Trigonometry. Contents. Syllabus subject matter

Transcription

1 Trigonomtry 2 ontnts 2.1 ythagoras s Thorm 2.2 Shaow rkoning an triangulation 2.3 Th tangnt ratio 2.4 Th sin ratio 2.5 Th osin ratio 2.6 Using trigonomtri ratios haptr rviw Syllaus sujt mattr asi knowlg an prours alulation an stimation Mtri masurmnt Simpl algrai manipulations Elmnts of appli gomtry ppliations of trigonomtry using sin, osin an tangnt ratios an ythagoras s Thorm Syllaus gui haptr 2

2 Th shap of a right-angl triangl is fix y just on othr angl. ny othr straight-si shaps an ivi into triangls, an ths an always ivi into right-angl ons. This is why right-angl triangls hav always n us as th asis of uiling, survying, lan masurmnt an navigation. 2.1 ythagoras s Thorm Th Grk mathmatiian ythagoras ( ) is gnrally givn rit for first proving th rlationship twn th sis of a right-angl triangl, although it is known that th anint Egyptians, aylonian an hins wr familiar with som of th proprtis.! In a right-angl triangl th longst si is opposit th right angl an is all th hypotnus. Th squar of th hypotnus is qual to th sum of th squars of th othr two sis. 2 = a whr is th hypotnus an a th othr two sis ar a an. Th naming onvntion for triangls is us aov. Th vrtis (ornrs) ar nam using apital lttrs, an th opposit sis hav th sam lowr-as lttr. In th triangl aov, =, = an = a. Th right angl is shown y a small squar in th ornr. Tahr nots Invstigation ythagoras s Thorm Gomtrially, th ara of a squar is qual to x 2, whr x is qual to th lngth of th si. This mans w an work out th squar of a numr y ounting unit squars in a squar of that si lngth. = 16 squars = 9 whol squars + part squars = 12 squars a = 4 squars 38 NEW QMTHS 11

3 W an show that ythagoras s Thorm is tru y ounting th ara in squars rawn on th sis of th triangl. Th ara is ount y iviing th squars into unit squars as shown at th ottom of pag 38. Work in groups of two or thr popl to show ythagoras s Thorm using this mtho. Eah group shoul hav iffrnt-siz triangls an ompar th aras ount in th squars. You may hav to ount som parts of squars y aing thm togthr as shown. Exampl 1 Fin th unknown lngth in this triangl. 10 m 15 m Writ ythagoras s Thorm. 2 = a x is th hypotnus. x 2 = Work out th squars. = alulat th sum. = 325 Fin th squar root. x = 325 m Us on your alulator. = m Roun off m Writ th answr. Th unknown lngth is aout m. x You will n to us sutration to alulat an unknown si if th hypotnus is givn. Exampl 2 Fin th thir si of th following triangl. 11 m 18 m Writ ythagoras s Thorm. 2 = a is th hypotnus = Work out th squars. 324 = Now us sutration. 2 = = 203 Fin th squar root. = 203 m Us on your alulator. = m Roun off m Writ th answr. Th thir si is aout 14.2 m. TRIGONOMETRY 39 HTER 2

4 Exampl 3 Th fram for a wall is 2400 high an 4000 long. iagonal ra must a. How long will it? (uiling lngths in ustralia ar always in millimtrs.) Skth th fram (2.4 m y 4 m). 2.4 m 4 m Show th triangl. Writ a lttr () for th ra lngth. 2.4 m 4 m Writ ythagoras s Thorm. 2 = a is th hypotnus. 2 = Us your alulator. = = Fin th squar root. = m Us on your alulator. = m = mm 4665 (to narst mm) Writ th answr. Th ra will 4665 mm long. Tahr nots Invstigation ythagoran tripls ythagoras s Thorm an us to fin som stanar right-angl triangls whr th valus of th sis ar whol numrs. Ths ar all ythagoran tripls. Th 3, 4, 5 an 5, 12, 13 tripls an thir multipls ar th most ommon, ut thr ar many othrs. 1 Work in groups of two or thr to fin whih of th following ar ythagoran tripls. You shoul also hk whthr th tripls hav a ommon fator to fin whthr thy ar multipls of simplr ons. 5, 6, 7 6, 8, 10 9, 12, 15 12, 16, 20 10, 24, 26 8, 15, 17 10, 14, 18 7, 24, 25 9, 40, 41 12, 35, 37 16, 63, 65 20, 24, 32 2 an you fin othrs? ythagoran tripls an thir multipls an somtims us to alulat missing sis without squaring. To us thm, you must al to rognis th most ommon ons: 3, 4, 5 an 5, 12, 13. Othrs suh as 8, 15, 17 an 7, 24, 25 ar nountr mor rarly. 40 NEW QMTHS 11

5 Exampl 4 Fin th missing si in th following triangl. 9 m 15 m Th hypotnus is 3 5 an th shortst si is 3 3. Th sis ar 3 tims iggr than th 3, 4, 5 tripl. Writ th answr. x Thr is a 3, 4, 5 tripl. Missing si x = 3 4 = 12 m. Th missing si is 12 m long. ythagoras s Thorm is ommonly us in rvrs to fin whthr or not an angl is 90. This is a ommon us in uiling an survying appliations.! Th rvrs of ythagoras s Thorm says that, if th sis of a triangl satisfy 2 = a 2 + 2, thn th angl opposit si is a right angl. a In fat, th siz of angl an hk as follows: If 2 a 2 + 2, thn 90 ( is aut). If 2 = a 2 + 2, thn = 90 ( is right). If 2 a 2 + 2, thn 90 ( is otus). Exampl 5 Dtrmin whthr this triangl is aut, otus or right-angl m 12.2 m 17.6 m Fin th sum of squars of th shortr sis = = Fin th squar of th longst si = ompar th rsults ompar with ythagoran rsult. Th longr si is longr than a hypotnus. Writ th answr. Th triangl is otus-angl. TRIGONOMETRY 41 HTER 2

6 ythagoras s Thorm is somtims us in survying to work out a istan that annot irtly masur aus thr is an ostrution. This is on y offstting at right angls from on point until a istan an masur from th son point. Exampl 6 survyor wants to know th lngth of a small lak an maks th following masurmnts twn th points shown in th iagram. = 35 m D = 15 m E = 395 m DE = 224 m Fin th lngth of th lak. Draw th triangl. Writ th known sis. Lal th unknown si. 395 m E D 224 m Us ythagoras s Thorm. 2 = a = x = x Now us sutration. x 2 = = Fin th squar root. x = m = m Roun off to masurmnt auray. 325 m Now us th original iagram. 325 m 35 m 15 m = 275 m Writ th answr. Th lngth of th lak is aout 275 m. x itional xris 2.1 Exris 2.1 ythagoras s Thorm 1 Fin th hypotnus in ah of th following triangls. a 16 m 3.15 m a 11 m 2.95 m 16.8 m 74 km 325 m 22.3 m 56 km 154 m 42 NEW QMTHS 11

7 2 Fin th lngth of th iagonal in ah of th following rtangls. a 2700 mm 1 m 3200 mm 4.69 m 23 m mm 52 m mm 106 m 5 m 3 Fin th unknown si in ah of th following triangls. a 5.24 m 95.2 km a 2.68 m 76.1 km 267 mm 98 mm 425 m 312 m 58 m 43 m 4 Us ythagoras s Thorm to fin th unknown si in ah of th following triangls. a 15 m 5.6 m 25 mm 8 m a 28 km 16 mm 3.1 m 21 km 52 m 51 m f 9.3 m 16.7 m f 5 Us ythagoran tripls to fin th missing si in ah of th following triangls. a 36 m a 6 m 10 mm 26 mm 8 m 50 m 65 m 25 m 48 m 40 m TRIGONOMETRY 43 HTER 2

8 6 Dtrmin whthr ah of th following triangls is aut, otus or right-angl. a 12 m 225 m 154 m 10 m 9 m 46 km 52 km 325 m 47 m 52 m 10 mm mm 48 km 69 m mm Molling an prolm solving 7 wall fram 3500 long an 2100 high is to hav a iagonal ra fitt. How long must th ra? 8 powr point is 1 m from th ornr of a squar room. projtor is to pla in th ntr of th opposit wall. If th room is 6 m wi, what is th minimum lngth of th powr or rquir? 9 ak yar is 19.5 m wi an 15.2 m p. What is th longst strth-a-lin washing lin that an us? (Hint: Thr ar no ostals in th ak yar, an th lin an go on th iagonal.) 10 vaant ornr lok is 25 m y 30 m. Insta of going roun th footpath, many popl ut aross th vaant lok whn thy ar walking to th us stop aroun th ornr. How muh furthr will thy hav to walk whn th lok is fn? 11 survyor wants to know th lngth of a small lak, an maks th following masurmnts twn th points shown in th iagram. = 28 m D = 25 m E = 627 m DE = 452 m Fin th lngth of th lak. E D 12 survy annot masur irtly ovr a small hill, so th following masurmnts ar ma twn th points shown in th iagram, whr an ar on th sam ontour lin. = 45 m D = 55 m E = 248 m DE = 427 m Fin th horizontal istan. E D 44 NEW QMTHS 11

9 2.2 Shaow rkoning an triangulation Triangls that ar multipls of th 3, 4, 5 tripl ar xatly th sam shap. Th only iffrn twn thm is thir siz. W say that th triangls ar similar an th ratio of thir sis is th sal fator. Th sam applis to any othr tripls or vn non-rightangl triangls that ar th sam shap. W an us th sal fator to fin unknown sis in on of th triangls. Exampl 7 Fin th sal fator an unknown si in th iagram of similar triangls shown at right. E 72 mm g ompar orrsponing known sis. Kp xat numr on your alulator. Now multiply to gt th unknown si. Us th xat numr. Writ th answr. 40 mm D 96 mm Sal fator from DE to is: D = = g = 72 mm sal fator = 72 mm = 30 mm Th unknown si is of lngth 30 mm. Th similarity prinipl is appli to hight masurmnt with th ai of a shaow stik. To masur th hight of an ojt, a stik of known hight is pla in th shaow of th uiling. Th stik is put on th groun so that th top of th stik is just on th g of th shaow. Exampl 8 Fin th hight of th uiling shown in th following iagram, showing a shaow stik st up in th aftrnoon. T 2 m E 3.2 m S 14.4 m F TRIGONOMETRY 45 HTER 2

10 Th sunlight forms th hypotnus of th triangls EST an EF. Th shaow forms th sis ES an EF, an th stik an th uiling mak th thir sis. Th triangls ar similar. ompar orrsponing known sis. Sal fator from EF to EST is: EF = ES 3.2 Kp th numr on your alulator. = 4.5 Now multiply to gt th unknown si. F = 2 m sal fator = 2 m 4.5 = 9 m Writ th answr. Th uiling is 9 m high. Th shaow stik an us vn whn th stik is not in th shaow of th uiling. Howvr, th shaows must masur at th sam tim so that th triangls rmain similar. ftr all, th lngth of th shaow ast y th stik will th sam whrvr it is pla at a givn tim. Exampl 9 stik of hight 1.5 m asts a shaow of lngth 3.8 m. t th sam tim, a tr asts a shaow 31 m long. What is th hight of th tr? Draw a iagram to show th information. h 1.5 m 31 m Fin th sal fator. Kp xat numr on your alulator. Now multiply to gt th unknown si. Us th xat numr on your alulator. Roun an writ th answr. 3.8 m Sal fator (tr to stik) is: lngth of tr shaow = lngth of stik shaow 3.8 = Tr hight = 1.5 m sal fator = 1.5 m = m Th tr is aout 12.2 m high. 46 NEW QMTHS 11

11 Similar triangls an also us in survying to fin istans that annot masur xatly. Using triangls in this way to masur aspts of th Earth is all triangulation. Exampl 10 survyor ns to masur th istan aross a rivr. Thr ar two trs on th opposit ank. ftr paing along th rivr ank, th survyor fins that thy must 45 m apart. Sh thn movs ak 5 m from th ank, irtly opposit th first tr. Hr assistant has to mov 8.4 m along th ank to pla a stik irtly in hr lin of sight to th son tr. Fin th with of th rivr. Draw a iagram to show th information. Lal points. Writ lngths. T 2 45 m T 1 Rivr 45 m 8.4 m 5 m S Fin th sal fator. T 1 T 2 45 Sal fator is = Kp xat numr on your alulator. = ST 1 = S sal fator Us xat numr on your alulator. = 5 m Kp xat numr on your alulator. = m Work out sir istan. T 1 = ST 1 S = m = m Roun an writ th answr. Th rivr is aout 22 m wi. Invstigation Shaow stik us mtr rulr maks an xllnt shaow stik aus th alulations ar simplifi y having th stik xatly 1 m high. Us a mtr rulr an a long tap in groups of two or thr to masur th hights of uilings an trs aroun th shool. You will n to mak sur th mtr rulr is vrtial whn you ar masuring its shaow. This an on using a spirit lvl. If you on t hav a spirit lvl, you an hk that th stik is vrtial y lin of sight against somthing known to vrtial, suh as th ornr of a uiling or a tlphon or light pol. hk your masurmnts y omparing th rsults of your group with th rsults of othr groups. TRIGONOMETRY 47 HTER 2

12 Thnology Using a graphis alulator program You an us a graphis alulator program to alulat th hights of ojts using shaow stiks. Th following program an us for this purpos. Entr th program as shown for your graphis alulator an try it. Graphis alulators hav an LH ky, whih is prss for typing a lttr suh as,,, t. Thy may also lok onto alphati haratrs using -LOK, whih is otain using SHIFT, 2n or 2nF for th LH ky. -LOK is turn off y again prssing th LH ky. asio FX-9850G LUS Us th ursor kys to RGM in th main mnu. to go rss F3 EXE to go into th program mnu an prss = NEW to put in a nw program. -LOK is automatially on to ntr th nam of th program. Typ SHDOW as th nam of th program an prss EXE to stor th nam. To mak th program isplay SHDOW STIK, ntr this insi quots an thn ntr th isplay symol, using th RGM mnu utton: F6 F2 SHIFT LH SHDOW. STIK F2 SHIFT VRS F5 rss EXIT to go ak to th main programming mnu. To mak th program ask for th stik hight H, typ F6 F2 SHIFT LH STIK. HEIGHT F2 SHIFT VRS F5 rss EXIT to go ak to th main programming mnu. Typ F4 LH H EXE. rss EXIT to go ak to th main programming mnu. To mak th program ask for th stik shaow lngth S, typ F6 F2 SHIFT LH STIK. SHDOW F2 SHIFT VRS F5 rss EXIT to go ak to th main programming mnu. 48 NEW QMTHS 11

13 Typ F4 LH S EXE. rss EXIT to go ak to th main programming mnu. To mak th program ask for th ojt shaow lngth, typ F6 F2 SHIFT LH OJET. SHDOW F2 SHIFT VRS F5 rss EXIT to go ak to th main programming mnu. Typ F4 LH EXE. rss EXIT to go ak to th main programming mnu. To mak th program alulat th ojt hight, typ LH H LH LH S LH EXE To mak th program isplay th answr, put in OJET HEIGHT IS an thn. Typ F6 F2 SHIFT LH OJET. HEIGHT. IS F2 SHIFT VRS F5 LH F5 rss EXIT twi to rturn to th program List. Th program SHDOW shoul highlight. rss EXE to run th program. You will n to prss EXE aftr ah valu is ntr an aftr ah isplay. Try th program with a 3 m stik with a shaow 2 m long an an ojt shaow 8 m long. This givs an ojt hight of 12 m. Txas Instrumnts TI-83 rss th RGM ky an us th ursor kys to highlight NEW. rss ENTER, typ th nam SHDOW an prss ENTER again. rss RGM, highlight I/O, slt 3: Disp an prss ENTER. Typ in SHDOW STIK as 2n LH + SHDOW 0 STIK + ENTER LH To mak th program ask for th stik hight H, prss RGM, highlight I/O, slt 1: Input an prss ENTER. Typ in 2n LH + STIK 0 HEIGHT + 0 LH, LH H ENTER. TRIGONOMETRY 49 HTER 2

14 To mak th program ask for th stik shaow lngth S, prss RGM, highlight I/O, slt 1: Input an prss ENTER. Typ in 2n LH + STIK 0 SHDOW 0 + LH, LH S ENTER. To mak th program ask for th ojt shaow lngth, prss RGM, highlight I/O, slt 1: Input an prss ENTER. Typ in 2n LH + STIK 0 SHDOW + 0 LH, LH ENTER. To mak th program alulat th ojt hight, typ in LH H LH LH S STO LH ENTER To mak th program isplay th answr, prss RGM, highlight I/O, slt 3: Disp an prss ENTER. Typ in 2n LH + OJET 0 HEIGHT 0 IS 0 + LH, LH ENTER alulator instrutions rss 2n MODE Run th program y prssing (QUIT) to lav th program ntry stion. RGM, highlighting EXE an prssing ENTER. Try th program with a 3 m stik with a shaow 2 m long an an ojt shaow 8 m long. This givs an ojt hight of 12 m. Sharp EL-9650 S th instrutions givn on th D-ROM. Exris 2.2 Shaow rkoning an triangulation 1 Th following iagrams show similar triangls. Us sal fators to fin th unknown sis. a 3.1 m 2 m 8.4 m a 4 m 1.5 m 17.3 m 3 m 5.2 m 25.6 m f 2 m 2.5 m 8.5 m 1.5 m 2.2 m 7.4 m 3 m 2.5 m 7.2 m f 50 NEW QMTHS 11

15 Molling an prolm solving 2 stunt us a mtr rulr as a shaow stik to masur th hights of som shool uilings. t 2 pm th lngth of th shaow ast y th mtr rulr was 60 m an th lngth of th shaow of -lok was 4.86 m. Th stunt thn mov to th oval an masur th lngth of th shaow of th gym; it was 3.5 m long. ftr that th stunt mov to -lok an masur its shaow as 7.2 m long. It was thn 2:45 pm an th tahr insist that th lass mak a last hk of th stik s shaow lngth for moving insi. Th shaow of th stik was now 90 m long. Th stunts wr to alulat th hights of th uilings for homwork. a What i this stunt o inorrtly? What was th hight of -lok? What was th hight of -lok? What o you stimat th hight of th gym to? 3 In Hoart, th lat aftrnoon summr sun asts th shaow of th mountain aross th wstrn suurs, vn though th astrn shor is still in sunlight. Th shaow rahs onstitution Dok, 8 km from th summit of Mt Wllington, at 6:20 pm. t th sam tim a 2.5 m shaow stik asts a shaow 15.7 m long. How high is Mt Wllington? 4 Th shaow of a san un on Morton Islan is 35 m long. t th sam tim th shaow ast y a 2 m shaow stik is masur to 3.6 m long. How high is th san un? 5 Th faa of a histori uiling was prsrv whn th uiling was molish for a nw shopping ntr. In th morning th sun shon through th faa, asting a shaow aross th strt. Thr wr right paths in th shaow from th winows of th faa. If th whol shaow was 25 m long an th right path from a 1.8 m-high winow was 2.4 m long, how high was th faa? 6 Th istan aross a roa utting is to alulat using triangulation. Sightings ar takn of two points an on th opposit si of th roa that ar known to 120 m apart. Th sis of th roa utting ar paralll, an point on th survyor s si is irtly opposit point. From a point 8 m away from th roa g at, th survyor notis that a small tr on th g of th roa just osurs point. Th tr is foun to 6.7 m from on th survyor s si of th roa. What is th with of th roa utting? 7 Watr is rushing out th long, narrow ntran to a lagoon at low ti. Thr is a larg rok on th opposit ank of th ntran. fishrman is irtly opposit th rok. H walks 4 m along th ank of th ntran an puts a stik in th san. ftr walking a furthr 2 m along th ank, h thn has to mov 4.5 m ak from th ank in orr to lin up th stik an rok. How wi is th lagoon ntran? Roa Lagoon TRIGONOMETRY 51 HTER 2

16 8 From a point at th watr s g at low ti, two frins an s a san un irtly own th ah. On walks 30 m along th watr-lin towars th un an 3 m towars th high-watr mark. Th othr walks 4 m towars th high-watr mark to mak a straight lin with th first prson an th san un. Us triangulation mthos to fin th istan from th oupl s first position to th san un. High-watr mark 4 m 30 m 3 m Low ti 2.3 Th tangnt ratio Th trigonomtri ratios rlat th sis to th angls of right-angl triangls. for looking at th ratios w will xamin th naming of th sis. Th thr sis ar nam in rlation to th angl in whih you ar intrst.! Th longst si is opposit th right angl an is all th hypotnus. Th opposit si is furthst away from th angl. It is iagonally opposit th angl. Th ajant si is nxt to th angl an maks on arm of th angl. (Th hypotnus is th othr arm.) For angl θ θ Hypotnus jant Φ Opposit For angl φ θ Hypotnus Opposit Φ jant Exampl 11 Nam th opposit, ajant an hypotnus for th angls mark in th triangls low. a g F H S h f G R T a Th hypotnus is opposit th right angl. Th hypotnus is si f. Th si opposit G is g (naming onvntion). Th opposit to angl G is si g. Th si nxt to th angl is si h. Th ajant to angl G is si h. Th hypotnus is opposit th right angl. Th hypotnus is si SR. Th si opposit S is RT. Th opposit to angl S is si RT. Th si nxt to th angl is si ST. Th ajant to angl S is si ST. 52 NEW QMTHS 11

17 Two right-angl triangls with th sam ornr angl will always th sam shap. Th triangls will similar an so will rlat y a sal fator. Th triangls low hav a ornr angl of 28, an th son 2.7 tims as ig as th first on. 1.5 m 2.8 m m 7.7 m Th opposit an ajant sis in ah triangl will ivi to giv th sam amount an Th ratios of th opposit an ajant sis will atually th sam for any similar rightangl triangl. W an say that vry right-angl triangl with a 28 ornr angl will hav th sam ratio of opposit to ajant (aout 0.53). In th past ths ratios wr alulat for iffrnt angls an print in a look-up tal all th tangnt tal. Nowaays w us a alulator to look up th ratios.! Tangnt ratio In any right-angl triangl, th tangnt ratio of an angl is givn y opposit si Tangnt of th angl = ajant si W usually arviat th tangnt ratio as tan an writ opposit si a tan = = -- ajant si 28 a Exampl 12 Fin th valu of tan H for this triangl. H 8.1 m I 4.5 m 6.7 m opposit Tangnt uss opposit an ajant. tan H = ajant 6.7 Th opposit si is 6.7 an th ajant = si is Writ th answr. tan H is approximatly J TRIGONOMETRY 53 HTER 2

18 Whn using a alulator for this work, mak sur it is st to gr mo. Most sintifi alulators will show a small DEG on th isplay if thy ar st in this mo. If th isplay shows RD or GRD, it will not giv th orrt answrs. Th angl mo is st in graphis alulators using th SETU or MODE mnu. Exampl 13 Us a alulator to fin tan 34.8 orrt to 4 imal plas. Us th tan ky. tan 34.8 = Roun an writ th answr. tan 34.8 is aout Not: On som olr alulators you may n to ntr th numr an thn prss tan. Exampl 14 Using a alulator, fin M suh that tan M = Us tan 1. 2nF tan 1.89 = Roun an writ th answr. M is aout Not: Som alulators us th INV, 2n or SHIFT ky insta of 2nF, an on som olr alulators th numr must ntr for th kys ar prss. Exampl 15 Fin th angls in th following triangl. K M 746 mm 208 mm onsir sis in rlation to angl M. opposit tan M = ajant Opposit is 208 an ajant is = Kp th xat numr on your alulator. = Us tan 1 to fin M, thn roun. So M 15.6 Us sum of angls in a triangl. Thus K = 74.4 Writ th answrs. Th angls ar aout 15.6 an Not: W oul fin K using tangnt again, ut is is asir to us th sum of angls in a triangl. L Th tangnt ratio an us to fin on of th shortr sis from th othr. W us th angl opposit th missing si to mak th alulation asir. ythagoras s Thorm an thn us to fin th hypotnus, if sir. 54 NEW QMTHS 11

19 Exampl 16 Fin th mark sis in th following triangls. a a 28.3 m 53.6 Th missing si is opposit th known opposit tan = angl. ajant Th ajant is known. Th opposit is a tan 53.6 = rquir Multiply to gt a. a = 28.3 tan 53.6 Us your alulator, kping xat numr. = = m Roun an writ th answr. a is aout 38.4 m. Th missing si is not opposit th known angl. Th angl opposit th rquir si an work out using th sum of angls in a triangl. a S 48.5 E = = 41.5 Us th tangnt ratio. opposit tan E = ajant tan 41.5 = Multiply to gt. = 908 tan 41.5 Us your alulator, kping xat numr. = = mm Roun an writ th answr. is aout 803 mm. E 908 mm H F Exris 2.3 Th tangnt ratio itional 1 Nam th opposit, ajant an hypotnus for ah of th iniat angls in th iagrams low. a K f xris 2.3 g L M R v w Q x TRIGONOMETRY 55 HTER 2

20 2 Fin th tangnt of ah iniat angl in th following triangls. a 28 mm M 16.8 m 8.3 m 17 mm 33 mm N 14.6 m J M X 0.75 km 32 m 0.63 km 27 m R 0.41 km D 18 m M 98 mm L 285 mm 268 mm R 3 Fin th following valus orrt to 4 imal plas. a tan 54 tan 46.7 tan 21.8 tan 45 tan Fin th angl, orrt to 1 imal pla, for whih: a tan = tan V = 2.67 tan G = tan K = 1 tan M = Fin th angls in th following triangls orrt to 1 imal pla. a M K L S 46 m N 58 m 6 m R D 847 mm X 509 mm J 122 m 88 m V 5 m 5.3 m T V 4.8 m W 6 Fin th mark si in ah of th following triangls. a a m 8310 mm 56.8 m 646 mm km NEW QMTHS 11

21 Molling an prolm solving 7 stairas has stps that ar 265 mm wi an risrs that ar 165 mm high. It is to rpla y a ramp. a What will th angl of th ramp? Th maximum prmitt angl is 7. ommnt. Stp Risr 8 roof truss has a hight of 1.6 m an an ovrall with of 12.4 m. Th highst point of th truss is 1 m from th ntr. What is th slop of ah si? 9 survyor s assistant has a 3.5 m-high stik on th othr si of a stram. Th survyor fins that h has to turn his thoolit up an angl of 6 to s th top of th stik. Th thoolit is 1 m off th groun. How wi is th stram? 1.6 m 12.4 m 6 10 rig slops up from ah n at an angl of 6.4 an has a flat stion 20 m long in th mil. If th rivr it rosss is 860 m wi an th ns ar 15 m aov th watr, how high is th mil stion aov th watr? 20 m m m 11 Whn it is staning, th lgs of an ironing oar mak an angl of 42.5 with th floor. Th ross lgs just fit unr th oar, whih is 1.2 m long. How high is th ironing oar? 2.4 Th sin ratio Th ratio of th opposit an ajant is th sam for similar right-angl triangls. This is also tru for th ratio of th opposit an hypotnus in similar right-angl triangls. Evry right-angl triangl with a 28 ornr angl will hav th sam ratio of opposit to hypotnus (aout 0.47).! Sin ratio In any right-angl triangl, th sin ratio of an angl is givn y opposit si Sin of th angl = hypotnus W usually arviat th sin ratio as sin an writ opposit si a sin = = -- hypotnus a TRIGONOMETRY 57 HTER 2

22 Exampl 17 Fin th valu of sin Q for this triangl. Q Sin uss opposit an hypotnus. opposit sin Q = hypotnus Th opposit si is 5.2 an th hypotnus 5.2 is 6.1. = = Roun an writ th answr. sin Q is approximatly R s with th tangnt ratio, w an us a alulator to fin th sin ratio of an angl. gain, mak sur th alulator is st in gr mo. Exampl 18 Us a alulator to fin sin 54 orrt to 4 imal plas. Us th sin ky. sin 54 = Roun an writ th answr. sin 54 is aout Not: On som olr alulators you may n to ntr th numr an thn prss sin. Exampl 19 Us a alulator to fin orrt to 1 imal pla, if sin = Us sin 1. 2nF sin 0.66 = Roun an writ th answr. is aout Not: Som alulators us th INV, 2n or SHIFT ky insta of 2nF, an on som olr alulators th numr must ntr for th kys ar prss. Exampl 20 Fin th unknown angls in this triangl. 9.5 E F 12.6 G 58 NEW QMTHS 11

23 onsir angl G so that th sis ar opposit an hypotnus. Opposit is 9.5 an hypotnus is sin G = Kp th xat numr on your alulator. = Us sin 1 to fin G, thn roun. So G 48.9 Us sum of angls in a triangl. Thus F = 41.1 Writ th answrs. Th angls ar aout 48.9 an = opposit hypotnus Th masurmnt of many nginring, survying an onstrution projts is oftn inomplt. Th sin ratio an us to fin th missing masurmnts. In a right-angl triangl, on si an on angl must known. Exampl 21 Fin th unknown angl an sis in th following triangl. Q 18.5 m p 63.4 R q Us th angl sum of a triangl Q + 90 = 180 Q = 26.6 opposit To fin p, us th angl opposit th si. sin = hypotnus p sin 63.4 = Multiply to gt p. p = 18.5 sin 63.4 Us your alulator, kping auray. = = Roun to auray of qustion m opposit To fin q, us th angl opposit th si. sin Q = hypotnus q sin 26.6 = Multiply to gt q. q = 18.5 sin 26.6 Us your alulator, kping auray. = = Roun to 3 figurs, th auray of th 8.28 m qustion. Writ th answrs. Th unknown angl Q is 26.6, an th unknown sis ar aout 16.5 m an 8.28 m. TRIGONOMETRY 59 HTER 2

24 Th sin ratio an us to alulat th hypotnus from an angl an th opposit si. Exampl 22 Fin th unknown sis in th following triangl. M 35.6 Z 4.28 m U MU is th opposit an UZ th hypotnus, opposit sin Z = so us sin. hypotnus 4.28 sin 35.6 = UZ Multiply y UZ. UZ sin 35.6 = Now ivi y sin 35.6 to gt UZ. UZ = sin Us your alulator, kping auray. = You may want to us th mmory. = Roun, ut kp auray on alulator m Fin th othr angl. U = 180 U = 54.4 opposit Us sin again. sin U = hypotnus MZ sin 54.4 = UZ Multiply to gt MZ. MZ = UZ sin 54.4 Us th aurat numr kpt on your alulator. = = Roun to 3 figurs, th qustion auray m Writ th answrs. Th unknown sis ar UZ 7.35 m an MZ 5.98 m. Not: ythagoras s Thorm or tangnt oul us to work out th thir si insta of using th sin ratio again. Exampl 23 train travls 2 km along a sloping trak. Th trak is at an angl of 8. Through what hight os th train ris? Draw a iagram, put in th information an mark th si w want with a lttr m h 8 60 NEW QMTHS 11

25 Opposit an hypotnus ar involv, so us sin. h sin 8 = Multiply to gt h. h = 2000 sin 8 = = Roun to auray of qustion. 300 m Writ th answr. Th train riss approximatly 300 m. Exris 2.4 Th sin ratio itional 1 Fin th sin ratios of th angls iniat in th following triangls. a 9.8 K 6.1 L L I N Us ythagoras s Thorm to fin th thir si an thn fin th sin ratio of th angl iniat in ah of th following triangls. a F G Y D M 5.2 R 4.2 X X Q xris 2.4 H F E 3 Us a alulator to fin th following orrt to 4 imal plas. a sin 63 sin 15 sin 87 sin 17 sin 45 f sin 31.8 g sin 22.1 h sin 30.7 i sin 59.5 j sin Fin th angl (orrt to 1 imal pla) for whih: a sin Q = 0.58 sin H = 0.95 sin J = 0.06 sin Y = 0.36 sin D = 0.83 Z TRIGONOMETRY 61 HTER 2

26 5 Fin th unknown angls in th following triangls. a 140 J K L F M Q H T R Fin th unknown angl an sis in ah of th following triangls. a M R I 15.3 m 62 T D mm 62.5 m 41 J R G 1960 m mm 22.6 V 7 Fin th unknown sis in ah of th following triangls. a J L 8.63 m M V 4.5 m V K m U 17.4 R 19 m O V 21 L 28 mm Molling or prolm solving 8 slot-ar nthusiast has 7 straight stions of trak to mak a rig with. Sh plans to us 3 to slop up an 3 to slop ak own, with on flat in th mil. Th rig an hav a maximum slop of 15, an th trak stions ar ah 30 m long. What is th maximum hight of th rig? 9 luminium planks 6 m long ar us on a uiling sit. Th maximum saf slop to run whlarrows of onrt up to th son floor is 12. a What is th hight ris for on plank? What is th horizontal istan ovr? How many planks must us to go up 2.4 m? 62 NEW QMTHS 11

27 10 wir 30 m long is us to hol a raio mast in pla. Th wir is at an angl of 72 to th groun an is ti 2 m from th top of th mast. How high is th raio mast? 11 Th mast of a yaht is 5.4 m from th ow. taut lin from th top of th mast to th ow maks an angl of 74.8 with th k. How long is th lin an how high is th mast? jt lims at an angl of 26 to th groun. Th ruising hight for th jt is 8000 m, an th liming sp is 300 km/h. a How far os th jt travl (sloping istan) to rah its ruising hight? How far os it travl in 1 minut? How long os it tak to rah ruising hight? What horizontal istan os it ovr whil liming? 13 t a sout rally a flying fox is st up to travl twn a trhous an a rais platform. Th platform is 2 m off th groun, an th trhous is 6.5 m from th groun. To work proprly th rop must at an angl of at last 8. What ar th maximum lngth of th rop an th maximum istan of th platform from th trhous? 2.5 Th osin ratio Th osin ratio is an altrnativ to th sin ratio. It is similar, xpt that th ajant si is us insta of th opposit.! osin ratio In any right-angl triangl, th osin ratio of an angl is givn y ajant si osin of th angl = hypotnus W usually arviat th osin ratio as os an writ ajant os = = -- hypotnus a TRIGONOMETRY 63 HTER 2

28 Exampl 24 Fin th valu of os H from th triangl shown. M 62 S os uss ajant an hypotnus. ajant os H = hypotnus Th ajant si is 42 an th hypotnus is = = 0.56 Writ th answr. os H is H Exampl 25 Us a alulator to fin os 46.9, orrt to 4 imal plas. Mak sur th alulator is in gr mo. Us th os ky. os 46.9 = Roun an writ th answr. os 46.9 is aout Not: On som olr alulators you may n to ntr th numr an thn prss os. Exampl 26 If os G = , what is G? Us os 1. 2nF os = Roun an writ th answr. G is aout Not: Som alulators us th INV, 2n or SHIFT ky insta of 2nF, an on som olr alulators th numr must ntr for th kys ar prss. Exampl 27 Us osin to fin th angls in this triangl. V J 64 NEW QMTHS 11

29 onsir V so that th sis ar th ajant an hypotnus: ajant = 13.5, hypotnus = Kp th xat valu on your alulator. ajant 13.5 os V = = hypotnus 16.9 = Us os 1 to fin V an roun. So V 37.0 Us sum of angls in a triangl to fin J. Thus J = 53.0 Writ th answrs, showing orrt auray. Th angls ar aout 37.0 an osin an us to fin missing sis, in a similar way to sin. osin is most usful whn th ajant si is to work out from th hypotnus or vi vrsa. Exampl 28 Fin th mark si in th following triangl mm jant an hypotnus ar involv, so ajant x os 65 = = us osin. hypotnus 2895 Multiply to gt x. x = 2895 os 65 Us your alulator, kping auray. = = Roun to auray of qustion mm Writ th answr. Th mark si is aout 1223 mm long. x 65 Exampl 29 Fin th hypotnus in th following triangl. y m jant is known an hypotnus is ajant 29.6 os 34.6 = = rquir, so us osin. hypotnus y Multiply y y. y os 34.6 = 29.6 Now ivi y os 34.6 to gt y y = os 34.6 Us your alulator, kping auray = You may want to us th mmory. = Roun to auray of qustion m Writ th answr. Th hypotnus is aout 36.0 m long. Not: W oul us sin insta of osin y fining th thir angl first. TRIGONOMETRY 65 HTER 2

30 Exampl 30 Th slop of a suuran roa up a stp hill is 28. Th survy map shows that th roa ovrs a horizontal istan of 807 m to rah th top of th hill. What is th atual lngth of roa surfa that must us for alulations of guttring an saling rquirmnts? Draw a iagram, put in th information an mark th want si with a lttr. r 807 m jant an hypotnus ar involv, 807 os 28 = so us osin. r Multiply y r. r os 28 = 807 Now ivi y os 28 to gt r. 807 r = os 28 Us your alulator, kping auray. 807 = You may want to us th mmory. = Roun to auray of qustion. 914 m Writ th answr. Th atual lngth of roa surfa is aout 914 m. 28 itional xris 2.5 Exris 2.5 Th osin ratio 1 Fin th osins of th angls iniat in th triangls low. a D R N 54 W 16 M I V S T T S 2 Us your alulator to fin th following orrt to 4 imal plas. a os 58 os 27.5 os 72.9 os 47.2 os Fin th angl (orrt to 1 imal pla) for whih: a os R = os Y = os = 0.4 os T = os = NEW QMTHS 11

31 4 Us osin to fin th angls in th following triangls. a Z L M 23.5 Q M K F R Y T L Fin th mark sis in th following triangls. a a m 28.5 m m 8.15 km m Fin th hypotnus in ah of th following. a a mm 86.2 m 42.8 m mm m 7 Fin th mark si in ah of th following triangls. a a m 6134 mm km 2096 mm m 71 Molling an prolm solving 8 arpntr ns to ut a right-angl wg out of timr. Th wg has to 25 m long, an th sloping g must 28 m. t what angl os th timr n to ut? TRIGONOMETRY 67 HTER 2

32 9 whlhair ramp up to an ntran has a horizontal lngth of 7555 mm an a sloping lngth of 7580 mm. Th maximum angl allow for suh ramps y ustralian Stanars is mm a What is th angl of th ramp? Dos th ramp mt th stanar? 7555 mm 10 long slop for a train lin is inlin at an angl of 8.4. To lay th trak for th slop, 248 stions of 20 m trak wr us. What horizontal istan is travll y th train in going up th slop? 11 tnt rop is ti to a pg that is 4825 mm away from th tnt pol. Whn strth tight from th pg to th top of th pol it maks an angl of 27.3 with th groun. How long is th rop? 12 fir hos is ti to th top of a pol to ry. It is pull 12.5 m away from th pol so that non of it lis on th groun, an it maks an angl of 65.4 with th groun. How long is th fir hos? 2.6 Using trigonomtri ratios In solving trigonomtri prolms, you must rmmr th thr ratios. On of th mnmonis that popl us to hlp to rmmr th ratios is shown low.! Signals of hlp sin opposit = hypotnus aus all hans os ajant = hypotnus to offr assistan tan opposit = ajant To solv a triangl, w fin th unknown sis an angls. Solving a right-angl triangl pns on hoosing th orrt trigonomtri ratio. Th orrt ratio is hosn y onsiring th sis involv as hypotnus, opposit an ajant to th known or rquir angl. In ass whr two sis ar known, ythagoras s Thorm may us to fin th thir. 68 NEW QMTHS 11

33 Exampl 31 Solv this triangl m K p 67.2 Q k For angl, p is opposit an q is ajant, opposit tan = so us tangnt. ajant ut in valus. p tan 67.2 = Multiply y 28.6 to fin p. p = 28.6 tan 67.2 = = Roun to auray of qustion m For angl, k is hypotnus an q is ajant, ajant os = so us osin. hypotnus ut in valus os 67.2 = k Multiply y k. k os 67.2 = Divi y os 67.2 to fin k. k = os = = Roun to auray of qustion m Us sum of angls in a triangl to fin Q. Q = = 22.8 Writ th answrs. p 68.0 m, k 73.8 m an Q = Exampl 32 Fin th unknown si an angls for this triangl. M 28.3 m 12.8 m For angl M, z is ajant an is hypotnus, so us osin. Kp xat valu on your alulator. Us os 1 to fin M. ajant os M = hypotnus 12.8 = = M = Roun, ut kp auray on alulator Us sum of angls to fin Z. Z = M 26.9 Z m TRIGONOMETRY 69 HTER 2

34 For angl M, m is opposit an is hypotnus, opposit sin M = so us sin. hypotnus m = Multiply y 28.3 to fin m. m = 28.3 sin M Us xat valu for M from alulator. = 28.3 sin = = Roun m Writ th answrs to th auray of th qustion. M 63.1, Z 26.9 an m 25.2 m. Not: You oul fin m using ythagoras s Thorm insta of sin. Exampl 33 In FGH, F = 90, G = 33.9 an g = 77.1 m. Solv FGH. Draw a iagram. H 77.1 m f F 33.9 h G Us sum of angls to fin H. H = = 56.1 opposit tan H = ajant Us angl H to fin h, so th unknown is opposit. Thn h is opposit an g is ajant, so us tangnt. ut in valus. h tan 56.1 = Multiply y 77.1 to fin h. h = 77.1 tan 56.1 = = Roun, ut kp auray on alulator. 115 m Us ythagoras to fin f. f 2 = h 2 + g 2 Sustitut valus, using xat valu of h on alulator. = ( ) = = Us th ky. f = = Roun to 3 figur auray. 138 m Writ answrs to th auray of th qustion. H = 56.1, h 115 m an f 138 m. In survying an mapping, angls may n to spifi vry xatly. This has traitionally n on using th smallr units of minuts an sons of ar. 70 NEW QMTHS 11

35 ! ngl units Eah gr is ivi into 60 minuts of ar. Eah minut of ar is ivi into 60 sons of ar. Th symols us for grs, minuts an sons ar, an rsptivly. 1 = 60 (1 gr = 60 minuts) 1 = 60 (1 minut = 60 sons) Morn alulators work out th trigonomtri ratios in imal grs. for w an us a alulator for sin, osin or tangnt, w must hang grs, minuts an sons to grs only. Exampl 34 hang to grs, orrt to 4 imal plas. 24 hang 24 to minuts y iviing y = = Writ th angl = hang 18.4 to grs y iviing y = = Roun an writ th angl On most sintifi an graphis alulators, 10 DMS 18 DMS 24 DMS th DMS or ky an us Not: On som alulators, you will n to us th 2nF ky to hang to grs only. Exampl 35 hang to grs, minuts an sons, orrt to th narst son. Multiply th imal part y 60 to hang to minuts = = Writ th angl = Multiply th imal part y 60 to hang to sons = = 2.64 Roun an writ th answr On most sintifi an graphis alulators, INV DMS th DMS or ky an us in rvrs Not: On som alulators, th DMS utton is us to hang to grs, minuts an sons, so th INV or 2nF ky is not n. ngls ar sri in a varity of ways. In orr to apply trigonomtry fftivly, you must familiar with th ommon mthos. TRIGONOMETRY 71 HTER 2

36 ! n angl of lvation is form whn you hav to look up from a horizontal lin to s a point. Th angl you hav to look up is th angl of lvation. Lin of sight ngl of lvation Horizontal lin oint n angl of prssion is form whn you hav to look own from a horizontal lin to s a point. Th angl you hav to look own is th angl of prssion. Horizontal lin ngl of prssion Lin of sight oint Exampl 36 From th top of a liff 20 m high, th angl of prssion of a surf-oat is How far is th oat from th foot of th liff? Draw a iagram, lal points an show th information T m Fin an angl in th triangl. FT = = 78.5 Us tangnt as opposit an ajant ar opposit tan 78.5 = involv. ajant ut in valus. x tan 78.5 = Multiply y 20 to fin x. x = 20 tan 78.5 = = Roun to th auray of th qustion. 98 m Writ th answr. Th oat is aout 98 m from th liff. x F Exampl 37 plan flying at a hight of 1500 m is spott y an osrvr. Sh nots that th plan has an angl of lvation of How far away (on th groun) is th plan? hang th angl to grs = 5.8 Draw a iagram, lal points an show th information. O 5.8 x G 1500 m 72 NEW QMTHS 11

37 Us tangnt as opposit an ajant ar involv. tan 5.8 = opposit ajant ut in valus tan 5.8 = x Multiply y x. x tan 5.8 = 1500 Divi y tan 5.8 to fin x x = tan = = Roun to auray of qustion m or 15 km Writ th answr. Th plan is aout 15 km away. Dirtions on th Earth ar givn in trms of th asi ompass points north, south, ast an wst.! Th aring of an ojt is givn as th 3-igit lokwis angl th osrvr turns from u north in orr to fa th ojt. n ojt u wst has a aring of 270. aring an also givn as th angl from north or south towars ithr ast or wst. Th irtion shown in th iagram an givn as ithr S37 W or 217. N N W 37 E 217 S Exampl 38 oat sails at a aring of 107 for 200 km. How far south an ast os it travl (to th narst kilomtr)? Draw a iagram, lal points an show th information. N 107 S 73 y 200 km x Fin an angl in th triangl. SF = = 73 For x, opposit an hypotnus ar involv. opposit x sin 73 = = hypotnus 200 F TRIGONOMETRY 73 HTER 2

38 Multiply y 200 to fin x. x = 200 sin 73 = = Roun. 191 km ajant y For y, ajant an hypotnus ar involv. os 73 = = hypotnus 200 Multiply y 200 to fin y. y = 200 os 73 = = Roun. 58 km Writ th answrs. Th oat travls aout 58 km south an 191 km ast. itional xris 2.6 Exris 2.6 Using trigonomtri ratios 1 Solv ths triangls. a K M K m 205 m mm V L m V V m N Fin th unknown sis an angls in ths triangls. a M D R mm 145 mm Q S V m D 8.6 mm O L m 0.96 m D 1.35 m 70.6 m 3 Solv ths triangls. a, = 90, = 28, = 8 m DG, D = 90, p = 5 m, g = 7 m KLM, M = 90, L = 34, l = 6 mm RMS, S = 90, R = 57.2, m = 5.3 km EFG, E = 90, = 76 m, f = 57 m 4 hang th following to imal grs, orrt to 4 imal plas. a hang th following to grs, minuts an sons, orrt to th narst son. a R 74 NEW QMTHS 11

39 NQM11 h02.fm ag 75 Friay, ugust 31, :21 M Molling an prolm solving 6 Th pilot of a plan osrvs that th angl of prssion of th ontrol towr at th airport is 3.7. If th altimtr shows th plan is at a hight of 2400 m, fin th horizontal istan to th ontrol towr. 7 From th othr si of th strt, th angl of lvation of th top of a uiling is 72. If th strt is 35 m wi, fin th hight of th uiling. 8 From th top of a lighthous 47 m high, th angl of prssion of a oat at sa is t sa-lvl, how far from th lighthous is th oat? 9 Sharon knows that sh taks 12 pas vry 10 m. Sh pas out a istan of 60 pas from th foot of a tr an looks up at th top. Its angl of lvation is thn 38. Sharon s ys ar 1.5 m from th groun. What is th hight of th tr? 10 Th hight of a mountain is shown on th map as 1700 m. From th top of a hill 370 m high, th angl of lvation of th top of th mountain is Fin th horizontal istan twn th hill an th mountain. 11 ship travls 150 km on a aring of 134. How far south an ast os it travl? 12 ftr travlling for 2 hours in a irtion S36 E an aroplan is 600 km south of its starting point. Fin th atual istan travll an its sp. 13 train is travlling at 60 km/h on a trak running wst to ast. n osrvr south of th trak nots that th train is irtly north of him. Half an hour latr th train is at a aring of 020. What is th istan of th osrvr south of th trak? 14 ylist travlling on a roa running north south notis that a hill is irtly wst of hr. ftr yling at 20 km/h for a furthr 2 hours, sh notis that th irtion of th hill is now S75 W. How far is th hill from th roa, an how far is it from th nw position of th ylist? 15 On town is irtly north of anothr. From a point 20 km to th wst of th roa joining thm, th towns ar at arings of 050 an 120. How far apart ar th towns? TRIGONOMETRY 75 haptr 2 summary HTER 2

40 haptr Rviw Ex 2.1 Ex 2.1 Ex 2.2 Ex 2.2 Ex 2.3 Ex 2.3 Ex 2.4 Ex 2.5 Ex 2.6 Ex 2.6 Ex 2.1 ommuniation an justifiation 1 triangl is nam GHK. Skth th triangl an show th usual nams of th sis. 2 Th triangl QR is suh that Q = 90. Stat ythagoras s Thorm for this triangl. 3 How ar similar triangls rlat to ah othr? 4 What is mant y shaow rkoning? 5 In th triangl, th opposit an ajant for angl θ ar rsptivly: a an a an an a D an a E an a 6 tan θ = opposit hypotnus ajant opposit D E ajant opposit hypotnus hypotnus 7 sin θ = opposit hypotnus ajant opposit D E ajant opposit hypotnus hypotnus 8 os θ = opposit hypotnus ajant opposit D E ajant opposit hypotnus hypotnus 9 Stat what is mant y an angl of prssion. 10 How an irtions on th Earth stat? Knowlg an prours 11 Us ythagoras s Thorm to fin th missing sis in th following triangls. a 12.6 m θ 2 km ajant opposit ajant opposit ajant opposit m 18.4 m 7.9 m 6.3 m 8 km m 8000 mm mm Ex Us ythagoras s Thorm to work out whthr triangls with th following imnsions ar aut, right-angl or otus. a 3, 5, 7 8, 10, 12 8, 15, 17 13, 15, NEW QMTHS 11

41 13 shaow stik 2.5 m high asts a shaow of lngth 4 m. t th sam tim a uiling asts a shaow of lngth 30 m. How high is th uiling? 14 Nam th opposit, ajant an hypotnus for th angl mark in ah of th following triangls. a L R Ex 2.2 Ex 2.3 K M D Q S 15 Us th iagrams low to fin: a sin Q sin V sin sin os Q f os R g os X h os i tan Q j tan 9 m Q 12 m 15 m R V 8.1 m W 10.3 m 13.1 m X 6000 mm 6500 mm 2500 mm Ex Us your alulator to fin, orrt to 4 imal plas: a sin 47 os 3.8 tan 85 tan 45 os 72 f sin 54.9 g os 19.6 h sin 45 i os j tan Fin th valu (orrt to 1 imal pla) of th angl for whih: a sin T = 0.55 os R = tan E = 3.5 tan K = 0.34 os = f sin W = Fin th unknown angls in th following triangls. a 6.3 m D F 7.9 m 8.2 m K N f 82 m L 45 m E 5105 m 4.5 m M O 6214 m Q J H 65 m 2800 mm 4600 mm R I 48 m S Ex Ex Ex TRIGONOMETRY 77 HTER 2

42 Ex Ex 2.1 Ex 2.1 Ex 2.6 Ex 2.6 Ex 2.6 Ex Solv th following triangls. a 28 m L 16 m K 30 m R M V f 14.2 m 8.9 m 22 T U M E I R g h Q i F mm G 4.6 mm L Molling an prolm solving 20 Fin th lngth of stl raing n to fit a iagonal ra to a wall fram 6.2 m long an 2.4 m high. 21 triangular ours for a yaht ra has a north lg an an ast lg, an th final lg rturns to th start. Th final lg is 7 km long an th first lg is 4.5 km long. How long is th son lg? 22 From a point 100 m from th as of a liff th angl of lvation of th top of th liff is 16. Th angl of lvation of th top of a lighthous uilt on th liff is 30. Fin th hight of th liff an th hight of th lighthous. 23 aint-makr has splay th lgs of a tal out at an angl of 15 to th vrtial. If th hight of th tal surfa is 850 mm an th tal-top is 45 mm thik, fin th lngth of th lgs. 24 kit lin is lt out 70 m whil th kit riss 45 m. If th lin is tight, fin th angl of lvation of th kit from th oy holing th string. 25 n airraft is 370 km north of its original position. Its aring is 342 from th original position. Fin th istan atually travll mm 17.5 mm K G 19.6 mm 23 m km J H S 78 NEW QMTHS 11