

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

This question in 0580/41 Oct/Nov 2020

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

8 marks

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

This question in 0580/42 Feb/March 2021

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

17 marks

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

This question in 0580/41 May/June 2022

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

16 marks

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

This question in 0580/41 May/June 2022

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

10 marks

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

This question in 0580/42 May/June 2022

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

18 marks

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

This question in 0580/42 May/June 2023

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

9 marks

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

This question in 0580/42 May/June 2024

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

10 marks

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

This question in 0580/42 May/June 2024

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

5 marks

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

This question in 0580/41 Oct/Nov 2025