11.2 The Law of Sines

Transcription

1 894 Applitions of Trigonometry 11. The Lw of Sines Trigonometry literlly mens mesuring tringles nd with Chpter 10 under our belts, we re more thn prepred to do just tht. The min gol of this setion nd the next is to develop theorems whih llow us to solve tringles tht is, find the length of eh side of tringle nd the mesure of eh of its ngles. In Setions 10., 10.3 nd 10.6, we ve hd some experiene solving right tringles. The following exmple reviews wht we know. Exmple Given right tringle with hypotenuse of length 7 units nd one leg of length 4 units, find the length of the remining side nd the mesures of the remining ngles. Express the ngles in deiml degrees, rounded to the nerest hundreth of degree. Solution. For definitiveness, we lbel the tringle below. = 7 α β b = 4 To find the length of the missing side, we use the Pythgoren Theorem to get + 4 = 7 whih then yields = 33 units. Now tht ll three sides of the tringle re known, there re severl wys we n find α using the inverse trigonometri funtions. To derese the hnes of propgting error, however, we stik to using the dt given to us in the problem. In this se, the lengths 4 nd 7 were given, so we wnt to relte these to α. Aording to Theorem 10.4, os(α) = 4 7. Sine α is n ute ngle, α = ros ( 4 7) rdins. Converting to degrees, we find α Now tht we hve the mesure of ngle α, we ould find the mesure of ngle β using the ft tht α nd β re omplements so α + β = 90. One gin, we opt to use the dt given to us in the problem. Aording to Theorem 10.4, we hve tht sin(β) = 4 7 so β = rsin ( 4 7) rdins nd we hve β A few remrks bout Exmple re in order. First, we dhere to the onvention tht lower se Greek letter denotes n ngle 1 nd the orresponding lowerse English letter represents the side opposite tht ngle. Thus, is the side opposite α, b is the side opposite β nd is the side opposite γ. Tken together, the pirs (α, ), (β, b) nd (γ, ) re lled ngle-side opposite pirs. Seond, s mentioned erlier, we will strive to solve for quntities using the originl dt given in the problem whenever possible. While this is not lwys the esiest or fstest wy to proeed, it 1 s well s the mesure of sid ngle s well s the length of sid side

2 11. The Lw of Sines 895 minimizes the hnes of propgted error. 3 Third, sine mny of the pplitions whih require solving tringles in the wild rely on degree mesure, we shll dopt this onvention for the time being. 4 The Pythgoren Theorem long with Theorems 10.4 nd llow us to esily hndle ny given right tringle problem, but wht if the tringle isn t right tringle? In ertin ses, we n use the Lw of Sines to help. Theorem 11.. The Lw of Sines: Given tringle with ngle-side opposite pirs (α, ), (β, b) nd (γ, ), the following rtios hold or, equivlently, sin(α) = sin(β) b = sin(γ) sin(α) = b sin(β) = sin(γ) The proof of the Lw of Sines n be broken into three ses. For our first se, onsider the tringle ABC below, ll of whose ngles re ute, with ngle-side opposite pirs (α, ), (β, b) nd (γ, ). If we drop n ltitude from vertex B, we divide the tringle into two right tringles: ABQ nd BCQ. If we ll the length of the ltitude h (for height), we get from Theorem 10.4 tht sin(α) = h nd sin(γ) = h so tht h = sin(α) = sin(γ). After some rerrngement of the lst eqution, we get sin(α) using the tringles ABQ nd ACQ to get sin(β) α B β γ = sin(γ). If we drop n ltitude from vertex A, we n proeed s bove = sin(γ), ompleting the proof for this se. α b A C A C b Q A C b For our next se onsider the tringle ABC below with obtuse ngle α. Extending n ltitude from vertex A gives two right tringles, s in the previous se: ABQ nd ACQ. Proeeding s before, we get h = b sin(γ) nd h = sin(β) so tht sin(β) b = sin(γ). B h γ β h B Q γ B β α γ B β Q h γ A b C A b C 3 Your Siene tehers should thnk us for this. 4 Don t worry! Rdins will be bk before you know it!

3 896 Applitions of Trigonometry Dropping n ltitude from vertex B lso genertes two right tringles, ABQ nd BCQ. We know tht sin(α ) = h so tht h = sin(α ). Sine α = 180 α, sin(α ) = sin(α), so in ft, we hve h = sin(α). Proeeding to BCQ, we get sin(γ) = h so h = sin(γ). Putting this together with the previous eqution, we get sin(γ) B = sin(α), nd we re finished with this se. β h α α γ Q A b C The remining se is when ABC is right tringle. In this se, the Lw of Sines redues to the formuls given in Theorem 10.4 nd is left to the reder. In order to use the Lw of Sines to solve tringle, we need t lest one ngle-side opposite pir. The next exmple showses some of the power, nd the pitflls, of the Lw of Sines. Exmple Solve the following tringles. Give ext nswers nd deiml pproximtions (rounded to hundredths) nd sketh the tringle. 1. α = 10, = 7 units, β = 45. α = 85, β = 30, = 5.5 units 3. α = 30, = 1 units, = 4 units 4. α = 30, = units, = 4 units 5. α = 30, = 3 units, = 4 units 6. α = 30, = 4 units, = 4 units Solution. 1. Knowing n ngle-side opposite pir, nmely α nd, we my proeed in using the Lw of Sines. Sine β = 45 b, we use sin(45 ) = 7 sin(10 ) so b = 7 sin(45 ) sin(10 ) = units. Now tht we hve two ngle-side pirs, it is time to find the third. To find γ, we use the ft tht the sum of the mesures of the ngles in tringle is 180. Hene, γ = = 15. To find, we hve no hoie but to used the derived vlue γ = 15, yet we n minimize the propgtion of error here by using the given ngle-side opposite pir (α, ). The Lw of Sines gives us sin(15 ) = 7 sin(10 ) so tht = 7 sin(15 ) sin(10 ).09 units.5. In this exmple, we re not immeditely given n ngle-side opposite pir, but s we hve the mesures of α nd β, we n solve for γ sine γ = = 65. As in the previous exmple, we re fored to use derived vlue in our omputtions sine the only 5 The ext vlue of sin(15 ) ould be found using the differene identity for sine or hlf-ngle formul, but tht beomes unneessrily messy for the disussion t hnd. Thus ext here mens 7 sin(15 ) sin(10 ).

4 11. The Lw of Sines 897 ngle-side pir vilble is (γ, ). The Lw of Sines gives sin(85 ) = 5.5 sin(65 ). After the usul rerrngement, we get = 5.5 sin(85 ) sin(65 ) 5.77 units. To find b we use the ngle-side pir (γ, ) b whih yields sin(30 ) = 5.5 sin(65 ) hene b = 5.5 sin(30 ) sin(65 ).90 units. β = 30 β = 45 = 7 = α = 10 γ = 15 α = 85 γ = 65 b 5.7 b.90 Tringle for number 1 Tringle for number 3. Sine we re given (α, ) nd, we use the Lw of Sines to find the mesure of γ. We strt with sin(γ) 4 = sin(30 ) 1 nd get sin(γ) = 4 sin (30 ) =. Sine the rnge of the sine funtion is [ 1, 1], there is no rel number with sin(γ) =. Geometrilly, we see tht side is just too short to mke tringle. The next three exmples keep the sme vlues for the mesure of α nd the length of while vrying the length of. We will disuss this se in more detil fter we see wht hppens in those exmples. 4. In this se, we hve the mesure of α = 30, = nd = 4. Using the Lw of Sines, we get sin(γ) 4 = sin(30 ) so sin(γ) = sin (30 ) = 1. Now γ is n ngle in tringle whih lso ontins α = 30. This mens tht γ must mesure between 0 nd 150 in order to fit inside the tringle with α. The only ngle tht stisfies this requirement nd hs sin(γ) = 1 is γ = 90. In other words, we hve right tringle. We find the mesure of β to be β = = 60 nd then determine b using the Lw of Sines. We find b = sin(60 ) sin(30 ) = units. In this se, the side is preisely long enough to form unique right tringle. = 4 = 1 = 4 β = 60 = α = 30 α = 30 b 3.46 Digrm for number 3 Tringle for number 4 5. Proeeding s we hve in the previous two exmples, we use the Lw of Sines to find γ. In this se, we hve sin(γ) 4 = sin(30 ) 3 or sin(γ) = 4 sin(30 ) 3 = 3. Sine γ lies in tringle with α = 30,

5 898 Applitions of Trigonometry we must hve tht 0 < γ < 150. There re two ngles γ tht fll in this rnge nd hve sin(γ) = 3 : γ = rsin ( ) 3 rdins nd γ = π rsin ( 3) rdins At this point, we puse to see if it mkes sense tht we tully hve two vible ses to onsider. As we hve disussed, both ndidtes for γ re omptible with the given ngle-side pir (α, ) = (30, 3) in tht both hoies for γ n fit in tringle with α nd both hve sine of 3. The only other given piee of informtion is tht = 4 units. Sine >, it must be true tht γ, whih is opposite, hs greter mesure thn α whih is opposite. In both ses, γ > α, so both ndidtes for γ re omptible with this lst piee of given informtion s well. Thus hve two tringles on our hnds. In the se γ = rsin ( 3) rdins 41.81, we find 6 β = Using the Lw of Sines with the ngle-side opposite pir (α, ) nd β, we find b 3 sin( ) sin(30 ) 5.70 units. In the se γ = π rsin ( 3) rdins , we repet the ext sme steps nd find β nd b 1.3 units. 7 tringles re drwn below. Both β = 4 α = 30 β = 3 γ b 5.70 = 4 = 3 α = 30 γ b For this lst problem, we repet the usul Lw of Sines routine to find tht sin(γ) 4 = sin(30 ) 4 so tht sin(γ) = 1. Sine γ must inhbit tringle with α = 30, we must hve 0 < γ < 150. Sine the mesure of γ must be stritly less thn 150, there is just one ngle whih stisfies both required onditions, nmely γ = 30. So β = = 10 nd, using the Lw of Sines one lst time, b = 4 sin(10 ) sin(30 ) = units. = 4 β = 10 = 4 α = 30 γ = 30 b 6.93 Some remrks bout Exmple 11.. re in order. We first note tht if we re given the mesures of two of the ngles in tringle, sy α nd β, the mesure of the third ngle γ is uniquely 6 To find n ext expression for β, we onvert everything bk to rdins: α = 30 = π rdins, γ = rsin ( ) 6 3 rdins nd 180 = π rdins. Hene, β = π π rsin ( ) 6 3 = 5π rsin ( ) 6 3 rdins An ext nswer for β in this se is β = rsin ( ) 3 π rdins

6 11. The Lw of Sines 899 determined using the eqution γ = 180 α β. Knowing the mesures of ll three ngles of tringle ompletely determines its shpe. If in ddition we re given the length of one of the sides of the tringle, we n then use the Lw of Sines to find the lengths of the remining two sides to determine the size of the tringle. Suh is the se in numbers 1 nd bove. In number 1, the given side is djent to just one of the ngles this is lled the Angle-Angle-Side (AAS) se. 8 In number, the given side is djent to both ngles whih mens we re in the so-lled Angle-Side-Angle (ASA) se. If, on the other hnd, we re given the mesure of just one of the ngles in the tringle long with the length of two sides, only one of whih is djent to the given ngle, we re in the Angle-Side-Side (ASS) se. 9 In number 3, the length of the one given side ws too short to even form tringle; in number 4, the length of ws just long enough to form right tringle; in 5, ws long enough, but not too long, so tht two tringles were possible; nd in number 6, side ws long enough to form tringle but too long to swing bk nd form two. These four ses exemplify ll of the possibilities in the Angle-Side-Side se whih re summrized in the following theorem. Theorem Suppose (α, ) nd (γ, ) re intended to be ngle-side pirs in tringle where α, nd re given. Let h = sin(α) If < h, then no tringle exists whih stisfies the given riteri. If = h, then γ = 90 so extly one (right) tringle exists whih stisfies the riteri. If h < <, then two distint tringles exist whih stisfy the given riteri. If, then γ is ute nd extly one tringle exists whih stisfies the given riteri Theorem 11.3 is proved on se-by-se bsis. If < h, then < sin(α). If tringle were to exist, the Lw of Sines would hve sin(γ) = sin(α) so tht sin(γ) = sin(α) > = 1, whih is impossible. In the figure below, we see geometrilly why this is the se. h = sin(α) = h = sin(α) α α < h, no tringle = h, γ = 90 Simply put, if < h the side is too short to onnet to form tringle. This mens if h, we re lwys gurnteed to hve t lest one tringle, nd the remining prts of the theorem 8 If this sounds fmilir, it should. From high shool Geometry, we know there re four ongruene onditions for tringles: Angle-Angle-Side (AAS), Angle-Side-Angle (ASA), Side-Angle-Side (SAS) nd Side-Side-Side (SSS). If we re given informtion bout tringle tht meets one of these four riteri, then we re gurnteed tht extly one tringle exists whih stisfies the given riteri. 9 In more reputble books, this is lled the Side-Side-Angle or SSA se.

7 900 Applitions of Trigonometry tell us wht kind nd how mny tringles to expet in eh se. If = h, then = sin(α) nd the Lw of Sines gives sin(α) = sin(γ) so tht sin(γ) = sin(α) = = 1. Here, γ = 90 s required. Moving long, now suppose h < <. As before, the Lw of Sines 10 gives sin(γ) = sin(α). Sine h <, sin(α) < or sin(α) < 1 whih mens there re two solutions to sin(γ) = sin(α) : n ute ngle whih we ll ll γ 0, nd its supplement, 180 γ 0. We need to rgue tht eh of these ngles fit into tringle with α. Sine (α, ) nd (γ 0, ) re ngle-side opposite pirs, the ssumption > in this se gives us γ 0 > α. Sine γ 0 is ute, we must hve tht α is ute s well. This mens one tringle n ontin both α nd γ 0, giving us one of the tringles promised in the theorem. If we mnipulte the inequlity γ 0 > α bit, we hve 180 γ 0 < 180 α whih gives (180 γ 0 ) + α < 180. This proves tringle n ontin both of the ngles α nd (180 γ 0 ), giving us the seond tringle predited in the theorem. To prove the lst se in the theorem, we ssume. Then α γ, whih fores γ to be n ute ngle. Hene, we get only one tringle in this se, ompleting the proof. α γ 0 γ 0 h h < <, two tringles h α γ, one tringle One lst omment before we use the Lw of Sines to solve n pplition problem. In the Angle- Side-Side se, if you re given n obtuse ngle to begin with then it is impossible to hve the two tringle se. Think bout this before reding further. Exmple Ssquth Islnd lies off the ost of Ippizuti Lke. Two sightings, tken 5 miles prt, re mde to the islnd. The ngle between the shore nd the islnd t the first observtion point is 30 nd t the seond point the ngle is 45. Assuming stright ostline, find the distne from the seond observtion point to the islnd. Wht point on the shore is losest to the islnd? How fr is the islnd from this point? Solution. We sketh the problem below with the first observtion point lbeled s P nd the seond s Q. In order to use the Lw of Sines to find the distne d from Q to the islnd, we first need to find the mesure of β whih is the ngle opposite the side of length 5 miles. To tht end, we note tht the ngles γ nd 45 re supplementl, so tht γ = = 135. We n now find β = γ = = 15. By the Lw of Sines, we hve d sin(30 ) = 5 sin(15 ) whih gives d = 5 sin(30 ) sin(15 ) 9.66 miles. Next, to find the point on the ost losest to the islnd, whih we ve lbeled s C, we need to find the perpendiulr distne from the islnd to the ost Remember, we hve lredy rgued tht tringle exists in this se! 11 Do you see why C must lie to the right of Q?

8 11. The Lw of Sines 901 Let x denote the distne from the seond observtion point Q to the point C nd let y denote the distne from C to the islnd. ( Using Theorem 10.4, we get sin (45 ) = y d. After some rerrnging, ) we find y = d sin (45 ) miles. Hene, the islnd is pproximtely 6.83 miles from the ost. To find the distne from Q to C, we note tht β = = 45 so by symmetry, 1 we get x = y 6.83 miles. Hene, the point on the shore losest to the islnd is pproximtely 6.83 miles down the ost from the seond observtion point. Ssquth Islnd Ssquth Islnd β β d 9.66 miles d 9.66 miles y miles P 30 γ Q 45 Shoreline Q 45 C 5 miles x miles We lose this setion with new formul to ompute the re enlosed by tringle. Its proof uses the sme ses nd digrms s the proof of the Lw of Sines nd is left s n exerise. Theorem Suppose (α, ), (β, b) nd (γ, ) re the ngle-side opposite pirs of tringle. Then the re A enlosed by the tringle is given by A = 1 b sin(α) = 1 sin(β) = 1 b sin(γ) Exmple Find the re of the tringle in Exmple 11.. number 1. Solution. From our work in Exmple 11.. number 1, we hve ll three ngles nd ll three sides to work with. However, to minimize propgted error, we hoose A = 1 sin(β) from Theorem 11.4 beuse it uses the most piees of given informtion. We re ( given ) = 7 nd β = 45, nd we lulted = 7 sin(15 ) sin(10 ). Using these vlues, we find A = 1 (7) 7 sin(15 ) sin(10 ) sin (45 ) = 5.18 squre units. The reder is enourged to hek this nswer ginst the results obtined using the other formuls in Theorem Or by Theorem 10.4 gin...

9 90 Applitions of Trigonometry Exerises In Exerises 1-0, solve for the remining side(s) nd ngle(s) if possible. As in the text, (α, ), (β, b) nd (γ, ) re ngle-side opposite pirs. 1. α = 13, β = 17, = 5. α = 73., β = 54.1, = α = 95, β = 85, = α = 95, β = 6, = α = 117, = 35, b = 4 6. α = 117, = 45, b = 4 7. α = 68.7, = 88, b = 9 8. α = 4, = 17, b = α = 68.7, = 70, b = α = 30, = 7, b = α = 4, = 39, b = γ = 53, α = 53, = α = 6, = 57, b = γ = 74.6, = 3, = β = 10, b = 16.75, = β = 10, b = 16.75, = β = 10, γ = 35, b = β = 9.13, γ = 83.95, b = γ = 10, β = 61, = 4 0. α = 50, = 5, b = Find the re of the tringles given in Exerises 1, 1 nd 0 bove. (Another Clssi Applition: Grde of Rod) The grde of rod is muh like the pith of roof (See Exmple ) in tht it expresses the rtio of rise/run. In the se of rod, this rtio is lwys positive beuse it is mesured going uphill nd it is usully given s perentge. For exmple, rod whih rises 7 feet for every 100 feet of (horizontl) forwrd progress is sid to hve 7% grde. However, if we wnt to pply ny Trigonometry to story problem involving rods going uphill or downhill, we need to view the grde s n ngle with respet to the horizontl. In Exerises - 4, we first hve you hnge rod grdes into ngles nd then use the Lw of Sines in n pplition.. Using right tringle with horizontl leg of length 100 nd vertil leg with length 7, show tht 7% grde mens tht the rod (hypotenuse) mkes bout 4 ngle with the horizontl. (It will not be extly 4, but it s pretty lose.) 3. Wht grde is given by 9.65 ngle mde by the rod nd the horizontl? I hve friends who live in Pifi, CA nd their rod is tully this steep. It s not nie rod to drive.

10 11. The Lw of Sines Along long, stright streth of mountin rod with 7% grde, you see tll tree stnding perfetly plumb longside the rod. 14 From point 500 feet downhill from the tree, the ngle of inlintion from the rod to the top of the tree is 6. Use the Lw of Sines to find the height of the tree. (Hint: First show tht the tree mkes 94 ngle with the rod.) (Another Clssi Applition: Berings) In the next severl exerises we introdue nd work with the nvigtion tool known s berings. Simply put, bering is the diretion you re heding ording to ompss. The lssi nomenlture for berings, however, is not given s n ngle in stndrd position, so we must first understnd the nottion. A bering is given s n ute ngle of rottion (to the est or to the west) wy from the north-south (up nd down) line of ompss rose. For exmple, N40 E (red 40 est of north ) is bering whih is rotted lokwise 40 from due north. If we imgine stnding t the origin in the Crtesin Plne, this bering would hve us heding into Qudrnt I long the terminl side of θ = 50. Similrly, S50 W would point into Qudrnt III long the terminl side of θ = 0 beuse we strted out pointing due south (long θ = 70 ) nd rotted lokwise 50 bk to 0. Counter-lokwise rottions would be found in the berings N60 W (whih is on the terminl side of θ = 150 ) nd S7 E (whih lies long the terminl side of θ = 97 ). These four berings re drwn in the plne below. N60 W 60 N 40 N40 E W E S50 W 50 S 7 S7 E The rdinl diretions north, south, est nd west re usully not given s berings in the fshion desribed bove, but rther, one just refers to them s due north, due south, due est nd due west, respetively, nd it is ssumed tht you know whih qudrntl ngle goes with eh rdinl diretion. (Hint: Look t the digrm bove.) 5. Find the ngle θ in stndrd position with 0 θ < 360 whih orresponds to eh of the berings given below. () due west (b) S83 E () N5.5 E (d) due south 14 The word plumb here mens tht the tree is perpendiulr to the horizontl.

11 904 Applitions of Trigonometry (e) N31.5 W (f) S W 15 (g) N45 E (h) S45 W 6. The Colonel spots mpfire t of bering N4 E from his urrent position. Srge, who is positioned 3000 feet due est of the Colonel, rekons the bering to the fire to be N0 W from his urrent position. Determine the distne from the mpfire to eh mn, rounded to the nerest foot. 7. A hiker strts wlking due west from Ssquth Point nd gets to the Chupbr Trilhed before she relizes tht she hsn t reset her pedometer. From the Chupbr Trilhed she hikes for 5 miles long bering of N53 W whih brings her to the Muffin Ridge Observtory. From there, she knows bering of S65 E will tke her stright bk to Ssquth Point. How fr will she hve to wlk to get from the Muffin Ridge Observtory to Ssquh Point? Wht is the distne between Ssquth Point nd the Chupbr Trilhed? 8. The ptin of the SS Bigfoot sees signl flre t bering of N15 E from her urrent lotion. From his position, the ptin of the HMS Ssquth finds the signl flre to be t bering of N75 W. If the SS Bigfoot is 5 miles from the HMS Ssquth nd the bering from the SS Bigfoot to the HMS Ssquth is N50 E, find the distnes from the flre to eh vessel, rounded to the nerest tenth of mile. 9. Crl spies potentil Ssquth nest t bering of N10 E nd rdios Jeff, who is t bering of N50 E from Crl s position. From Jeff s position, the nest is t bering of S70 W. If Jeff nd Crl re 500 feet prt, how fr is Jeff from the Ssquth nest? Round your nswer to the nerest foot. 30. A hiker determines the bering to lodge from her urrent position is S40 W. She proeeds to hike miles t bering of S0 E t whih point she determines the bering to the lodge is S75 W. How fr is she from the lodge t this point? Round your nswer to the nerest hundredth of mile. 31. A wthtower spots ship off shore t bering of N70 E. A seond tower, whih is 50 miles from the first t bering of S80 E from the first tower, determines the bering to the ship to be N5 W. How fr is the bot from the seond tower? Round your nswer to the nerest tenth of mile. 3. Skippy nd Slly deide to hunt UFOs. One night, they position themselves miles prt on n bndoned streth of desert runwy. An hour into their investigtion, Skippy spies UFO hovering over spot on the runwy diretly between him nd Slly. He reords the ngle of inlintion from the ground to the rft to be 75 nd rdios Slly immeditely to find the ngle of inlintion from her position to the rft is 50. How high off the ground is the UFO t this point? Round your nswer to the nerest foot. (Rell: 1 mile is 580 feet.) 15 See Exmple in Setion 10.1 for review of the DMS system.

12 11. The Lw of Sines The ngle of depression from n observer in n prtment omplex to grgoyle on the building next door is 55. From point five stories below the originl observer, the ngle of inlintion to the grgoyle is 0. Find the distne from eh observer to the grgoyle nd the distne from the grgoyle to the prtment omplex. Round your nswers to the nerest foot. (Use the rule of thumb tht one story of building is 9 feet.) 34. Prove tht the Lw of Sines holds when ABC is right tringle. 35. Disuss with your lssmtes why knowing only the three ngles of tringle is not enough to determine ny of the sides. 36. Disuss with your lssmtes why the Lw of Sines nnot be used to find the ngles in the tringle when only the three sides re given. Also disuss wht hppens if only two sides nd the ngle between them re given. (Sid nother wy, explin why the Lw of Sines nnot be used in the SSS nd SAS ses.) 37. Given α = 30 nd b = 10, hoose four different vlues for so tht () the informtion yields no tringle (b) the informtion yields extly one right tringle () the informtion yields two distint tringles (d) the informtion yields extly one obtuse tringle Explin why you nnot hoose in suh wy s to hve α = 30, b = 10 nd your hoie of yield only one tringle where tht unique tringle hs three ute ngles. 38. Use the ses nd digrms in the proof of the Lw of Sines (Theorem 11.) to prove the re formuls given in Theorem Why do those formuls yield squre units when four quntities re being multiplied together?

13 906 Applitions of Trigonometry 11.. Answers 1. α = 13 β = 17 γ = 150 = 5 b α = 73. β = 54.1 γ = 5.7 = 117 b Informtion does not produe tringle 4. α = 95 β = 6 γ = 3 = b Informtion does not produe tringle 6. α = 117 β 56.3 γ 6.7 = 45 b = α = 68.7 β 76.9 γ 34.4 = 88 b = α = 4 β γ = 17 b = α = 68.7 β γ 8. = 88 b = α = 4 β γ 5.66 = 17 b = Informtion does not produe tringle 10. α = 30 β = 90 γ = 60 = 7 b = 14 = α = 4 β 3.78 γ 114. = 39 b = α = 53 β = 74 γ = 53 = 8.01 b = α = 6 β γ 4.57 = 57 b = α β 6.81 γ = 74.6 = 3.05 b 1.40 = 3 α = 6 β γ = 57 b = α β 3.99 γ = 74.6 = 3.05 b 0.17 = α 8.61 β = 10 γ b = = Informtion does not produe tringle 17. α = 43 β = 10 γ = b = α = 66.9 β = 9.13 γ = b = Informtion does not produe tringle 0. α = 50 β.5 γ = 5 b = The re of the tringle from Exerise 1 is bout 8.1 squre units. The re of the tringle from Exerise 1 is bout squre units. The re of the tringle from Exerise 0 is bout 149 squre units.. rtn ( ) rdins, whih is equivlent to About 17% 4. About 53 feet

14 11. The Lw of Sines () θ = 180 (b) θ = 353 () θ = 84.5 (d) θ = 70 (e) θ = 11.5 (f) θ = (g) θ = 45 (h) θ = 5 6. The Colonel is bout 3193 feet from the mpfire. Srge is bout 55 feet to the mpfire. 7. The distne from the Muffin Ridge Observtory to Ssquh Point is bout 7.1 miles. The distne from Ssquth Point to the Chupbr Trilhed is bout.46 miles. 8. The SS Bigfoot is bout 4.1 miles from the flre. The HMS Ssquth is bout.9 miles from the flre. 9. Jeff is bout 34 feet from the nest. 30. She is bout 3.0 miles from the lodge 31. The bot is bout 5.1 miles from the seond tower. 3. The UFO is hovering bout 9539 feet bove the ground. 33. The grgoyle is bout 44 feet from the observer on the upper floor. The grgoyle is bout 7 feet from the observer on the lower floor. The grgoyle is bout 5 feet from the other building.

15 908 Applitions of Trigonometry 11.3 The Lw of Cosines In Setion 11., we developed the Lw of Sines (Theorem 11.) to enble us to solve tringles in the Angle-Angle-Side (AAS), the Angle-Side-Angle (ASA) nd the mbiguous Angle-Side-Side (ASS) ses. In this setion, we develop the Lw of Cosines whih hndles solving tringles in the Side-Angle-Side (SAS) nd Side-Side-Side (SSS) ses. 1 We stte nd prove the theorem below. Theorem Lw of Cosines: Given tringle with ngle-side opposite pirs (α, ), (β, b) nd (γ, ), the following equtions hold = b + b os(α) b = + os(β) = + b b os(γ) or, solving for the osine in eh eqution, we hve os(α) = b + b os(β) = + b os(γ) = + b b To prove the theorem, we onsider generi tringle with the vertex of ngle α t the origin with side b positioned long the positive x-xis. B = ( os(α), sin(α)) α A = (0, 0) b C = (b, 0) From this set-up, we immeditely find tht the oordintes of A nd C re A(0, 0) nd C(b, 0). From Theorem 10.3, we know tht sine the point B(x, y) lies on irle of rdius, the oordintes 1 Here, Side-Angle-Side mens tht we re given two sides nd the inluded ngle - tht is, the given ngle is djent to both of the given sides.

16 11.3 The Lw of Cosines 909 of B re B(x, y) = B( os(α), sin(α)). (This would be true even if α were n obtuse or right ngle so lthough we hve drwn the se when α is ute, the following omputtions hold for ny ngle α drwn in stndrd position where 0 < α < 180.) We note tht the distne between the points B nd C is none other thn the length of side. Using the distne formul, Eqution 1.1, we get = ( os(α) b) + ( sin(α) 0) ( = ( os(α) b) + sin (α)) = ( os(α) b) + sin (α) = os (α) b os(α) + b + sin (α) = ( os (α) + sin (α) ) + b b os(α) = (1) + b b os(α) Sine os (α) + sin (α) = 1 = + b b os(α) The remining formuls given in Theorem 11.5 n be shown by simply reorienting the tringle to ple different vertex t the origin. We leve these detils to the reder. Wht s importnt bout nd α in the bove proof is tht (α, ) is n ngle-side opposite pir nd b nd re the sides djent to α the sme n be sid of ny other ngle-side opposite pir in the tringle. Notie tht the proof of the Lw of Cosines relies on the distne formul whih hs its roots in the Pythgoren Theorem. Tht being sid, the Lw of Cosines n be thought of s generliztion of the Pythgoren Theorem. If we hve tringle in whih γ = 90, then os(γ) = os (90 ) = 0 so we get the fmilir reltionship = +b. Wht this mens is tht in the lrger mthemtil sense, the Lw of Cosines nd the Pythgoren Theorem mount to pretty muh the sme thing. Exmple Solve the following tringles. Give ext nswers nd deiml pproximtions (rounded to hundredths) nd sketh the tringle. 1. β = 50, = 7 units, = units. = 4 units, b = 7 units, = 5 units Solution. 1. We re given the lengths of two sides, = 7 nd =, nd the mesure of the inluded ngle, β = 50. With no ngle-side opposite pir to use, we pply the Lw of Cosines. We get b = 7 + (7)() os (50 ) whih yields b = 53 8 os (50 ) 5.9 units. In order to determine the mesures of the remining ngles α nd γ, we re fored to used the derived vlue for b. There re two wys to proeed t this point. We ould use the Lw of Cosines gin, or, sine we hve the ngle-side opposite pir (β, b) we ould use the Lw of Sines. The dvntge to using the Lw of Cosines over the Lw of Sines in ses like this is tht unlike the sine funtion, the osine funtion distinguishes between ute nd obtuse ngles. The osine of n ute is positive, wheres the osine of n obtuse ngle is negtive. Sine the sine of both ute nd obtuse ngles re positive, the sine of n ngle lone is not This shouldn t ome s too muh of shok. All of the theorems in Trigonometry n ultimtely be tred bk to the definition of the irulr funtions long with the distne formul nd hene, the Pythgoren Theorem.

17 910 Applitions of Trigonometry enough to determine if the ngle in question is ute or obtuse. Sine both uthors of the textbook prefer the Lw of Cosines, we proeed with this method first. When using the Lw of Cosines, it s lwys best to find the mesure of the lrgest unknown ngle first, sine this will give us the obtuse ngle of the tringle if there is one. Sine the lrgest ngle is opposite the longest side, we hoose to find α first. To tht end, we use the formul os(α) = b + b nd substitute = 7, b = 53 8 os (50 ) nd =. We get 3 os(α) = 7 os (50 ) 53 8 os (50 ) Sine α is n ngle in tringle, we know the rdin mesure of α must lie between 0 nd π rdins. This mthes the rnge of the rosine funtion, so we hve ( ) 7 os (50 ) α = ros rdins os (50 ) At this point, we ould find γ using γ = 180 α β = 15.01, tht is if we trust our pproximtion for α. To minimize propgtion of error, however, we ould use the Lw of Cosines gin, 4 in this se using os(γ) = +b b. Plugging in = 7, b = ( ) 53 8 os (50 7 os(50 ) nd =, we get γ = ros ) rdins We sketh the tringle below os(50 ) β = 50 = 7 = α γ b 5.9 As we mentioned erlier, one we ve determined b it is possible to use the Lw of Sines to find the remining ngles. Here, however, we must proeed with ution s we re in the mbiguous (ASS) se. It is dvisble to first find the smllest of the unknown ngles, sine we re gurnteed it will be ute. 5 In this se, we would find γ sine the side opposite γ is smller thn the side opposite the other unknown ngle, α. Using the ngle-side opposite pir (β, b), we get sin(γ) sin(50 ). The usul lultions produes γ 53 8 os(50 ) nd = α = 180 β γ = Sine ll three sides nd no ngles re given, we re fored to use the Lw of Cosines. Following our disussion in the previous problem, we find β first, sine it is opposite the longest side, b. We get os(β) = + b = 1 5, so we get β = ros ( 1 5) rdins As in 3 fter simplifying... 4 Your instrutor will let you know whih proedure to use. It ll boils down to how muh you trust your lultor. 5 There n only be one obtuse ngle in the tringle, nd if there is one, it must be the lrgest.

18 11.3 The Lw of Cosines 911 the previous problem, now tht we hve obtined n ngle-side opposite pir (β, b), we ould proeed using the Lw of Sines. The Lw of Cosines, however, offers us rre opportunity to find the remining ngles using only the dt given to us in the sttement of the problem. Using this, we get γ = ros ( 5 7 ) rdins 44.4 nd α = ros ( 9 35) rdins β = 5 = 4 α γ 44.4 We note tht, depending on how mny deiml ples re rried through suessive lultions, nd depending on whih pproh is used to solve the problem, the pproximte nswers you obtin my differ slightly from those the uthors obtin in the Exmples nd the Exerises. A gret exmple of this is number in Exmple , where the pproximte vlues we reord for the mesures of the ngles sum to , whih is geometrilly impossible. Next, we hve n pplition of the Lw of Cosines. Exmple A reserher wishes to determine the width of vernl pond s drwn below. From point P, he finds the distne to the estern-most point of the pond to be 950 feet, while the distne to the western-most point of the pond from P is 1000 feet. If the ngle between the two lines of sight is 60, find the width of the pond. b = feet feet P Solution. We re given the lengths of two sides nd the mesure of n inluded ngle, so we my pply the Lw of Cosines to find the length of the missing side opposite the given ngle. Clling this length w (for width), we get w = (950)(1000) os (60 ) = from whih we get w = feet.

19 91 Applitions of Trigonometry In Setion 11., we used the proof of the Lw of Sines to develop Theorem 11.4 s n lternte formul for the re enlosed by tringle. In this setion, we use the Lw of Cosines to derive nother suh formul - Heron s Formul. Theorem Heron s Formul: Suppose, b nd denote the lengths of the three sides of tringle. Let s be the semiperimeter of the tringle, tht is, let s = 1 ( + b + ). Then the re A enlosed by the tringle is given by A = s(s )(s b)(s ) We prove Theorem 11.6 using Theorem Using the onvention tht the ngle γ is opposite the side, we hve A = 1 b sin(γ) from Theorem In order to simplify omputtions, we strt by mnipulting the expression for A. A = ( ) 1 b sin(γ) = 1 4 b sin (γ) = b 4 ( 1 os (γ) ) sine sin (γ) = 1 os (γ). The Lw of Cosines tells us os(γ) = +b b, so substituting this into our eqution for A gives A = b ( 1 os (γ) ) 4 [ = b ( + b ) ] 1 4 b [ ( = b + b ) ] b [ = b 4 b ( + b ) ] 4 4 b = 4 b ( + b ) 16 = (b) ( + b ) 16 ( [ b + b ]) ( b + [ + b ]) = = 16 ( + b b ) ( + b + b ) 16 differene of squres.

20 11.3 The Lw of Cosines 913 A = = = = = ( [ b + b ]) ([ + b + b ] ) 16 ( ( b) ) ( ( + b) ) 16 ( ( b))( + ( b))(( + b) )(( + b) + ) 16 (b + )( + b)( + b )( + b + ) 16 (b + ) ( + b) ( + b ) ( + b + ) perfet squre trinomils. differene of squres. At this stge, we reognize the lst ftor s the semiperimeter, s = 1 +b+ ( + b + ) =. To omplete the proof, we note tht (s ) = + b + = + b + = b + Similrly, we find (s b) = + b nd (s ) = +b. Hene, we get A = (b + ) ( + b) = (s )(s b)(s )s ( + b ) ( + b + ) so tht A = s(s )(s b)(s ) s required. We lose with n exmple of Heron s Formul. Exmple Find the re enlosed of the tringle in Exmple number. Solution. We re given = 4, b = 7 nd = 5. Using these vlues, we find s = 1 ( ) = 8, (s ) = 8 4 = 4, (s b) = 8 7 = 1 nd (s ) = 8 5 = 3. Using Heron s Formul, we get A = s(s )(s b)(s ) = (8)(4)(1)(3) = 96 = squre units.

21 914 Applitions of Trigonometry Exerises In Exerises 1-10, use the Lw of Cosines to find the remining side(s) nd ngle(s) if possible. 1. = 7, b = 1, γ = α = 104, b = 5, = = 153, β = 8., = = 3, b = 4, γ = α = 10, b = 3, = 4 6. = 7, b = 10, = = 1, b =, = 5 8. = 300, b = 30, = = 5, b = 5, = = 5, b = 1, ; = 13 In Exerises 11-16, solve for the remining side(s) nd ngle(s), if possible, using ny pproprite tehnique. 11. = 18, α = 63, b = 0 1. = 37, b = 45, = = 16, α = 63, b = =, α = 63, b = α = 4, b = 117, = β = 7, γ = 170, = Find the re of the tringles given in Exerises 6, 8 nd 10 bove. 18. The hour hnd on my ntique Seth Thoms shoolhouse lok in 4 inhes long nd the minute hnd is 5.5 inhes long. Find the distne between the ends of the hnds when the lok reds four o lok. Round your nswer to the nerest hundredth of n inh. 19. A geologist wnts to mesure the dimeter of rter. From her mp, it is 4 miles to the northern-most point of the rter nd miles to the southern-most point. If the ngle between the two lines of sight is 117, wht is the dimeter of the rter? Round your nswer to the nerest hundredth of mile. 0. From the Pedimxus Interntionl Airport tour heliopter n fly to Cliffs of Insnity Point by following bering of N8. E for 19 miles nd it n fly to Bigfoot Flls by following bering of S68.5 E for 07 miles. 6 Find the distne between Cliffs of Insnity Point nd Bigfoot Flls. Round your nswer to the nerest mile. 1. Cliffs of Insnity Point nd Bigfoot Flls from Exerise 0 bove both lie on stright streth of the Gret Ssquth Cnyon. Wht bering would the tour heliopter need to follow to go diretly from Bigfoot Flls to Cliffs of Insnity Point? Round your ngle to the nerest tenth of degree. 6 Plese refer to Pge 903 in Setion 11. for n introdution to berings.

22 11.3 The Lw of Cosines 915. A nturlist sets off on hike from lodge on bering of S80 W. After 1.5 miles, she hnges her bering to S17 W nd ontinues hiking for 3 miles. Find her distne from the lodge t this point. Round your nswer to the nerest hundredth of mile. Wht bering should she follow to return to the lodge? Round your ngle to the nerest degree. 3. The HMS Ssquth leves port on bering of N3 E nd trvels for 5 miles. It then hnges ourse nd follows heding of S41 E for miles. How fr is it from port? Round your nswer to the nerest hundredth of mile. Wht is its bering to port? Round your ngle to the nerest degree. 4. The SS Bigfoot leves hrbor bound for Nessie Islnd whih is 300 miles wy t bering of N3 E. A storm moves in nd fter 100 miles, the ptin of the Bigfoot finds he hs drifted off ourse. If his bering to the hrbor is now S70 W, how fr is the SS Bigfoot from Nessie Islnd? Round your nswer to the nerest hundredth of mile. Wht ourse should the ptin set to hed to the islnd? Round your ngle to the nerest tenth of degree. 5. From point 300 feet bove level ground in firetower, rnger spots two fires in the Yeti Ntionl Forest. The ngle of depression 7 mde by the line of sight from the rnger to the first fire is.5 nd the ngle of depression mde by line of sight from the rnger to the seond fire is 1.3. The ngle formed by the two lines of sight is 117. Find the distne between the two fires. Round your nswer to the nerest foot. (Hint: In order to use the 117 ngle between the lines of sight, you will first need to use right ngle Trigonometry to find the lengths of the lines of sight. This will give you Side-Angle-Side se in whih to pply the Lw of Cosines.) fire 117 fire firetower 6. If you pply the Lw of Cosines to the mbiguous Angle-Side-Side (ASS) se, the result is qudrti eqution whose vrible is tht of the missing side. If the eqution hs no positive rel zeros then the informtion given does not yield tringle. If the eqution hs only one positive rel zero then extly one tringle is formed nd if the eqution hs two distint positive rel zeros then two distint tringles re formed. Apply the Lw of Cosines to Exerises 11, 13 nd 14 bove in order to demonstrte this result. 7. Disuss with your lssmtes why Heron s Formul yields n re in squre units even though four lengths re being multiplied together. 7 See Exerise 78 in Setion 10.3 for the definition of this ngle.

23 916 Applitions of Trigonometry Answers 1. α β γ = 59.3 = 7 b = α = 104 β 9.40 γ b = 5 = α β = 8. γ = 153 b 1.88 = α β γ = 90 = 3 b = 4 = 5 5. α = 10 β 5.8 γ 34.7 = 37 b = 3 = 4 6. α 3.31 β γ 98.1 = 7 b = 10 = Informtion does not produe tringle 8. α β γ 9.14 = 300 b = 30 = α = 60 β = 60 γ = 60 = 5 b = 5 = α.6 β γ = 90 = 5 b = 1 = α = 63 β γ = 18 b = α β γ = 37 b = 45 = 6 α = 63 β γ = 18 b = Informtion does not produe tringle 14. α = 63 β 54.1 γ 6.9 = b = α = 4 β 89.3 γ b = 117 = α 3 β = 7 γ = b 69. = The re of the tringle given in Exerise 6 is 100 = squre units. The re of the tringle given in Exerise 8 is squre units. The re of the tringle given in Exerise 10 is extly 30 squre units. 18. The distne between the ends of the hnds t four o lok is bout 8.6 inhes. 19. The dimeter of the rter is bout 5. miles. 0. About 313 miles 1. N31.8 W. She is bout 3.9 miles from the lodge nd her bering to the lodge is N37 E. 3. It is bout 4.50 miles from port nd its heding to port is S47 W. 4. It is bout 9.61 miles from the islnd nd the ptin should set ourse of N16.4 E to reh the islnd. 5. The fires re bout feet prt. (Try to void rounding errors.)