5Trigonometri rtios nd their pplitions 5.1 Kik off with CS 5.2 Trigonometry of right-ngled tringles 5.3 Elevtion, depression nd erings 5.4 The sine rule 5.5 The osine rule 5.6 rs, setors nd segments 5.7 Review
5.1 Kik off with CS Defining rules Using CS tehnology it is possile to define ommonly used mthemtil rules nd sve them so tht they n e used t ny stge to solve prolem. 1 Use CS tehnology to define the rule for Pythgors theorem s = 2 + 2 or = 2 2 nd hene nswer the following questions. If = 3 nd = 5, determine the vlue of. If = 9.1 nd = 24.3, determine the vlue of. If = 78.86 nd = 155.32, determine the vlue of. In this topi, you will study the sine rule nd the osine rule, whih n e used to lulte ngles nd side lengths in non right-ngled tringles. Sine rule sin() = sin() = sin(c) Cosine rule 2 = 2 + 2 2 os() 2 fter studying the sine rule, use CS tehnology to rewrite nd define it s: = sin() or = sin 1 sin() nd nswer the following questions sin() regrding C. Clulte if = 15 m, = 48 nd = 74. Clulte if = 7 m, = 32 nd = 86. Clulte if = 12.7 m, = 16.3 m nd = 45. 3 fter studying the osine rule, use CS tehnology to rewrite nd define it s: = 2 + 2 2 os() or = os 1 2 + 2 2 nd nswer the following 2 questions regrding C. Clulte if = 10 m, = 8 m nd = 30. Clulte if = 260 m, = 120 m nd = 115. Clulte if = 20 m, = 12 m nd = 13 m. d Clulte if = 2 m, = 3.5 m nd = 2.5 m. Plese refer to the Resoures t in the Prelims setion of your eookplus for omprehensive step-y-step guide on how to use your CS tehnology.
5.2 Units 1 & 2 OS 4 Topi 3 Conept 1 Trigonometry of right-ngled tringles Conept summry Prtie questions Trigonometry of right-ngled tringles Trigonometry, derived from the Greek words trigon (tringle) nd metron (mesurement), is the rnh of mthemtis tht dels with the reltionship etween the sides nd ngles of tringle. It involves finding unknown ngles, side lengths nd res of tringles. The priniples of trigonometry re used in mny prtil situtions suh s uilding, surveying, nvigtion nd engineering. In previous yers you will hve studied the trigonometry of right-ngled tringles. We will review this mteril efore onsidering non right-ngled tringles. sin (θ) opposite side = hypotenuse, whih is revited to sin (θ) = O H djent side os (θ) = hypotenuse, whih is revited to os (θ) = Opposite (O) H C opposite side tn (θ) = djent side, whih is revited to tn (θ) = O The symol θ (thet) is one of the mny letters of the Greek lphet used to represent the ngle. Other symols inlude α (lph), β (et) nd γ (gmm). Non-Greek letters my lso e used. Writing the mnemoni SOH CH TO eh time we perform trigonometri lultions will help us to rememer the rtios nd solve the prolem. Pythgors theorem For speifi prolems it my e neessry to determine the side lengths of right-ngled tringle efore lulting the trigonometri rtios. In these situtions, Pythgors theorem is used. Pythgors theorem sttes: In ny right-ngled tringle, 2 = 2 + 2. Proof of Pythgors theorem Strt with squre with side length +. re 1 = ( + ) 2 Divide this squre into four ongruent right-ngled tringles, Δ, nd squre of length,. re 2 = re of four tringles + one squre = 4 1 + 2 2 = 2 + 2 Hypotenuse (H) () djent + θ + 180 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2
WorKEd EXMPLE 1 think re 1= re 2 ( + ) 2 = 2 + 2 2 + 2 + 2 = 2 + 2 2 + 2 = 2 Pythgors theorem sttes tht the squre of the hypotenuse is equl to the sum of the squres of the other two sides. Determine the vlue of the pronumerls, orret to 2 deiml ples. 4 x 7 24 25 50 h 1 Lel the sides, reltive to the mrked ngles. WritE/drW 2 Write wht is given. Hve: ngle nd hypotenuse (H) 3 Write wht is needed. Need: opposite (O) side 4 Determine whih of the trigonometri rtios is required, using SOH CH TO. x O 4 H 50 sin (θ) = O H 5 Sustitute the given vlues into the pproprite rtio. sin (50 ) = x 4 6 Trnspose the eqution nd solve for x. 4 sin (50 ) = x x = 4 sin (50 ) 7 Round the nswer to 2 deiml ples. = 3.06 1 Lel the sides, reltive to the mrked ngles. 7 24 25 h H 2 Write wht is given. Hve: ngle nd djent () side 3 Write wht is needed. Need: hypotenuse (H) 4 Determine whih of the trigonometri rtios is required, using SOH CH TO. os (θ) = H 5 Sustitute the given vlues into the pproprite rtio. os (24 25 ) = 7 h 7 h = os (24 25 ) 6 Round the nswer to 2 deiml ples. = 7.69 Topi 5 TrIgonoMETrIC rtios nd THEIr PPLICTIonS 181
WorKEd EXMPLE 2 think Find the ngle θ, giving the nswer in degrees nd minutes. 1 Lel the sides, reltive to the mrked ngles. Ext vlues Most of the trigonometri vlues tht we will del with in this topi re only pproximtions. However, ngles of 30, 45 nd 60 hve ext vlues of sine, osine nd tngent. Consider n equilterl tringle, C, of side length 2 units. If the tringle is perpendiulrly iseted, then two ongruent tringles, D nd CD, re otined. From tringle D it n e seen tht D retes right-ngled tringle with ngles of 60 nd 30 nd se length (D) of 1 unit. The length of D is otined using Pythgors theorem. Using tringle D nd the three trigonometri rtios, the following ext vlues re otined: sin (30 ) = 1 2 os (30 ) = 3 2 WritE/drW 2 Write wht is given. Hve: opposite (O) nd djent () sides 3 Write wht is needed. Need: ngle 4 Determine whih of the trigonometri rtios is required, using SOH CH TO. 5 Sustitute the given vlues into the pproprite rtio. O 18 θ tn (30 ) = 1 3 or 3 3 12 tn (θ) = O tn (θ) = 18 12 6 Write the nswer to the nerest minute. θ = tn 1 18 12 = 56 19 sin (60 ) = 3 2 os (60 ) = 1 2 60 tn (60 ) = 3 1 or 3 30 2 2 3 18 θ D 2 12 C 182 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2
WorKEd EXMPLE 3 Consider right-ngled isoseles tringle EFG whose equl sides re of 1 unit. The hypotenuse EG is otined y using Pythgors theorem. (EG) 2 = (EF) 2 + (FG) 2 = 1 2 + 1 2 = 2 EG = 2 Using tringle EFG nd the three trigonometri rtios, the following ext vlues re otined: sin (45 ) = 1 2 or 2 2 os (45 ) = 1 2 or 2 2 tn (45 ) = 1 1 or 1 Determine the height of the tringle shown in surd form. think 1 Lel the sides reltive to the mrked ngle. WritE/drW 60 8 m 2 Write wht is given. Hve: ngle nd djent () side 3 Write wht is needed. Need: opposite (O) side 4 Determine whih of the trigonometri rtios is required, using SOH CH TO. h O tn (θ) = O 5 Sustitute the given vlues into the pproprite rtio. tn (60 ) = h 8 6 Sustitute ext vlues where pproprite. 3 = h 8 7 Trnspose the eqution to find the required vlue. h = 8 3 8 Stte the nswer. The tringle s height is 8 3 m. E 2 45 1 h G 1 F 60 8 m Topi 5 TrIgonoMETrIC rtios nd THEIr PPLICTIonS 183
Exerise 5.2 PRtise Trigonometry of right-ngled tringles 1 WE1 Find the vlue of the pronumerls, orret to 2 deiml ples. 10 7.5 47 8 d x x 17 40 684 x 32 14 2 Find the vlue of x nd y, orret to 2 deiml ples. 1.03 27 47 x 78 x x 504 3.85 14 25 3 WE2 Find the ngle θ, giving the nswer in degrees nd minutes. θ 10 7 5 12 4 Find the ngle θ, giving the nswer in degrees nd minutes. 11.7 θ 4.2 48 θ 30 θ 28 53.2 θ 20 78.1 5 WE3 n isoseles tringle hs se of 12 m nd equl ngles of 30. Find, in the simplest surd form: the height of the tringle the re of the tringle the perimeter of the tringle. 6 Find the perimeter of the omposite shpe elow, in surd form. The length mesurements re in metres. 14 26 60 θ d d d x θ 1.74 θ 17 x y 38 48 6.8 30 30 12 m Consolidte 62 38 7 ldder 6.5 m long rests ginst vertil wll nd mkes n ngle of 50 to the horizontl ground. How high up the wll does the ldder reh? If the ldder needs to reh 1 m higher, wht ngle should it mke to the ground, to the nerest minute? 3.26 2.1 184 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2
Mster 5.3 Units 1 & 2 OS 4 Topi 3 8 rod 400 m long goes stright up slope. If the rod rises 50 m vertilly, wht is the ngle tht the rod mkes with the horizontl? 9 n ie-rem one hs dimeter of 6 m nd sloping edge of 15 m. Find the ngle t the ottom of the one. 10 vertil flgpole is supported y wire tthed from the top of the pole to the horizontl ground, 4 m from the se of the pole. Jonne mesures the ngle the wire mkes with the ground s 65. How tll is the flgpole? 11 stepldder stnds on floor, with its feet 1.5 m prt. If the ngle formed y the legs is 55, how high ove the floor is the top of the ldder? 12 In the figure t right, find the vlue of the pronumerls, orret to 2 deiml ples. 13 The ngle formed y the digonl of retngle nd one of its shorter sides is 60. If the digonl is 8 m long, find the dimensions of the retngle, in surd form. 14 n dvertising lloon is tthed to rope 120 m long. The rope mkes n ngle of 75 to level ground. How high ove the ground is the lloon? 15 In the figure t right, find the vlue of the pronumerl x, orret to 2 deiml ples. 16 n isoseles tringle hs sides of 17 m, 20 m nd 20 m. Find the mgnitude of the ngles. 17 grden ed in the shpe of trpezium is shown t right. Wht volume of grden mulh is needed to over it to depth of 15 m? 18 ldder 10 m long rests ginst vertil wll t n ngle 120 of 55 to the horizontl. It slides down the wll, so tht it 4 m now mkes n ngle of 48 with the horizontl. Through wht vertil distne did the top of the ldder slide? Does the foot of the ldder move through the sme distne? Justify your nswer. Elevtion, depression nd erings Trigonometry is espeilly useful for mesuring distnes nd heights tht re diffiult or imprtil to ess. For exmple, two importnt pplitions of right-ngled tringles re: 1. ngles of elevtion nd depression 2. erings. ngles of elevtion nd depression ngles of elevtion nd depression re employed when deling with diretions tht require us to look up nd down respetively. n ngle of elevtion is the ngle etween the horizontl nd n ojet tht is higher thn the oserver (for exmple, the top of mountin or flgpole). Conept 2 Elevtion nd depression Conept summry Prtie questions 6 x 14 12 m 33 58 Line of sight θ 48 70 ngle of elevtion Topi 5 Trigonometri rtios nd their pplitions 185
WorKEd EXMPLE 4 n ngle of depression is the ngle etween the horizontl nd n ojet tht is lower thn the oserver (for exmple, ot t se when the oserver is on liff). Unless otherwise stted, the ngle of elevtion or depression is mesured nd drwn from the horizontl. ngles of elevtion nd depression re eh mesured from the horizontl. When solving prolems involving ngles of elevtion nd depression, it is lwys est to drw digrm. The ngle of elevtion is equl to the ngle of depression euse they re lternte Z ngles. erings erings mesure the diretion of one ojet from nother. There re two systems used for desriing erings. True erings re mesured in lokwise diretion, strting from north (0 T). Conventionl or ompss erings re mesured: first, reltive to north or south, nd seond, reltive to est or west. ngle of depression E Line of sight D θ D nd E re lternte ngles. D = E From liff 50 metres high, the ngle of depression to ot t se is 12. How fr is the ot from the se of the liff? think 1 Drw digrm nd lel ll the given informtion. Inlude the unknown length, x, nd the ngle of elevtion, 12. WritE/drW 12 x 12 50 m 2 Write wht is given. Hve: ngle nd opposite side 3 Write wht is needed. Need: djent side 4 Determine whih of the trigonometri rtios is required (SOH CH TO). 5 Sustitute the given vlues into the pproprite rtio. tn (θ) = O tn (12 ) = 50 x 6 Trnspose the eqution nd solve for x. x tn (12 ) = 50 50 x = tn (12 ) 7 Round the nswer to 2 deiml ples. = 235.23 8 nswer the question. The ot is 235.23 m from the se of the liff. W N 150 T E S Compss ering equivlent is S30 E. 186 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2
N N Units 1 & 2 OS 4 Topi 3 Conept 5 erings Conept summry Prtie questions WorKEd EXMPLE 5 W S N20 W 20 True ering equivlent is 340 T. The two systems re interhngele. For exmple, ering of 240 T is the sme s S60 W. When solving questions involving diretion, lwys strt with digrm showing the si ompss points: north, south, est nd west. E W S S70 E 20 True ering equivlent is 110 T. W N S E E W 240 T 60 N S S60 W ship sils 40 km in diretion of N52 W. How fr west of the strting point is it? think 1 Drw digrm of the sitution, lelling eh of the ompss points nd the given informtion. WritE/drW N x S 2 Write wht is given for the tringle. Hve: ngle nd hypotenuse 3 Write wht is needed for the tringle. Need: opposite side W 40 km 4 Determine whih of the trigonometri rtios is sin (θ) = O required (SOH CH TO). H 5 Sustitute the given vlues into the pproprite rtio. sin (52 ) = x 40 6 Trnspose the eqution nd solve for x. 40 sin (52 ) = x 7 Round the nswer to 2 deiml ples. x = 31.52 8 nswer the question. The ship is 31.52 km west of the strting point. WorKEd EXMPLE 6 52 ship sils 10 km est, then 4 km south. Wht is its ering from its strting point? think 1 Drw digrm of the sitution, lelling eh of the ompss points nd the given informtion. E WritE/drW N 10 km 4 km S θ E Topi 5 TrIgonoMETrIC rtios nd THEIr PPLICTIonS 187
2 Write wht is given for the tringle. Hve: djent nd opposite sides 3 Write wht is needed for the tringle. Need: ngle 4 Determine whih of the trigonometri rtios is required (SOH CH TO). 5 Sustitute the given vlues into the pproprite rtio. 6 Trnspose the eqution nd solve for θ, using the inverse tn funtion. Exerise 5.3 PRCTISE CONSOLIDTE tn (θ) = O tn (θ) = 4 10 θ = tn 1 4 10 7 Convert the ngle to degrees nd minutes. = 21.801 409 49 = 21 48 8 Express the ngle in erings form. The ering of the ship ws initilly 0 T; it hs sine rotted through n ngle of 90 nd n dditionl ngle of 21 48. To otin the finl ering these vlues re dded. ering = 90 + 21 48 = 111 48 T 9 nswer the question. The ering of the ship from its strting point is 111 48 T. Elevtion, depression nd erings 1 WE4 From vertil fire tower 60 m high, the ngle of depression to fire is 6. How fr wy, to the nerest metre, is the fire? 2 person stnds 20 m from the se of uilding nd mesures the ngle of elevtion to the top of the uilding s 55. If the person is 1.7 m tll, how high, to the nerest metre, is the uilding? 3 WE5 pir of kykers pddle 1800 m on ering of N20 E. How fr north of their strting point re they, to the nerest metre? 4 ship sils 230 km on ering of S20 W. How fr west of its strting point hs it trvelled, orret to the nerest kilometre? 5 WE6 ship sils 20 km south, then 8 km west. Wht is its ering from the strting point? 6 ross-ountry ompetitor runs 2 km west, then due north for 3 km. Wht is the true ering of the runner from the strting point? 7 Express the following onventionl erings s true erings, nd the true erings in onventionl form. N35 W S47 W N58 E d S17 E e 246 T f 107 T g 321 T h 074 T 188 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2
MSTER 8 ering of S30 E is the sme s: 030 T 120 T C 150 T D 210 T E 240 T ering of 280 T is the sme s: N10 W S10 W C S80 W D N80 W E N10 E 9 n oserver on liff top 57 m high oserves ship t se. The ngle of depression to the ship is 15. The ship sils towrds the liff, nd the ngle of depression is then 25. How fr, to the nerest metre, did the ship sil etween sightings? 10 new skysrper is proposed for the Melourne Doklnds region. It is to e 500 m tll. Wht would e the ngle of depression, in degrees nd minutes, from the top of the uilding to the islnd on lert Prk Lke, whih is 4.2 km wy? 11 From resue heliopter 2500 m ove the oen, the ngles of depression to two shipwrek survivors re 48 (survivor 1) nd 35 (survivor 2). Drw lelled digrm tht represents the sitution. Clulte how fr prt the two survivors re. 12 lookout tower hs een ereted on top of mountin. t distne of 5.8 km, the ngle of elevtion from the ground to the se of the tower is 15.7, nd the ngle of elevtion to the oservtion dek (on the top of the tower) is 15.9. How high, to the nerest metre, is the oservtion dek ove the top of the mountin? 13 From point on level ground, the ngle of elevtion to the top of uilding 50 m high is 45. From point on the ground nd in line with nd the foot of the uilding, the ngle of elevtion to the top of the uilding is 60. Find, in simplest surd form, the distne from to. 14 yht re onsists of four legs. The first three legs re 4 km due est, then 5 km south, followed y 2 km due west. How long is the finl leg, if the re finishes t the strting point? On wht ering must the finl leg e siled? 15 Two hikers set out from the sme mpsite. One wlks 7 km in the diretion 043 T nd the other wlks 10 km in the diretion 133 T. Wht is the distne etween the two hikers? Wht is the ering of the first hiker from the seond? 16 ship sils 30 km on ering of 220 T, then 20 km on ering of 250 T. Find: how fr south of the originl position it is how fr west of the originl position it is the true ering of the ship from its originl position, to the nerest degree. 17 The town of rknw is due west of rley. Chris, in n ultrlight plne, strts t third town, Chmpton, whih is due north of rknw, nd flies diretly towrds rley t speed of 40 km/h in diretion of 110 T. She rehes rley in 3 hours. Find: the distne etween rley nd rknw the time to omplete the journey from Chmpton to rknw, vi rley, if she inreses her speed to 45 km/h etween rley nd rknw. Topi 5 Trigonometri rtios nd their pplitions 189
5.4 Units 1 & 2 OS 4 Topi 3 Conept 3 The sine rule Conept summry Prtie questions 18 ird flying t 50 m ove the ground ws oserved t noon from my front door t n ngle of elevtion of 5. Two minutes lter its ngle of elevtion ws 4. If the ird ws flying stright nd level, find the horizontl distne of the ird: i from my doorwy t noon ii from my doorwy t 12.02 pm. Hene, find: i the distne trvelled y the ird in the two minutes ii its speed of flight in km/h. The sine rule When working with non right-ngled tringles, it is usul to lel the ngles, nd C, nd the sides, nd, so tht side is the side opposite ngle, side is opposite ngle nd side is opposite ngle C. In non right-ngled tringle, perpendiulr line, h, n e drwn from the ngle to side. Using tringle D, we otin sin () = h. Using tringle CD, we otin sin (C) = h. Trnsposing eh eqution to mke h the sujet, we otin h = sin () nd h = sin (C). Equte to get sin () = sin (C). Trnspose to get sin (C) = sin () In similr wy, if perpendiulr line is drwn from ngle to side, we get sin () = sin (C) From this, the sine rule n e stted. In ny tringle C: sin () = sin () = sin (C) Notes 1. When using this rule, depending on the vlues given, ny omintion of the two equlities my e used to solve prtiulr tringle. 2. To solve tringle mens to find ll unknown side lengths nd ngles. The sine rule n e used to solve non right-ngled tringles if we re given: 1. two ngles nd one side length 2. two side lengths nd n ngle opposite one of these side lengths. h C C D h = sin () nd h = sin (C) C 190 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2
WorKEd EXMPLE 7 In the tringle C, = 4 m, = 7 m nd = 80. Find, C nd. think 1 Drw lelled digrm of the tringle C nd fill in the given informtion. WritE/drW = 7 C 2 Chek tht one of the riteri for the sine rule hs The sine rule n e used sine two side een stisfied. lengths nd n ngle opposite one of these side lengths hve een given. 3 Write the sine rule to find. To find ngle : sin () = sin () 4 Sustitute the known vlues into the rule. 4 sin () = 7 sin (80 ) 5 Trnspose the eqution to mke sin () the sujet. 4 sin (80 ) = 7 sin () 80 = 4 4 sin (80 ) sin () = 7 6 Evlute. = sin 1 4 sin (80 ) 7 = sin 1 (0.562 747 287) = 34.246 004 71 7 Round the nswer to degrees nd minutes. = 34 15 8 Determine the vlue of ngle C using the ft tht C = 180 (80 + 34 15 ) the ngle sum of ny tringle is 180. = 65 45 9 Write the sine rule to find. To find side length : sin (C) = sin () 10 Sustitute the known vlues into the rule. sin (65 45 ) = 7 sin (80 ) 7 sin (65 45 ) 11 Trnspose the eqution to mke the sujet. = sin (80 ) 12 Evlute. Round the nswer to 2 deiml ples nd inlude the pproprite unit. 7 0.911 762 043 = 0.984 807 753 6.382 334 305 = 0.984 807 753 = 6.480 792 099 = 6.48 m Topi 5 TrIgonoMETrIC rtios nd THEIr PPLICTIonS 191
WorKEd EXMPLE 8 think The miguous se When using the sine rule there is one importnt issue to onsider. If we re given two side lengths nd n ngle opposite one of these side lengths, then sometimes two different tringles n e drwn. For exmple, if = 10, = 6 nd C = 30, two possile tringles ould e reted. In the first se, ngle is n ute ngle, while in the seond se, ngle is n otuse ngle. The two vlues for will dd to 180. The miguous se does not work for every exmple. It would e useful to know, efore ommening question, whether or not the miguous se exists nd, if so, to then find oth sets of solutions. = 6 = 10 The miguous se exists if C is n ute ngle nd > > sin (C), or ny equivlent sttement; for exmple, if is n ute ngle nd > > sin (), nd so on. In the tringle C, = 10 m, = 6 m nd C = 30. Show tht the miguous se exists. = 6 = 10 Find two possile vlues of, nd hene two possile vlues of nd. 1 Chek tht the onditions for n miguous se exist, i.e. tht C is n ute ngle nd tht > > sin (C). WritE/drW C = 30 so C is n ute ngle. sin (C) = sin (30 ) = 0.5 > > sin (C) 10 > 6 > 10 sin (30 ) 10 > 6 > 5 This is orret. 2 Stte the nswer. This is n miguous se of the sine rule. Cse 1 1 Drw lelled digrm of the tringle C nd fill in the given informtion. = 6 = 10 2 Write the sine rule to find. To find ngle : sin () = sin (C) 3 Sustitute the known vlues into the rule. 10 sin () = 6 sin (30 ) 30 C 30 30 C C 192 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2
4 Trnspose the eqution to mke sin () the sujet. 10 sin (30 ) = 6 sin () 10 sin (30 ) sin () = 6 5 Evlute ngle, in degrees nd minutes. = sin 1 10 sin (30 ) 6 6 Determine the vlue of ngle, using the ft tht the ngle sum of ny tringle is 180. 7 Write the sine rule to find. To find side length : sin () = sin (C) = 56 27 = 180 (30 + 56 27 ) = 93 33 8 Sustitute the known vlues into the rule. sin (93 33 ) = 6 sin (30 ) 6 sin (93 33 ) 9 Trnspose the eqution to mke the = sujet nd evlute. sin (30 ) = 11.98 m Cse 2 1 Drw lelled digrm of the tringle C nd fill in the given informtion. 2 Write the lterntive vlue for ngle. Sutrt the vlue otined for in Cse 1 from 180. 3 Determine the lterntive vlue of ngle, using the ft tht the ngle sum of ny tringle is 180. = 6 = 10 30 To find the lterntive ngle : If sin () = 0.8333, then ould lso e: = 180 56 27 = 123 33 = 180 (30 + 123 33) = 26 27 4 Write the sine rule to find the lterntive. To find side length : sin () = sin (C) 5 Sustitute the known vlues into the rule. sin (26 27 ) = 6 sin (30 ) 6 Trnspose the eqution to mke the sujet nd evlute. C 6 sin (26 27 ) = sin (30 ) = 5.35 m Topi 5 Trigonometri rtios nd their pplitions 193
Exerise 5.4 PRtise CONSOLIDTE Hene, for Worked exmple 8 there were two possile solutions s shown y the digrms elow. = 6 123 33ʹ 26 27ʹ The sine rule = 10 = 6 30 C 56 27ʹ = 5.35 = 11.98 93 33ʹ = 10 1 WE7 In the tringle C, = 10, = 12 nd = 58. Find, C nd. 2 In the tringle C, = 17.35, = 26.82 nd = 101 47. Find C, nd. 3 WE8 In the tringle C, = 10, = 8 nd C = 50. Find two possile vlues of nd hene two possile vlues of. 4 In the tringle C, = 20, = 12 nd = 35. Find two possile vlues for the perimeter of the tringle. 5 In the tringle C, = 27, C = 42 nd = 105. Find, nd. 6 In the tringle C, = 7, = 5 nd = 68. Find the perimeter of the tringle. 7 Find ll unknown sides nd ngles for the tringle C, given = 32, = 51 nd = 28. 8 Find the perimeter of the tringle C if = 7.8, = 6.2 nd = 50. 9 In tringle C, = 40, C = 80 nd = 3. The vlue of is: 2.64 2.86 C 14 D 4.38 E 4.60 10 Find ll unknown sides nd ngles for the tringle C, given = 27, = 43 nd = 6.4. 11 Find ll unknown sides nd ngles for the tringle C, given = 25, = 17 nd = 13. 12 To lulte the height of uilding, Kevin mesures the ngle of elevtion to the top s 48. He then wlks 18 m loser to the uilding nd mesures the ngle of elevtion s 64. How high is the uilding? 13 river hs prllel nks tht run diretly est west. From one nk Kylie tkes ering to tree on the oppsite nk. The ering is 047 T. She then wlks 10 m due est nd tkes seond ering to the tree. This is 305 T. Find: her distne from the seond mesuring point to the tree the width of the river, to the nerest metre. 14 ship sils on ering of S20 W for 14 km, then hnges diretion nd sils for 20 km nd drops nhor. Its ering from the strting point is now N65 W. How fr is it from the strting point? On wht ering did it sil the 20 km leg? 15 ross-ountry runner runs t 8 km/h on ering of 150 T for 45 minutes, then hnges diretion to ering of 053 T nd runs for 80 minutes until he is due est of the strting point. i How fr ws the seond prt of the run? ii Wht ws his speed for this setion? iii How fr does he need to run to get k to the strting point? MSTER 30 C 194 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2
5.5 Units 1 & 2 OS 4 Topi 3 Conept 5 The osine rule Conept summry Prtie questions From fire tower,, fire is spotted on ering of N42 E. From seond tower,, the fire is on ering of N12 W. The two fire towers re 23 km prt, nd is N63 W of. How fr is the fire from eh tower? 16 liff is 37 m high. The rok slopes outwrd t n ngle of 50 to the horizontl, then uts k t n ngle of 25 to the vertil, meeting the ground diretly elow the top of the liff. Crol wishes to seil from the top of the liff to the ground s shown in the digrm elow right. Her liming rope is 45 m long, nd she needs 2 m to seure it to tree t the top of the liff. Will the rope e long enough to llow her to reh the ground? The osine rule Rope 25 50 Rok 37 m In ny non right-ngled tringle C, perpendiulr line n e drwn from ngle to side. Let D e the point where the perpendiulr line meets side, nd the length of the perpendiulr line e h. Let the length D = x units. The perpendiulr line retes two right-ngled tringles, D nd CD. Using tringle D nd Pythgors theorem, we otin: 2 = h 2 + x 2 [1] Using tringle CD nd Pythgors theorem, we otin: h 2 = h 2 + ( x) 2 [2] D Expnding the rkets in eqution [2]: x x 2 = h 2 + 2 2x + x 2 Rerrnging eqution [2] nd using 2 = h 2 + x 2 from eqution [1]: 2 = h 2 + x 2 + 2 2x = 2 + 2 2x = 2 + 2 2x From tringle D, x = os (), therefore 2 = 2 + 2 2x eomes 2 = 2 + 2 2 os () This is lled the osine rule nd is generlistion of Pythgors theorem. C Topi 5 Trigonometri rtios nd their pplitions 195
WorKEd EXMPLE 9 In similr wy, if the perpendiulr line ws drwn from ngle to side or from ngle C to side, the two right-ngled tringles would give 2 = 2 + 2 2 os (C) nd 2 = 2 + 2 2 os () respetively. From this, the osine rule n e stted: In ny tringle C 2 = 2 + 2 2 os () 2 = 2 + 2 2 os () 2 = 2 + 2 2 os (C) The osine rule n e used to solve non right-ngled tringles if we re given: 1. three sides of the tringle 2. two sides of the tringle nd the inluded ngle (the ngle etween the given sides). Find the third side of tringle C orret to 2 deiml ples given = 6, = 10 nd = 76. think 1 Drw lelled digrm of the tringle C nd fill in the given informtion. 2 Chek tht one of the riteri for the osine rule hs een stisfied. 3 Write the pproprite osine rule to find side. Note: One the third side hs een found, the sine rule ould e used to find other ngles if neessry. If three sides of tringle re known, n ngle ould e found y trnsposing the osine rule to mke os, os or os C the sujet. WritE/drW = 10 76 os () = 2 + 2 2 2 os () = 2 + 2 2 2 os (C) = 2 + 2 2 2 = 6 C Yes, the osine rule n e used sine two side lengths nd the inluded ngle hve een given. To find side : 2 = 2 + 2 2 os () 4 Sustitute the given vlues into the rule. = 6 2 + 10 2 2 6 10 os (76 ) 5 Evlute. = 36 + 100 120 0.241 921 895 = 106.969 372 5 = 106.969 372 5 6 Round the nswer to 2 deiml ples. = 10.34 orret to 2 deiml ples C 196 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2
WorKEd EXMPLE 10 Find the smllest ngle in the tringle with sides 4 m, 7 m nd 9 m. think 1 Drw lelled digrm of the tringle, ll it C nd fill in the given informtion. Note: The smllest ngle will e opposite the smllest side. 2 Chek tht one of the riteri for the osine rule hs een stisfied. WritE/drW = 7 Let = 4 = 7 = 9 = 9 C = 4 The osine rule n e used sine three side lengths hve een given. 3 Write the pproprite osine rule to find ngle. os () = 2 + 2 2 4 Sustitute the given vlues into the rerrnged rule. = 72 + 9 2 4 2 2 2 7 9 5 Evlute. = 49 + 81 16 126 = 114 126 6 Trnspose the eqution to mke the sujet y tking the inverse os of oth sides. = os 1 114 126 = 25.208 765 3 7 Round the nswer to degrees nd minutes. = 25 13 WorKEd EXMPLE 11 think Two rowers set out from the sme point. One rows N70 E for 2000 m nd the other rows S15 W for 1800 m. How fr prt re the two rowers? 1 Drw lelled digrm of the tringle, ll it C nd fill in the given informtion. WritE/drW 15 1800 m N 2000 m C 70 Topi 5 TrIgonoMETrIC rtios nd THEIr PPLICTIonS 197
2 Chek tht one of the riteri for the osine rule hs een stisfied. The osine rule n e used sine two side lengths nd the inluded ngle hve een given. 3 Write the pproprite osine rule to To find side : find side. 2 = 2 + 2 2 os (C) 4 Sustitute the given vlues into the rule. = 2000 2 + 1800 2 2 2000 1800 os (125 ) 5 Evlute. = 40 000 000 + 3 240 000 7 200 000 0.573 576 436 = 11 369 750.342 = 11 369 750.342 = 3371.906 04 6 Round the nswer to 2 deiml ples. = 3371.91 7 nswer the question. The rowers re 3371.91 m prt. Exerise 5.5 PRCTISE CONSOLIDTE The osine rule 1 WE9 Find the third side of tringle C given = 3.4, = 7.8 nd C = 80. 2 In tringle C, = 64.5 m, = 38.1 m nd = 58 34. Find the third side,. 3 WE10 Find the smllest ngle in the tringle with sides 6 m, 4 m nd 8 m. 4 In tringle C, = 356, = 207 nd = 296. Find the smllest ngle. 5 WE11 Two rowers set out from the sme point. One rows N30 E for 1500 m nd the other rows S40 E for 1200 m. How fr prt re the two rowers? 6 Two rowers set out from the sme point. One rows 16.2 km on ering of 053 T nd the other rows 31.6 km on ering of 117 T. How fr prt re the two rowers? 7 In tringle C, = 17, = 10 nd = 115. Find, nd hene find nd C. 8 In tringle C, = 23.6, = 17.3 nd = 26.4. Find the size of ll the ngles. 9 In tringle DEF, d = 3 m, e = 7 m nd F = 60. Find f in ext form. 10 Mri yles 12 km in diretion of N68 W, then 7 km in diretion of N34 E. How fr is she from her strting point? Wht is the ering of the strting point from her finishing point? 11 grden ed is in the shpe of tringle with sides of length 3 m, 4.5 m nd 5.2 m. Clulte the smllest ngle. Hene, find the re of the grden. (Hint: Drw digrm with the longest length s the se of the tringle.) 12 hokey gol is 3 m wide. When Sophie is 7 m from one post nd 5.2 m from the other, she shoots for gol. Within wht ngle, to the nerest degree, must the shot e mde if it is to sore gol? 198 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2
MSTER 5.6 Units 1 & 2 OS 4 Topi 3 Conept 5 rs, setors nd segments Conept summry Prtie questions 13 Three irles of rdii 5 m, 6 m nd 8 m re positioned 5 m so tht they just touh one nother. Their entres form the 6 m verties of tringle. Find the lrgest ngle in the tringle. 14 n dvertising lloon is tthed to two ropes 120 m nd 100 m long. The ropes re nhored to level ground 35 m prt. How high n the lloon fly? 15 plne flies N70 E for 80 km, then on ering of 8 m S10 W for 150 km. How fr is the plne from its strting point? Wht diretion is the plne from its strting point? 16 plne tkes off t 10:00 m from n irfield nd flies t 120 km/h on ering of N35 W. seond plne tkes off t 10:05 m from the sme irfield nd flies on ering of S80 E t speed of 90 km/h. How fr prt re the plnes t 10:25 m? 17 For the given shpe t right, determine: 8 the length of the digonl 150 x the mgnitude (size) of ngle 7 the length of x. 60 18 From the top of vertil liff 68 m high, n oserver 10 noties yht t se. The ngle of depression to the yht is 47. The yht sils diretly wy from the liff, nd fter 10 minutes the ngle of depression is 15. How fst does the yht sil? rs, setors nd segments Rdin mesurement In ll of the trigonometry tsks overed so fr, the unit for mesuring ngles hs een the degree. There is nother ommonly used mesurement for ngles: the rdin. This is used in situtions involving length nd res ssoited with irles. Consider the unit irle, irle with rdius of 1 unit. OP is the rdius. If OP is rotted θ ntilokwise, the point P tres pth long the irumferene of the irle to new point, P 1. The r length PP 1 is rdin mesurement, symolised y θ. Note: 1 is equivlent to the ngle in degrees formed when the length of PP 1 is 1 unit; in other words, when the r is the sme length s the rdius. O O P OP = 1 unit P 1 θ θ P OP = 1 unit Topi 5 Trigonometri rtios nd their pplitions 199
WorKEd EXMPLE 12 think If the length OP is rotted 180, the point P tres out hlf the irumferene. Sine the irle hs rdius of 1 unit, nd C = 2πr, the r PP 1 hs length of π. The reltionship etween degrees nd rdins is thus estlished. 180 = π This reltionship will e used to onvert from one system to nother. Rerrnging the si onversion ftor gives: 180 = π 1 = π 180 π To onvert n ngle in degrees to rdin mesure, multiply y 180. lso, sine π = 180, it follows tht 1 = 180 π. To onvert n ngle in rdin mesure to degrees, multiply y 180 π. Where possile, it is ommon to hve rdin vlues with π in them. It is usul to write rdins without ny symol, ut degrees must lwys hve symol. For exmple, n ngle of 25 must hve the degree symol written, ut n ngle of 1.5 is understood to e 1.5 rdins. P 1 O 1 2 irumferene 180 Convert 135 to rdin mesure, expressing the nswer in terms of π. Convert the rdin mesurement 4π to degrees. 5 WritE 1 To onvert n ngle in degrees to rdin mesure, multiply the ngle y π 180. 135 = 135 π 180 = 135π 180 2 Simplify, leving the nswer in terms of π. = 3π 4 1 To onvert rdin mesure to n ngle in degrees, multiply the ngle y 180 π. 2 Simplify. Note: π nels out. 4π 5 = 4π 5 180 π = 720 5 = 144 P 200 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2
WorKEd EXMPLE 13 If the lultion does not simplify esily, write the nswers in degrees nd minutes, or rdins to 4 deiml ples. If ngles re given in degrees nd minutes, onvert to degrees only efore onverting to rdins. r length n r is setion of the irumferene of irle. The length of n r is proportionl to the ngle sutended t the entre. For exmple, n ngle of 90 will rete n r tht is 1 4 the irumferene. We hve lredy defined n r length s equivlent to θ rdins if the irle hs rdius of 1 unit. Therefore, simple diltion of the unit irle will enle us to lulte the r length for ny sized irle, s long s the ngle is expressed in rdins. If the rdius is dilted y ftor of r, the r length is lso dilted y ftor of r. Therefore, l = r θ, where l represents the r length, r represents the rdius nd θ represents n ngle mesured in rdins. Note: In order to use the formul for the length of the r, the ngle must e in rdin mesure. O O θ r = 1 θ r θ rθ Diltion y ftor of r Find the length of the r tht sutends n ngle of 75 t the entre of irle with rdius 8 m. think 1 Drw digrm representing the sitution nd lel it with the given vlues. WritE/drW O 75 r = 8 l = rθ 2 Convert the ngle from 75 to rdin mesure 75 = 75 π π y multiplying the ngle y 180. 180 = 75π 180 3 Evlute to 4 deiml ples. = 1.3090 4 Write the rule for the length of the r. l = r θ 5 Sustitute the vlues into the formul. = 8 1.3090 6 Evlute to 2 deiml ples nd inlude the = 10.4720 pproprite unit. = 10.47 m Topi 5 TrIgonoMETrIC rtios nd THEIr PPLICTIonS 201
WorKEd EXMPLE 14 Find the ngle sutended y 17 m r in irle of rdius 14 m: in rdins in degrees. think WritE 1 Write the rule for the length of the r. l = r θ 2 Sustitute the vlues into the formul. 17 = 14 θ 3 Trnspose the eqution to mke θ the sujet. θ = 17 14 4 Evlute to 4 deiml ples nd inlude the = 1.214 285 714 pproprite unit. = 1.2143 1 To onvert rdin mesure to n ngle in 1.2143 = 1.2143 180 degrees, multiply the ngle y 180 π π. 2 Evlute. = 69.573 446 55 3 Convert the ngle to degrees nd minutes. = 69 34 WorKEd EXMPLE 15 think re of setor In the digrm t right, the shded re is the minor setor O, nd the unshded re is the mjor setor O. The re of the setor is proportionl to the r length. For exmple, n re of 1 4 of the irle ontins n r tht is 1 4 of the irumferene. Thus, in ny irle: re of setor re of irle = r length irumferene of irle πr = rθ where θ is mesured in rdins. 2 2πr rθ πr2 = 2πr The re of setor is: = 1 2 r2 θ = 1 2 r 2 θ setor hs n re of 157 m 2 nd sutends n ngle of 107. Wht is the rdius of the irle? 1 Convert the ngle from 107 to rdin mesure y π multiplying the ngle y 180. WritE 107 = 107 π 180 = 107π 180 O Mjor setor Minor setor 202 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2
2 Evlute to 4 deiml ples. = 1.8675 3 Write the rule for the re of setor. 4 Sustitute the vlues into the formul. 5 Trnspose the eqution to mke r 2 the sujet. re of segment segment is tht prt of setor ounded y the r nd the hord. s n e seen from the digrm t right: re of segment = re of setor re of tringle = 1 O 2 r 2 1 θ 2 r 2 θ sin (θ ) r θ = 1 2 r 2 (θ sin (θ )) Note: θ is in rdins nd θ is in degrees. The re of segment: = 1 2 r 2 (θ sin (θ )) = 1 2 r2 θ 157 = 1 2 r2 1.8675 r 2 = 2 157 1.8675 r 2 = 168.139 016 5 6 Tke the squre root of oth sides of the eqution. r = 12.966 842 97 7 Evlute to 2 deiml ples nd inlude the pproprite unit. WorKEd EXMPLE 16 = 12.97 m Segment Find the re of the segment in irle of rdius 5 m, sutended y n ngle of 40. think WritE 1 Convert the ngle from 40 to rdin mesure y 40 = 40 π 180 π multiplying the ngle y 180. = 40π 180 2 Evlute to 4 deiml ples. = 0.6981 3 Write the rule for the re of segment. = 1 2 r 2 (θ sin (θ )) 4 Identify eh of the vriles. r = 5, θ = 0.6981, θ = 40 5 Sustitute the vlues into the formul. = 1 2 52 (0.6981 sin (40 )) 6 Evlute. 7 Round to 2 deiml ples nd inlude the pproprite unit. = 1 25 0.0553 2 = 0.69125 = 0.69 m 2 Topi 5 TrIgonoMETrIC rtios nd THEIr PPLICTIonS 203
Exerise 5.6 PRtise CONSOLIDTE rs, setors nd segments 1 WE12 Convert the following ngles to rdin mesure, expressing nswers in terms of π. 30 60 120 d 150 e 225 f 270 g 315 h 480 i 72 j 200 2 Convert the following rdin mesurements into degrees. π 3π 7π d 5π e 7π 4 2 6 3 12 f 17π g π h 13π i 11π j 8π 6 12 10 8 3 WE13 Find the length of the r tht sutends n ngle of 65 t the entre of irle of rdius 14 m. 4 Find the length of the r tht sutends n ngle of 153 t the entre of irle of rdius 75 mm. 5 WE14 Find the ngle sutended y 20 m r in irle of rdius 75 m: in rdins in degrees. 6 Find the ngle sutended y n 8 m r in irle of rdius 5 m: in rdins in degrees. 7 WE15 setor hs n re of 825 m 2 nd sutends n ngle of 70. Wht is the rdius of the irle? 8 setor hs n re of 309 m 2 nd sutends n ngle of 106. Wht is the rdius of the irle? 9 WE16 Find the re of the segment in irle of rdius 25 m sutended y n ngle of 100. 10 Find the re of the segment of irle of rdius 4.7 m tht sutends n ngle of 85 20 t the entre. 11 Convert the following ngles in degrees to rdins, giving nswers to 4 deiml ples. 27 109 243 d 351 e 7 f 63 42 g 138 21 h 274 8 i 326 53 j 47 2 12 Convert the following rdin mesurements into degrees nd minutes. 2.345 0.6103 1 d 1.61 e 3.592 f 7.25 g 0.182 h 5.8402 i 4.073 j 6.167 13 Find the length of the r tht sutends n ngle of 135 t the entre of irle of rdius 10 m. Leve the nswer in terms of π. 14 n r of irle is 27.8 m long nd sutends n ngle of 205 t the entre of the irle. Wht is the rdius of the irle? 15 n r of length 8 m is mrked out on the irumferene of irle of rdius 13 m. Wht ngle does the r sutend t the entre of the irle? 16 The minute hnd of lok is 35 m long. How fr does the tip of the hnd trvel in 20 minutes? 204 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2
MstEr 17 hild s swing is suspended y rope 3 m long. Wht is the length of the r it trvels if it swings through n ngle of 42? 18 Find the re of the setor of irle of rdius 6 m with n ngle of 100. Write your nswer in terms of π. 19 grden ed is in the form of setor of irle of rdius 4 m. The r of the setor is 5 m long. Find: the re of the grden ed the volume of mulh needed to over the ed to depth of 10 m. 20 setor whose ngle is 150 is ut from irulr piee of rdord whose rdius is 12 m. The two stright edges of the setor re joined so s to form one. Wht is the surfe re of the one? Wht is the rdius of the one? 21 Two irrigtion sprinklers spred wter in irulr pths with rdii of 7 m nd 4 m. If the sprinklers re 10 m prt, find the re of rop tht reeives wter from oth sprinklers. 22 Two irles of rdii 3 m nd 4 m hve their entres 5 m prt. Find the re of the intersetion of the two irles. Topi 5 TrIgonoMETrIC rtios nd THEIr PPLICTIonS 205
ONLINE ONLY 5.7 Review the Mths Quest review is ville in ustomisle formt for you to demonstrte your knowledge of this topi. the review ontins: Multiple-hoie questions providing you with the opportunity to prtise nswering questions using CS tehnology short-nswer questions providing you with the opportunity to demonstrte the skills you hve developed to effiiently nswer questions using the most pproprite methods studyon is n intertive nd highly visul online tool tht helps you to lerly identify strengths nd weknesses prior to your exms. You n then onfidently trget res of gretest need, enling you to hieve your est results. www.jplus.om.u Extended-response questions providing you with the opportunity to prtise exm-style questions. summry of the key points overed in this topi is lso ville s digitl doument. REVIEW QUESTIONS Downlod the Review questions doument from the links found in the Resoures setion of your eookplus. Units 1 & 2 Trigonometri rtios nd their pplitions Sit topi test 206 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2
5 nswers Exerise 5.2 1 6.43 11.89 24.99 d 354.05 2 4.14 18.11 445.90 d x = 21.14, y = 27.13 3 44 26 67 23 44 25 d 17 10 4 68 58 38 41 47 4 d 61 55 5 2 3 m 12 3 m 2 12 + 8 3 m 6 26 3 + 54 m 7 4.98 m 66 56 8 7 11 9 23 4 10 8.58 m 11 1.44 m 12 = 14.90, = 20.05 13 4 nd 4 3 14 115.91 m 15 x = 13.39 16 64 51, 64 51, 50 18 17 10.91 m 3 18 0.76 m No, the foot of the ldder moves through distne of 0.95 m. Exerise 5.3 1 571 m 2 30 m 3 1691 m 4 79 km 5 201 48 T 6 326.3 T 7 325 T 227 T 058 T d 163 T e S66 W f S73 E g N39 W h N74 E 8 C D 9 90 m 10 6 47 11 Heliopter 35 48 2500 m 1319.36 m 12 22 m 13 50 50 3 m 3 14 5.39 km N21 48 W 15 12.2 km 348 T or N12 W 16 29.82 km 38.08 km 232 T 17 112.76 km 5 hours 30 minutes 18 i 571.5 m ii 715 m i 143.5 m ii 4.31 km/h Exerise 5.4 1 44 58, 77 2, 13.79 2 39 18, 38 55, 17.21 3 = 73 15, = 8.83; or = 106 45, = 4.12 4 51.9 or 44.86 5 33, 38.98, 21.98 6 19.12 7 = 48 26, C = 103 34, = 66.26; or = 131 34, C = 20 26, = 23.8 8 24.17 9 10 C = 110, = 3.09, = 4.64 11 = 33 33, C = 121 27, = 26.24; or = 146 27, C = 8 33, = 4.57 12 43.62 m 13 6.97 m 4 m 14 13.11 km N20 47 W 15 i 8.63 km ii 6.48 km/h iii 9.90 km 22.09 km from nd 27.46 km from 16 Yes, she needs 43 m ltogether. Exerise 5.5 1 7.95 2 55.22 m 3 28 57 4 35 32 5 2218 m 6 28.5 km 7 23.08, 41 53, 23 7 8 = 61 15, = 40, C = 78 45 9 37 m 10 12.57 km S35 1 E S 2 S 1 Topi 5 Trigonometri rtios nd their pplitions 207
11 35 6 6.73 m 2 12 23 13 70 49 14 89.12 m 15 130 km S22 12 E 16 74.3 km 17 8.89 m 76 59 x = 10.07 m 18 1.14 km/h Exerise 5.6 π 1 6 5π d 6 e π 3 5π 4 f 2π 3 3π 2 g 7π 8π h 2π i 4 3 5 j 10π 9 2 45 270 210 d 300 e 105 f 510 g 15 h 234 i 247.5 j 1440 3 15.88 m 4 200.28 mm 5 0.2667 15 17 6 1.6 91 40 7 36.75 m 8 18.28 m 9 237.66 m 2 10 5.44 m 2 11 0.4712 1.9024 4.2412 d 6.1261 e 0.1222 f 1.1118 g 2.4147 h 4.7845 i 5.7052 j 0.8209 12 134 22 34 58 57 18 d 92 15 e 205 48 f 415 24 g 10 26 h 334 37 i 233 22 j 353 21 13 15π 2 14 7.77 m 15 35 16 16 73.3 m 17 2.20 m 18 = 10π m 2 19 10 m 2 1 m 3 20 188.5 m 2 5 m 21 2.95 m 2 22 6.64 m 2 208 MTHS QUEST 11 SPECILIST MTHEMTICS VCE Units 1 nd 2