The unit circle and radian measure

Transcription

1 hter 8 The unit irle nd rdin mesure Sllus referene:.,.,. ontents: Rdin mesure r length nd setor re The unit irle nd the trigonometri rtios D litions of the unit irle E Multiles of ¼ 6 nd ¼ F The eqution of stright line

2 90 THE UNIT IRLE ND RDIN MESURE (hter 8) OPENING PROLEM onsider n equilterl tringle with sides m long. isets side [] nd the vertil ngle. ltitude [N] Things to think out: n ou use this figure to elin wh sin 0 ± =? Use our lultor to find the vlues of sin 0 ±, sin 50 ±, sin 90 ±, sin 0 ±, nd sin( 0 ± ). n ou elin our results even though the ngles re not etween 0 ± nd 90 ±? 0 0 m N m RDIN MESURE DEGREE MESUREMENT OF NGLES We hve seen reviousl tht one full revolution mkes n ngle of 60 ±, nd the ngle on stright line is 80 ±. Hene, one degree, ±, n e defined s 60th of one full revolution. This mesure of ngle is ommonl used surveors nd rhitets. For greter ur we define one minute, 0,s 60 th of one degree nd one seond, 00,s 60th of one minute. Oviousl minute nd seond re ver smll ngles. Most grhis lultors n onvert frtions of ngles mesured in degrees into minutes nd seonds. This is lso useful for onverting frtions of hours into minutes nd seonds for time mesurement, s one minute is 60th of one hour, nd one seond is 60th of one minute. RDIN MESUREMENT OF NGLES rdius=r r length =r n ngle is sid to hve mesure of rdin ( ) if it is sutended t the entre of irle n r equl in length to the rdius. The smol is used for rdin mesure ut is usull omitted. ontrst, the degree smol is lws used when the mesure of n ngle is given in degrees. 60 From the digrm to the right, it n e seen tht is slightl smller thn 60 ±. In ft, ¼ 57: ±. The word rdin is n revition of rdil ngle. DEGREE-RDIN ONVERSIONS r GRPHIS LULTOR INSTRUTIONS r If the rdius of irle is r, then n r of length ¼r, or hlf the irumferene, will sutend n ngle of ¼ rdins. 80 or ¼ Therefore, ¼ rdins 80 ±.

3 THE UNIT IRLE ND RDIN MESURE (hter 8) 9 So, = 80 ± ¼ ¼ 57: ± nd ± = ¼ 80 ¼ 0:075. To onvert from degrees to rdins, we multil ¼ 80. To onvert from rdins to degrees, we multil 80 ¼. ¼ 80 We indite degrees with smll. To indite rdins we use smll or else use no smol t ll. Degrees Rdins 80 ¼ Emle onvert 5 ± to rdins in terms of ¼. 5 ± = (5 ¼ 80 ) rdins or 80 ± = ¼ rdins = ¼ rdins ) 80 ± = ¼ rdins ) 5 ± = ¼ rdins ngles in rdins m e eressed either in terms of ¼ or s deimls. Emle onvert 6:5 ± to rdins. 6:5 ± = (6:5 ¼ 80 ) rdins ¼ : rdins EXERISE 8 onvert to rdins, in terms of ¼: 90 ± 60 ± 0 ± d 8 ± e 9 ± f 5 ± g 5 ± h 70 ± i 60 ± j 70 ± k 5 ± l 50 ± m 6 ± n 80 ± o 0 ± onvert to rdins, orret to signifint figures: 6:7 ± 7: ± 7:9 ± d 9:6 ± e 96:7 ± Emle onvert 5¼ 6 to degrees. 5¼ 6 = 5¼ 6 80 ¼ = 50 ± ±

4 9 THE UNIT IRLE ND RDIN MESURE (hter 8) onvert the following rdin mesures to degrees: ¼ 5 ¼ 5 ¼ d ¼ 8 e ¼ 9 f 7¼ ¼ 9 g 0 h ¼ ¼ 0 i 6 j 8 7¼ Emle onvert 0:68 rdins to degrees. 0:68 rdins =(0:68 80 ¼ )± ¼ 6:6 ± onvert the following rdins to degrees. Give our nswers orret to deiml les. :5 0:867 d :79 e 5:67 5 o nd omlete, giving nswers in terms of ¼: Degrees Rdins Degrees Rdins THEORY OF KNOWLEDGE There re severl theories for wh one omlete turn ws divided into 60 degrees: ² 60 is roimtel the numer of ds in er. ² The lonins used ounting sstem in se 60. If the drew 6 equilterl tringles within irle s shown, nd divided eh ngle into 60 sudivisions, then there were 60 sudivisions in one turn. The division of n hour into 60 minutes, nd minute into 60 seonds, is from this se 60 ounting sstem. ² 60 hs divisors, inluding ever integer from to 0 eet The ide of mesuring n ngle the length of n r dtes to round 00 nd the Persin mthemtiin l-kshi. The onet of rdin is generll redited to Roger otes, however, who desried it s we know it tod. Wht other mesures of ngle re there? Whih is the most nturl unit of ngle mesure?

5 THE UNIT IRLE ND RDIN MESURE (hter 8) 9 R LENGTH ND SETOR RE You should e fmilir with these terms relting to the rts of irle: setor r (rt of irle) entre n r, setor, or segment is desried s: ² minor if it involves less thn hlf the irle ² mjor if it involves more thn hlf the irle. hord segment rdius For emle: minor segment minor r (lk) mjor segment mjor r (red) R LENGTH l In the digrm, the r length is l. ngle µ is mesured in rdins. We use rtio to otin: O r r length irumferene = µ ¼ ) l ¼r = µ ¼ ) l = µr For µ in rdins, r length l = µr. For µ in degrees, r length l = µ 60 ¼r. RE OF SETOR X r O r Y In the digrm, the re of minor setor XOY is shded. µ is mesured in rdins. We use rtio to otin: re of setor re of irle = µ ¼ ) ¼r = µ ¼ ) = µr For µ in rdins, re of setor = µr. For µ in degrees, re of setor = µ 60 ¼r.

6 9 THE UNIT IRLE ND RDIN MESURE (hter 8) Emle 5 setor hs rdius m nd ngle rdins. Find its: r length re r length = µr = =6m re = µr = = 6 m EXERISE 8 Use rdins to find the r length nd re of setor of irle of: rdius 9 m nd ngle 7¼ rdius :9 m nd ngle :67 rdins. setor hs n ngle of 07:9 ± nd n r length of 5:9 m. Find its: rdius re. setor hs n ngle of :9 rdins nd n re of 0:8 m. Find its: rdius erimeter. Emle 6 Find the re of setor with rdius 8: m nd r length : m. l = µr fµ in rdinsg ) µ = l r = : 8: ) re = µr = : 8: 8: ¼ 5:5 m Find, in rdins, the ngle of setor of: rdius : m nd r length :95 m rdius 0 m nd re 0 m. 5 Find µ (in rdins) for eh of the following, nd hene find the re of eh figure:. 7 m 6 m 8 m 5 m 8. m 8 m 6 Find the r length nd re of setor of rdius 5 m nd ngle rdins. 7 If setor hs rdius m nd r length m, show tht its re is m.

7 8 The one is mde from this setor: 0 m s m THE UNIT IRLE ND RDIN MESURE (hter 8) 95 m r m Find orret to signifint figures: the slnt length s m the vlue of r the r length of the setor d the setor ngle µ in rdins. 9 The end wll of uilding hs the she illustrted, where the entre of r is t. Find: 0 m to signifint figures 5 m µ to signifint figures the re of the wll. 0 T [T] is tngent to the given irle. O =m nd the irle hs rdius 5 m. Find the erimeter of the shded region. O nutil mile (nmi) is the distne on the Erth s surfe tht sutends n ngle of minute (or 60 of degree) of the Gret irle r mesured from the entre of the Erth. knot is seed of nutil mile er hour. Given tht the rdius of the Erth is 670 km, show tht nmi is roimtel equl to :85 km. lulte how long it would tke lne to fl from Perth to delide ( distne of 0 km) if the lne n fl t 80 knots. nutil mile ( nmi) N S ' P Q fene shee is tethered to ost whih is 6 m from long fene. The length of roe is 9 m. Find the re whih 6 m the shee n feed on. ost shee

8 96 THE UNIT IRLE ND RDIN MESURE (hter 8) THE UNIT IRLE ND THE TRIGONOMETRI RTIOS The unit irle is the irle with entre (0, 0) nd rdius unit. - - IRLES WITH ENTRE (0, 0) r (0, 0) P (, ) onsider irle with entre (0, 0) nd rdius r units. Suose P(, ) is n oint on this irle. Sine OP = r, ( 0) +( 0) = r ) + = r fdistne formulg + = r is the eqution of irle with entre (0, 0) nd rdius r. The eqution of the unit irle is + =. NGLE MESUREMENT Suose P lies nwhere on the unit irle nd is (, 0). Let µ e the ngle mesured from [O] on the ositive -is. µ is ositive for ntilokwise rottions nd negtive for lokwise rottions. P Positive diretion For emle, µ = 0 ± nd Á = 50 ±. Negtive diretion 0 You n elore ngle mesurement further liking on the ion. DYNMI NGLES

9 DEFINITION OF SINE ND OSINE ² os µ is the -oordinte of P ² sin µ is the -oordinte of P THE UNIT IRLE ND RDIN MESURE (hter 8) 97 onsider oint P(, ) whih lies on the unit irle in the first qudrnt. [OP] mkes n ngle µ with the -is s shown. Using right ngled tringle trigonometr: os µ = DJ HYP = = sin µ = OPP HYP = = tn µ = OPP DJ = = sin µ os µ In generl, for oint P nwhere on the unit irle: P( os, sin) We n hene find the oordintes of n oint on the unit irle with given ngle µ mesured from the ositive -is. For emle: ( os75,sin75) ( os65,sin65) ( os55,sin55) Sine the unit irle hs eqution + =, (os µ) + (sin µ) = for ll µ. We ommonl write this s os µ + sin µ =. For ll oints on the unit irle, 6 6 nd So, 6 os µ 6 nd 6 sin µ 6 for ll µ P (, ) - P( os, sin) ( os7, sin7) or ( os( -), sin( -)) DEFINITION OF TNGENT Suose we etend [OP] to meet the tngent t (, 0). The intersetion etween these lines ours t Q, nd s P moves so does Q. The osition of Q reltive to is defined s the tngent funtion. Notie tht s ONP nd OQ re equingulr nd therefore similr. Q onsequentl O = NP ON nd hene Q = sin µ os µ. Under the definition tht Q = tn µ, tn µ = sin µ os µ. - P sin N os Q (, tn) tn ( 0, ) - tngent

10 98 THE UNIT IRLE ND RDIN MESURE (hter 8) INVESTIGTION THE TRIGONOMETRI RTIOS In this investigtion we elore the signs of the trigonometri rtios in eh qudrnt of the unit irle. Wht to do: lik on the ion to run the Unit irle softwre. Drg the oint P slowl round the irle. Note the sign of eh trigonometri rtio in eh qudrnt. Qudrnt os µ sin µ tn µ ositive nd rd st th THE UNIT IRLE Hene note down the trigonometri rtios whih re ositive in eh qudrnt. From the Investigtion ou should hve disovered tht: ² sin µ, os µ, nd tn µ re ositive in qudrnt ² onl sin µ is ositive in qudrnt ² onl tn µ is ositive in qudrnt ² onl os µ is ositive in qudrnt. We n use letter to show whih trigonometri rtios re ositive in eh qudrnt. The stnds for ll of the rtios. You might like to rememer them using nd rd S T st th ll Sill Turtles rwl. Emle 7 Use unit irle digrm to find the vlues of os( 70 ± ) nd sin( 70 ± ). (0, ) os( 70 ± )=0 sin( 70 ± )= fthe -oordinteg fthe -oordinteg

11 PERIODIITY OF TRIGONOMETRI RTIOS THE UNIT IRLE ND RDIN MESURE (hter 8) 99 Sine there re ¼ rdins in full revolution, if we dd n integer multile of ¼ to µ (in rdins) then the osition of P on the unit irle is unhnged. For µ in rdins nd k Z, os (µ +k¼) = os µ nd sin (µ +k¼) = sin µ. (, ) µ + ¼ µ tn(µ + ¼) = = = tn µ (-, -) For µ in rdins nd k Z, tn(µ + k¼) = tn µ. This eriodi feture is n imortnt roert of the trigonometri funtions. EXERISE 8 For eh unit irle illustrted: i stte the et oordintes of oints,, nd in terms of sine nd osine ii use our lultor to give the oordintes of,, nd orret to signifint figures With the id of unit irle, omlete the following tle: µ (degrees) 0 ± 90 ± 80 ± 70 ± 60 ± 50 ± µ (rdins) sine osine tngent

12 00 THE UNIT IRLE ND RDIN MESURE (hter 8) Use our lultor to evlute: i ii o nd omlete the following tle. Use our lultor to evlute the trigonometri rtios, then to write them etl. µ (degrees) 0 ± 5 ± 60 ± 5 ± 50 ± 0 ± 5 ± µ (rdins) sine osine tngent Use our lultor to evlute: i sin 00 ± ii sin 80 ± iii sin 0 ± iv sin 60 ± v sin 50 ± vi sin 0 ± vii sin 5 ± viii sin 5 ± Use the results from to o nd omlete: sin(80 ± µ) =:::::: Justif our nswer using the digrm longside: P( os, sin) d Find the otuse ngle with the sme sine s: i 5 ± ii 5 ± ¼ ¼ iii iv 6 5 Use our lultor to evlute: i os 70 ± ii os 0 ± iii os 60 ± iv os 0 ± v os 5 ± vi os 55 ± vii os 80 ± viii os 00 ± Use the results from to o nd omlete: os(80 ± µ) =:::::: Justif our nswer using the digrm longside: P( os, sin) d Find the otuse ngle whih hs the negtive osine of: i 0 ± ii 9 ± ¼ ¼ iii 5 iv 5 6 Without using our lultor, find: sin 7 ± if sin ± ¼ 0:680 sin 59 ± if sin ± ¼ 0:857 os ± if os 7 ± ¼ 0:7986 d os ± if os 56 ± ¼ 0:95 e sin 5 ± if sin 65 ± ¼ 0:906 f os ± if os 8 ± ¼ 0:669 7 o nd omlete: Qudrnt Degree mesure Rdin mesure os µ sin µ tn µ 0 ± <µ<90 ± 0 <µ< ¼ ositive ositive

13 THE UNIT IRLE ND RDIN MESURE (hter 8) 0 In whih qudrnts re the following true? i os µ is ositive. ii os µ is negtive. iii os µ nd sin µ re oth negtive. iv os µ is negtive nd sin µ is ositive. 8 If OP = µ nd OQ = µ lso, wht is the mesure of OQ? o nd omlete: [OQ] is refletion of [OP] in the... nd so Q hs oordintes... Wht trigonometri formule n e dedued from nd? - Q P( os, sin) 9 o nd omlete: µ sin µ sin( µ) os µ os( µ) 0:75 :77 : 6:5 :7 Wht do ou suset is true from for generl ngle µ? The oordintes of P in the figure re (os µ, sin µ). P i finding the oordintes of Q in terms of µ in two different ws, rove tht our susiion in is orret. - - ii Hene elin wh os(¼ µ) = os µ. Q D PPLITIONS OF THE UNIT IRLE The identit os µ + sin µ = is essentil for finding trigonometri rtios. Emle 8 Find etl the ossile vlues of os µ for sin µ =. Illustrte our nswers. os µ + sin µ = ) os µ + = ) os µ = 5 9 ) os µ = We

14 0 THE UNIT IRLE ND RDIN MESURE (hter 8) EXERISE 8D. Find the ossile et vlues of os µ for: sin µ = sin µ = sin µ =0 d sin µ = Find the ossile et vlues of sin µ for: os µ = 5 os µ = os µ = d os µ =0 Emle 9 If sin µ = nd ¼<µ<¼, find os µ nd tn µ. Give et vlues. Now os µ + sin µ = ) os µ = S T -Er ) os µ = 7 6 ) os µ = 7 ut ¼<µ< ¼, so µ is qudrnt ngle ) os µ is negtive. ) os µ = 7 nd tn µ = sin µ os µ = = 7 7 Without using lultor, find: sin µ if os µ = nd 0 <µ< ¼ os µ if sin µ = 5 nd ¼ <µ<¼ os µ if sin µ = 5 nd ¼ <µ<¼ d sin µ if os µ = 5 nd ¼<µ<¼. Emle 0 If tn µ = nd ¼ <µ<¼, find sin µ nd os µ. Give et nswers. tn µ = sin µ os µ = ) sin µ = os µ S T Now sin µ + os µ = ) ( os µ) + os µ = ) os µ + os µ = ) 5 os µ = ) os µ = 5 ut ¼ <µ<¼, so µ is qudrnt ngle. ) os µ is ositive nd sin µ is negtive. ) os µ = 5 nd sin µ = 5.

15 THE UNIT IRLE ND RDIN MESURE (hter 8) 0 If sin = ¼ nd <<¼, find tn etl. If os = ¼ 5 nd <<¼, find tn etl. If sin = nd ¼<< ¼, find tn etl. d If os = ¼ nd <<¼, find tn etl. 5 Find et vlues for sin nd os given tht: tn = nd 0 << ¼ tn = nd ¼ <<¼ tn = 5 nd ¼<< ¼ d tn = 5 nd ¼ <<¼ 6 Suose tn = k where k is onstnt nd ¼<< ¼. Write eressions for sin nd os in terms of k. FINDING NGLES WITH PRTIULR TRIGONOMETRI RTIOS From Eerise 8 ou should hve disovered tht: For µ in rdins: ² sin(¼ µ) = sin µ ² os(¼ µ) = os µ ² os(¼ µ) = os µ We need results suh s these, nd lso the eriodiit of the trigonometri rtios, to find ngles whih hve rtiulr sine, osine, or tngent. Emle Find the two ngles µ on the unit irle, with 0 6 µ 6 ¼, suh tht: os µ = sin µ = tn µ = Using tehnolog, os µ = os ( ) ¼ : Using tehnolog, sin µ = sin ( ) ¼ 0:88 Er Using tehnolog, tn µ = tn () ¼ : ) µ ¼ : or ¼ : ) µ ¼ : or 5:05 Qe - ) µ ¼ 0:88 or ¼ 0:88 ) µ ¼ 0:88 or :9 - ) µ ¼ : or ¼ +: ) µ ¼ : or :5

16 0 THE UNIT IRLE ND RDIN MESURE (hter 8) Emle Find two ngles µ on the unit irle, with 0 6 µ 6 ¼, suh tht: sin µ = 0: os µ = tn µ = sin µ = 0: Using tehnolog, sin ( 0:) ¼ 0: ut 0 6 µ 6 ¼ ) µ ¼ ¼ +0: or ¼ 0: ) µ ¼ :55 or 5:87-0. os µ = Using tehnolog, os ( ) ¼ :0 - -We -.0 ut 0 6 µ 6 ¼ ) µ ¼ :0 or ¼ :0 ) µ ¼ :0 or :98 tn µ = Using tehnolog, tn ( ) ¼ 0: - - ut 0 6 µ 6 ¼ ) µ ¼ ¼ 0: or ¼ 0: ) µ ¼ :8 or 5:96-0. The green rrow shows the ngle tht the lultor gives. EXERISE 8D. Find two ngles µ on the unit irle, with 0 6 µ 6 ¼, suh tht: tn µ = os µ =0:8 sin µ = 5 d os µ =0 e tn µ =: f os µ =0:786 g sin µ = h tn µ =0: i sin µ = 9 0 Find two ngles µ on the unit irle, with 0 6 µ 6 ¼, suh tht: os µ = sin µ =0 tn µ = : d sin µ = 0: e tn µ = 6:67 f os µ = 7 g tn µ = 5 h os µ = i sin µ = 5

17 THE UNIT IRLE ND RDIN MESURE (hter 8) 05 INVESTIGTION Usull we write funtions in the form = f(). For emle: = +7, = 6 +8, = sin However, sometimes it is useful to eress oth nd in terms of nother vrile t, lled the rmeter. In this se we s we hve rmetri equtions. Wht to do: PRMETRI EQUTIONS GRPHIS LULTOR INSTRUTIONS Use our grhis lultor to lot f(, ) j = os t, = sin t, 0 ± 6 t 6 60 ± g. Use the sme sle on oth es. Note: Your lultor will need to e set to degrees. Desrie the resulting grh. Is it the grh of funtion? Evlute +. Hene determine the eqution of this grh in terms of nd onl. Use our grhis lultor to lot: f(, ) j = os t, = sin(t), 0 ± 6 t 6 60 ± g f(, ) j = os t, = sin(t), 0 ± 6 t 6 60 ± g f(, ) j = os t, = os t sin t, 0 ± 6 t 6 60 ± g d f(, ) j = os t + sin t, = os t, 0 ± 6 t 6 60 ± g e f(, ) j = os t, = sin t, 0 ± 6 t 6 60 ± g E MULTIPLES OF ¼ 6 ND ¼ ngles whih re multiles of ¼ 6 nd ¼ trigonometri rtios etl. our frequentl, so it is imortnt for us to write their MULTIPLES OF ¼ OR 5± Tringle OP is isoseles s ngle OP lso mesures 5 ±. Letting O = P =, + = fpthgorsg ) = ) = ) = s >0 So, P is (, ) where ¼ 0: P (, ) os ¼ = nd sin ¼ =

18 06 THE UNIT IRLE ND RDIN MESURE (hter 8) You should rememer these vlues. If ou forget, drw right ngled isoseles tringle with side length. 5 For multiles of ¼, the numer is the imortnt thing to rememer. The signs of the oordintes re determined whih qudrnt the ngle is in. ³, (0, ) ³, (-, 0) 5¼ ¼ 7¼ ¼ (, 0) ³, ³, (0, -) MULTIPLES OF ¼ 6 OR 0± Sine O = OP, tringle OP is isoseles. The remining ngles re therefore lso 60 ±, nd so tringle OP is equilterl. P ( Qw, k) The ltitude [PN] isets se [O], so ON = If P is (, k), then ( ) + k = fpthgorsg k ) k = ) k = fs k>0g 60 Qw N ( 0,\\ ) So, P is (, ) where ¼ 0:866. Now os ¼ = nd sin ¼ = N PO = ¼ 6 =0±. Hene os ¼ 6 = nd sin ¼ 6 = 60 O 0 P N You should rememer these vlues. If ou forget, divide in two n equilterl tringle with side length. 0 60

19 THE UNIT IRLE ND RDIN MESURE (hter 8) 07 For multiles of ¼ 6, the numers nd re imortnt. The et oordintes of eh oint re found smmetr. ³, ³, (-,0) ³, ³, 7¼ 6 ( 0, ) ¼ ( 0, -) ³, ³ ( 0, ) ³,, ³, SUMMRY ² For multiles of ¼, the oordintes of the oints on the unit irle involve 0 nd. ² For other multiles of ¼, the oordintes involve. ² For other multiles of ¼ 6, the oordintes involve nd. You should e le to use this summr to find the trigonometri rtios for ngles whih re multiles of ¼ 6 nd ¼. For emle, onsider the ngles: ² 5 ± = 5¼ ³, 5¼ 5¼ is in qudrnt, so the signs re oth negtive nd oth hve size. ² 00 ± = 5¼ whih is multile of ¼ 6. ¼ ¼ Qw 5¼ is in qudrnt, so the signs re (+, ) nd from the digrm the -vlue is. ³, Emle Find the et vlues of sin, os, nd tn for = ¼. ³, ¼ ¼ 0.7 sin( ¼ )= os( ¼ )= ¼-0.7 tn( ¼ )=

20 08 THE UNIT IRLE ND RDIN MESURE (hter 8) Emle Find the et vlues of sin ¼, os ¼, nd tn ¼. -Qw ¼ sin( ¼ )= os( ¼ )= ³ -, - ¼ tn( ¼ )= = EXERISE 8E Use unit irle digrm to find et vlues for sin µ, os µ, nd tn µ for µ equl to: ¼ ¼ 7¼ d ¼ e ¼ Use unit irle digrm to find et vlues for sin, os, nd tn for equl to: ¼ 6 ¼ 7¼ 6 d 5¼ e ¼ 6 Find the et vlues of: os 0 ±, sin 0 ±, nd tn 0 ± os( 5 ± ), sin( 5 ± ), nd tn( 5 ± ) Find the et vlues of os 90 ± nd sin 90 ±. Wht n ou s out tn 90 ±? Emle 5 Without using lultor, show tht 8 sin( ¼ ) os( 5¼ 6 )= 6. ³, 5¼ 6 ¼ ³, sin( ¼ )= nd os( 5¼ 6 )= ) 8 sin( ¼ ) os( 5¼ 6 )=8( )( ) =( ) = 6 5 Without using lultor, evlute: sin 60 ± sin 0 ± os 60 ± sin 60 ± os 0 ± d os ( ¼ 6 ) e sin ( ¼ ) f os ( ¼ ) sin( 7¼ 6 ) g sin( ¼ ) os( 5¼ ) h sin ( 7¼ 6 ) i os ( 5¼ 6 ) sin ( 5¼ 6 ) j tn ( ¼ ) sin ( ¼ ) k tn( 5¼ ) sin( ¼ ) l tn 50 ± tn 50 ± hek ll nswers using our lultor.

21 THE UNIT IRLE ND RDIN MESURE (hter 8) 09 Emle 6 Find ll ngles 0 6 µ 6 ¼ with osine of. 5¼ ¼ Qw Sine the osine is, we drw the vertil line =. euse is involved, we know the required ngles re multiles of ¼ 6. =Qw The re ¼ nd 5¼. 6 Find ll ngles etween 0 ± nd 60 ± with: sine of sine of osine of d osine of e osine of f sine of 7 Find ll ngles etween 0 nd ¼ (inlusive) whih hve: tngent of tngent of tngent of d tngent of 0 e tngent of f tngent of 8 Find ll ngles etween 0 nd ¼ with: osine of sine of sine of 9 Find µ if 0 6 µ 6 ¼ nd: os µ = sin µ = os µ = d sin µ = e os µ = f sin µ = g os µ = h os µ = i tn µ = j tn µ = 0 Find ll vlues of µ for whih tn µ is: zero undefined. F THE EQUTION OF STRIGHT LINE If stright line mkes n ngle of µ with the ositive -is then its grdient is m = tn µ. Proof: -ste -ste Grdient m = -ste -ste = tn µ

22 0 THE UNIT IRLE ND RDIN MESURE (hter 8) Emle 7 Find the eqution of the given line: ¼ 6 The line hs grdient m = tn ¼ 6 = nd -interet. ) the line hs eqution = +. EXERISE 8F Find the eqution of eh line: 60 ¼ ¼ 6 Find the eqution of eh line: ¼ 6 ¼ ~` ¼ REVIEW SET 8 NON-LULTOR onvert these to rdins in terms of ¼: 0 ± 5 ± 50 ± d 50 ± Find the ute ngles tht would hve the sme: sine s ¼ sine s 65 ± osine s 76 ±. Find: sin 59 ± if sin ± ¼ 0:58 os 9 ± if os 88 ± ¼ 0:05 os 75 ± if os 05 ± ¼ 0:59 d sin( ± ) if sin 7 ± ¼ 0:7. Use unit irle digrm to find: os 60 ± nd sin 60 ± os( ¼) nd sin( ¼). 5 Elin how to use the unit irle to find µ when os µ = sin µ, 0 6 µ 6 ¼. 6 Find et vlues for sin µ, os µ, nd tn µ for µ equl to: ¼ 8¼

23 THE UNIT IRLE ND RDIN MESURE (hter 8) 7 If sin = nd ¼<< ¼, find tn etl. 8 If os µ = find the ossile vlues of sin µ. 9 Evlute: sin( ¼ ) os( ¼ ) tn ( ¼ ) os ( ¼ 6 ) sin ( ¼ 6 ) 0 Given tn = ¼ nd <<¼, find: sin os. Find the erimeter nd re of the setor. Suose os µ = REVIEW SET 8 7 nd µ is ute. Find the et vlue of tn µ. LULTOR Determine the oordintes of the oint on the unit irle orresonding to n ngle of: 0 ± 6 ± onvert to rdins to signifint figures: 7 ± :6 ± ± onvert these rdin mesurements to degrees, to deiml les: :6 0:5 d 5:7 Determine the re of setor of ngle 5¼ nd rdius m. M 5 Find the oordintes of the oints M, N, nd P on the unit irle. 0 7 N -5 P 6 Find the ngle [O] mkes with the ositive -is if the -oordinte of the oint on the unit irle is 0:. 7 Find ll ngles etween 0 ± nd 60 ± whih hve: osine of sine of tngent of 8 Find µ for 0 6 µ 6 ¼ if: os µ = sin µ = 9 Find the otuse ngles whih hve the sme: sine s 7 ± sine s ¼ 5 osine s 86 ± 0 Find the erimeter nd re of setor of rdius m nd ngle 6 ±. Find the rdius nd re of setor of erimeter 6 m with n ngle of ¼.

24 THE UNIT IRLE ND RDIN MESURE (hter 8) Find two ngles on the unit irle with 0 6 µ 6 ¼, suh tht: os µ = sin µ = tn µ = REVIEW SET 8 onvert these rdin mesurements to degrees: ¼ 5 5¼ 7¼ 9 d ¼ 6 Illustrte the regions where sin µ nd os µ hve the sme sign. Use unit irle digrm to find: os( ¼ ) nd sin( ¼ ) os( ¼ ) nd sin( ¼ ) Suose m = sin, where is ute. Write n eression in terms of m for: sin(¼ ) sin( + ¼) os d tn 5 P Stte the vlue of µ in: i degrees ii rdins. Stte the r length P. O Stte the re of the minor setor OP. 6 Without lultor, evlute tn ( ¼ ). 7 Show tht os( ¼ ) sin( ¼ )=. 8 If os µ =, ¼ <µ<¼ find the et vlue of: sin µ tn µ sin(µ + ¼) 9 Without using lultor, evlute: tn 60 ± sin 5 ± os ( ¼ ) + sin( ¼ ) os( 5¼ ) tn( 5¼ ) 0 Simlif: sin(¼ µ) sin µ os µ tn µ Find the eqution of the line drwn. Find the et vlue of k given the oint (k, ) lies on the line. 0 Three irles with rdius r re drwn s shown, eh with its entre on the irumferene of the other two irles., nd re the entres of the three irles. Prove tht n eression for the re of the shded region is = r (¼ ):

25 hter 9 Non-right ngled tringle trigonometr Sllus referene:.6 ontents: res of tringles The osine rule The sine rule D Using the sine nd osine rules

26 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) OPENING PROLEM tringulr sil is to e ut from setion of loth. Two of the sides must hve lengths m nd 6 m s illustrted. The totl re for the sil must e :6 m, the mimum llowed for the ot to re in its lss. Things to think out: n ou find the size of the ngle µ etween the two sides of given length? n ou find the length of the third side of the sil? 6 m m RES OF TRINGLES height DEMO se se se If we know the se nd height mesurements of tringle we n lulte the re using re = se height. However, ses rise where we do not know the height ut we n use trigonometr to lulte the re. These ses re: ² knowing two sides nd the inluded ngle etween them ² knowing ll three sides 8 m 8 m 9 m 9 0 m 0 m USING THE INLUDED NGLE Tringle hs ngles of size,, nd. The sides oosite these ngles re lelled,, nd resetivel. Using trigonometr, we n develo n lterntive re formul tht does not deend on erendiulr height. n tringle tht is not right ngled must e either ute or otuse. We will onsider oth ses: D h h D ( 80 -)

27 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) 5 In oth tringles the ltitude h is onstruted from to D on [] (etended if neessr). sin = h ) h = sin ut sin(80 ± ) = h ) h = sin(80 ± ) sin(80 ± ) = sin ) h = sin So, sine re = h, we now hve re = sin. Using different ltitudes we n show tht the re is lso sin or sin. Given the lengths of two sides of tringle, nd the size of the inluded ngle etween them, the re of the tringle is hlf of the rodut of two sides nd the sine of the inluded ngle. side inluded ngle side Emle Find the re of tringle. m 8 5 m re = sin 5 sin 8± = ¼ 8:7 m If we rerrnge the re formul to find the inluded ngle etween two sides, we need to use the inverse sine funtion denoted sin. For hel with this nd the other inverse trigonometri funtions ou should onsult the kground Knowledge hter on the D. LINK Emle tringle hs sides of length 0 m nd m nd n re of 50 m. Determine the two ossile mesures of the inluded ngle. Give our nswers urte to deiml le. If the inluded ngle is µ, then 0 sin µ =50 ) sin µ = Now sin ( ) ¼ 65:± ) µ ¼ 65: ± or 80 ± 65: ± ) µ ¼ 65: ± or :6 ± The two different ossile ngles re 65: ± nd :6 ±. 0 m 65. m 0 m. 6 m

28 6 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) EXERISE 9 Find the re of: km 9 m km 0 m 0. m ¼ 6. m If tringle hs re 50 m, find the vlue of : 7 m 68 m rllelogrm hs two djent sides with lengths m nd 6 m resetivel. If the inluded ngle mesures 5 ±, find the re of the rllelogrm. rhomus hs sides of length m nd n ngle of 7 ±. Find its re. 5 Find the re of regulr hegon with sides of length m. 6 rhomus hs n re of 50 m nd n internl ngle of size 6 ±. Find the length of its sides. 7 regulr entgonl grden lot hs entre of smmetr O nd n re of 8 m. Find the distne O. 8 Find the ossile vlues of the inluded ngle of tringle with: sides of length 5 m nd 8 m, nd re 5 m sides of length 5 km nd 5 km, nd re 800 km. 9 The ustrlin 50 ent oin hs the she of regulr dodegon, whih is olgon with sides. Eight of these 50 ent oins will fit etl on n ustrlin $0 note s shown. Wht frtion of the $0 note is not overed? 0 Find the shded re in: m 5. m O m P 5 7 mm

29 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) 7 D is n r of the irle with entre nd rdius 7: m. E is n r of the irle with entre F nd rdius 8:7 m. Find the shded re. D E F THE OSINE RULE The osine rule involves the sides nd ngles of tringle. In n : = + os or = + os or = + os We will develo the first formul for oth n ute nd n otuse tringle. Proof: h h (80 - ) - D D In oth tringles drw the ltitude from down to [] (etended if neessr), meeting it t D. Let D = nd let D = h. l the theorem of Pthgors in D: = h +( ) = h +( + ) ) = h + + ) = h In oth ses, ling Pthgors to D gives h + =. ) h =, nd we sustitute this into the equtions ove. ) = + ) = + + In D, os = os(80 ± ) = ) os = ) = + ) os = os ) os = ) = + os The other vritions of the osine rule ould e develoed rerrnging the verties of.

30 8 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) Note tht if =90 ± then os =0 nd = + os redues to = +, whih is the Pthgoren Rule. The osine rule n e used to solve rolems involving tringles given: ² two sides nd n inluded ngle ² three sides. If we re given two sides nd non-inluded ngle, then when we tr to find the third side we otin qudrti eqution. This is n miguous se where there m e two lusile solutions. We m not e le to solve for the length uniquel if there re two ositive, lusile solutions to the qudrti eqution. Emle Find, orret to deiml les, the length of []. m m the osine rule: = + os ± ) ¼ ( + os ± ) ) ¼ 8:80 ) [] is 8:80 m in length. Rerrngement of the originl osine rule formule n e used for finding ngles if we know ll three sides. The formule for finding the ngles re: os = + os = + os = + We then need to use the inverse osine funtion os to evlute the ngle. Emle In tringle, =7m, =5m, nd =8m. Find the mesure of. Find the et re of tringle. 7 m 8 m The re of = 8 5 sin 60± 5 m =0 =0 m f sin 60 ± = the osine rule: os = ( ) ( 5 8) µ ) = os ) = os ( ) ) =60 ± So, mesures 60 ±. g

31 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) 9 EXERISE 9 Find the length of the remining side in the given tringle: 5 m Q km m R 6. m K P 6. km L 7. 8 m M Find the mesure of ll ngles of: Find the mesure of otuse P QR. Hene find the re of PQR. P m m 0 m 5 m m 7 m Q R Find the smllest ngle of tringle with sides m, m, nd 7 m. Find the lrgest ngle of tringle with sides m, 7 m, nd 9 m. The smllest ngle is oosite the shortest side. 5 Find os µ ut not µ. Find the vlue of. 5 m m m m m 6 Find the et vlue of in eh of the following digrms: 6 m 7 m 60 m m 5 m 0 m m 60 m 5 m 7 Solve the Oening Prolem on ge. 8 Find in eh of the following digrms: m 70 m m 5 m 0 8 m m

32 0 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) 9 Show tht there re two lusile vlues for in this tringle: m 5 m 0 6 m THE SINE RULE The sine rule is set of equtions whih onnets the lengths of the sides of n tringle with the sines of the ngles of the tringle. The tringle does not hve to e right ngled for the sine rule to e used. In n tringle with sides,, nd units in length, nd oosite ngles,, nd resetivel, sin = sin = sin or sin = sin = sin. Proof: The re of n tringle is given sin = sin = sin. Dividing eh eression sin gives = sin = sin. The sine rule is used to solve rolems involving tringles, given: ² two ngles nd one side ² two sides nd non-inluded ngle. FINDING SIDE LENGTHS Emle 5 Find the length of [] orret to two deiml les. m 58 9 m 58 m 9 Using the sine rule, sin 58 ± = sin 9 ± ) = sin 58± sin 9 ± ) ¼ 6:70 7 ) [] is out 6:7 m long.

33 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) EXERISE 9. Find the vlue of : m m 5 m m 5 8. km 80 km onsider tringle. Given =6 ±, =9 ±, nd =8m, find. Given =8 ±, =5 ±, nd =m, find. Given = ±, =8 ±, nd =6: m, find. FINDING NGLES The rolem of finding ngles using the sine rule is more omlited euse there m e two ossile nswers. For emle, if sin µ = then µ =60 ± or 0 ±. We ll this sitution n miguous se. You n lik on the ion to otin n intertive demonstrtion of the miguous se, or else ou n work through the following investigtion. DEMO INVESTIGTION THE MIGUOUS SE You will need lnk sheet of er, ruler, rotrtor, nd omss for the tsks tht follow. In eh tsk ou will e required to onstrut tringles from given informtion. You ould lso do this using omuter kge suh s The Geometer s Skethd. Wht to do: Tsk : Drw =0m. t onstrut n ngle of 0 ±. Using s the entre, drw n r of irle of rdius 6 m. Let denote the oint where the r intersets the r from. How mn different ossile oints re there, nd therefore how mn different tringles m e onstruted? Tsk : Tsk : Tsk : s efore, drw =0m nd onstrut 0 ± ngle t. This time drw n r of rdius 5 m entred t. How mn different tringles re ossile? Reet, ut this time drw n r of rdius m entred t. How mn different tringles re ossile? Reet, ut this time drw n r of rdius m entred t. How mn different tringles re ossile now? You should hve disovered tht when ou re given two sides nd non-inluded ngle there re numer of different ossiilities. You ould get two tringles, one tringle, or it m e imossile to drw n tringles t ll for some given dimensions. Now onsider the lultions involved in eh of the ses of the investigtion.

34 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) Tsk : Given: =0m, =6m, =0 ± euse sin = sin ) sin = sin ) sin = 0 sin 0± 6 ¼ 0:8 sin µ = sin(80 ± µ) there re two ossile ngles: ¼ 56: ± or 80 ± 56: ± = :56 ± Tsk : Given: =0m, =5m, =0 ± sin = sin ) sin = sin 0 sin 0± ) sin = = 0 m 5 There is onl one ossile solution for in the rnge from 0 ± to 80 ±, nd tht is =90 ±. Onl one tringle is therefore ossile. omlete the solution of the tringle ourself. Tsk : Given: =0m, =m, =0 ± sin = sin ) sin = sin 0 sin 0± ) sin = ¼ : m There is no ngle tht hs sine vlue >. Therefore there is no solution for this given dt, nd no tringles n e drwn to mth the informtion given. Tsk : Given: =0m, =m, =0 ± sin = sin ) sin = sin 0 sin 0± ) sin = ¼ 0:67 Two ngles hve sine rtio of 0:67 : ¼ :6 ± or 80 ± :6 ± = 55:8 ± However, in this se onl one of these two ngles is vlid. If =0 ± then nnot ossil equl 55:8 ± euse 0 ± + 55:8 ± > 80 ±. Therefore, there is onl one ossile solution, ¼ :6 ±. onlusion: Eh sitution using the sine rule with two sides nd non-inluded ngle must e emined ver refull. 0 0 m m 6 m m 5 m m 6 m

35 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) Emle 6 Find the mesure of ngle in tringle if =7m, =m, nd ngle mesures 5 ±. 5 7 m m sin ) sin = sin = sin 5± 7 fsine ruleg sin 5± ) sin = 7 ³ ) = sin sin 5 ± ) ¼ :6 ± or 80 ± :6 ± ) ¼ :6 ± or 8: ± 7 or its sulement fs m e otuseg ) mesures :6 ± if ngle is ute, or 8: ± if ngle is otuse. In this se there is insuffiient informtion to determine the tul she of the tringle. There re two ossile tringles. Sometimes there is informtion in the question whih enles us to rejet one of the nswers. Emle 7 Find the mesure of ngle L in tringle KLM given tht ngle K mesures 56 ±, LM =6:8 m, nd KM =:5 m. L K m. 5 m sin L :5 We rejet L ¼ 8: ±, sine 8: ± +56 ± > 80 ± ) L ¼ :8 ±, unique solution in this se. M = sin 56± 6:8 f the sine ruleg :5 sin 56± ) sin L = 6:8 ³ ) L = sin :5 sin 56 ± or its sulement 6:8 ) L ¼ :8 ± or 80 ± :8 ± ) L ¼ :8 ± or 8: ± whih is imossile in tringle. EXERISE 9. Tringle hs ngle =0 ±, =8m, nd =m. Find the two ossile mesures of ngle. onsider tringle. Given =:6 m, =7: m, nd =65 ±, find the mesure of. Given =:8 m, =: m, nd = ±, find the mesure of. Given =6:5 km, =:8 km, nd =7 ±, find the mesure of.

36 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) Is it ossile to hve tringle with the mesurements shown? Elin our nswer. 98. m m Given D =0m, find the mgnitude of nd hene the length D. 0 m 78 D 5 Find nd in the given figure. m 95 m 0 8 m 6 Tringle hs =58 ±, =5: m, =8m, nd =6:8 m. Find orret to the nerest degree using the sine rule. Find orret to the nerest degree using the osine rule. o nd omlete: When fed with using either the sine rule or the osine rule it is etter to use the... s it voids... 7 In tringle, =0 ±, =9m, nd =7m. Find the re of the tringle. 8 Find the et vlue of, giving our nswer in the form + where, Q. 5 ( + ) m 0 ( - 5) m D USING THE SINE ND OSINE RULES If we re given rolem involving tringle, we must first deide whih rule is est to use. If the tringle is right ngled then the trigonometri rtios or Pthgors Theorem n e used. For some rolems we n dd n etr line or two to the digrm to rete right ngled tringle. However, if we do not hve right ngled tringle then we usull hve to hoose etween the sine nd osine rules. In these ses the following heklist m e helful:

37 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) 5 Use the osine rule when given: ² three sides ² two sides nd n inluded ngle. Use the sine rule when given: ² one side nd two ngles ² two sides nd non-inluded ngle, ut ewre of n miguous se whih n our when the smller of the two given sides is oosite the given ngle. Emle 8 The ngles of elevtion to the to of mountin re mesured from two eons nd t se. T The mesurements re shown on the digrm. If the eons re 7 m rt, how high is the mountin? N m. 5. m T N h m Let T e m nd NT e h m. T =: ± 9:7 ± feterior ngle of g =:5 ± We find in T using the sine rule: sin 9:7 ± = 7 sin :5 ± ) = 7 sin 9:7± sin :5 ± ¼ 660:6 Now, in NT, sin : ± = h ¼ h 660:6 ) h ¼ sin : ± 660:6 ¼ 0 The mountin is out 0 m high. EXERISE 9D Rodrigo wishes to determine the height of flgole. He tkes sighting to the to of the flgole from oint P. He then moves further w from the flgole 0 metres to oint Q nd tkes seond sighting. The informtion is shown in the digrm longside. How high is the flgole? Q 8 P 5 0 m

38 6 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) Q To get from P to R, rk rnger hd to wlk long 75 m 6 m th to Q nd then to R s shown. Wht is the distne in stright line from P to R? P lke R golfer led his tee shot distne of 0 m to oint. He then led 65 m si iron to the green. If the distne from tee to green is 0 m, determine the ngle the golfer ws off line with his tee shot. 0 m 65 m 0 m G T ommunitions tower is onstruted on to of tower uilding s shown. Find the height of the tower m uilding 5 Hikers Ritv nd Esko leve oint P t the sme time. Ritv wlks km on ering of 00 ±, then further 6 km on ering of 55 ±. Esko hikes diretl from P to the m site. N 55 How fr does Esko hike? In whih diretion does Esko hike? Ritv hikes t 0 km h nd Esko hikes t 6 km h. i Who will rrive t the m site first? ii How long will this erson need to wit efore the other erson rrives? N P 0 km 6 km d On wht ering should the hikers wlk from the m site to return to P? msite 6 footll gol is 5 metres wide. When ler is 6 metres from one gol ost nd metres from the other, he shoots for gol. Wht is the ngle of view of the gols tht the ler sees? gol osts ngle of view ler 7 tower metres high stnds on to of hill. From oint some distne from the se of the hill, the ngle of elevtion to the to of the tower is : ± nd the ngle of elevtion to the ottom of the tower is 8: ±. Find the height of the hill.

39 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) 7 8 From the foot of uilding I hve to look ± uwrds to sight the to of tree. From the to of the uilding, 50 metres ove ground level, I hve to look down t n ngle of 50 ± elow the horizontl to sight the tree to. How high is the tree? How fr from the uilding is this tree? m Emle 9 Find the mesure of R PV. S R P U T 6 m V Q m W 5 m In RVW, RV = 5 + = m. fpthgorsg In PUV, PV = 6 + = 5 m. fpthgorsg In PQR, PR = 6 +5 = 6 m. fpthgorsg P 5 m 6 m V R m rerrngement of the osine rule, os µ = ( 6) +( 5) ( ) = = 6 5 µ ) µ = os 6 ¼ 6:6 ± 6 5 ) R PV mesures out 6:6 ±. 9 Find the mesure of P QR in the retngulr o shown. P 8 m Q m R 7 m 0 Two oservtion osts nd re km rt. third oservtion ost is loted suh tht is ± nd is 67 ±. Find the distne of from nd from.

40 8 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) Stn nd Olg re onsidering uing shee frm. surveor hs sulied them with the given urte sketh. Find the re of the roert, giving our nswer in: km hetres. Q 8 km 70 P 0 km 0 km R S hetre is equivlent to 00 m 00 m. Tho nd Ples strt t oint. The eh wlk in stright line t n ngle of 0 ± to one nother. Tho wlks t 6 km h nd Ples wlks t 8 km h. How fr rt re the fter 5 minutes? The ross-setion design of the kering for driverless-us rodw is shown oosite. The metl stri is inlid into the onrete nd is used to ontrol the diretion nd seed of the us. Find the width of the metl stri. 00 mm 0 8 n orienteer runs for km, then turns through n ngle of ± nd runs for nother 6 km. How fr is she from her strting oint? 5 Sm nd Mrkus re stnding on level ground 00 metres rt. lrge tree is due north of Mrkus nd on ering of 065 ± from Sm. The to of the tree ers t n ngle of elevtion of 5 ± to Sm nd 5 ± to Mrkus. Find the height of the tree. 6 helioter oserves two shis nd. is :8 km from the helioter nd is :9 km from it. The ngle of view from the helioter to nd, is 8:6 ±. How fr re the shis rt? 0 D metl stri REVIEW SET 9 Determine the re of the tringle. 7 km 0 8 km NON-LULTOR You re given enough detils of tringle so tht ou ould use either the osine rule or the sine rule to find n unknown. Whih rule should ou use? Elin our nswer. Kd ws sked to drw the illustrted tringle etl. Use the osine rule to find. 8 m 7 m Wht should Kd s resonse e? 60 m tringle hs sides of length 7 m nd m, nd its re is m. Find the sine of the inluded ngle.

41 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) 9 5 ot is ment to e siling diretl from to km However, it trvels in stright line to efore the tin relises he is off ourse. The ot is turned through n ngle of 60 ±, then trvels nother 0 km to. The tri would hve een km shorter if the ot hd gone stright from to. How fr did the ot trvel? 6 Show tht the ellow shded re is given = 9 ¼ 8 sin( ¼ 8 ). ¼ 8 7 m REVIEW SET 9 Determine the vlue of : m 7 5 km LULTOR m 9 m 7 km km Find the unknown side nd ngles: Find the re of qudrilterl D: m m m 0 7 m D 0 6 m vertil tree is growing on the side of hill with grdient 0 ± to the horizontl. From oint 50 m downhill from the tree, the ngle of elevtion to the to of the tree is 8 ±. Find the height of the tree. 5 From oint, the ngle of elevtion to the to of tll uilding is 0 ±. On wlking 80 m towrds the uilding, the ngle of elevtion is now ±. How tll is the uilding? 6 Peter, Sue, nd li re se-kking. Peter is 0 m from Sue on ering of ±. li is on ering of 0 ± nd is 0 m from Sue. Find the distne nd ering of Peter from li.

42 0 NON-RIGHT NGLED TRINGLE TRIGONOMETRY (hter 9) REVIEW SET 9 Find the vlue of : m m m m 9 m 7 m Find the vlue of if the tringle hs re 80 m. 9. m. m Find the mesure of E DG: E F 6 m H D m G m nke nd Lus re onsidering uing lok of lnd. The lnd gent sulies them with the given urte sketh. Find the re of the roert, giving our nswer in: m hetres. 5 m 75 0 m 0 90 m D 5 fmil in Germn drives t 0 km h for 5 minutes on the ering 0 ±, nd then 80 km h for 0 minutes on the ering 7 ±. Find the distne nd ering of the r from its strting oint. 6 Soil ontrtor Frnk ws given the following dimensions over the telehone: The tringulr grden lot hs mesuring ±, [] is 8 m long, nd [] is 6 m long. Soil to deth of 0 m is required. Elin wh Frnk needs etr informtion from his lient. Wht is the mimum volume of soil needed if his lient is unle to sul the neessr informtion?