4.4. Applications of Trigonometry to Triangles. Introduction. Prerequisites. Learning Outcomes

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1 pplitions of Trigonometry to Tringles 4.4 Introdution We originlly introdued trigonometry using right-ngled tringles. However, the sujet hs pplitions in deling with ny tringles suh s those tht might rise in surveying, nvigtion or mehnisms. In this Setion we show how, given ertin informtion out tringle, we n use pproprite rules, lled the sine rule nd the osine rule, to fully solve the tringle i.e. otin the lengths of ll the sides nd the size of ll the ngles of tht tringle. Prerequisites efore strting this Setion you should. Lerning Outomes fter ompleting this Setion you should e le to. 1 hve knowledge of the sis of trigonometry e wre of the stndrd trigonometri identities use trigonometry in everydy situtions fully determine ll the sides nd ngles nd the re of ny tringle from prtil informtion

2 1. pplitions of trigonometry to tringles re of tringle Clulte n expression for the re of the right-ngled tringle C shown in Figure 1. θ C Figure 1 Your solution The re S of ny tringle is given y S = 1 (se) (perpendiulr height) where perpendiulr height mens the perpendiulr distne from the side lled the se to the opposite vertex. Thus for the right-ngled tringle shown S = 1. Now onsider the tringle C shown. (This is n otuse ngled tringle ut the formul otined will hold for ny tringle.) h µ C C Figure Using the onstrution shown in Figure, the re of tringle C is S = 1 h D HELM (VERSION 1: Mrh 18, 004): Workook Level 1 4.4: pplitions of Trigonometry to Tringles

3 If we use C to denote the ngle C then sin(180 C) = h (tringle CD is right-ngled). h = sin(180 C) = sin C (see the grph of the sine wve or expnd sin(180 )). S = 1 sin C y other similr onstrutions we ould demonstrte tht S = 1 sin nd S = 1 (1) sin (1, ) Note the pttern here: in eh formul for the re the ngle involved is the one etween the sides whose lengths our in tht expression. Clerly if C is right-ngle (so sin C =1)then s you dedued in the previous guided exerise. S = 1 The Sine Rule The sine rule is formul whih, if we re given ertin informtion out tringle, enles us to fully solve the tringle i.e. otin the lengths of ll three sides nd the vlue of ll three ngles. To show the rule we note tht from the formule (1) nd (1) for the re S of the tringle C in Figure we hve sin C = sin or sin = sin C Similrly using (1) nd (1) sin = sin or sin = sin Key Point For ny tringle C where is the length of the side opposite ngle, the side length opposite ngle nd the side length opposite ngle C the sine rule sttes sin = sin = sin C () 3 HELM (VERSION 1: Mrh 18, 004): Workook Level 1 4.4: pplitions of Trigonometry to Tringles

4 Use of the sine rule: To e le to fully determine ll the ngles nd sides of tringle it follows from the sine rule tht we must know either two ngles nd one side : (knowing two ngles of tringle relly mens tht ll three re known sine the sum of the ngles is 180 ) or two sides nd n ngle opposite one of those two sides. Exmple Solve the tringle C given tht =3m =46m nd ngle = =63.5. Solution Using the first pir of equtions in the sine rule () we hve 3 sin = 46. sin = 3 sin sin 63.5 =0.61 so = sin 1 (0.61) = 38.4 (y lultor) You should, however, note refully tht euse of the form of the grph of the sine funtion there re two ngles etween 0 nd 180 whih hve the sme vlue for their sine viz. x nd (180 x). See Figure 3. sin θ x 180 x θ In our exmple or Figure 3 = sin 1 (0.61) = 38.4 = = However sine we re given tht ngle is 63.5, the vlue of for ngle is lerly impossile. To omplete the prolem we simply note tht C = 180 ( )=78.35 The remining side is lulted from the sine rule, using either nd sin or nd sin. HELM (VERSION 1: Mrh 18, 004): Workook Level 1 4.4: pplitions of Trigonometry to Tringles 4

5 Complete this prolem y finding the length of side. Your solution Using, for exmple, we hve sin = sin C = sin C sin sin = = = m. The miguous se When, s the ove exmple, we re given two sides nd the non-inluded ngle of tringle, prtiulr re is required. Suppose, to e definite, tht sides nd nd the ngle re given. Then the ngle C is given y the sine rule s sin C = sin C Vrious ses n rise: (i) sin > This implies tht thn 1. sin > 1 in whih se no tringle exists sine sin C nnot e greter (ii) sin = In this se sin C = sin =1 soc =90. (Clerly this se n only rise if the given ngle is less thn 90!) (iii) sin < 5 HELM (VERSION 1: Mrh 18, 004): Workook Level 1 4.4: pplitions of Trigonometry to Tringles

6 Hene sin C = sin < 1. s mentioned erlier there re two possile vlues of ngle C in the rnge 0 to 180, one ute ngle (< 90 ) nd one otuse (etween 90 nd 180.) These ngles re C 1 = x nd C = 180 x. See Figure 4. If the given ngle is greter thn 90 then the otuse ngle C is not possile solution euse, of ourse, tringle nnot possess two otuse ngles. C C 1 C C 1 Figure 4 For less thn 90 there re still two possiilities. If the given side is greter thn the given side, the otuse ngle solution C is not possile euse then the lrger ngle would e opposite the smller side. (This ws the sitution in the worked exmple erlier.) The finl se <, <90 does give rise to two possile vlues C 1,C of the ngle C nd is referred to s the miguous se. In this se there will e two possile vlues 1 nd for the third side of the tringle orresponding to the two ngle vlues 1 = 180 ( + C 1 ) = 180 ( + C ) Show tht two tringles fit the following dt for tringle C. =4.5 m =7m =35 Otin the sides nd ngle of oth possile tringles. Your solution HELM (VERSION 1: Mrh 18, 004): Workook Level 1 4.4: pplitions of Trigonometry to Tringles 6

7 We hve, y the sine rule sin = sin = 7 sin =0.89 So = sin (y lultor) or = In this se, oth vlues of re indeed possile sine oth vlues re lrger thn ngle. (Side is longer thn side.) This is indeed the miguous se with two possile tringles. = 1 =63.15 = = C = C 1 =81.85 C = C =8.15 = 1 where = where 1 sin = 4.5 sin 35 sin 8.15 = 4.5 sin 35 1 = = = m = m You n lerly see tht we hve one ute ngled tringle 1 C 1 nd one otuse ngled C orresponding to the given dt. The Cosine Rule The osine rule is n lterntive formul for solving tringle C. It is prtiulrly useful for the se where the sine rule nnot e used, viz. when two sides of the tringle re known together with the ngle etween these two sides. Consider the two tringles C shown in Figure 5. C D C θ D () () Figure 5 In oth ses, using the right-ngled tringle D D = sin 7 HELM (VERSION 1: Mrh 18, 004): Workook Level 1 4.4: pplitions of Trigonometry to Tringles

8 In Figure 5() D = os CD = os In Figure 5() D = os(180 ) = os CD = + D = os In oth ses, in the right-ngled tringle DC (C) =(CD) +(D) So, using the ove results, = ( os ) + (sin ) = os + (os + sin ) giving = + os (3) Eqution (3) is one form of the osine rule. Clerly it n e used, s we stted ove, to lulte the side if the sides nd nd the inluded ngle re known. Note tht if =90, os =0nd (3) redues to Pythgors Theorem. Two similr formule to (3) for the osine rule n e similrly derived. One is = + os (3) Write down the third form of the osine rule, with s the sujet. Your solution = + os C (3) HELM (VERSION 1: Mrh 18, 004): Workook Level 1 4.4: pplitions of Trigonometry to Tringles 8

9 Exmple Solve the tringle where =7.00 m, =3.59 m, =47. Solution Sine two sides nd the ngle etween these sides is given we must first use the osine rule in the form (3): = (7.00) +(3.59) (7.00)(3.59) os 47 = = so = =5.55 m. We n now most esily use the sine rule to solve one of the remining ngles: 7.00 sin = 5.55 sin 47 so sin = 7.00 sin =0.974 from whih = 1 =76.96 or = = t this stge it is not ovious whih vlue is orret or whether this is the miguous se nd oth vlues of re possile. The two possile vlues for the remining ngle C re C 1 = 180 ( )=56.04 C = 180 ( ) =9.96 Sine for the sides of this tringle >> then similrly for the ngles we must hve >>C so the vlue C =9.96 is the orret one for the third side. The osine rule n lso e pplied to some tringles where the lengths, nd of the three sides re known nd the only lultions needed re finding the ngles. tringle C hs sides = 7m =11m =1m. Otin the vlues of ll the ngles of the tringle. (You will need to trnspose one of the osine rule formule (3) to otin the osine of n ngle s the sujet.) 9 HELM (VERSION 1: Mrh 18, 004): Workook Level 1 4.4: pplitions of Trigonometry to Tringles

10 Your solution Suppose, for the ske of exmple, we find ngle first using the trnsposed form of (3): os = +. os = + ( generl result) In this exmple os = = so = os 1 (0.818) = 35.1 (There is no other possiility etween 0 nd 180 for. No miguous se rises using the osine rule!) nother ngle or C ould now e otined using the sine rule or the osine rule. We shll use the ltter to otin. Trnsposing (3). os = + so here os = = 0.49 so = os 1 (0.49) = 64.6 fter whih C =80.3 HELM (VERSION 1: Mrh 18, 004): Workook Level 1 4.4: pplitions of Trigonometry to Tringles 10

11 Exerises 1. Determine the remining ngles nd sides for the following tringles: () C () C C () The tringles C with =50,=5,=6.(Tke speil re here!) (d) C. Determine ll the ngles of the tringles C where the sides hve lengths =7,=66 nd =9 3. () Use the osine rule nd hlf ngle formule to show tht, for tringle with sides of length,, os 1 s(s ) = where s = + + sin 1 = (s )(s ) (the semi-perimeter of the tringle.) () Hene show tht the re of tringle is given y s(s )(s )(s ) 4. Two ships leve port t 8.00 m, one trvelling t 1 knots (nutil miles per hour) the other t 10 knots. The fster ship mintins ering of N47 W, the slower one ering S0 W. Clulte the seprtion of the ships t middy. (Hint: Drw n pproprite digrm.) 5. The rnk mehnism shown elow hs n rm O of length 30 mm rotting ntilokwise out 0 nd onneting rod of length 60 mm. moves long the horizontl line O. Wht is the length O when O hs rotted y 1 of omplete revolution from 8 the horizontl? O 11 HELM (VERSION 1: Mrh 18, 004): Workook Level 1 4.4: pplitions of Trigonometry to Tringles

12 1. () Using the sine rule =6 sin 130 sin sin 130 = 6 sin 0 = sin C. From the two left-hnd equtions Then, sine C =30 sin 30, the right hnd pir of equtions give =6 sin () gin using the sine rule sin = 4 sin 80 = 3 sin C so sin C = 3 4 sin 80 = there re two possile ngles stisfying sin C = or C = sin 1 (0.7386). These re nd = However the otuse ngle vlue is impossile here euse the ngle is 80 nd the sum of the ngles would then exeed 180 Hene =47.01 so = 180 ( )=5.39. Then, sin 5.39 = 4 so =4 sin 80 sin sin 80 () y the sine rule sin = 5 sin 50 = 6 sin C. sin C =6 sin 50 5 = Then C = sin 1 (0.9193) = 66.8 (lultor) or = In this se oth vlues of C sy C 1 =66.8 nd C = re possile nd there re two possile tringles stisfying the given dt. Continued use of the sine rule produes i. with C 1 =66.8 (ute ngle tringle) = 1 = 180 ( )=63.18 = 1 =5.83 ii. with C = = =16.8 = =1.89 (d) In this se sine two sides nd the inluded ngle re given we must use the osine rule. The pproprite form is = + os =10 +1 ()(10)(1) os 6 =8.894 so = = 5.3 Continuing we use the osine rule gin to determine sy ngle C where = + os C tht is 10 =1 +(5.3) (1.)(5.3) os C from whih os C = nd C =55.51 (There is no other possiility for C etween 0 nd 180. Rell tht the osine of n ngle etween 90 nd 180 is negtive.) Finlly, = 180 ( )= HELM (VERSION 1: Mrh 18, 004): Workook Level 1 4.4: pplitions of Trigonometry to Tringles 1

13 . We use the osine rule firstly to find the ngle opposite the longest side. This will tell us whether the tringle ontins n otuse ngle. Hene we solve for using = + os C 81 = os C from whih 84 os C =4 os C = 4 84 from whih C =87.7. So there is no otuse ngle in this tringle nd we n use the sine rule knowing tht there is only one possile tringle fitting the dt. (We ould ontinue to use the osine rule if we wished of ourse.) Choosing to find the ngle we hve 6 sin = 9 sin 87.7 from whih sin = giving = (The otuse se for is not possile, s explined ove.) Finlly = 180 ( )= () y the osine rule ( ) ) + = os = os ( 1 ( ) from whih 4 os ( ) = + + =( + ) =( + + )( + ) If s = os (, then + = + + =(s ) ) =(s)(s ) from whih os ( ) = s(s ) Similrly using the first eqution of ( ) ut putting os =1 sin ( ) we otin ( ) 4 sin = + = ( ) = ( + )( + ). sin ( ) = (s )(s ) (s )(s ) = 13 HELM (VERSION 1: Mrh 18, 004): Workook Level 1 4.4: pplitions of Trigonometry to Tringles

14 () We showed t the egining of this setion tht the re of tringle is given y re= 1 sin = 1 ( ) ( ) () sin os Using the ove results we otin re = (s )(s ) s(s ) = s(s )(s )(s ) s required. N O 4. S t middy (4 hours trvelling) ships nd re respetively 48 nd 40 nutil miles from the port O. In tringle O we hve O = 180 (47 +0 )=113. We must use the osine rule to otin the required distne prt of the ships. Denoting the distne y, susul, = (48)(40) os 113 from whih = nd =73.5 nutil miles. 30mm 60mm (Position fter 1 8 revolution) 5. O y the sine rule sin = 60. sin = 30 sin sin 45 =0.353 so = (The otuse vlue of sin 1 (0.353) is impossile. Hene, = 180 ( )= Using the sine rule gin = O sin from whih O =77.5 mm. HELM (VERSION 1: Mrh 18, 004): Workook Level 1 4.4: pplitions of Trigonometry to Tringles 14