HPTR 10 Pythgors theorem nd trigonometry (2) 31 HPTR Liner equtions In hpter 19, Pythgors theorem nd trigonometry were used to find the lengths of sides nd the sizes of ngles in right-ngled tringles. These methods will now be used with three-dimensionl shpes. 31.1 Problems in three dimensions In cuboid ll the edges re perpendiculr to ech other. Problems with cuboids nd other 3- shpes involve identifying suitble right-ngled tringles nd pplying Pythgors theorem nd trigonometry to them. xmple 1 FGH is cuboid with length, bredth 6 cm nd height 9 cm. i lculte the length of. ii lculte the length of G. Give your nswer correct to 3 significnt figures. b lculte the size of ngle G. Give your nswer correct to the nerest degree. Solution 1 i 6 cm 6 cm Look for right-ngled tringle where is one side nd the lengths of the other two sides re known. is suitble tringle. So drw tringle mrking the known lengths. H F G 9 cm ii 2 2 2 8 2 6 2 64 36 100 10 cm G Use Pythgors theorem for this tringle. Look for right-ngled tringle where G is one side nd the lengths of the other two sides re known. G is suitble tringle. G So drw tringle G mrking the known lengths. 10 cm 9 cm G 2 2 G 2 G 2 10 2 9 2 100 81 181 G 181 13.4536 G 13.5 cm (to 3 s.f.) Use Pythgors theorem for this tringle. 498
31.1 Problems in three dimensions HPTR 31 b G 9 cm For ngle G. 9 cm is the opposite side. 10 cm is the djcent side. tn (ngle G) 10 0.9 ngle G 41.987 ngle G 42 (to the nerest degree) 9 tn o p p dj 10 cm xercise 31 Where necessry give lengths correct to 3 significnt figures nd ngles correct to one deciml plce. 1 FGH is cuboid of length, bredth 4 cm nd H height 13 cm. G lculte the length of F iii ii G iii F iv G. 13 cm b lculte the size of i ngle F ii ngle G iii ngle G. 2 F is tringulr prism. In tringle ngle 90, 5 cm nd 12 cm. In rectngle the length of 15 cm. lculte the length of. b lculte the length of i ii F. c lculte the size of i ngle F ii ngle F. 3 The digrm shows squre-bsed pyrmid. The lengths of sides of the squre bse,,re 10 cm nd the bse is on horizontl plne. The centre of the bse is the point M nd the vertex of the pyrmid is, so tht M is verticl. The point is the midpoint of the side. 15 cm. lculte the length of i ii M. b lculte the length of M. c lculte the size of ngle M. d Hence find the size of ngle. e lculte the length of. f lculte the size of ngle. ngle between line nd plne 12 cm Imgine light shining directly bove onto the plne. N is the shdow of on the plne. line drwn from point perpendiculr to the plne will meet the line N nd form right ngle with this line. ngle N is the ngle between the line nd the plne. 5 cm 10 cm F 15 cm M 15 cm N 4 cm 499
HPTR 31 Pythgors theorem nd trigonometry (2) xmple 2 The digrm shows pyrmid. The bse,, is horizontl rectngle in which 12 cm nd 9 cm. The vertex,, is verticlly bove the midpoint of the bse nd 1. lculte the size of the ngle tht mkes with the horizontl plne. Give your nswer correct to one deciml plce. Solution 2 9 cm 12 cm 1 9 cm M 12 cm 1 The bse,, of the pyrmid is horizontl so the ngle tht mkes with the horizontl plne is the ngle tht mkes with the bse. Let M be the midpoint of the bse which is directly below. Join to M nd M to. s M is perpendiculr to the bse of the pyrmid the ngle M is the ngle between nd the bse nd so is the required ngle. 1 rw tringle M mrking 1. To find the size of ngle M find the length of either M or M. lculte the length of M which is 1 2. 9 cm rw the right-ngled tringle mrking the known lengths. 12 cm M 2 9 2 12 2 81 144 2 225 225 15 M 1 2 7.5 Use Pythgors theorem to clculte the length of. For ngle M,18cm is the hypotenuse, 7.5 cm is the djcent side. 1 M 7.5 cm cos (ngle M) 7. 5 18 ngle M 65.37 dj cos h yp The ngle between nd the horizontl plne is 65.4 (to one d.p.) 500
31.1 Problems in three dimensions HPTR 31 xercise 31 Where necessry give lengths correct to 3 significnt figures nd ngles correct to one deciml plce. 1 The digrm shows pyrmid. The bse,, is horizontl rectngle in which 15 cm nd. The vertex,, is verticlly bove the centre of the bse nd 24 cm. lculte the size of the ngle tht mkes with the horizontl plne. 24 cm 15 cm 2 FGH is cuboid with rectngulr bse in which 12 cm nd 5 cm. The height,, of the cuboid is 15 cm. lculte the size of the ngle between F nd b between G nd c between nd H d Write down the size of the ngle between H nd F. 15 cm H 12 cm G F 5 cm 3 F is tringulr prism. In tringle, ngle 90, nd 10 cm. In rectngle, the length of 5 cm. lculte the size of the ngle between nd b nd c nd d nd F. 10 cm F 5 cm 4 The digrm shows squre-bsed pyrmid. The lengths of sides of the squre bse,, re nd the bse is on horizontl plne. The centre of the bse is the point M nd the vertex of the pyrmid is so tht M is verticl. The point is the midpoint of the side. 20 cm lculte the size of the ngle between nd the bse. M 20 cm 501
HPTR 31 Pythgors theorem nd trigonometry (2) 5 is horizontl rectngulr lwn in grden nd T is verticl pole. Ropes run from the top of the pole, T, to the corners,, nd, of the lwn. lculte the length of the rope T. b lculte the size of the ngle mde with the lwn by i the rope T ii the rope T iii the rope T. T 6 m 12 m 8 m 6 The digrm shows lerner s ski slope,, of length,, 500 m. Tringles F nd re congruent right-ngled tringles nd, F nd F re rectngles. The rectngle F is horizontl nd the rectngle F is verticl. The ngle between nd F is 20 nd the ngle between nd F is 10. 500 m lculte the length of F c the distnce F b the height of bove F d the width,, of the ski slope. 7 igrm 1 shows squre-bsed pyrmid. ch side of the squre is of length 60 cm nd 50 cm. 60 cm 60 cm 50 cm igrm 1 igrm 2 shows cube, FGH, in which ech edge is of length 60 cm. solid is mde by plcing the pyrmid on top of the cube so tht the bse,, of the pyrmid is on the top,, of the cube. The solid is plced on horizontl tble with the fce, FGH, on the tble. lculte the height of the vertex bove the tble. b lculte the size of the ngle between nd the horizontl. H 60 cm G 60 cm 60 cm F igrm 2 502
31.2 Trigonometric rtios for ny ngle HPTR 31 31.2 Trigonometric rtios for ny ngle The digrm shows circle, centre the origin nd rdius 1 unit. Imgine line, P, of length 1 unit fixed t, rotting in n nticlockwise direction bout, strting from the x-xis. The digrm shows P when it hs rotted through 40. y 1 0.8 0.6 P 0.4 1 0.2 40 Q 1.2 1 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1 0.2 x 0.4 0.6 0.8 1 1.2 The right-ngled tringle PQ hs hypotenuse P 1 Reltive to ngle PQ, side PQ is the opposite side nd side Q is the djcent side. This mens tht Q cos 40 nd PQ sin 40 For P, x cos 40 nd y sin 40 so the coordintes of P re (cos 40, sin 40 ). In generl when P rottes through ny ngle, the position of P on the circle, rdius 1 is given by x cos, y sin. The coordintes of P re (cos, sin ). So when P rottes through 400 the coordintes of P re (cos 400, sin 400 ). rottion of 400 is 1 complete revolution of 360 plus further rottion of 40. The position of P is the sme s in the previous digrm so (cos 400, sin 400 ) is the sme point s (cos 40, sin 40 ), therefore cos 400 cos 40 nd sin 400 sin 40. If P rottes through 40 this mens P rottes through 40 in clockwise direction. 503
HPTR 31 Pythgors theorem nd trigonometry (2) For 136, 225, 304 nd 40 the position of P is shown on the digrm. y 1.2 1 P 0.8 0.6 0.4 136 1.2 1 0.2 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1 1.2 x 0.2 40 225 0.4 P 0.6 0.8 1 304 P P 1.2 For P when 136, x cos 136 nd y sin 136. From the digrm, cos 136 0 nd sin 136 0 For P when 225, x cos 225 nd y sin 225. From the digrm, cos 225 0 nd sin 225 0 For P when 304, x cos 304 nd y sin 304. From the digrm, cos 304 0 nd sin 304 0 For P when 40, x cos 40 nd y sin 40. From the digrm, cos 40 0 nd sin 40 0 The digrm shows for ech qudrnt whether the sine nd cosine of ngles in tht qudrnt re positive or negtive. sin cos sin cos 2nd 1st 3rd 4th sin cos sin cos The sine nd cosine of ny ngle cn be found using your clcultor. The following tble shows some of these vlues corrected where necessry to 3 deciml plces. 504 0 30 40 45 60 90 136 180 225 270 304 360 sin 0 0.5 0.643 0.707 0.866 1 0.695 0 0.707 1 0.829 0 cos 1 0.866 0.766 0.707 0.5 0 0.719 1 0.707 0 0.559 1 Using these vlues nd others from clcultor the grphs of y sin nd y cos cn be drwn. grphicl clcultor would be useful here.
31.2 Trigonometric rtios for ny ngle HPTR 31 Grph of y sin y 1 0.5 180 180 360 540 θ 0.5 Notice tht the grph: cuts the -xis t, 180, 0, 180, 360, 540, repets itself every 360, tht is, it hs period of 360 hs mximum vlue of 1 t, 90, 450, hs minimum vlue of 1 t, 90, 270, 1 Grph of y cos y 1 0.5 180 180 360 540 θ 0.5 Notice tht the grph: cuts the -xis t 90, 90, 270, 450, repets itself every 360, tht is it hs period of 360 hs mximum vlue of 1 t, 0, 360, hs minimum vlue of 1 t, 180, 180, 540, 1 Notice lso tht the grph of y sin nd the grph of y cos re horizontl trnsltions of ech other. sin To find the vlue of the tngent of ny ngle, use tn c os From the grph of y cos it cn be seen tht cos 0 t 90, 270, 450, for exmple. s it is not possible to divide by 0 there re no vlues of tn t 90, 270, 450, tht is, the grph is discontinuous t these vlues of. Grph of y tn y 8 6 4 2 180 2 180 360 540 θ 4 6 8 Notice lso tht tn cn tke ny vlue. Notice tht the grph: cuts the -xis where tn 0, tht is, t 180, 0, 180, 360, 540 repets itself every 180, tht is it hs period of 180 does not hve vlues t 90, 270, 450, does not hve ny mximum or minimum points. 505
HPTR 31 Pythgors theorem nd trigonometry (2) xmple 3 For vlues of in the intervl 180 to 360 solve the eqution ii sin 0.7 ii 5 cos 2 Give ech nswer correct to one deciml plce. Solution 3 i sin 0.7 Use clcultor to find one vlue of. 44.4 y 1 y 0.7 To find the other solutions drw sketch of y sin for from 180 to 360 180 180 360 θ 1 The sketch shows tht there re two vlues of in the intervl 180 to 360 for which sin 0.7 ne solution is 44.4 nd by symmetry the other solution is 180 44.4 44.4, 180 44.4 44.4, 135.6 ii 5 cos 2 ivide ech side of the eqution by 5 cos 2 5 0.4 66.4 y 1 y 0.4 180 180 360 θ Use clcultor to find one vlue of. To find the other solutions drw sketch of y cos for from 180 to 360 The sketch shows tht there re three vlues of in the intervl 180 to 360 for which cos 0.4 1 66.4, 66.4, 360 66.4 66.4, 66.4, 293.6 ne solution is 66.4 nd by symmetry nother solution is 66.4 Using the period of the grph the other solution is 360 66.4 xercise 31 1 For 360 360 sketch the grph of y sin b y cos c y tn. 2 Find ll vlues of in the intervl 0 to 360 for which sin 0.5 b cos 0.1 c tn 1 3 Show tht one solution of the eqution 3 sin 1 is 19.5, correct to 1 deciml plce. b Hence solve the eqution 3 sin 1 for vlues of in the intervl 0 to 720 506
31.3 re of tringle HPTR 31 4 Show tht one solution of the eqution 10 cos 3 is 107.5 correct to 1 deciml plce. b Hence find ll vlues of in the intervl 360 to 360 for which 10 cos 3 5 Solve 4 tn 3 for vlues of in the intervl 180 to 360 31.3 re of tringle Lbelling sides nd ngles The vertices of tringle re lbelled with cpitl letters. The tringle shown is tringle. c b The sides opposite the ngles re lbelled so tht is the length of the side opposite ngle, b is the length of the side opposite ngle nd c is the length of the side opposite ngle. re of tringle 1 2 bse height re of tringle 1 2 bh In the right-ngled tringle N h sin h c So re of tringle 1 2 b sin tht is re of tringle 1 2 b sin N b The ngle is the ngle between the sides of length nd b nd is clled the included ngle. Theformul for the re of tringle mens tht re of tringle 1 2 product of two sides sine of the included ngle. For tringle there re other formule for the re. re of tringle 1 2 b sin 1 2 bc sin 1 2 c sin. These formule give the re of tringle whether the included ngle is cute or obtuse. xmple 4 Find the re of ech of the tringles correct to 3 significnt figures. b 7.3 cm 16.2 m 37 5. 7.4 m 118 507
HPTR 31 Pythgors theorem nd trigonometry (2) Solution 4 re 1 2 7.3 5.8 sin 37 re 12.74 re 12.7 cm 2 b re 1 2 7.4 16.2 sin 118 re 52.92 re 52.9 m 2 Substitute 7.3 cm, b 5., 37 into re 1 2 b sin Give the re correct to 3 significnt figures nd stte the units. Substitute into re of tringle 1 2 product of two sides sine of the included ngle. xmple 5 The re of this tringle is 20 cm 2. Find the size of the cute ngle x. Give your ngle correct to one deciml plce. 8.1 cm x 6.4 cm Solution 5 1 2 8.1 6.4 sin x 20 2 20 sin x 0.7716 8.1 6.4 x 50.49 x 50.5 xercise 31 Give lengths nd res correct to 3 significnt figures nd ngles correct to one deciml plce. 1 Work out the re of ech of these tringles. i ii iii 9.3 cm Use re of tringle 1 2 product of two sides sine of the included ngle. 9.2 cm 28 Find the vlue of sin x. Give the ngle correct to one deciml plce. 13.5 cm 10.6 cm 43 6.9 cm iv v vi 76.3 4.6 cm 4.6 cm 9.6 cm 137 4.7 cm 9.1 cm 34.7 8.6 cm 148.6 13.4 cm 2 is qudrilterl. Work out the re of the qudrilterl. 9.4 cm 57 12.6 cm 80 8.6 cm 508
31.3 re of tringle HPTR 31 3 The re of tringle is 15 cm 2 ngle is cute. Work out the size of ngle. 6.5 cm 8.4 cm 4 The re of tringle is 60.7 m 2 Work out the length of. 12.6 m 35 5 Tringle is such tht 6 cm, b 9cm nd ngle 25. Work out the re of tringle. b Tringle PQR is such tht p 6 cm, q 9cmnd ngle R 155. Work out the re of tringle PQR. c Wht do you notice bout your nswers? Why do you think this is true? 6 The digrm shows regulr octgon with centre. Work out the size of ngle. 6 cm. b Work out the re of tringle. c Hence work out the re of the octgon. 7 Work out the re of the prllelogrm. 5.7 cm 63 12. 8 n equilterl tringle hs sides of length 12 cm. lculte the re of the equilterl tringle. b regulr hexgon hs sides of length 12 cm. lculte the re of the regulr hexgon. 9 The digrm shows sector,, of circle, centre. The rdius of the circle is nd the size of ngle is 50. Work out the re of tringle. b Work out the re of the sector. c Hence work out the re of the segment shown shded in the digrm. 50 509
HPTR 31 Pythgors theorem nd trigonometry (2) 31.4 The sine rule c b The lst section showed tht 1 1 1 re of tringle 2 b sin 2 bc sin 2 c sin 1 2 b sin 1 2 bc sin nd 1 2 bc sin 1 2 c sin cncelling 1 2 nd b from both sides cncelling 1 2 nd c from both sides sin c sin nd b sin sin or or c b nd sin sin sin sin ombining these results b c sin sin sin This result is known s the sine rule nd cn be used in ny tringle. Using the sine rule to clculte length xmple 6 Find the length of the side mrked in the tringle. Give your nswer correct to 3 significnt figures. 74 Solution 6 8. 4 sin 37 sin 74 8.4 sin 37 sin 74 5.258 5.26 cm 8.4 cm 37 b Substitute 37, b 8.4, 74 into. sin sin Multiply both sides by sin 37. 510
31.4 The sine rule HPTR 31 xmple 7 Find the length of the side mrked x in the tringle. Give your nswer correct to 3 significnt figures. 18 Solution 7 Missing ngle 180 (18 124) 38 x 124 9.7 cm x 18 9.7 cm The ngle opposite 9.7 cm must be found before the sine rule cn be used. Use the ngle sum of tringle. 38 124 x 9. 7 sin 1 24 sin 38 x 9.7 sin 124 sin 38 x 13.06 Write down the sine rule with x opposite 124 nd 9.7 opposite 38. Multiply both sides by sin 124. x 13.1 cm Using the sine rule to clculte n ngle When the sine rule is used to clculte n ngle it is good ide to turn ech frction upside down (the reciprocl). This gives sin sin sin b c xmple 8 Find the size of the cute ngle x in the tringle. Give your nswer correct to one deciml plce. 74 7.9 cm Solution 8 s in x sin 74 7.9 8. 4 sin x 7.9 sin 74 8.4 sin x 0.904 x 8.4 cm Write down the sine rule with x opposite 7.9 nd 74 opposite 8.4 Multiply both sides by 7.9 Find the vlue of sin x. x 64.69 x 64.7 511
HPTR 31 Pythgors theorem nd trigonometry (2) xercise 31 Give lengths nd res correct to 3 significnt figures nd ngles correct to 1 deciml plce. 1 Find the lengths of the sides mrked with letters in these tringles. b c d e f d 76 64 11 cm 79 46 13.6 cm 17 b 27 134 e 6.1 cm 51 4.2 cm 62 f 58 6.1 cm 113 c 14.9 cm g 22 2 lculte the size of ech of the cute ngles mrked with letter. b c d 6 cm 17 cm 6. 18.4 cm 9.1 cm 32 21 104 7.6 cm 73 12.7 cm 3 The digrm shows qudrilterl nd its digonl. In tringle,work out the length of. b In tringle,work out the size of ngle. c Work out the size of ngle. 4 In tringle, 8.6 cm, ngle 52 nd ngle 63. lculte the length of. b lculte the length of. c lculte the re of tringle. 5 In tringle PQR ll the ngles re cute. PR 7. nd PQ 8.4 cm. ngle PQR 58. Work out the size of ngle PRQ. b Work out the length of QR. 102 18 8.6 cm 5.7 cm 46 6 The digrm shows the position of port (P), lighthouse (L) nd buoy (). The lighthouse is due est of the buoy. The lighthouse is on bering of 035 from the port nd the buoy is on bering of 312 from the port. Work out the size of i ngle PL ii ngle PL. The lighthouse is 8 km from the port. b Work out the distnce P. c Work out the distnce L. d Work out the shortest distnce from the port (P) to the line L. N P L 512
31.5 The cosine rule HPTR 31 31.5 The cosine rule The digrm shows tringle. The line N is perpendiculr to nd meets the line t N so tht N x nd N (b x). The length of N is h. In the right-ngled tringle N, x c cos Using 2 In tringle N Pythgors theorem gives c 2 x 2 h 2 1 c 2 b 2 c 2 2bc cos This result is known s the cosine rule nd cn be used in ny tringle. h x N (b x) b In tringle N Pythgors theorem gives 2 (b x) 2 h 2 2 b 2 2bx x 2 h 2 Using 1 substitute c 2 for x 2 h 2 2 b 2 2bx c 2 2 Similrly nd b 2 2 c 2 2c cos c 2 2 b 2 2b cos Using the cosine rule to clculte length xmple 9 Find the length of the side mrked with letter in ech tringle. Give your nswers correct to 3 significnt figures. b 12 cm Solution 9 2 12 2 8 2 2 12 8 cos 24 2 144 64 175.4007 2 32.599 27 32.599 27 5.709 577 5.71 cm 24 b x 2 7.3 2 5.8 2 2 7.3 5.8 cos 117 7.3 cm 117 x 5. Substitute b 12, c 8, 24 into 2 b 2 c 2 2bc cos. vlute ech term seprtely. Tke the squre root. Substitute the two given lengths nd the included ngle into the cosine rule. x 2 53.29 33.64 84.68 (0.4539 ) x 2 86.93 38.44 x 2 125.37 x 125.37 x 11.19 x 11.2 cm cos 117 0 Tke the squre root. 513
HPTR 31 Pythgors theorem nd trigonometry (2) Using the cosine rule to clculte n ngle To find n ngle using the cosine rule, when the lengths of ll three sides of tringle re known, rerrnge 2 b 2 c 2 2bc cos. 2bc cos b 2 c 2 2 cos b2 c2 2 2bc Similrly nd cos 2 c2 b 2 2c cos 2 b2 c 2 2b xmple 10 Find the size of ngle b ngle X. Give your nswers correct to one deciml plce. b 16 cm 13 cm 8.6 cm 12.7 cm X 11 cm 6.9 cm Solution 10 cos 11 2 16 2 13 2 11 2 16 cos 2 0 8 352 cos 0.590 909 53.77 53.8 b cos X 8.6 2 2 6. 9 12.7 2 2 8. 6 6.9 cos X 39. 72 118. 68 cos X 0.334 68 X 109.55 X 109.6 Substitute b 11, c 16, 13 into cos b2 c2 2. 2bc Substitute the three lengths into the cosine rule noting tht 12.7 cm is opposite the ngle to be found. The vlue of cos X is negtive so X is n obtuse ngle. 514
31.5 The cosine rule HPTR 31 xercise 31F Where necessry give lengths nd res correct to 3 significnt figures nd ngles correct to 1 deciml plce. 1 lculte the length of the sides mrked with letters in these tringles. b c 62 d e f d 9.6 cm 9 cm 9.6 cm 52 b 10.2 cm 9.2 cm 134 e 11.3 cm 75 6.3 cm 16.2 cm 8.4 cm c 18 8.4 cm 147 15.5 cm f 2 lculte the size of ech of the ngles mrked with letter in these tringles. b 7 cm 9 cm 15.3 cm 9.4 cm c 11 cm d 13.6 cm 8.6 cm 8.7 cm 8.7 cm 7.2 cm 14.4 cm 6. 3 The digrm shows the qudrilterl. Work out the length of. b Work out the size of ngle. c Work out the re of qudrilterl. 8.4 cm 26.4 cm 56 9. 4 Work out the perimeter of tringle PQR. R 16.3 cm P 8.6 cm 10.9 cm 5 In tringle, 10.1 cm, 9.4 cm nd 8.7 cm. lculte the size of ngle. 6 In tringle XYZ, XY 20.3 cm, XZ 14.5 cm nd ngle YXZ 38. lculte the length of YZ. 27 Q 515
HPTR 31 Pythgors theorem nd trigonometry (2) 7 is chord of circle with centre. The rdius of the circle is 7 cm nd the length of the chord is 11 cm. lculte the size of ngle. 7 cm 11 cm 8 The region is mrked on school field. The point is 70 m from on bering of 064. The point is 90 m from on bering of 132. Work out the size of ngle. b Work out the length of. N 70 m 9 hris rn 4 km on bering of 036 from P to Q. He then rn in stright line from Q to R where R is 7 km due st of P. hris then rn in stright line from R to P. lculte the totl distnce run by hris. 10 The digrm shows prllelogrm. Work out the length of ech digonl of the prllelogrm. 31.6 Solving problems using the sine rule, the cosine rule nd 1 2 b sin xmple 11 90 m 6 cm 65 The re of tringle is 12 cm 2 3. nd ngle 70. Find the length of i ii. Give your nswers correct to 3 significnt figures. b Find the size of ngle. Give your nswer correct to 1 deciml plce. Solution 11 i 1 2 3.8 sin 70 12 2 12 3.8 sin 70 6.721 6.72 cm 3. 70 Substitute c 3.8, 70 into re 1 2 c sin. ii b 2 6.721 2 3.8 2 2 6.721 3.8 cos 70 b 2 59.613 17.470 b 2 42.142 b 6.491 6.49 cm Substitute 6.721, c 3.8 nd 70 into b 2 2 c 2 2c cos. 516
31.6 Solving problems using the sine rule, the cosine rule nd 1 2 b sin HPTR 31 in sin 70 b 6ṣ 721 6. 491 sin 6.721 sin 70 6.491 sin 0.9728 76.62 ngle 76.6 Substitute 6.721, b 6.491 nd 70 into sin sin. b xercise 31G Where necessry give lengths nd res correct to 3 significnt figures nd ngles correct to 1 deciml plce, unless the question sttes otherwise. 1 tringle hs sides of lengths 9 cm, 10 cm nd 11 cm. lculte the size of ech ngle of the tringle. b lculte the re of the tringle. 2 In the digrm is stright line. lculte the length of. b lculte the size of ngle. c lculte the length of. 12.9 cm 12 cm 3 The re of tringle is 15 cm 2. 4.6 cm nd ngle 63. Work out the length of. b Work out the length of. c Work out the size of ngle. 4 is kite with digonl. lculte the length of. b lculte the size of ngle. c lculte the vlue of x. d lculte the length of. 5.4cm 65 47 5 Kultr wlked 9 km due South from point to point. He then chnged direction nd wlked 5 km to point. Kultr ws then 6 km from his strting point. Work out the bering of point from point. Give your nswer correct to the nerest degree. b Work out the bering of point from point.give your nswer correct to the nerest degree. 6 The digrm shows pyrmid. The bse of the pyrmid,, is rectngle in which 15 cm nd. The vertex of the pyrmid is where 20 cm. Work out the size of ngle correct to the nerest degree. x cm 50 30 x cm 20 cm 15 cm 517
HPTR 31 Pythgors theorem nd trigonometry (2) 7 The digrm shows verticl pole, PQ, stnding on hill. The hill is t n ngle of 8 to the horizontl. The point R is 20 m downhill from Q nd the line PR is t 12 to the hill. lculte the size of ngle RPQ. b lculte the length, PQ, of the pole. 8, nd re points on horizontl ground so tht 30 m, 24 m nd ngle 50. P nd Q re verticl posts, where P Q 10 m. Work out the size of ngle. b Work out the length of. c Work out the size of ngle PQ. d Work out the size of the ngle between Q nd the ground. hpter summry 10 m P 8 R 50 P 12 Q 20 m 30 m 24 m Q 10 m You should now be ble to: use Pythgors theorem to solve problems in 3 dimensions use trigonometry to solve problems in 3 dimensions work out the size of the ngle between line nd plne drw sketches of the grphs of y sin x, y cos x, y tn x nd use these grphs to solve simple trigonometric equtions use the formul re 1 2 b sin to clculte the re of ny tringle b c use the sine rule nd the cosine rule 2 b 2 c 2 2bc cos in sin sin sin tringles nd in solving problems. hpter 31 review questions 1 In the digrm, XY represents verticl tower on level ground. nd re points due West of Y. The distnce is 30 metres. The ngle of elevtion of X from is 30º. The ngle of elevtion of X from is 50º. lculte the height, in metres, of the tower XY. Give your nswer correct to 2 deciml plces. 30 30 m (1384 June 1996) 2 The digrm shows tringle. igrm NT ccurtely drwn 7.2 cm 8.35 cm ngle 74. 74 7.2 cm 8.35 cm lculte the re of tringle. Give your nswer correct to 3 significnt figures. b lculte the length of. Give your nswer correct to 3 significnt figures. (1385 June 2002) 50 X Y igrm NT ccurtely drwn 518
hpter 31 review questions HPTR 31 3 In tringle 8cm 15 cm ngle 70. lculte the re of tringle. Give your nswer correct to 3 significnt figures. 15 cm X is the point on such tht ngle X 90. b lculte the length of X. Give your nswer correct to 3 significnt figures. (1387 June 2003) 70 X igrm NT ccurtely drwn 4 The digrm shows cuboid.,,, nd re five vertices of the cuboid. 5 cm 3 cm. lculte the size of the ngle the digonl mkes with the plne. Give your nswer correct to 1 deciml plce. 5 cm 3 cm igrm NT ccurtely drwn 5 In tringle 15 cm ngle 70. lculte the length of. Give your nswer correct to 3 significnt figures. b lculte the size of ngle. Give your nswer correct to 1 deciml 15 cm plce. (1387 June 2003) 70 igrm NT ccurtely drwn 6 This is sketch of the grph of y cos x for vlues of x between 0 nd 360. Write down the coordintes of the point ii ii. y 360 x 7 ngle 150 60 m. The re of tringle is 450 m 2 lculte the perimeter of tringle. Give your nswer correct to 3 significnt figures. 150 igrm NT ccurtely drwn 60 m (1385 November 2000) 519
HPTR 31 Pythgors theorem nd trigonometry (2) 8 The digrm shows qudrilterl. 4.1 cm 4.1 cm 5.4 cm 7.6 cm 5.4 cm 62 117 ngle 117 ngle 62. lculte the length of. Give your nswer correct to 3 significnt figures. 7.6 cm b lculte the re of tringle. Give your nswer correct to 3 significnt figures. c lculte the re of the qudrilterl. Give your nswer correct to 3 significnt figures. (1385 June 2000) 9 This is grph of the curve y sin x for 0 x 180 1 y igrm NT ccurtely drwn 0.5 45 90 135 180 x 0.5 1 Using the grph or otherwise, find estimtes of the solutions in the intervl 0 x 360 of the eqution i sin x 0.2 ii sin x 0.6. cos x sin (x 90) for ll vlues of x. b Write down two solutions of the eqution cos x 0.2 (1385 November 2002) 10 In the digrm,, F nd F re rectngles. The plne F is horizontl nd the plne F is verticl. 10 cm 20 cm 20 cm. lculte the size of the ngle tht the line mkes with the plne F. 10 cm 20 cm F 20 cm 11 In tringle 10 cm 14 cm 16 cm. lculte the size of the smllest ngle in the tringle. Give your nswer correct to the nerest 0.1. b lculte the re of tringle. Give your nswer correct to 3 significnt figures. 10 cm 16 cm igrm NT ccurtely drwn 14 cm 520