Chap31 - 498-520.qxd 31/5/06 11:59 am Page 498 31 Pythagoras’ theorem and trigonometry (2) CHAPTER 10 Linear equations CHAPTER In Chapter 19, Pythagoras’ theorem and trigonometry were used to find the lengths of sides and the sizes of angles in right-angled triangles. These methods will now be used with three-dimensional shapes. 31.1 Problems in three dimensions In a cuboid all the edges are perpendicular to each other. Problems with cuboids and other 3-D shapes involve identifying suitable right-angled triangles and applying Pythagoras’ theorem and trigonometry to them. Example 1 H ABCDEFGH is a cuboid with length 8 cm, breadth 6 cm and height 9 cm. a i Calculate the length of AC. ii Calculate the length of AG. Give your answer correct to 3 significant figures. b Calculate the size of angle GAC. Give your answer correct to the nearest degree. C 6 cm A E 8 cm D C 6 cm 8 cm B Look for a right-angled triangle where AC is one side and the lengths of the other two sides are known. ABC is a suitable triangle. So draw triangle ABC marking the known lengths. B AC2 AB2 BC2 82 62 64 36 100 AC 10 cm ii F 9 cm A Solution 1 a i G Use Pythagoras’ theorem for this triangle. G Look for a right-angled triangle where AG is one side and the lengths of the other two sides are known. ACG is a suitable triangle. G So draw triangle ACG marking the known lengths. 9 cm C A 10 cm C AG2 AC2 CG2 AG2 102 92 100 81 181 AG 181 13.4536 … AG 13.5 cm (to 3 s.f.) 498 A Use Pythagoras’ theorem for this triangle. Chap31 - 498-520.qxd 31/5/06 11:59 am Page 499 CHAPTER 31 31.1 Problems in three dimensions b G 9 cm A 10 cm tan (angle GAC) 190 0.9 angle GAC 41.987 …° Angle GAC 42° (to the nearest degree) For angle GAC. 9 cm is the opposite side. 10 cm is the adjacent side. opp tan adj C Exercise 31A Where necessary give lengths correct to 3 significant figures and angles correct to one decimal place. 1 ABCDEFGH is a cuboid of length 8 cm, breadth 4 cm and height 13 cm. a Calculate the length of ii GB iii AC iii FA iv GA. b Calculate the size of ii angle GBC iii angle GAC. i angle FAB H G E F 13 cm D C 4 cm A B 8 cm F 2 ABCDEF is a triangular prism. In triangle ABC angle CAB 90°, AB 5 cm and AC 12 cm. In rectangle ABED the length of BE 15 cm. a Calculate the length of CB. b Calculate the length of ii AF. i CE c Calculate the size of ii angle FAD. i angle FED C D E 12 cm 15 cm A 5 cm B O 3 The diagram shows a square-based pyramid. The lengths of sides of the square base, ABCD, are 10 cm and the base is on a horizontal plane. The centre of the base is the point M and the vertex of the pyramid is O, so that OM is vertical. The point E is the midpoint of the side AB. OA OB OC OD 15 cm. D a Calculate the length of i AC ii AM. 10 cm b Calculate the length of OM. c Calculate the size of angle OAM. d Hence find the size of angle AOC. e Calculate the length of OE. f Calculate the size of angle OAB. 15 cm C M B E A B Angle between a line and a plane Imagine a light shining directly above AB onto the plane. AN is the shadow of AB on the plane. A line drawn from point B perpendicular to the plane will meet the line AN and form a right angle with this line. Angle BAN is the angle between the line AB and the plane. N A 499 Chap31 - 498-520.qxd 31/5/06 11:59 am Page 500 CHAPTER 31 Pythagoras’ theorem and trigonometry (2) O Example 2 The diagram shows a pyramid. The base, ABCD, is a horizontal rectangle in which AB 12 cm and AD 9 cm. The vertex, O, is vertically above the midpoint of the base and OB 18 cm. Calculate the size of the angle that OB makes with the horizontal plane. Give your answer correct to one decimal place. Solution 2 18 cm C D O B 9 cm A 18 cm C D M 9 cm B 12 cm A O The base, ABCD, of the pyramid is horizontal so the angle that OB makes with the horizontal plane is the angle that OB makes with the base ABCD. Let M be the midpoint of the base which is directly below O. Join O to M and M to B. As OM is perpendicular to the base of the pyramid the angle OBM is the angle between OB and the base and so is the required angle. Draw triangle OBM marking OB 18 cm. To find the size of angle OBM find the length of either MB or OM. Calculate the length of MB which is 12 DB. 18 cm D Draw the right-angled triangle ABD marking the known lengths. 9 cm A M 12 cm 12 cm B B DB2 92 122 81 144 DB2 225 DB 225 15 Use Pythagoras’ theorem to calculate the length of DB. MB 12 DB 7.5 O For angle OBM, 18 cm is the hypotenuse, 7.5 cm is the adjacent side. 18 cm M 7.5 cm B 7.5 cos (angle OBM) 18 angle OBM 65.37 …° The angle between OB and the horizontal plane is 65.4° (to one d.p.) 500 adj cos hyp Chap31 - 498-520.qxd 31/5/06 11:59 am Page 501 CHAPTER 31 31.1 Problems in three dimensions Exercise 31B Where necessary give lengths correct to 3 significant figures and angles correct to one decimal place. O 1 The diagram shows a pyramid. The base, ABCD, is a horizontal rectangle in which AB 15 cm and AD 8 cm. The vertex, O, is vertically above the centre of the base and OA 24 cm. Calculate the size of the angle that OA makes with the horizontal plane. 24 cm C D B 8 cm 15 cm A 2 ABCDEFGH is a cuboid with a rectangular base in which AB 12 cm and BC 5 cm. The height, AE, of the cuboid is 15 cm. Calculate the size of the angle a between FA and ABCD b between GA and ABCD c between BE and ADHE d Write down the size of the angle between HE and ABFE. H G E F 15 cm D C 5 cm A 3 ABCDEF is a triangular prism. In triangle ABC, angle CAB 90°, AB 8 cm and AC 10 cm. In rectangle ABED, the length of BE 5 cm. C Calculate the size of the angle between a CB and ABED b CD and ABED c CE and ABED 10 cm d BC and ADFC. B 12 cm F D E 5 cm A 4 The diagram shows a square-based pyramid. The lengths of sides of the square base, ABCD, are 8 cm and the base is on a horizontal plane. The centre of the base is the point M and the vertex of the pyramid is O so that OM is vertical. The point E is the midpoint of the side AB. OA OB OC OD 20 cm Calculate the size of the angle between OE and the base ABCD. B 8 cm O 20 cm C D M 8 cm A B E 501 Chap31 - 498-520.qxd 31/5/06 11:59 am Page 502 CHAPTER 31 Pythagoras’ theorem and trigonometry (2) 5 ABCD is a horizontal rectangular lawn in a garden and TC is a vertical pole. Ropes run from the top of the pole, T, to the corners, A, B and D, of the lawn. a Calculate the length of the rope TA. b Calculate the size of the angle made with the lawn by ii the rope TD iii the rope TA. i the rope TB T 6m D C 12 m A B 8m 6 The diagram shows a learner’s ski slope, ABCD, of length, AB, 500 m. Triangles BAF and CDE are congruent right-angled triangles and ABCD, AFED and BCEF are rectangles. The rectangle BCEF is horizontal and the rectangle AFED is vertical. The angle between AB and BCEF is 20° and the angle between AC and BCEF is 10°. D A E C 500 m F Calculate a the length of FB c the distance AC B b the height of A above F d the width, BC, of the ski slope. 7 Diagram 1 shows a square-based pyramid OABCD. Each side of the square is of length 60 cm and OA OB OC OD 50 cm. O C 50 cm D B 60 cm 60 cm A Diagram 2 shows a cube, ABCDEFGH, in which each edge is of length 60 cm. A solid is made by placing the pyramid on top of the cube so that the base, ABCD, of the pyramid is on the top, ABCD, of the cube. The solid is placed on a horizontal table with the face, EFGH, on the table. a Calculate the height of the vertex O above the table. b Calculate the size of the angle between OE and the horizontal. 502 Diagram 1 C 60 cm 60 cm D B A G 60 cm H F E Diagram 2 Chap31 - 498-520.qxd 31/5/06 11:59 am Page 503 CHAPTER 31 31.2 Trigonometric ratios for any angle 31.2 Trigonometric ratios for any angle The diagram shows a circle, centre the origin O and radius 1 unit. Imagine a line, OP, of length 1 unit fixed at O, rotating in an anticlockwise direction about O, starting from the x-axis. The diagram shows OP when it has rotated through 40°. y 1 0.8 P 0.6 1 0.4 0.2 40° 1.2 1 0.8 0.6 0.4 0.2 O 0.2 Q 0.4 0.6 0.8 1 x 0.2 0.4 0.6 0.8 1 1.2 The right-angled triangle OPQ has hypotenuse OP 1 Relative to angle POQ, side PQ is the opposite side and side OQ is the adjacent side. This means that OQ cos 40° and PQ sin 40° For P, x cos 40° and y sin 40° so the coordinates of P are (cos 40°, sin 40°). In general when OP rotates through any angle °, the position of P on the circle, radius 1 is given by x cos °, y sin °. The coordinates of P are (cos °, sin °). So when OP rotates through 400° the coordinates of P are (cos 400°, sin 400°). A rotation of 400° is 1 complete revolution of 360° plus a further rotation of 40°. The position of P is the same as in the previous diagram so (cos 400°, sin 400°) is the same point as (cos 40°, sin 40°), therefore cos 400° cos 40° and sin 400° sin 40°. If OP rotates through 40° this means OP rotates through 40° in a clockwise direction. 503 Chap31 - 498-520.qxd 31/5/06 11:59 am Page 504 CHAPTER 31 Pythagoras’ theorem and trigonometry (2) For 136, 225, 304 and 40 the position of P is shown on the diagram. y 1.2 1 0.8 P 0.6 0.4 136° 0.2 1.2 1 0.8 0.6 0.4 0.2 O 0.2 0.2 225° 0.4 0.6 40° 0.8 x 1.2 0.4 0.6 P 304° P 1 0.8 P 1 1.2 For P when 136, x cos 136° and y sin 136°. From the diagram, cos 136° 0 and sin 136° 0 For P when 225, x cos 225° and y sin 225°. From the diagram, cos 225° 0 and sin 225° 0 For P when 304, x cos 304° and y sin 304°. From the diagram, cos 304° 0 and sin 304° 0 For P when 40, x cos 40° and y sin 40°. From the diagram, cos 40° 0 and sin 40° 0 The diagram shows for each quadrant whether the sine and cosine of angles in that quadrant are positive or negative. sin cos sin cos 2nd 1st 3rd 4th sin cos sin cos The sine and cosine of any angle can be found using your calculator. The following table shows some of these values corrected where necessary to 3 decimal places. 0 30 40 45 60 90 136 180 sin ° 0 0.5 0.643 0.707 0.866 1 0.695 0 cos ° 1 0.866 0.766 0.707 0.5 0 225 270 304 0.707 1 0.829 0.719 1 0.707 0 0.559 Using these values and others from a calculator the graphs of y sin ° and y cos ° can be drawn. A graphical calculator would be useful here. 504 360 0 1 Chap31 - 498-520.qxd 31/5/06 11:59 am Page 505 CHAPTER 31 31.2 Trigonometric ratios for any angle Graph of y sin ° y 1 0.5 180 O 180 360 540 θ 0.5 Notice that the graph: ● cuts the -axis at … , 180, 0, 180, 360, 540, … ● repeats itself every 360°, that is, it has a period of 360° ● has a maximum value of 1 at … , 90, 450, … ● has a minimum value of 1 at … , 90, 270, … 1 Graph of y cos ° y 1 0.5 180 O 180 360 540 θ 0.5 Notice that the graph: ● cuts the -axis at … 90, 90, 270, 450, … ● repeats itself every 360°, that is it has a period of 360° ● has a maximum value of 1 at … , 0, 360, … ● has a minimum value of 1 at … , 180, 180, 540, … 1 Notice also that the graph of y sin ° and the graph of y cos ° are horizontal translations of each other. sin ° To find the value of the tangent of any angle, use tan ° cos ° From the graph of y cos ° it can be seen that cos ° 0 at 90, 270, 450, … for example. As it is not possible to divide by 0 there are no values of tan ° at 90, 270, 450, … that is, the graph is discontinuous at these values of . Graph of y tan ° y 8 6 4 2 180 O 2 4 6 8 180 360 540 θ Notice that the graph: ● cuts the -axis where tan ° 0, that is, at … 180, 0, 180, 360, 540 … ● repeats itself every 180°, that is it has a period of 180° ● does not have values at 90, 270, 450, … ● does not have any maximum or minimum points. Notice also that tan ° can take any value. 505 Chap31 - 498-520.qxd 31/5/06 11:59 am Page 506 CHAPTER 31 Pythagoras’ theorem and trigonometry (2) Example 3 For values of in the interval – 180 to 360 solve the equation ii sin ° 0.7 ii 5 cos ° 2 Give each answer correct to one decimal place. Solution 3 i sin ° 0.7 Use a calculator to find one value of . 44.4 y 1 180 y 0.7 O 180 360 θ To find the other solutions draw a sketch of y sin ° for from 180 to 360 The sketch shows that there are two values of in the interval 180 to 360 for which sin ° 0.7 One solution is 44.4 and by symmetry the other solution is 180 44.4 1 44.4, 180 44.4 44.4, 135.6 ii 5 cos ° 2 Divide each side of the equation by 5 cos ° 25 0.4 66.4 Use a calculator to find one value of . y To find the other solutions draw a sketch of y cos ° for from 180 to 360 1 y 0.4 180 O 180 The sketch shows that there are three values of in the interval 180 to 360 for which cos ° 0.4 360 θ 1 66.4, 66.4, 360 66.4 66.4, 66.4, 293.6 One solution is 66.4 and by symmetry another solution is 66.4 Using the period of the graph the other solution is 360 66.4 Exercise 31C 1 For 360 360 sketch the graph of a y sin ° b y cos ° 2 Find all values of in the interval 0 to 360 for which a sin ° 0.5 b cos ° 0.1 c y tan °. c tan ° 1 3 a Show that one solution of the equation 3 sin ° 1 is 19.5, correct to 1 decimal place. b Hence solve the equation 3 sin ° 1 for values of in the interval 0 to 720 506 Chap31 - 498-520.qxd 31/5/06 11:59 am Page 507 CHAPTER 31 31.3 Area of a triangle 4 a Show that one solution of the equation 10 cos ° 3 is 107.5 correct to 1 decimal place. b Hence find all values of in the interval 360 to 360 for which 10 cos ° 3 5 Solve 4 tan ° 3 for values of in the interval 180 to 360 31.3 Area of a triangle Labelling sides and angles The vertices of a triangle are labelled with capital letters. The triangle shown is triangle ABC. B c a C A b The sides opposite the angles are labelled so that a is the length of the side opposite angle A, b is the length of the side opposite angle B and c is the length of the side opposite angle C. Area of a triangle 12 base height B Area of triangle ABC 12 bh In the right-angled triangle BCN a h a sin C So area of triangle ABC 12 b a sin C that is C c h A N b area of triangle ABC ab sin C 1 2 The angle C is the angle between the sides of length a and b and is called the included angle. The formula for the area of a triangle means that Area of a triangle 12 product of two sides sine of the included angle. For triangle ABC there are other formulae for the area. Area of triangle ABC 12 ab sin C 12 bc sin A 12 ac sin B. These formulae give the area of a triangle whether the included angle is acute or obtuse. Example 4 Find the area of each of the triangles correct to 3 significant figures. a b B 7.3 cm C 16.2 m 37° 5.8 cm A 118° 7.4 m 507 Chap31 - 498-520.qxd 31/5/06 11:59 am Page 508 CHAPTER 31 Pythagoras’ theorem and trigonometry (2) Solution 4 a Area 12 7.3 5.8 sin 37° Substitute a 7.3 cm, b 5.8 cm, C 37° into area 12 ab sin C Area 12.74 … Area 12.7 cm2 Give the area correct to 3 significant figures and state the units. b Area 12 7.4 16.2 sin 118° Area 52.92 … Substitute into area of a triangle 12 product of two sides sine of the included angle. Area 52.9 m2 Example 5 x° 2 The area of this triangle is 20 cm . Find the size of the acute angle x°. Give your angle correct to one decimal place. Solution 5 1 8.1 6.4 sin x° 20 2 8.1 cm 6.4 cm Use area of a triangle 12 product of two sides sine of the included angle. 2 20 sin x° 0.7716 8.1 6.4 Find the value of sin x°. x° 50.49 …° x° 50.5° Give the angle correct to one decimal place. Exercise 31D Give lengths and areas correct to 3 significant figures and angles correct to one decimal place. 1 Work out the area of each of these triangles. i ii iii 28° 9.3 cm 10.6 cm 13.5 cm 9.2 cm 34.7° 43° 6.9 cm 9.1 cm iv v vi 8.6 cm 148.6° 13.4 cm 76.3° 4.6 cm 4.6 cm 9.6 cm 137° 4.7 cm 2 ABCD is a quadrilateral. Work out the area of the quadrilateral. D 57° 9.4 cm C 12.6 cm 8.6 cm 80° A 508 B Chap31 - 498-520.qxd 31/5/06 11:59 am Page 509 CHAPTER 31 31.3 Area of a triangle 3 The area of triangle ABC is 15 cm2 Angle A is acute. Work out the size of angle A. C 6.5 cm A B 8.4 cm C 4 The area of triangle ABC is 60.7 m2 Work out the length of BC. A 35° B 12.6 m 5 a Triangle ABC is such that a 6 cm, b 9 cm and angle C 25°. Work out the area of triangle ABC. b Triangle PQR is such that p 6 cm, q 9 cm and angle R 155°. Work out the area of triangle PQR. c What do you notice about your answers? Why do you think this is true? 6 The diagram shows a regular octagon with centre O. a Work out the size of angle AOB. OA OB 6 cm. b Work out the area of triangle AOB. c Hence work out the area of the octagon. A B O 7 Work out the area of the parallelogram. 5.7 cm 63° 12.8 cm 8 a An equilateral triangle has sides of length 12 cm. Calculate the area of the equilateral triangle. b A regular hexagon has sides of length 12 cm. Calculate the area of the regular hexagon. 9 The diagram shows a sector, AOB, of a circle, centre O. The radius of the circle is 8 cm and the size of angle AOB is 50°. a Work out the area of triangle AOB. b Work out the area of the sector AOB. c Hence work out the area of the segment shown shaded in the diagram. A 8 cm 50° O B 8 cm 509 Chap31 - 498-520.qxd 31/5/06 11:59 am Page 510 CHAPTER 31 Pythagoras’ theorem and trigonometry (2) 31.4 The sine rule B a c C A b The last section showed that 1 1 1 Area of triangle ABC 2 ab sin C 2 bc sin A 2 ca sin B 1 2 ab sin C 12 bc sin A and cancelling 12 and b from both sides a sin C c sin A 1 2 bc sin A 12 ca sin B cancelling 12 and c from both sides and or b sin A a sin B or a c sin A sin C and b a sin B sin A Combining these results a b c sin A sin B sin C This result is known as the sine rule and can be used in any triangle. Using the sine rule to calculate a length Example 6 Find the length of the side marked a in the triangle. Give your answer correct to 3 significant figures. 74° a 37° Solution 6 a 8.4 sin 74° sin 37° 8.4 sin 37° a sin 74° a 5.258 … a 5.26 cm 510 8.4 cm a b Substitute A 37°, b 8.4, B 74° into . sin A sin B Multiply both sides by sin 37°. Chap31 - 498-520.qxd 31/5/06 11:59 am Page 511 CHAPTER 31 31.4 The sine rule Example 7 Find the length of the side marked x in the triangle. Give your answer correct to 3 significant figures. 18° x Solution 7 Missing angle 180 (18 124) 38° 9.7 cm 124° 18° The angle opposite 9.7 cm must be found before the sine rule can be used. Use the angle sum of a triangle. x 9.7 cm 124° 38° x 9.7 sin 124° sin 38° 9.7 sin 124° x sin 38° Write down the sine rule with x opposite 124° and 9.7 opposite 38°. Multiply both sides by sin 124°. x 13.06 … x 13.1 cm Using the sine rule to calculate an angle When the sine rule is used to calculate an angle it is a good idea to turn each fraction upside down (the reciprocal). This gives sin A sin B sin C a b c Example 8 Find the size of the acute angle x in the triangle. Give your answer correct to one decimal place. 74° 7.9 cm x Solution 8 sin x sin 74° 8.4 7.9 8.4 cm Write down the sine rule with x opposite 7.9 and 74° opposite 8.4 7.9 sin 74° sin x 8.4 Multiply both sides by 7.9 sin x 0.904 … Find the value of sin x. x 64.69 …° x 64.7° 511 Chap31 - 498-520.qxd 31/5/06 11:59 am Page 512 CHAPTER 31 Pythagoras’ theorem and trigonometry (2) Exercise 31E Give lengths and areas correct to 3 significant figures and angles correct to 1 decimal place. 1 Find the lengths of the sides marked with letters in these triangles. a b c a 79° b 27° 46° 58° 4.2 cm 51° 62° 11 cm 6.1 cm d e 64° d c f 17° 13.6 cm 113° f e g 22° 76° 134° 14.9 cm 6.1 cm 2 Calculate the size of each of the acute angles marked with a letter. a b c 6 cm A 8 cm 32° 17 cm 21° 6.8 cm d C 18.4 cm 73° 9.1 cm B E 104° 7.6 cm 3 The diagram shows quadrilateral ABCD and its diagonal AC. a In triangle ABC, work out the length of AC. b In triangle ACD, work out the size of angle DAC. c Work out the size of angle BCD. D D 12.7 cm 8.6 cm C 102° 5.7 cm 46° 18° B A 4 In triangle ABC, BC 8.6 cm, angle BAC 52° and angle ABC 63°. a Calculate the length of AC. b Calculate the length of AB. c Calculate the area of triangle ABC. 5 In triangle PQR all the angles are acute. PR 7.8 cm and PQ 8.4 cm. Angle PQR 58°. a Work out the size of angle PRQ. b Work out the length of QR. 6 The diagram shows the position of a port (P), a lighthouse (L) and a buoy (B). The lighthouse is due east of the buoy. B The lighthouse is on a bearing of 035° from the port and the buoy is on a bearing of 312° from the port. ii angle PLB. a Work out the size of i angle PBL The lighthouse is 8 km from the port. b Work out the distance PB. c Work out the distance BL. d Work out the shortest distance from the port (P) to the line BL. 512 N L P Chap31 - 498-520.qxd 31/5/06 11:59 am Page 513 CHAPTER 31 31.5 The cosine rule 31.5 The cosine rule The diagram shows triangle ABC. The line BN is perpendicular to AC and meets the line AC at N so that AN x and NC (b x). The length of BN is h. In triangle ANB Pythagoras’ theorem gives c2 x2 h2 1 In triangle BNC Pythagoras’ theorem gives a2 (b x)2 h2 a2 b2 2bx x2 h2 Using 1 substitute c2 for x2 h2 a2 b2 2bx c2 2 B c a h x (b x) N A b C In the right-angled triangle ANB, x c cos A Substituting this into 2 a2 b2 c2 2bc cos A This result is known as the cosine rule and can be used in any triangle. b2 a2 c2 2ac cos B c2 a2 b2 2ab cos C Similarly and Using the cosine rule to calculate a length Example 9 Find the length of the side marked with a letter in each triangle. Give your answers correct to 3 significant figures. a b B a x 8 cm 7.3 cm 117° 24° C 12 cm Solution 9 a a2 122 82 2 12 8 cos 24° a 144 64 175.4007 … 2 5.8 cm A Substitute b 12, c 8, A 24° into a2 b2 c2 2bc cos A. Evaluate each term separately. a2 32.599 27 … a 32.599 27 … a 5.709 577 … a 5.71 cm b x2 7.32 5.82 2 7.3 5.8 cos 117° x2 53.29 33.64 84.68 (0.4539 …) x2 86.93 38.44 … x2 125.37 … x 125.37 … x 11.19 … x 11.2 cm Take the square root. Substitute the two given lengths and the included angle into the cosine rule. cos 117° 0 Take the square root. 513 Chap31 - 498-520.qxd 31/5/06 11:59 am Page 514 CHAPTER 31 Pythagoras’ theorem and trigonometry (2) Using the cosine rule to calculate an angle To find an angle using the cosine rule, when the lengths of all three sides of a triangle are known, rearrange a2 b2 c2 2bc cos A. 2bc cos A b2 c2 a2 b2 c2 a2 cos A 2bc Similarly a2 c2 b2 cos B 2ac and a2 b2 c2 cos C 2ab Example 10 Find the size of a angle BAC b angle X. Give your answers correct to one decimal place. a b B 12.7 cm 16 cm 13 cm 8.6 cm X A 11 cm C Solution 10 112 162 132 a cos A 2 11 16 208 cos A 352 cos A 0.590 909 … A 53.77 …° A 53.8° 8.62 6.92 12.72 b cos X 2 8.6 6.9 39.72 cos X 118.68 cos X 0.334 68 … X 109.55 …° X 109.6° 514 6.9 cm b2 c2 a2 Substitute b 11, c 16, a 13 into cos A . 2bc Substitute the three lengths into the cosine rule noting that 12.7 cm is opposite the angle to be found. The value of cos X is negative so X is an obtuse angle. Chap31 - 498-520.qxd 31/5/06 11:59 am Page 515 CHAPTER 31 31.5 The cosine rule Exercise 31F Where necessary give lengths and areas correct to 3 significant figures and angles correct to 1 decimal place. 1 Calculate the length of the sides marked with letters in these triangles. a b c a 8 cm 11.3 cm b 18° 16.2 cm 62° 15.5 cm 75° 9 cm 9.2 cm c d e d f 10.2 cm 9.6 cm 8.4 cm e 147° 134° 52° 9.6 cm f 8.4 cm 6.3 cm 2 Calculate the size of each of the angles marked with a letter in these triangles. a b 7 cm 9 cm 9.4 cm 15.3 cm A B 11 cm 13.6 cm c d 8.6 cm C D 8.7 cm 8.7 cm 7.2 cm 14.4 cm 6.8 cm 3 The diagram shows the quadrilateral ABCD. a Work out the length of DB. b Work out the size of angle DAB. c Work out the area of quadrilateral ABCD. C 26.4 cm D 56° 8.4 cm 9.8 cm A 16.3 cm B 4 Work out the perimeter of triangle PQR. R 8.6 cm Q 27° P 10.9 cm 5 In triangle ABC, AB 10.1 cm, AC 9.4 cm and BC 8.7 cm. Calculate the size of angle BAC. 6 In triangle XYZ, XY 20.3 cm, XZ 14.5 cm and angle YXZ 38°. Calculate the length of YZ. 515 Chap31 - 498-520.qxd 31/5/06 11:59 am Page 516 CHAPTER 31 Pythagoras’ theorem and trigonometry (2) 7 AB is a chord of a circle with centre O. The radius of the circle is 7 cm and the length of the chord is 11 cm. Calculate the size of angle AOB. O 7 cm B 11 cm A 8 The region ABC is marked on a school field. The point B is 70 m from A on a bearing of 064°. The point C is 90 m from A on a bearing of 132°. a Work out the size of angle BAC. b Work out the length of BC. N B 70 m A 90 m C 9 Chris ran 4 km on a bearing of 036° from P to Q. He then ran in a straight line from Q to R where R is 7 km due East of P. Chris then ran in a straight line from R to P. Calculate the total distance run by Chris. 10 The diagram shows a parallelogram. Work out the length of each diagonal of the parallelogram. 6 cm 65° 8 cm 31.6 Solving problems using the sine rule, the cosine rule and 1 ab sin C 2 Example 11 The area of triangle ABC is 12 cm2 AB 3.8 cm and angle ABC 70°. ii AC. a Find the length of i BC Give your answers correct to 3 significant figures. b Find the size of angle BAC. Give your answer correct to 1 decimal place. A 3.8 cm 70° B Solution 11 a i 12 3.8 BC sin 70° 12 C Substitute c 3.8, B 70° into area 12 ac sin B. 2 12 BC 3.8 sin 70° BC 6.721 … BC 6.72 cm ii b2 6.721 …2 3.82 2 6.721… 3.8 cos 70° b2 59.613 … 17.470 … b2 42.142 … b 6.491 … AC 6.49 cm 516 Substitute a 6.721 …, c 3.8 and B 70° into b2 a2 c2 2ac cos B. Chap31 - 498-520.qxd 31/5/06 11:59 am Page 517 1 31.6 Solving problems using the sine rule, the cosine rule and 2ab sin C sin A sin 70° b 6.721 … 6.491 … 6.721 … sin 70° sin A 6.491 … CHAPTER 31 sin A sin B Substitute a 6.721 …, b 6.491 … and B 70° into . a b sin A 0.9728 … A 76.62 …° Angle BAC 76.6° Exercise 31G Where necessary give lengths and areas correct to 3 significant figures and angles correct to 1 decimal place, unless the question states otherwise. 1 A triangle has sides of lengths 9 cm, 10 cm and 11 cm. a Calculate the size of each angle of the triangle. b Calculate the area of the triangle. 2 In the diagram ABC is a straight line. a Calculate the length of BD. b Calculate the size of angle DAB. c Calculate the length of AC. D 12.9 cm A 5.4cm B 12 cm 65° 47° C 2 3 The area of triangle ABC is 15 cm . AB 4.6 cm and angle BAC 63˚. a Work out the length of AC. b Work out the length of BC. c Work out the size of angle ABC. 4 ABCD is a kite with diagonal DB. a Calculate the length of DB. b Calculate the size of angle BDC. c Calculate the value of x. d Calculate the length of AC. D x cm 8 cm A 50° C 30° 8 cm x cm B 5 Kultar walked 9 km due South from point A to point B. He then changed direction and walked 5 km to point C. Kultar was then 6 km from his starting point A. a Work out the bearing of point C from point B. Give your answer correct to the nearest degree. b Work out the bearing of point C from point A. Give your answer correct to the nearest degree. 6 The diagram shows a pyramid. The base of the pyramid, ABCD, is a rectangle in which AB 15 cm and AD 8 cm. The vertex of the pyramid is O where OA OB OC OD 20 cm. Work out the size of angle DOB correct to the nearest degree. O 20 cm C D 8 cm A 15 cm B 517 Chap31 - 498-520.qxd 31/5/06 11:59 am Page 518 CHAPTER 31 Pythagoras’ theorem and trigonometry (2) 7 The diagram shows a vertical pole, PQ, standing on a hill. The hill is at an angle of 8° to the horizontal. The point R is 20 m downhill from Q and the line PR is at 12° to the hill. a Calculate the size of angle RPQ. b Calculate the length, PQ, of the pole. 12° R P Q 20 m 8° 8 A, B and C are points on horizontal ground so that AB 30 m, BC 24 m and angle CAB 50°. AP and BQ are vertical posts, where AP BQ 10 m. a Work out the size of angle ACB. b Work out the length of AC. c Work out the size of angle PCQ. d Work out the size of the angle between QC and the ground. P Q 10 m 10 m 30 m A B 50° 24 m C Chapter summary You should now be able to: use Pythagoras’ theorem to solve problems in 3 dimensions use trigonometry to solve problems in 3 dimensions work out the size of the angle between a line and a plane draw sketches of the graphs of y sin x °, y cos x °, y tan x ° and use these graphs to solve simple trigonometric equations use the formula area 12 ab sin C to calculate the area of any triangle a b c use the sine rule and the cosine rule a2 b2 c2 2bc cos A in sin A sin B sin C triangles and in solving problems. Chapter 31 review questions 1 In the diagram, XY represents a vertical tower on level ground. A and B are points due West of Y. The distance AB is 30 metres. The angle of elevation of X from A is 30º. The angle of elevation of X from B is 50º. Calculate the height, in metres, of the tower XY. Give your answer correct to 2 decimal places. A X Diagram NOT accurately drawn 50° 30° 30 m B Y (1384 June 1996) 2 The diagram shows triangle ABC. AC 7.2 cm BC 8.35 cm Angle ACB 74°. a Calculate the area of triangle ABC. Give your answer correct to 3 significant figures. b Calculate the length of AB. Give your answer correct to 3 significant figures. 518 C Diagram NOT accurately drawn 74° 7.2 cm A 8.35 cm B (1385 June 2002) Chap31 - 498-520.qxd 31/5/06 11:59 am Page 519 CHAPTER 31 Chapter 31 review questions 3 In triangle ABC AC 8 cm CB 15 cm Angle ACB 70°. a Calculate the area of triangle ABC. Give your answer correct to 3 significant figures. X 8 cm 70° C 4 The diagram shows a cuboid. A, B, C, D and E are five vertices of the cuboid. AB 5 cm BC 8 cm CE 3 cm. B 15 cm X is the point on AB such that angle CXB 90°. b Calculate the length of CX. Give your answer correct to 3 significant figures. (1387 June 2003) Diagram NOT accurately drawn E 3 cm D C Calculate the size of the angle the diagonal AE makes with the plane A ABCD. Give your answer correct to 1 decimal place. 5 In triangle ABC AC 8 cm BC 15 cm Angle ACB 70°. a Calculate the length of AB. Give your answer correct to 3 significant figures. b Calculate the size of angle BAC. Give your answer correct to 1 decimal place. Diagram NOT accurately drawn A 8 cm 5 cm B Diagram NOT accurately drawn A 8 cm 70° B 6 This is a sketch of the graph of y cos x° for values of x between 0 and 360. Write down the coordinates of the point ii A ii B. C 15 cm (1387 June 2003) y A 360 x O B 7 Angle ACB 150° BC 60 m. The area of triangle ABC is 450 m2 Calculate the perimeter of triangle ABC. Give your answer correct to 3 significant figures. C Diagram NOT accurately drawn 150° 60 m A B (1385 November 2000) 519 Chap31 - 498-520.qxd 31/5/06 11:59 am Page 520 CHAPTER 31 Pythagoras’ theorem and trigonometry (2) 8 The diagram shows a quadrilateral ABCD. AB 4.1 cm BC 7.6 cm AD 5.4 cm Angle ABC 117° Angle ADC 62°. a Calculate the length of AC. Give your answer correct to 3 significant figures. b Calculate the area of triangle ABC. Give your answer correct to 3 significant figures. c Calculate the area of the quadrilateral ABCD. Give your answer correct to 3 significant figures. Diagram NOT accurately drawn A 5.4 cm 4.1 cm 62° B 117° D 7.6 cm C (1385 June 2000) 9 This is a graph of the curve y sin x° for 0 x 180 y 1 0.5 O 45 90 135 x 180 0.5 1 a Using the graph or otherwise, find estimates of the solutions in the interval 0 x 360 of the equation ii sin x° 0.6. i sin x° 0.2 cos x° sin (x 90)° for all values of x. b Write down two solutions of the equation cos x° 0.2 (1385 November 2002) 10 In the diagram, ABCD, ABFE and EFCD are rectangles. The plane EFCD is horizontal and the plane ABFE is vertical. EA 10 cm DC 20 cm ED 20 cm. B A 10 cm 20 cm Calculate the size of the angle that the line AC makes with the plane EFCD. 11 In triangle ABC AB 10 cm AC 14 cm BC 16 cm. a Calculate the size of the smallest angle in the triangle. Give your answer correct to the nearest 0.1°. b Calculate the area of triangle ABC. Give your answer correct to 3 significant figures. 520 F E D C Diagram NOT accurately drawn A 14 cm 10 cm B 20 cm 16 cm C

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