4 (a) The diagram shows a sketch of quadrilateral ABCD. A C B D 4.5 cm 8 cm 10 cm 6 cm NOT TO SCALE 50° (i) Construct an accurate drawing of ABCD. AC has been drawn for you. A C [3] (ii) Measure ADC | . … [1] (iii) By taking a suitable measurement from your diagram, find the perimeter of quadrilateral ABCD. … cm [1] (b) P T X Q U R V W S 8 10 15 The diagram shows a cuboid. TW = 15 cm, WV = 10 cm and RV = 8 cm. (i) Show that TR = 19.7 cm, correct to 1 decimal place. [3] (ii) X is the midpoint of PQ. Calculate TRX t . TRX = t … [5]

4(a)(i) Correct quadrilateral with construction arcs 3 B2 for triangle ABC correct with construction arcs or triangle ADC correct or B1 for ABC correct with no/incorrect arcs or ˆ DAC = 50o 4(a)(ii) their ˆ ADC 1 FT their diagram 4(a)(iii) 18.5 + their DC 1 FT their diagram 4(b)(i) 82 + 102 + 152 oe M2 M1 for any correct 2D Pythag pair e.g. 82 + 102 19.72… A1 4(b)(ii) 28.[0] to 28.2 5 M1 for 82 + ( ) 2 10 2 M1 for 152 + ( ) 2 10 2 M2 for cos…= 2 2 2 19.7 2 19.7 + − × × RX TX RX or M1 for TX2 = 19.72 + RX2 – 2 × 19.7 × RX cos… (a)(i) To earn marks, you must accurately draw the quadrilateral using a ruler and compass, showing all construction arcs. (a)(ii) To earn marks, you must measure the angle ADC from your constructed diagram. (a)(iii) To earn marks, you must measure the length of DC from your diagram and add it to the given lengths to find the perimeter. (b)(i) To earn marks, you must use the Pythagorean theorem in three dimensions (or two successive 2D applications) to calculate the length of TR and show it rounds to 19.7 cm. (b)(ii) To earn marks, you must calculate the lengths of RX and TX using the Pythagorean theorem and then use the cosine rule to find angle TRX.

4 (a) NOT TO SCALE North 7.5 3.8 140° P Q R The diagram shows the positions of three villages, P, Q and R. R is due west of P and Q is on a bearing of 140° from P. PR = 3.8 km and QR = 7.5 km. (i) Calculate angle PRQ. Angle PRQ = … [4] (ii) Work out the bearing of R from Q. … [2] (b) The distance by road from village P to village T is 16.5 km. Kwesi leaves village P at 10 30 and drives to village T at an average speed of 45 km/h. He stops in village T for 15 minutes. He then drives back to village P and arrives there at 11 35. Calculate Kwesi’s average speed, in km/h, for the journey back from village T to village P. … km/h [4]

4(a)(i) 27.1 to 27.2 4 B1 for ∠RPQ = 130° soi M2 for [ ] ( ) 3.8sin 130 sin 7.5 = their oe or M1 for ( ) [ ] 7.5 3.8 sin 130 sin = their oe 4(a)(ii) 297.1 to 297.2 2 FT 270 + their 27.2, where 0 ⩽ their 27.2 ⩽ 90 or for 320 – their ∠PQR where 0 ⩽ their ∠PQR ⩽ 90 M1 for 180 + 90 + their 27.2 oe or for 360 – (40 + 180 – 130 – their 27.2) or for 360 – 40 – their ∠PQR 4(b) 35.3 to 35.4 4 M1 for 16.5 45 M1 for [time =] 65 – 15 – their 22 oe M1 for [ ] 16.5 60 28 × their

6 (a) NOT TO SCALE 2x 3x + 5 8 The diagram shows a right-angled triangle, with dimensions given in centimetres. (i) Show that x x 5 30 39 0 2 + - = . [3] (ii) Solve the equation x x 5 30 39 0 2 + - = . Show your working and give your answers correct to 2 decimal places. x = … or x = … [3] (iii) Calculate the area of the triangle. … cm2 [2] (b) NOT TO SCALE 35° A B C The diagram shows a different right-angled triangle. The length of the hypotenuse is 12 cm. Calculate the shortest distance from A to BC. … cm [4]

6(a)(i) ( ) ( ) 2 2 2 8 2 3 5 + = + x x M1 2 9 15 15 25 + + + x x x B1 Expansion of brackets 2 2 64 4 9 15 15 25 + = + + + x x x x leading to 2 5 30 39 0 + − = x x A1 6(a)(ii) 2 30 30 4 5 39 2 5 − ± −× × − × oe or 2 30 30 39 10 10 5 − −   ± −     oe B2 B1 for 2 30 4 5 39 − × × − oe or for 30 1680 2 5 − ± × their oe or 2 30 10   +     x oe 1.10 and –7.10 B1 6(a)(iii) 8.78 to 8.8[0] 2 M1 for 1 8 2 1.10 2 × × ×their 6(b) 5.63 to 5.64 nfww 4 M1 for correct method to find AB or AC [ ] cos35 12 = AC oe or ( ) [ ] cos 90 35 12 − = AB oe M2 for [ ] sin35 = × their AC oe or [ ] sin(90 35) = × − their AB oe or [ ] 1 12 sin35 2 1 12 2 × × × = × their AC oe or [ ] 1 12 sin(90 35) 2 1 12 2 × × × − = × their AB oe or M1 for [ ] sin35 = their AC oe or ( ) [ ] sin 90 35 − = their AB oe or for 1 1 12 12 sin35 2 2 × × = × × × d theirAC oe or 1 1 12 12 sin(90 35) 2 2 × × = × × × − d theirAB oe

7 (a) A rectangular field measures 30 m by 45 m. (i) Calculate the perimeter. … m [1] (ii) Calculate the length of a diagonal. … m [2] (b) Q a P R 11 a + 4 60° NOT TO SCALE The diagram shows a sketch of triangle PQR. All lengths are given in centimetres. Calculate the length a. a = … cm [4] (c) The diagram shows a sketch of quadrilateral ABCD. All lengths are given in centimetres. A B C D 8 9 10 NOT TO SCALE The area of quadrilateral ABCD is 70 cm2. Calculate DAB t . DAB = t … [6]

7(a)(i) 150 1 7(a)(ii) 54.1 or 54.08… 2 M1 for 302 + 452 7(b) 5 10 19 or 5.53 or 5.526… nfww 4 M1 for 2 2 2 ( 4) 11 2 11cos60 + = + − × × a a a oe B1 for a2 + 4a + 4a + 16 oe M1 for correct isolation of their terms in a to form a linear equation with use of cos 60 = 0.5 After 0 scored, SC1 for a + 4 = a2 + 121 – 11a oe 7(c) 107.8 or 107.78 to 107.79… 6 M2 for 1 10 tan 8 −      oe or M1 for tan … = 10 8 oe B1 for 1 10 8 2 × × soi M2 for 1 8 9sin 70 40 2 × × = − x their oe or M1 for 1 8 9sin 2 × × x (a)(i) To earn marks, you must correctly calculate the perimeter of the rectangle (2 × (length + width)). (a)(ii) To earn marks, you must use the Pythagorean theorem (length² + width² = diagonal²) to find the length of the diagonal. (b) To earn marks, you must apply the cosine rule to triangle PQR, setting up the equation correctly with 'a' and 'a+4', and then solve for 'a'. (c) To earn marks, you must first calculate the area of triangle ABD using the formula (1/2)ab sin(DAB), and then use the given total area to find the area of triangle BCD. You will then need to use the area of triangle BCD to find the length of BD, and finally use the sine rule or cosine rule in triangle ABD to find angle DAB.

8 (a) P O R Q x° 8x° 24° NOT TO SCALE P, Q and R are points on the circumference of a circle, centre O. Angle POQ = 8xo, angle RPO = xo and angle OQR = 24o. Calculate angle PQO. Angle PQO = … [4] (b) J N K 10 10.5 6 M L y° NOT TO SCALE Triangle KLM is similar to triangle KNJ. JKM and NKL are straight lines. KLM KNJ = t t and LKM t = y o. KL = 6 cm, KM = 10.5 cm and KN = 10 cm. The area of triangle JKN is 75 cm2. Calculate y. y = … [5]

8(a) 58° 4 B2 for x = 8 or B1 for ˆ PRQ = 4x or for ˆ POR = 180 – 2x M1 for a correct equation involving ˆ PQO 8(b) 58.99… to 59[.0] 5 B2 for JK = 17.5 or M1 for 6 10.5 10 = JK oe M2 for sin y = 75 0.5 10 17.5 × ×their oe or M1 for 0.5 × 10 × their 17.5 × sin y = 75 oe Alternative: B2 for area MKL = 27 or M1 for 2 6 10       or 2 10 6       M2 for sin y = ( ) 2 6 10 75 0.5 6 10.5 × × × their oe or M1 for 0.5 × 6 × 10.5 × sin y = 75 × their ( ) 2 6 10 oe

Cambridge O Level Mathematics (Syllabus D) · 4024

O-Level Mathematics (Syllabus D) (4024) Trigonometry past paper questions

Revise O-Level Mathematics (Syllabus D) (4024) — Trigonometry with real Cambridge past-paper questions grouped by topic, each shown with its marking scheme. Working one topic at a time targets exactly the skill each question tests.

47 free questions from 2016–2021, each with its marking scheme. Open the full topical bank to filter by subtopic, paper, and year.