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 the study of 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 On ompletion you should e le to... 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 HELM (2008): Setion 4.4: pplitions of Trigonometry to Tringles 53

2 1. pplitions of trigonometry to tringles re of tringle The re S of ny tringle is given y S = 1 (se) (perpendiulr height) where perpendiulr 2 height mens the perpendiulr distne from the side lled the se to the opposite vertex. Thus for the right-ngled tringle shown in Figure 33() S = 1. For the otuse-ngled tringle 2 shown in Figure 33() the re is S = 1 2 h. h θ () θ () D Figure 33 If we use to denote the ngle in Figure 33() then sin(180 ) = h (tringle D is right-ngled)... h = sin(180 ) = sin (see the grph of the sine wve or expnd sin(180 ))... S = 1 2 sin 1() y other similr onstrutions we ould demonstrte tht S = 1 2 sin 1() nd S = 1 2 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. lerly if is right-ngle (so sin = 1) then S = 1 2 s for Figure 33(). Note: from now on we will not generlly write ut use the more usul =. 54 HELM (2008): Workook 4: Trigonometry

3 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 in Figure 33 we hve sin = sin or Similrly using (1) nd (1) sin = sin or sin = sin = sin sin Key Point 18 The Sine Rule For ny tringle where is the length of the side opposite ngle, the side length opposite ngle nd the side length opposite ngle sttes sin = sin = sin 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 3 Solve the tringle given tht = 32 m, = 46 m nd ngle = Solution Using the first pir of equtions in the Sine rule (Key Point 18) we hve 32 sin = sin = 32 sin sin = so = sin 1 (0.6212) = 38.4 (y lultor) HELM (2008): Setion 4.4: pplitions of Trigonometry to Tringles 55

4 Solution (ontd.) 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 i.e. x nd (180 x). See Figure 34. sin θ x 180 x θ In our exmple or = sin 1 (0.6212) = 38.4 = = Figure 34 However sine we re given tht ngle is 63.25, the vlue of for ngle is lerly impossile. To omplete the prolem we simply note tht = 180 ( ) = The remining side is lulted from the Sine rule, using either nd sin or nd sin. Tsk Find the length of side in Exmple 3. Your solution nswer Using, for exmple, we hve = sin sin sin = sin = 32 sin = = m. 56 HELM (2008): Workook 4: Trigonometry

5 The miguous se When, s in Exmple 3, we re given two sides nd the non-inluded ngle of tringle, prtiulr re is required. Suppose tht sides nd nd the ngle re given. Then the ngle is given y the Sine rule s Vrious ses n rise: (i) sin > This implies tht (ii) sin = In this se (iii) sin < sin sin = sin sin = sin Figure 35 > 1 in whih se no tringle exists sine sin nnot exeed 1. = 1 so = 90. Hene sin = sin < 1. s mentioned erlier there re two possile vlues of ngle in the rnge 0 to 180, one ute ngle (< 90 ) nd one otuse (etween 90 nd 180.) These ngles re 1 = x nd 2 = 180 x. See Figure 36. If the given ngle is greter thn 90 then the otuse ngle 2 is not possile solution euse, of ourse, tringle nnot possess two otuse ngles Figure 36 For less thn 90 there re still two possiilities. If the given side is greter thn the given side, the otuse ngle solution 2 is not possile euse then the lrger ngle would e opposite the smller side. (This ws the sitution in Exmple 3.) The finl se <, < 90 does give rise to two possile vlues 1, 2 of the ngle nd is referred to s the miguous se. In this se there will e two possile vlues 1 nd 2 for the third side of the tringle orresponding to the two ngle vlues 1 = 180 ( + 1 ) 2 = 180 ( + 2 ) HELM (2008): Setion 4.4: pplitions of Trigonometry to Tringles 57

6 Tsk Show tht two tringles fit the following dt for tringle : = 4.5 m = 7 m = 35 Otin the sides nd ngle of oth possile tringles. Your solution nswer We hve, y the Sine rule, sin = sin = 7 sin = 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 the miguous se with two possile tringles. = 1 = = 2 = = 1 = = 2 = = 1 where 1 sin = = sin 35 2 where sin = 4.5 sin 35 1 = = = m = m You n lerly see tht we hve one ute ngled tringle 1 1 nd one otuse ngled 2 2 orresponding to the given dt. 58 HELM (2008): Workook 4: Trigonometry

7 The osine rule The osine rule is n lterntive formul for solving tringle. It is prtiulrly useful for the se where the Sine rule nnot e used, i.e. when two sides of the tringle re known together with the ngle etween these two sides. onsider the two tringles shown in Figure 37. D D () () Figure 37 In Figure 37() using the right-ngled tringle D, D = sin. In Figure 37() using the right-ngled tringle D, D = sin(π ) = sin. In Figure 37() D = os... D = os In Figure 37() D = os(180 ) = os... D = + D = os In oth ses, in the right-ngled tringle D () 2 = (D) 2 + (D) 2 So, using the ove results, giving 2 = ( os ) (sin ) 2 = 2 2 os + 2 (os 2 + sin 2 ) 2 = os (3) Eqution (3) is one form of the osine rule. lerly 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 = 0 nd (3) redues to Pythgors theorem. Two similr formule to (3) for the osine rule n e similrly derived - see following Key Point: HELM (2008): Setion 4.4: pplitions of Trigonometry to Tringles 59

8 Key Point 19 osine Rule For ny tringle with sides,, nd orresponding ngles,, 2 = os os = = os os = = os os = Exmple 4 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): 2 = (7.00) 2 + (3.59) 2 2(7.00)(3.59) os 47 = = so = = m. We n now most esily use the Sine rule to solve one of the remining ngles: 7.00 sin = sin 47 so sin = = sin from whih = 1 = or = 2 = 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 re 1 = 180 ( ) = = 180 ( ) = Sine for the sides of this tringle > > then similrly for the ngles we must hve > > so the vlue 2 = 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. 60 HELM (2008): Workook 4: Trigonometry

9 Tsk tringle hs sides = 7m = 11 m = 12 m. Otin the vlues of ll the ngles of the tringle. (Use Key Point 19.) Your solution nswer Suppose we find ngle first using the following formul from Key Point 19 os = Here 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 ould now e otined using the Sine rule or the osine rule. Using the following formul from Key Point 19: os = = Sine + + = 180 we n dedue = 80.3 = so = os 1 (0.429) = 64.6 HELM (2008): Setion 4.4: pplitions of Trigonometry to Tringles 61

10 Exerises 1. Determine the remining ngles nd sides for the following tringles: () () () (d) The tringles with = 50, = 5, = 6. (Tke speil re here!) 2. Determine ll the ngles of the tringles where the sides hve lengths = 7, = 66 nd = 9 3. Two ships leve port t 8.00 m, one trvelling t 12 knots (nutil miles per hour) the other t 10 knots. The fster ship mintins ering of N47 W, the slower one ering S20 W. lulte the seprtion of the ships t middy. (Hint: Drw n pproprite digrm.) 4. 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 the 8 horizontl? O 62 HELM (2008): Workook 4: Trigonometry

11 nswers 1. () Using the Sine rule sin 130 = 6 sin 20 = sin. sin 130 = 6 sin From the two left-hnd equtions Then, sine = 30 sin 30, the right hnd pir of equtions give = 6 sin () gin using the Sine rule sin = 4 sin 80 = 3 sin so sin = 3 4 sin 80 = there re two possile ngles stisfying sin = or = 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 = so = 180 ( ) = Then, sin = 4 sin so = sin 80 sin 80 () In this se sine two sides nd the inluded ngle re given we must use the osine rule. The pproprite form is 2 = os = (2)(10)(12) os 26 = so = = 5.32 ontinuing we use the osine rule gin to determine sy ngle where 2 = os tht is 10 2 = (5.32) 2 2(1.2)(5.32) os from whih os = nd = (There is no other possiility for etween 0 nd 180. Rell tht the osine of n ngle etween 90 nd 180 is negtive.) Finlly, = 180 ( ) = (d) y the Sine rule sin = 5 sin 50 = 6 sin.. sin 50. sin = 6 5 = Then = sin 1 (0.9193) = (lultor) or = In this se oth vlues of sy 1 = nd 2 = re possile nd there re two possile tringles stisfying the given dt. ontinued use of the Sine rule produes (i) with 1 = (ute ngle tringle) = 1 = 180 ( ) = = 1 = 5.83 (ii) with 2 = = 2 = = 2 = 1.89 HELM (2008): Setion 4.4: pplitions of Trigonometry to Tringles 63

12 nswers ontinued 2. 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 2 = os 81 = os from whih 84 os = 4 os = 4/84 giving = 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.) hoosing to find the ngle we hve 6 sin = 9 sin from whih sin = giving = (The otuse se for is not possile, s explined ove.) Finlly = 180 ( ) = N O 3. 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 ( ) = 113. We must use the osine rule to otin the required distne prt of the ships. Denoting the distne y, s usul, 2 = (48)(40) os 113 from whih 2 = nd = 73.5 nutil miles. 4. y the Sine rule 30mm 30 sin = 60 sin 45 60mm... sin = sin 45 = so = (Position fter 1 8 revolution) O 45 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. 64 HELM (2008): Workook 4: Trigonometry