Tringle pplitions Syllus Guide Chpter 1 Contents 1.1 Using right-ngled tringles 1.2 pplying erings nd diretion 1.3 The sine rule 1.4 The osine rule 1.5 Trig pplitions 1.6 Working in three dimensions Chpter summry Chpter review Syllus sujet mtter Periodi funtions nd pplitions Trigonometry inluding the definition nd prtil pplitions of the sine, osine nd tngent rtios Simple prtil pplitions of the sine nd osine rules (the miguous se is not essentil) Quntittive onepts nd skills Metri mesurement inluding mesurement of mss, length, re nd volume in prtil ontets Clultion nd estimtion with nd without instruments si lgeri mnipultions
Tringle pplitions The word trigonometry omes from Greek nd mens Erth mesurement. Even tody, trigonometry is still used to hek property oundries nd to survey for new ridges, rods nd other ivil engineering. However, the importne of trigonometry now etends fr eyond the originl purposes of the nient Greeks. It is used in stronomy, nvigtion, optis, sound engineering, uilding nd onstrution, nd mny other res of modern life. 1.1 Using right-ngled tringles When working with tringles it is usul to nme n ngle y single pitl letter. The opposite side is usully nmed y the sme lower-se letter. C In the tringle on the right, we ould stte the length of side s = 44 mm or s = 44 mm. We ould stte the size of the top ngle s C = 79 or s C = 79. Greek letters suh s or α re lso used for ngles. ngles my e mesured in degrees (of r), shown y the symol. There re 360 degrees in one full turn (revolution). If greter ury is needed, minutes ( ) nd seonds ( ) of r my lso e used. There re 60 minutes of r in one degree nd 60 seonds of r in one minute.! 1 revolution = 360 1 = 60 1 = 60 Nowdys, it is eoming more ommon to epress ngles in degrees only, for emple, s 36.5 rther thn 36 30. Most sientifi lultors hve funtion mrked s or DMS to hnge from deiml degrees to degrees, minutes nd seonds of r, nd vie vers. Mke sure tht you know how to use your lultor to onvert etween deiml degrees nd degrees, minutes nd seonds. In right-ngled tringle, the longest side is lled the hypotenuse. It is lso the side opposite the right ngle. You my rememer tht Pythgors s Theorem reltes the lengths of the sides of right-ngled tringle.! Pythgors s Theorem In ny right-ngled tringle, the squre of the length of the hypotenuse is equl to the sum of the squres of the lengths of the other two sides. 2 = 2 + 2 2 New QMths 11
Emple 1 Find the unknown side in the tringle on the right. 73 m 27 m vertil post stnds 3.7 m ove the ground. tut wire joins the top of the post to n nhor point 5.6 m from the se of the post. How long is the wire? Wire Solution The 73 m side is the hypotenuse. 73 2 = 27 2 + 2 Evlute squres. 5329 = 729 + 2 Sutrt to get 2 nd reverse the eqution. 2 = 5329 729 = 4600 Find the squre root. = 4600 = 67.8232 Round nd write the nswer s sentene. is out 68 m. Re-drw the digrm with the known informtion. w 3.7 m Post 5.6 m The wire is the hypotenuse of the tringle. w 2 = 3.7 2 + 5.6 2 Evlute. = 13.69 + 31.36 = 45.05 Find the squre root. w = 45.05 = 6.7119 Round nd write the nswer s sentene. The wire is out 6.8 m long. The sides other thn the hypotenuse re nmed s opposite nd djent (net to) to the ngle we re onsidering.! For ngle For ngle φ Hypotenuse djent φ Opposite Hypotenuse Opposite φ djent Chpter 1 Tringle pplitions 3
Tringle pplitions We use the trigonometri rtios sine, osine nd tngent for right-ngled tringles. They re revited to sin, os nd tn.! Trigonometri rtios opposite sin X = ------------------------------ side = -- hypotenuse y djent os X = ------------------------------ side = -- z hypotenuse y opposite tn X = ------------------------------ side = -- djent side z X z y Z Y The rtios n e rememered using the following mnemoni.! SOH CH TO (sok-h-tow) Sin = O/H Cos = /H Tn = O/ Emple 2 Clulte the vlue of the pronumerl in eh of the following tringles. 16.4 mm 52 7 26 m 34.6 g Solution First hoose the pproprite trig rtio. The known side is the hypotenuse nd the side we wnt is the opposite, so use sin. sin 52 7 = Multiply y 26 to otin. = 26 sin 52 7 Use your lultor to otin sin. = 26 0.7892 = 20.5208 Round ppropritely. 20.5 Write the nswer. is pproimtely 20.5 m. The side we wnt is the hypotenuse. os 34.6 = --------- 16.4 The known side is the djent, so use os. g Rerrnge. gos 34.6 = 16.4 Isolte g. g = ----------------------- 16.4 os 34.6 When using lultor, don t round off until the lst step. = 19.9237 19.9 Write in words. g is out 19.9 mm. ----- 26 4 New QMths 11
Emple 3 Find the mgnitude of the ngle. 49 m 16 m Solution The osine rtio should e used here. 16 os = ----- 49 = 0.3265 Use the inverse os funtion os 1 on your lultor. = 70.9416 Round off. 70.9 Write in words. The ngle is out 71. Sometimes we use si geometry to find right-ngled tringles in whih to pply trigonometri rtios. Emple 4 To find the pproimte length of the following lered re we need to find the length of MN in the figure shown here. M 43 m P 30 m 40 N O Solution First drw perpendiulr from O to meet MN t Q s shown on the right. M 30 m Q 40 N 43 m 43 m P 30 m O Chpter 1 Tringle pplitions 5
Tringle pplitions MQOP is retngle, so MQ = PO nd OQ = PM. MQ = 30 m nd OQ = 43 m Now find QN using QNO. tn 40 = -------- 43 QN Rerrnge to isolte QN. QNtn 40 = 43 QN = ----------------- 43 tn 40 Evlute nd round off. 51 m Now use the digrm to help lulte MN. MN = MQ + QN = 30 + 51 = 81 m Write in words. MN is out 81 m. dditionl Eerise 1.1 Eerise 1.1 Using right-ngled tringles 1 Chnge the following ngles to degrees only. 42 36 67 21 28 17 24 d 16 24 36 e 7 11 19 2 Chnge the following ngles to degrees, minutes nd seonds. 19.2 49.67 56.97 d 87.25 e 22.16 3 Find the unknown side or ngle mrked in eh of the following tringles. 25 17.9 21 48 28 p d e f m 28 12 4 Find the vlue of the unknown length in eh of the following. 65 m φ 29 47 35 17 m 8 2 m 92 54 48.1 78 42 36 u d e f 7 m 31 m 39 74 d 58 e 47 16 8 m 3 2 m 800 m 64 f 6 New QMths 11
g h i h g 26 4 7 m 14 m 41 i 29 17 4 m 5 Find the ngle mrked in eh of the following tringles. 16 48 k 24 β 2.3 18 1.8 d e f g 1860 m 5.1 m 9 2 m 3744 m 4 7 m f 2 2 m 7 6 m 6 In PMQ, M = 90, Q = 60 nd PQ = 40 m. Find the length of: MQ PM 7 In MNO, MN = 24 mm, MO = 16 mm nd M = 60. OQ is drwn perpendiulr to MN so tht it meets MN t Q. Find the length of: MQ QO NO 8 In the digrm shown, C = 14 m. Find the length of: C D d CD D C 15 14 m 30 9 In the digrm, lulte the length of: MO NO MN d OL L 30 O 60 N Modelling nd prolem solving 10 Clulte the length of. 40 mm 49 m 73 C M 31 Chpter 1 Tringle pplitions 7
Tringle pplitions 11 Find the length of MN. O 1.9 m N 70 m L 116 12 PQRS is retngle with PS = 97 m nd SR = 64 m. If PT = 35 m, find QTR. Q M R 13 ivil engineering ompny is using tunnel-oring mhine to onstrut tunnel into hill tht is inlined n ngle of 15 to the horizontl tunnel. fter the tunnel hs een drilled 90 m into the hill, ventiltion shft needs to e driven vertilly up to the surfe. Wht will e the depth of the shft? 14 hest of drwers 75 m high is pled in n tti in whih the roof slopes down towrds the horizontl floor. The losest tht the hest of drwers n stnd to the orner formed y the roof nd the floor is 28 m. Find the ngle of inline of the roof to the floor. Y 15 Find the perimeter of WXYZ shown on the right. P T 64 2 S X 32 m 16 uilder wnts to onstrut roof s shown on the right with pith of 32 30. How long should the ems e ut? W 78 m em em 32 30 32 30 11.8 m Z 17 irle of rdius 30 m hs hord drwn 20 m from its entre, O. Clulte: the mgnitude of POM the length of the hord PM. P Chord O M 8 New QMths 11
18 The roof of new home is designed so tht the solr pnel n ollet the mimum energy from the sun. Clulte: the height of the roof the length of the em. Solr pnel Roof 24 m C 34 28 62 12 19 This digrm shows prt of ontour mp. The ontours re t 20 m intervls nd the mp is drwn to sle of 1 : 25 000. n ll-terrin vehile n mnge slopes inlined t less thn 18 to the horizontl. Would it e possile for the vehile to proeed diretly from X to Y on the route shown? Y X 400 420 460 Contour intervl: 20 m 0 250 500 750 1000 m Sle 1 : 25 000 20 pendulum 1.2 m long is designed for grndfther lok. It swings through n ngle of 12. How wide should the se of the lok e mde? 1.2 pplying erings nd diretion Surveyors nd nvigtors use diretions nd erings in their work. ering gives the diretion in the horizontl plne of n ojet from n oservtion point.! The ering of the ojet is given s the 3-digit lokwise ngle the oserver turns from due north in order to fe the ojet. So n ojet due west hs ering of 270 T (T for true ering). ering n lso e given s the ngle from north or south towrds either est or west. So ering of 160 T n e epressed s S20 E. s shown in the digrm on the right, the diretion of from n oserver t O n e stted s N40 E or 040 T. N The diretion of from O n e written s S20 W or 200 T. The diretion of C from O n e written s S20 E or 160 T. W 20 40 O 20 E S C Chpter 1 Tringle pplitions 9
Tringle pplitions Emple 5 Two ots, nd, re situted so tht is 26 km due est of. ot sets off for n islnd in the diretion 113 T, nd tkes the ourse 203 T for the sme islnd. fter they meet, how fr hs trvelled? Wht is the ering from the islnd to s strting point? Solution Drw digrm showing the strting positions nd diretions tken y nd. Put ompss diretion points NSWE t the ruil positions. Lel ngles nd show ll given informtion. N N W 113 23 26 km 67 203 E S W Islnd N C β S E The distne trvelled y is C. Write down the informtion from C. C = 23 nd C = 67 Clulte the other ngle in C. C = 90 Use C to write n eqution with C. sin 67 = ------- C 26 Isolte C. C = 26 sin 67 Evlute nd round off. 23.9 Write the nswer in words. hs trvelled pproimtely 23.9 km. Lote the islnd nd s strting point on the digrm. We wnt to know the ering from C to. We know tht β nd form right ngle. Now we stte the ering. S β is lternte to 67. β= 67 = 90 67 = 23 The ering from the islnd to s strting point is 023 T. dditionl Eerise 1.2 Eerise 1.2 pplying erings nd diretion 1 The ering from C to is N50 E. The ering from C to is N80 W. D is due north of. Wht is the ering of: C from? C from? from D? d D from? D N C 10 New QMths 11
2 X is 45 m due west of P, nd the ering of Y from P is 120 T. Find: the ering of P from X the ering of P from Y the distne from X to Y d the ering of X from Y. X 45 m N 120 P 60 m Y 3 fishing ot leves port nd trvels on ering of N60 E for 1.6 km until it is due north of uoy. The uoy is due est of the port. Wht is the ering of the port from the ot? How fr does the ot hve to trvel to reh the uoy? How fr is it from the uoy to the port? N 60 1 6 km ot 60 Port uoy 4 motorist trvels 8 km due north from petrol sttion nd then 6 km due west. Wht is the ering of the motorist from the petrol sttion? 5 Two towns, X nd Y, re 25 km prt, with Y eing due north of X. Town Z is due est of X nd 60 km wy. Find the distne nd ering of Y from Z. Modelling nd prolem solving 6 Julio nd Mri leve shool t the sme time. Julio yles due west t 18 km/h nd Mri yles due south t 24 km/h. Clulte the distne tht seprtes them, fter they hve yled for 2 hours. How long did it tke for them to e 15 km prt? Clulte the ering from Mri to Julio t ny time. 7 trithlete leves the strt of re nd runs 12 km on ering of 030 T. She then yles for 15 km on ering of 060 T. Wht re the ering nd distne of her strting point from her position t the end of the yle leg? 8 yht trvels on ering of 052 T until it is 3.2 km north of its strting point. It then trvels on ering of 155 T until it is due est of its strting point. How fr is the yht from the ple where it strted? 9 n orienteer egins from strting point, sttion 1, nd runs on ering of 350 T until he rehes sttion 2, whih is 3.3 km north of sttion 1. He then runs due est for 5.2 km to reh sttion 3. He then hnges diretion to run roughly south-west until he rehes sttion 4, whih is 2.1 km due est of sttion 1. Wht is the ering from sttion 3 to sttion 4? 10 n irrft flies due south for 100 km nd then due west for nother 250 km. Find the distne nd ering of the irrft from its strting point. 11 P nd Q re two ushfire oservtion sttions. Q is 10 km due west of P. fire is sighted t ering of 322 T from P nd 052 T from Q. Clulte the distne of the fire from eh oservtion sttion. Chpter 1 Tringle pplitions 11
Tringle pplitions Etr Mteril 1.3 The sine rule The re of ny C my e written s: re of tringle 1 -- 2 1 2 sin C or -- sin or -- sin So -- sin C = -- sin = -- sin Multiplying y 2 gives sin C = sin = sin sin C sin sin Dividing y, we hve ------------ = ------------ = ------------ Inverting gives the sine rule:! Sine rule In ny C ------------ = ------------ = ------------ sin sin sin C 1 2 1 2 1 2 1 2 C Detiled Proof forml proof of the sine rule is given on the CD-ROM. We use the sine rule to solve non-right-ngled tringles when we know two ngles nd one side or two sides nd n opposite ngle. Emple 6 In XYZ, X = 50, Y = 34 nd = 12. Find Z, y nd z. Solution Sketh tringle to show the informtion. y Z 12 50 34 z Write the sine rule for, X, y nd Y. ------------ = ------------ y sin X sin Y Sustitute. ----------------- 12 = ----------------- y sin 50 sin 34 Isolte y nd evlute. 12 y = ----------------------- sin 34 8.8 sin 50 Now find Z. Z + 50 + 34 = 180 Z = 96 Write the sine rule for, X, z nd Z. ------------ = ------------ z sin X sin Z Sustitute. ----------------- 12 = ----------------- z sin 50 sin 96 Isolte z nd evlute. 12 z = ----------------------- sin 96 15.6 sin 50 Write the nswer. Z = 96, y 8.8 nd z 15.6. X Y 12 New QMths 11
The sine rule does not lwys give unique nswer for the ngle in tringle. Sine sin = sin (180 ), you must try two possile nswers to hek whether one or oth nswers re possile. Your lultor gives only the ute ngle, so you must work out the other ngle. For emple, if sin = 0.6, your lultor will give sin 1 0.6 36.87. Now 180 36.87 = 143.13. Cheking, you find tht sin 36.87 0.6 nd sin143.13 0.6. If oth nswers re possile, we sy it is n emple of the miguous se. Emple 7 In C, = 6 m, = 7.5 m nd = 40. Clulte the remining ngles nd side nd sketh the possile tringles. Solution Write the sine rule for, nd,. ----------- = ----------- sin sin Sustitute relevnt vlues. ---------------- 6 = ----------- 7.5 sin 40 sin 7.5 Rerrnge nd evlute. sin = ------------------------ sin40 6 = 0.8034 Use sin 1 on the lultor. So 53.46 or 126.54 Keep et vlues in your lultor. Find the possile totls of nd. + 40 + 53.46 or 40 + 126.54 oth re less thn 180, so oth re possile. 93.46 or 166.54 Now find C. C 180 93.46 or 180 166.54 = 86.54 or 13.46 Use the sine rule to find the possile ----------- = ----------- vlues of side. sin sinc ---------------- 6 = ----------- sin 40 sinc 6 sinc = --------------- sin40 6 Use the et vlues in your lultor. = --------------------------- sin86.54 6 or --------------------------- sin13.46 sin 40 sin 40 9.32 or 2.17 This is n miguous se. The possile tringles re shown elow. 53.46, C 86.54 nd 9.32 m or 126.54, C 13.46, nd 2.17 m 86.54 7.5 m 6 m C 13.46 7.5 m 6 m C 40 9 32 m 53.46 40 2.17 m 126.54 Chpter 1 Tringle pplitions 13
Tringle pplitions Emple 8 Jermine sees tht the top of snd dune t the end of eh is t n ngle of elevtion of 16. On wlking 40 m loser, she finds tht the ngle of elevtion inreses to 22. Wht is the height of the snd dune? Solution Mke sketh of the informtion. T Lel sides nd ngles s needed. The height is side T. h Side TS is in etween the known nd 22 16 unknown sides. Cll it nd find it first. S 40 m F Use the eterior ngle of SFT to find STF. 22 = 16 + STF STF = 6 Use the sine rule in SFT to find. ---------------- = ------------- 40 sin 16 sin6 Don t other to lulte. 40 = ------------------------ sin16 sin6 Now use ST nd the sine rtio to find h. h sin 22 = -- Isolte h. h = sin 22 Evlute using the unlulted vlue of. 40 = ------------------------------------------ sin16 sin22 = 39.5129 sin6 Round nd write the nswer. The dune is out 39.5 m high. dditionl Eerise 1.3 Eerise 1.3 The sine rule 1 Clulte the lengths of the unknown sides in these tringles orret to 1 deiml ple. k 62 48 107 18 h 14 q 25 58 37 32 p d e f 40 29 43 27.2 76 m n y 108 52 d 8 115.5 14 New QMths 11
2 Clulte the vlue of in eh of the following tringles orret to 1 deiml ple. 62.5 d e f 91 16 9 110 96 3 Find the length of the longest side of PQR if P = 31 42, Q = 28 50 nd PR = 42 mm. 4 Solve eh of the following tringles. (Clulte the sizes of ll unknown ngles nd sides, inluding oth possiilities in miguous ses.) XYZ, where Z = 65, y = 11 m nd z = 16.2 m PQR, where P = 35.3, Q = 52.8 nd q = 67.5 m DEF, where F = 111, f = 12.5 km nd d = 8.96 km d C, where = 124.1, C = 18.7 nd = 94.6 m e XYZ, where X = 68.41, Y = 54.23 nd = 12.75 m f KLM, where K = 71.83, M = 42.57 nd l = 2.614 km 5 In DEF, f = 30 m, d = 35 m nd F = 18. Find oth possile vlues for e nd D. 6 In C, C = 18 m, C = 15 m nd = 38. Find oth possile vlues of nd mke sketh of eh possile C. Modelling nd prolem solving 7 The folding hir shown on the right hs set whih is 36 m deep. The legs of the hir re inlined t n ngle of 55 to the set nd join t n ngle of 70. The set is 51 m ove the floor. Find the distne from the set, d, t whih the legs re joined. 8 52 127.6 12 135.5 82.9 51 m 48.3 5.8 6.29 d 23.6 36 m 55 70 12 55 16 8 Three gers re rrnged s shown on the right. The rdii of the gers re 7.2 m, 5.4 m nd 3.2 m. Clulte. 5.4 m 38 3.2 m 7.2 m Chpter 1 Tringle pplitions 15
Tringle pplitions 9 heliopter is sighted t the sme time y two ground oservers who re 4.8 km prt, s shown in the digrm on the right. The ngles from the oservers to the heliopter (mesured to the horizontl) re 21.4 nd 28.6. How high is the heliopter? 28.6 21.4 4.8 km 10 nne is flying from risne to Dly t speed of 150 km/h. She thinks her ltimeter, whih shows height of 15 000 feet (4600 m), is fulty. Looking hed, she sees the top of hill mrked s 405 m on her mp t n ngle of depression of 13. Two minutes lter, the ngle of depression hs inresed to 20.6. Find her height ove the hill nd thus her ltitude. Wht should the rdio opertor tell her out her ltimeter? 1.4 The osine rule The sine rule n e used in tringles where n ngle nd its opposite side re known. However, if the known ngle is inluded etween two sides or only the sides re known, the sine rule nnot e pplied. In these ses, the osine rule is used insted.! Cosine rule In ny C 2 = 2 + 2 2 os or os = ----------------------------- 2 + 2 2 2 2 = 2 + 2 2 os or os = ----------------------------- 2 + 2 2 2 2 = 2 + 2 2 os C or os C = ----------------------------- 2 + 2 2 2 where the ngles nd sides re nmed in the usul wy. C Detiled Proof Forml proofs of the osine rule for ute-ngled, right-ngled nd otuse-ngled tringles re shown on the CD-ROM. 16 New QMths 11
Emple 9 In the tringle shown t right, lulte: Y 2.1 m Z 3.9 m Solution Write the osine rule for XYZ. 2 = y 2 + z 2 2yz os X Sustitute vlues. = 2.1 2 + 3.9 2 2 2.1 3.9 os 50 Evlute. = 9.0911 = 3.0151 Round nd write the nswer. is pproimtely 3.02 m. You n now use the sine rule. ----------- = ----------- y sin X siny Trnsform. sin Y = ---------------- y sin X 2.1 = ------------------------------ sin50 Use the et vlue of from prt. 3.0151 = 0.5335 Find Y nd round off. Y = 32.2. Write the nswer in words. Y is out 32.2. X 50 Y Emple 10 Clulte the lrgest ngle in the tringle with sides 12 m, 14 m nd 20 m. Solution Mke rough sketh. In tringles, the lrgest ngle is opposite the longest side. So the tringle n e drwn s shown. 14 m 12 m 20 m It s esier to use the seond form of the os = ----------------------------- 2 + 2 2 osine rule. 2 Sustitute using the informtion from the = -------------------------------------- 14 2 + 12 2 20 2 digrm. 2 14 12 Keep et numer on lultor. 60 = -------- 336 is otuse euse os is negtive. = 0.1785 Most lultors will give the orret ngle. 100.29 Write the nswer in words. Lrgest ngle is out 100.29 or 100 17. C Chpter 1 Tringle pplitions 17
Tringle pplitions Emple 11 tower in diretion N52 E is found to e t rnge of 4.86 km. The rnge finder shows tht nother tower in diretion 106 T is t rnge of 7.96 km. Find the distne etween the towers. Solution Drw sketh nd nme the points. Nme the oservtion point P. Nme the tower positions F nd S. Drw in the erings. The ngle etween the towers must e 54. We wnt to find the distne etween the towers, FS. Lel FS s. Write the osine rule for. FS 2 = PF 2 + PS 2 2 PF PS os P Sustitute. = 4.86 2 + 7.96 2 2 4.86 7.96 os 54 Keep et vlue on lultor. = 41.5035 Use the et vlue for the squre root. FS = 41.5035 Evlute nd round off. 6.44 Write the nswer in words. Distne etween the towers is out 6.44 km. P N 52 54 4.86 km 106 7.96 km F S dditionl Eerise 1.4 Eerise 1.4 The osine rule 1 Clulte the unknown side lengths in the following tringles. 3 m 6 m d e f 30 m 150 2 Find in eh of the following tringles. 8 12.6 20 16 18 New QMths 11 61 4 m 27 m 20 9.6 10.4 12 121 30 m 18 5 m 15 m 42.25 d e f 20 10 24 15 25.6 100 m 55 90 m 35 m 100 21 m 21 27 54.2 33.1
3 Find the third side of eh of the following tringles. MNO, where M = 41.1, n = 27.8 nd o = 39.2 XYZ, where Z = 28.3, y = 571 nd = 421 DEF, where F = 45.6, d = 72.3 nd e = 89.4 d GHI, where G = 67.3, h = 37.9 nd i = 40.8 e STU, where S = 112.8, t = 62.8 nd u = 122 f PQR, where Q = 74.8, p = 891.9 nd r = 642.7 4 Find ll unknown ngles nd sides of eh of the following tringles. DG, where D = 121, = 3 m nd g = 5 m UVW, where W = 55, u = 45 m nd v = 50 m XYZ, where = 7.2 m, y = 12.5 m nd z = 8.3 m d C, where = 8.5 m, = 20 m nd = 15.2 m e LMN, where L = 42.3, m = 15.4 km nd n = 12.9 km f REF, where E = 168.2, f = 192 mm nd r = 151 mm 5 In C, = 12, = 9 nd = 8. Clulte the gretest ngle. 6 In XYZ, = 8.8, y = 11 nd z = 17.6. Clulte the smllest ngle. 7 In PQR, p = 5, q = 1 nd r = 2. Clulte the gretest ngle. Modelling nd prolem solving 8 The digrm on the right shows rne with two rms, eh 15 m long. The rms of the rne re seprted y n ngle of 128. The lod is tthed t point X nd the ounterweight is tthed t point Z. Find XZ. Z 15 m 128 15 m Y X 9 weight is supported y les tthed to horizontl em s shown in the digrm on the right. Find the ngle tht eh le mkes with the em. 12 m 18 m 9 m 10 hiker enounters swmp t point X s shown in the digrm on the right. To trvel round the swmp, the hiker wlks 3.4 km from X on ering of 325 T nd then wlks 1.8 km to point Y on the other side of the swmp. Find the ering of X from Y. 1.8 km 3.4 km Y Swmp 4.2 km N X 325 T Chpter 1 Tringle pplitions 19
Tringle pplitions 1.5 Trig pplitions When n ojet is ove or elow eye level, n oserver must look up or down to see the ojet. The ngles up nd down hve speil nmes.! ngles of elevtion nd depression Horizontl Line of sight ngle of elevtion ngle of depression Line of sight Horizontl The ngle of elevtion of n ojet is the ngle n oserver must look upwrds from the horizontl to see the ojet. The ngle of depression of n ojet is mesured downwrds from the horizontl. Emple 12 stellite reeiver on top of tower is t n ngle of elevtion of 38 from point M on the ground tht is 72 m from the entre of the se of the tower. There is wrning sign on the tower, with its top positioned one-third of the vertil distne up the tower. Clulte the height of the tower. Clulte the ngle of elevtion from the point M to the top of the wrning sign. 72 m M Solution Drw digrm showing ll the relevnt informtion. Then show just the tringles nd nme the verties. Q R 38 P 72 M 20 New QMths 11
From the digrm, we wnt to find QP. QP tn 38 = ------- Form n eqution involving QP. 72 Isolte QP. QP = 72 tn 38 Retin the et nswer in your lultor. = 56.2526 Round off for the nswer. 56.3 Stte the nswer in words. The tower is out 56.3 m high. First form n eqution to find RP. Use the QP RP = ------- et nswer in your lultor from prt. 3 Retin the et nswer in your lultor. = 18.7508 Now we n use tn to find. tn = ------- RP 72 Use the et figure on your lultor. 18.7508 = ------------------------- 72 = 0.2604 Use tn 1 to find. = 14.5972 Now round off. 14.6 Write the nswer in words. The ngle of elevtion is out 14.6. lterntive Method Emple 13 To n oserver on top of liff, 120 m ove se level, ot ppers to e t n ngle of depression of 17. fter the ot sils diretly towrds the liff, the ngle of depression inreses to 37. How fr did the ot sil? 17 120 m Solution Drw the tringles, nme the verties nd put ll the informtion on the digrm. 37 Z 17 73 53 120 m lterntive Method W There re two right-ngled tringles, WYZ nd XYZ. The length we wnt is WX, the differene of the ses of these tringles. X Y Chpter 1 Tringle pplitions 21
Tringle pplitions First find WY using the tn rtio in WYZ. In WYZ, WZY = 73. WY tn 73 = -------- 120 WY = 120 tn 73 Keep the et figure in your lultor memory. = 392.5023 Now find XY using the tn rtio in XYZ. In XYZ, XZY = 53. tn 53 = -------- XY 120 Use the et figures in your lultor in the lst XY = 120 tn 53 step of the prolem. = 159.2453 The length we wnt is the differene etween Now WX = WY XY WY nd XY. = 233.2569 Round nd write the nswer in words. The ot siled pproimtely 233 m. Teher Notes Investigtion Viewing distnes nd ngles This digrm shows inem. The sreen is 2.5 m high nd is mounted so tht the ottom of the sreen is 3 m from the floor. Eye level for the udiene is pproimtely 120 m. Sreen 2.5 m 1.2 m 3 m Work in groups to determine the distne (to the nerest 10 m) tht will give the viewer the mimum viewing ngle. Use tle of vlues of suh s tht elow, or spredsheet, to find the est distne. (m) 1 2 3 4 5 6 7 8 Modern inems hve sloping floor to mke it esier for viewers to see the sreen. Keep the sme dimensions for the sreen, ut hnge the seting so tht the floor slopes up towrds the k. Mke nother tle or spredsheet to do your investigtions. Look t the hnges of ngle with flt nd sloping floors of different ngles. Spredsheet Spredsheet 22 New QMths 11
When solving trigonometri prolems, you must deide whih rules to use. The following provides some ides to guide you in hoosing the orret (or esiest) method.! When solving tringles: Look for solutions using right-ngled tringles first. Use the sine rule where you know one ngle nd the opposite side. e reful of miguous ses. Use the osine rule where the known ngle is enlosed y two known sides or the three sides re known. fter using the osine rule one, the sine rule my e useful for other ngles nd sides. If the known side is in different tringle from n unknown side, the side ommon to the tringles should usully e found s n intermedite step. Emple 14 Find the mrked sides nd ngles elow. 58 28 m 32 16 m 95 Solution The ngles in tringle dd up to 180. So the third ngle must e 90. The hypotenuse is 28 m. is opposite 58. Use sin = -------------------------- opposite. sin 58 = ----- hypotenuse 28 Rerrnge to find. = 28 sin 58 = 23.7453 Round nd write nswer. is out 23.7 m. You wnt the third side. Use the osine 2 = 2 + 2 2 os rule. Sustitute. = 16 2 + 19 2 2 16 19 os 95 = 256 + 361 + 52.9906 Keep the et numer on your = 669.9906 lultor. Use. = 25.8841 Round nd write nswer. is out 25.9 m. 19 m 58 32 m 69 28 m 75 71 32 36 m Chpter 1 Tringle pplitions 23
Tringle pplitions The first tringle hs three things known. is in seond tringle. The ommon side n e found. Lel this side s. 32 m 69 75 71 36 m Use the sine rule for. ---------------- = ---------------- 32 sin 69 sin 71 Rerrnge to find. 32 = ------------------------ sin69 sin 71 Keep the et nswer on your lultor. = 31.5959 Now use the osine rule for. 2 = 2 + 36 2 2 36 os 75 Keep et figures on the lultor. = 1705.5151 = 41.2978 Round nd write nswer. is out 41.3 m. Emple 15 Two points, P nd Q, re 100 m prt on level ground. vertil tower stnds somewhere etween P nd Q suh tht the ngles of elevtion to the top of the tower from P nd Q re 60 nd 35 respetively. How high is the tower? Solution Drw sketh. R The ngles dd up to 180. 85 So PRQ = 85. You wnt to find RS. It is in the right-ngled tringle QRS. 60 35 You need to find RQ first. P S Q 100 m It is in PQR s well s QRS. Use the sine rule to find RQ in PQR. ---------------- RQ = ---------------- 100 sin 60 sin 85 Rerrnge to get RQ. 100 RQ = --------------------------- sin60 Don t other to lulte. sin 85 Now use QRS. sin 35 = ------- RS RQ Rerrnge to get RS. RS = RQ sin 35 Put in the epression for RQ. 100 = --------------------------------------------- sin60 sin35 sin 85 Use your lultor. = 49.8629 Round nd write nswer. The tower is pproimtely 50 m high. 24 New QMths 11
Eerise 1.5 Trig pplitions 1 Find the mrked ngles nd sides in the following tringles. d 2.8 m 49 18 mm 22 mm 36 17 m 109 54 14 m 69 d 48 m dditionl Eerise 1.5 2 Find the mrked sides in the following. d 105 31 22 45 m 42 24 36 24 m 71 42 247 m 75 d 69 32 m Modelling nd prolem solving 3 n eletriity pole is stilised y wire. One end of the wire is onneted to the pole 11.2 m ove the ground. The other end is fied to the ground y spike. From the point where it is tthed to the pole, the ngle of depression of the wire is 65. How fr is the spike from the se of the pole? 65 11.2 m 4 Some workers re doing repirs to n eletriity pole. The worker who is t ground level is stnding so tht his eye is 1.8 m ove the ground. His feet re t the sme level s the se of the eletriity pole, 13.2 m wy from the pole. When he looks up, he sees the top of the helmet of his workmte in the herry-piker t n ngle of elevtion of 40. Wht is the vertil distne from the se of the pole to the helmet of the worker in the herry-piker? 40 1.8 m 13.2 m Chpter 1 Tringle pplitions 25
Tringle pplitions 5 uilding in the entrl usiness distrit is 85 m tll. The ngle of depression from the top of this uilding to fire lrm on the fe of uilding on the opposite side of the street is 58 20. The street is 15 m wide. Clulte: the height of the fire lrm ove the ground the ngle of elevtion from the se of the first uilding to the fire lrm. 6 ot leves jetty t the se of lighthouse tht is 80 m high. It sils diretly wy from the lighthouse nd fter 5 minutes is t n ngle of depression of 7.67 from n oserver t the top of the lighthouse. t wht speed did the ot sil? 7 n oserver sees tht the ngle of elevtion to the top of tower, 80 m high nd due south of her position, is 37. The top of nother tower, 60 m high nd due west of her position, is t n ngle of elevtion of 28. Clulte the horizontl distne etween the towers. 8 From point P, level with the se of vertil flgpole nd 50 m from it, the ngle of elevtion to the top of the pole is 48. Clulte the height of the pole. Clulte the ngle of elevtion from P to point hlfwy up the pole. 9 From the okpit of n irrft ruising 1520 m ove the ground, the ngles of depression to the tops of two identil (sme height) strutures re 50 nd 70 respetively. How fr prt re the 20 m high strutures? 10 n oserver in lighthouse, 120 m ove se level, is wthing fishing trwler move towrd the lighthouse. The ngle of depression of the trwler from the oserver is 18. How fr is the trwler from the lighthouse? Some time lter the ngle of depression is mesured s 28. How fr hs the trwler trvelled in this time? 11 While nn is wlking on stright rod running est west from to, whih re 10 km prt, she sees hurh in the distne. The ering from to the hurh is N48 E nd from the hurh is t ering of N67 W. Find the distne from to the hurh. 12 The top of lighthouse is t n ngle of elevtion of 35 from n oserver some distne from point on its se, s shown on the right. The top of the lighthouse is 38 m ove the point on the se. This point ppers to e t n ngle of elevtion of 8 from the oserver. How fr is the oserver from the se of the lighthouse? 38 m 35 8 26 New QMths 11
13 n irrft flies on ourse of 322 T for 250 km nd then hnges ourse to 244 T to n irstrip where it lnds. If the irstrip is 480 km from the strting position, lulte: the distne trvelled from the time tht the ourse ws hnged until the plne lnded the ering from the irstrip to the strting position. 14 Two ylists leve from the sme shopping entre. The first trvels long stright rod due west t 18 km/h. Thirty minutes lter, the seond leves nd proeeds long stright rod in south-westerly diretion t 24 km/h. How fr prt re the two ylists 30 minutes fter the seond deprts from the shops? 15 girl is flying kite nd hs etended 250 m of string when n updrft mkes the kite rise suddenly nd she hs to let out nother 60 m of string. If the ngle of elevtion to the kite ws 38 efore the updrft, how fr hs the kite een driven vertilly? 310 m 250 m 38 16 Two points, nd, on the sme nk of river re 120 m prt. Point C is on the opposite nk. C nd C re mesured to e 55 nd 33 respetively. Find the width of the river. 55 33 120 m 17 Sophie nd Will set out from point t the sme time. Sophie trvels t 30 km/h long stright rod in the diretion 042 T. Will trvels t 24 km/h long nother stright rod in the diretion 142 T. Find their distne prt fter 4 hours. 18 From distne, the ngle of elevtion to the top of tree is 25.13. Ten metres loser, the ngle of elevtion is 31.6. How high is the tree? C 31 6 25 13 Chpter 1 Tringle pplitions 27
Tringle pplitions 19 tower is ereted t the se of rod tht is inlined t onstnt ngle of 8 to the horizontl. ngles of elevtion to the top of the tower re mesured t two points tht re 80 m prt, some distne up the slope wy from the tower. The ngles re found to e 25 nd 8 respetively. Wht is the height of the tower? P 8 Q 80 m R 20 low-impt moile phone tower is on top of uilding. The ngles of elevtion of the top nd ottom of the tower re 16 nd 10 respetively. Thirty metres loser to the uilding, the ngle of elevtion of the ottom of the tower hs inresed to 17. Find the height of the tower ove the uilding nd the uilding s height ove the ground. Csio Clultor Progrm TI Clultor Progrm Shrp Clultor Progrm Tehnology You n use smll progrm on grphis lultor to solve tringles. The progrm TRISOLVE n e used to find the unknown sides or ngles in ny tringle, given ny set of three inputs (prt from ngle-ngle-ngle). The progrm is given in full on the CD-ROM. Enter the progrm (or lod it from the CD-ROM) nd try it with vrious tringles. right-ngled tringle is entered y giving right ngle s one of the ngles. The lultor must e set in degree mode. 1.6 Working in three dimensions Mny trigonometri prolems n e solved using tringles in only two dimensions, suh s the height nd distne pplitions erlier in this hpter. pplitions involving tringles in three dimensions (3D) must e simplified efore they n e solved with trigonometry.! Solution of 3D prolems To solve three-dimensionl prolems: line o drwing my help to visulise the prolem. Mke 3D drwing inluding tringles nd ll relevnt informtion. Lel points on the drwing. Unfold the digrm nd drw the tringles flt. Lel the unknowns nd intermedite sides nd ngles. Use the methods of right-ngled tringles, sine nd osine rules to find the unknowns. Write the solution to the prolem in n pproprite form. 28 New QMths 11
Emple 16 From ertin point,, mountin pek due north hs n ngle of elevtion of 20. From nother point,, 2 km est of nd on the sme level s, the ering of the pek is N40 W. Find the height of the pek ove the level of nd. Solution Mke sketh inside o to mke 3D drwing. Cll the foot of the mountin M nd the pek P. Show the diretions of north nd est. P M MN = 40 nd N = 90, so M = 50. Cll the height of the mountin h. Cll the distne from to the foot of the mountin. Redrw the digrm with the sides nd ngles mrked, ignoring the prts of the o you don t need. 20 2 km 40 20 N E P h M 50 2 km Then unfold the digrm to mke flt tringles. In this se they re joined long M. P h M P h Unfold M 20 20 50 50 2 km 2 km Use tn in M to find. tn 50 = -- 2 Rerrnge, ut don t lulte. = 2 tn 50 Now use tn in PM to find h. h tn 20 = -- Rerrnge. h = tn 20 Sustitute. = 2 tn 50 tn 20 Clulte the nswer. = 0.8675 km Round nd write the nswer in pproprite form. The pek is out 868 m high. Chpter 1 Tringle pplitions 29
Tringle pplitions Emple 17 Hssn oserves tht the top of trnsmission tower t ering of 038 T is t n ngle of elevtion of 12. Ftim is 575 m due est of Hssn, nd she sys the ering of the tower is 295 T. Find the height of the tower. Solution Mke sketh inside o. T Cll Hssn s position H. Cll Ftim s position F. Show north nd est. N W Cll the ottom of the tower W. 38 12 Cll the top of the tower T. 295 WHN = 38 nd FHN = 90, so FHW = 52. H 575 m F The other ering shows tht WFH = 25. T h TWH = 90. W Unfold nd drw the tringles flt. Cll the height h. Cll the ommon side. 12 The ngles in WHF dd up to 180. 52 25 H Thus HWF = 103. 575 m Use the sine rule in WHF to find. ---------------- = ------------------- 575 sin 25 sin 103 Rerrnge ut don t lulte. 575 = --------------------------- sin25 sin 103 Now use tn in THW to find h. h tn 12 = -- Rerrnge. h = tn 12 Sustitute nd lulte. = ---------------------------------------------- 575 sin25 tn12 sin 103 = 53.0110 Round nd write the nswer. The trnsmission tower is out 53 m high. F E In the net emple, one side hs to e duplited to unfold the digrm to mke it flt. In ddition, the required side nnot e found diretly. fter pplying trigonometry you must solve simple eqution to find the height. 30 New QMths 11
Emple 18 Mri nd Jk re stnding on level rod 150 m prt. hurh spire is due north of Mri nd on ering of 070 T from Jk. The top of the spire ppers t n ngle of elevtion of 27 to Jk nd 18 to Mri. Find the height of the spire ove the rod. Solution Mke sketh inside o. N Cll Jk s position J. T Cll Mri s position M. 70 h Show north nd est. Cll the ottom of the spire S. 27 J S Cll the top of the spire T. Cll the height of the spire h. Cll the distne of the spire from Jk nd its distne from Mri y. 18 NJS nd JSM re lternte, so JSM = 70. M Unfold the digrm to mke it flt, showing the side TS on two of the flt tringles. y the ngle sum of tringle, JTS = 63 nd MTS = 72. J 27 70 T h S J 150 m T h 63 S 27 70 h 72 T E 150 m M 18 y 150 m Use tn in TSJ to find h. tn 63 = -- h Use tn in TSM to find h. tn 72 = -- y h Now use the osine rule in JMS. 150 2 = 2 + y 2 2y os 70 To proeed further you need nd y. = h tn 63 nd y = h tn 72 Rerrnge the previous epressions. Now sustitute nd y. 150 2 = (h tn63 ) 2 + (htn72 ) 2 2(htn63 )(htn72 ) os 70 Note: tn 2 = (tn ) 2. = h 2 tn 2 63 + h 2 tn 2 72 2h 2 tn 63 tn 72 os 70 Ftorise. = h 2 (tn 2 63 + tn 2 72 2 tn 63 tn 72 os 70 ) Clulte. 22 500 = h 2 9.1921 Rerrnge nd lulte. h 2 = 22 500 9.1921 = 2447.7350 h = 49.4745 Round nd write the nswer. The height of the spire is out 49.5 m. M y 18 Chpter 1 Tringle pplitions 31
Tringle pplitions dditionl Eerise 1.6 Eerise 1.6 Working in three dimensions Modelling nd prolem solving 1 retngulr-sed pyrmid with se mesuring 16 m y 12 m nd height of 10 m is to e onstruted using perspe sheeting. Clulte the ngle tht the tringulr fes YZ nd ZW mke with the horizontl se. 10 m X Y W 16 m O Z C 12 m 2 flgpole 12 m high is temporrily ereted for n ustrli Dy eremony. It is held in ple y two wires: one 15 m long fstened due west of the pole nd the other 17 m long fstened due south of the pole. oth supporting wires re tthed to the top of the flgpole. Clulte: the distne tht eh wire is from the pole t the point where it meets the ground the ngle of inlintion tht eh wire mkes with the horizontl the distne etween the points where the wires meet the ground d the ngle etween the wires. 3 From lighthouse 150 m ove se level, the lighthouse keeper oserves ot due est t n ngle of depression of 25 nd nother ot due south t n ngle of depression of 32. Find the distne tht seprtes the ots. Find the ering of the seond ot from the first. 4 n irrft flying due est t 385 km/h t onstnt height of 1500 m psses over villge t prtiulr instnt. One minute lter, person who is 1.6 km south of the villge sees the plne. In wht diretion nd t wht ngle of elevtion does the person see the plne? 5 sulptor hs removed setion from opper lok with dimensions s shown in the digrm on the right. Clulte: QR y d the ngle tht the tringulr fe PQR mkes with the horizontl se. 6 lrge tent is onstruted in the shpe of retngulr-sed pyrmid CDE with the verte E vertilly ove G, the midpoint of C, s shown in the digrm on the right. Clulte: the ngle tht the tringulr fe ED mkes with the se of the pyrmid the ngle etween the lines DE nd D. Q 15 m F 48 m E G y S P 40 m 45 D R 9 m C 10 m 32 New QMths 11
7 stright stik 2 m long is stuk in the ground nd is inlined t 60 to the horizontl, pointing in n esterly diretion. Clulte the length nd diretion of the shdow of the stik when the sun is due north t n elevtion of 30. (Hint: Imgine piee of string hnging vertilly from the top of the stik to the ground. Consider the shdow of the string nd its reltion to the shdow of the stik.) 8 The digrm on the right shows the frme for doule-hipped roof. Clulte: the ngle tht the tringulr fe RSU mkes with the horizontl T 8.4 m W U the length of the rfter VW Q 6 m the ngle the sloping roof PTUS R mkes with the horizontl X Y 9 m Z d the length of the rfter UY e the length of the rfter UR. P V 12 m S 9 hillside is onsidered to hve uniform slope of 22 with the horizontl s shown in the digrm on the right. hiker, strting from point E, wlks due north up the hillside for 120 m to point C, then wlks due west to point. The hiker then wlks 170 m k to the strting point. Clulte the ngle of inlintion of E to the horizontl nd the totl distne overed y the hiker. 170 m 120 m 22 C D N 22 E 10 Two groups of people set off to spotlight possums t night. One group spots possum. The tree tht hs the possum is t ering of N28 W. The ngle of elevtion of their spotlight em is 38. t this instnt the other group, whih is 100 m west of the first group, sights the sme tree t ering of N29 E. How high up the tree is the possum? 11 Elen sees tht the ngle of elevtion to the top of tower, 80 m high, whih is due south of her position is 3l7. The top of nother tower, 60 m high, whih is due west of her is t n ngle of elevtion of 28. Clulte the horizontl distne etween the two towers. 12 Miguel is 90 m due west of Consuell on level rod. He sees tht the ngle of elevtion of kite flown y their son is 30, while Consuell sees tht it is due north t n ngle of elevtion of 38. Wht is the height of the kite? Chpter 1 Tringle pplitions 33
Chpter summry The side of tringle is usully nmed y lower-se letter, nd the opposite verte is nmed y the sme pitl letter. ngles re mesured in degrees, minutes nd seonds. There re 360 in full turn, 60 in 1 nd 60 in 1. We use sin, os nd tn to solve right-ngled tringles. The side opposite the right ngle is lled the hypotenuse. The side etween the ngle we wnt nd the right ngle is the djent side, nd the side opposite the ngle we wnt is the opposite side. opposite djent opposite sin = -------------------------- os = -------------------------- tn = ------------------- hypotenuse hypotenuse djent The ering of the ojet is given s the 3-digit lokwise ngle the oserver turns from due north in order to fe the ojet. ering n lso e given s the ngle from north or south towrds either est or west. The sine rule my e used to solve non-right-ngled tringles where side nd n opposite ngle re known. In ny C ----------- = ----------- = ----------- sin sin sinc The osine rule my e used to solve tringles if ll three sides re known, or in ses where two sides nd the enlosed ngle re known. In ny C 2 = 2 + 2 2 os or os = ----------------------------- 2 + 2 2 2 The ngle of elevtion of n ojet is the ngle n oserver must look upwrds from the horizontl to see the ojet. The ngle of depression of n ojet is mesured downwrds from the horizontl. When solving tringles: Look for solutions using right-ngled tringles first. Use the sine rule where you know one ngle nd the opposite side. e reful of miguous ses. Use the osine rule where the known ngle is enlosed y two known sides or the three sides re known. fter using the osine rule one, the sine rule my e useful for other ngles nd sides. If the known side is in different tringle from n unknown side, the side ommon to the tringles should usully e found s n intermedite step. To solve three-dimensionl prolems: 1 line o drwing my help to visulise the prolem. 2 Mke 3D drwing inluding tringles nd ll relevnt informtion. 3 Lel points on the drwing. 4 Unfold the digrm nd drw the tringles flt. 5 Lel the unknowns nd intermedite sides nd ngles. 6 Use the methods of right-ngled tringles, sine nd osine rules to find the unknowns. 7 Write the solution to the prolem in n pproprite form. 34 New QMths 11
Chpter review Knowledge nd proedures 1 Write down the trigonometri rtios in terms of the sides of right-ngled tringle. 2 Find the missing sides of the following tringles. 15.0 m 16.5 m E 1.1 E 1.1 22.5 m 21.3 m 3 Find the length of the unknown side or ngle mrked in eh of the following tringles. 27.9 42 24 29 47 53 α u 64 15 41 d e f f 72 67 m 98.1 52 42 55.2 E 1.1 4 lighthouse, L, is 40 km wy from hrour loted t K in the digrm on the right. The lighthouse is on ering of 225 T from the hrour. fishing ot trvels from the hrour on ering of 205 T until it is due south of the lighthouse. Clulte: how fr the fishing ot trvels how fr the fishing ot is from the lighthouse. L 40 km N K 205 225 E 1.2 5 The re of the tringle shown is equl to: 1 -- 1 sin -- 1 sin C -- os 2 1 2 2 1 2 D -- os E -- tn 6 Solve eh of the following tringles. C, = 42, = 36 m, = 28 m FEG, F = 58.2, f = 42.5 m, E = 28.4 KMN, M = 35, m = 29 m, n = 19 m 7 Find the ngles in tringle with sides 8.2 m, 9.5 m nd 10.6 m. 2 M E 1.3 E 1.3 E 1.4 Chpter 1 Tringle pplitions 35
Chpter review E 1.4 E 1.5 8 Find the mrked ngles nd sides in the following tringles. 22.3 m 48 m 48.5 103 37.4 m 54 m β 9 Find the mrked side in eh of the following. φ 34 mm α 38 mm 44 mm E 1.1 61 Modelling nd prolem solving 10 roof truss is eing mde to the design shown on the right. Clulte the lengths of timer required, nd the lengths required to mke the uts t the orret ngles s shown in the detil drwings. 27 164 m y 90 90 28.1 17.6 m 54.2 z 57.3 w 90 18 19.5 m 90 6500 mm E 1.2 E 1.3 E 1.4 E 1.5 E 1.6 E 1.6 11 yht sils on tringulr ourse. The first leg is 18 km NE nd the seond leg is 15 km t 168 T. Wht is the third leg? 12 person stnding t point Z on the nk of X river oserves tree t point X on the opposite nk, t ering of 292.17 T. The person then wlks 885 m down the nk to Z Y point Y nd notes the tree is now t ering 885 m of 285.33 T. Find the distne etween the 292.17 T tree nd the initil position, Z, of the oserver. 13 ostl surveillne vessel Intereptor detets two suspet ots on the rdr sreen. The first is 60 km wy from the Intereptor on ering of 062.2 T. The seond is 42 km wy on ering of 296.58 T. How fr prt re the two suspet ots? 14 The top of hill is t n ngle of elevtion of 7 30. hundred metres loser to the hill, the ngle of elevtion is 8. Wht is the height of the hill? 15 surveyor notes tht the top of wter tower is t ering of 062 T. fter wlking 200 m diretly est she finds tht the tower is t ering of 059 10 T. The ngle of elevtion of the top of the tower is then 4 31. Wht is the height of the tower? 16 tle is 1 m wide, 2.5 m long nd 900 mm high. The 5 m thik top is supported on legs splyed outwrds long the digonls y 8 from the vertil. The legs meet the floor diretly under the orners of the tle. Find how fr from the ends nd sides the legs re tthed. 285.33 T 36 New QMths 11