Setion 11.: The Lw of Sines, from College Trigonometry: Correted Edition y Crl Stitz, Ph.D. nd Jeff Zeger, Ph.D. is ville under Cretive Commons Attriution-NonCommeril-ShreAlike 3.0 liense. 013, Crl Stitz.
896 Applitions of Trigonometry 11. The Lw of Sines Trigonometry literlly mens mesuring tringles nd with Chpter 10 under our elts, 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 11..1. 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 lel the tringle elow. = 7 = 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 using the inverse trigonometri funtions. To derese the hnes of propgting error, however, we stik to using the dt given to us in the prolem. In this se, the lengths 4 nd 7 were given, so we wnt to relte these to. Aording to Theorem 10.4, os() = 4 7. Sine is n ute ngle, = ros ( 4 7) rdins. Converting to degrees, we find 55.15. Now tht we hve the mesure of ngle, we ould find the mesure of ngle using the ft tht nd re omplements so + = 90. One gin, we opt to use the dt given to us in the prolem. Aording to Theorem 10.4, we hve tht sin() = 4 7 so = rsin ( 4 7) rdins nd we hve 34.85. A few remrks out Exmple 11..1 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, is the side opposite nd is the side opposite γ. Tken together, the pirs (, ), (, ) 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 prolem whenever possile. 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
11. The Lw of Sines 897 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 eing. 4 The Pythgoren Theorem long with Theorems 10.4 nd 10.10 llow us to esily hndle ny given right tringle prolem, ut 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 (, ), (, ) nd (γ, ), the following rtios hold or, equivlently, sin() = sin() = sin(γ) sin() = sin() = sin(γ) The proof of the Lw of Sines n e roken into three ses. For our first se, onsider the tringle ABC elow, ll of whose ngles re ute, with ngle-side opposite pirs (, ), (, ) 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() = h nd sin(γ) = h so tht h = sin() = sin(γ). After some rerrngement of the lst eqution, we get sin() using the tringles ABQ nd ACQ to get sin() B γ = sin(γ). If we drop n ltitude from vertex A, we n proeed s ove = sin(γ), ompleting the proof for this se. A C A C Q A C For our next se onsider the tringle ABC elow with otuse ngle. Extending n ltitude from vertex A gives two right tringles, s in the previous se: ABQ nd ACQ. Proeeding s efore, we get h = sin(γ) nd h = sin() so tht sin() = sin(γ). B h γ h B Q γ B γ B Q h γ A C A C 3 Your Siene tehers should thnk us for this. 4 Don t worry! Rdins will e k efore you know it!
898 Applitions of Trigonometry Dropping n ltitude from vertex B lso genertes two right tringles, ABQ nd BCQ. We know tht sin( ) = h so tht h = sin( ). Sine = 180, sin( ) = sin(), so in ft, we hve h = sin(). 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(), nd we re finished with this se. h γ Q A 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 11... Solve the following tringles. Give ext nswers nd deiml pproximtions (rounded to hundredths) nd sketh the tringle. 1. = 10, = 7 units, = 45. = 85, = 30, = 5.5 units 3. = 30, = 1 units, = 4 units 4. = 30, = units, = 4 units 5. = 30, = 3 units, = 4 units 6. = 30, = 4 units, = 4 units Solution. 1. Knowing n ngle-side opposite pir, nmely nd, we my proeed in using the Lw of Sines. Sine = 45, we use sin(45 ) = 7 sin(10 ) so = 7 sin(45 ) sin(10 ) = 7 6 3 5.7 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 = 180 10 45 = 15. To find, we hve no hoie ut to used the derived vlue γ = 15, yet we n minimize the propgtion of error here y using the given ngle-side opposite pir (, ). 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, ut s we hve the mesures of nd, we n solve for γ sine γ = 180 85 30 = 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 e found using the differene identity for sine or hlf-ngle formul, ut tht eomes unneessrily messy for the disussion t hnd. Thus ext here mens 7 sin(15 ) sin(10 ).
11. The Lw of Sines 899 ngle-side pir ville 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 we use the ngle-side pir (γ, ) whih yields sin(30 ) = 5.5 sin(65 ) hene = 5.5 sin(30 ) sin(65 ).90 units. = 30 = 45 = 7 = 5.5 5.77.09 = 10 γ = 15 = 85 γ = 65 5.7.90 Tringle for numer 1 Tringle for numer 3. Sine we re given (, ) 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 numer 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 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 = 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 = 30. This mens tht γ must mesure etween 0 nd 150 in order to fit inside the tringle with. 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 to e = 180 30 90 = 60 nd then determine using the Lw of Sines. We find = sin(60 ) sin(30 ) = 3 3.46 units. In this se, the side is preisely long enough to form unique right tringle. = 4 = 1 = 4 = 60 = = 30 = 30 3.46 Digrm for numer 3 Tringle for numer 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 = 30,
900 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 41.81 nd γ = π rsin ( 3) rdins 138.19. At this point, we puse to see if it mkes sense tht we tully hve two vile ses to onsider. As we hve disussed, oth ndidtes for γ re omptile with the given ngle-side pir (, ) = (30, 3) in tht oth hoies for γ n fit in tringle with nd oth hve sine of 3. The only other given piee of informtion is tht = 4 units. Sine >, it must e true tht γ, whih is opposite, hs greter mesure thn whih is opposite. In oth ses, γ >, so oth ndidtes for γ re omptile 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 6 180 30 41.81 = 108.19. Using the Lw of Sines with the ngle-side opposite pir (, ) nd, we find 3 sin(108.19 ) sin(30 ) 5.70 units. In the se γ = π rsin ( 3) rdins 138.19, we repet the ext sme steps nd find 11.81 nd 1.3 units. 7 tringles re drwn elow. Both 11.81 = 4 = 30 108.19 = 3 γ 41.81 5.70 = 4 = 3 = 30 γ 138.19 1.3 6. For this lst prolem, 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 inhit tringle with = 30, we must hve 0 < γ < 150. Sine the mesure of γ must e stritly less thn 150, there is just one ngle whih stisfies oth required onditions, nmely γ = 30. So = 180 30 30 = 10 nd, using the Lw of Sines one lst time, = 4 sin(10 ) sin(30 ) = 4 3 6.93 units. = 4 = 10 = 4 = 30 γ = 30 6.93 Some remrks out Exmple 11.. re in order. We first note tht if we re given the mesures of two of the ngles in tringle, sy nd, the mesure of the third ngle γ is uniquely 6 To find n ext expression for, we onvert everything k to rdins: = 30 = π rdins, γ = rsin ( ) 6 3 rdins nd 180 = π rdins. Hene, = π π rsin ( ) 6 3 = 5π rsin ( ) 6 3 rdins 108.19. 7 An ext nswer for in this se is = rsin ( ) 3 π rdins 6 11.81.
11. The Lw of Sines 901 determined using the eqution γ = 180. 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 numers 1 nd ove. In numer 1, the given side is djent to just one of the ngles this is lled the Angle-Angle-Side (AAS) se. 8 In numer, the given side is djent to oth 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 numer 3, the length of the one given side ws too short to even form tringle; in numer 4, the length of ws just long enough to form right tringle; in 5, ws long enough, ut not too long, so tht two tringles were possile; nd in numer 6, side ws long enough to form tringle ut too long to swing k nd form two. These four ses exemplify ll of the possiilities in the Angle-Side-Side se whih re summrized in the following theorem. Theorem 11.3. Suppose (, ) nd (γ, ) re intended to e ngle-side pirs in tringle where, nd re given. Let h = sin() ˆ ˆ ˆ ˆ 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-y-se sis. If < h, then < sin(). If tringle were to exist, the Lw of Sines would hve sin(γ) = sin() so tht sin(γ) = sin() > = 1, whih is impossile. In the figure elow, we see geometrilly why this is the se. h = sin() = h = sin() < 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 out 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 reputle ooks, this is lled the Side-Side-Angle or SSA se.
90 Applitions of Trigonometry tell us wht kind nd how mny tringles to expet in eh se. If = h, then = sin() nd the Lw of Sines gives sin() = sin(γ) so tht sin(γ) = sin() = = 1. Here, γ = 90 s required. Moving long, now suppose h < <. As efore, the Lw of Sines 10 gives sin(γ) = sin(). Sine h <, sin() < or sin() < 1 whih mens there re two solutions to sin(γ) = sin() : 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. Sine (, ) nd (γ 0, ) re ngle-side opposite pirs, the ssumption > in this se gives us γ 0 >. Sine γ 0 is ute, we must hve tht is ute s well. This mens one tringle n ontin oth nd γ 0, giving us one of the tringles promised in the theorem. If we mnipulte the inequlity γ 0 > it, we hve 180 γ 0 < 180 whih gives (180 γ 0 ) + < 180. This proves tringle n ontin oth of the ngles nd (180 γ 0 ), giving us the seond tringle predited in the theorem. To prove the lst se in the theorem, we ssume. Then γ, whih fores γ to e n ute ngle. Hene, we get only one tringle in this se, ompleting the proof. γ 0 γ 0 h h < <, two tringles h γ, one tringle One lst omment efore we use the Lw of Sines to solve n pplition prolem. In the Angle- Side-Side se, if you re given n otuse ngle to egin with then it is impossile to hve the two tringle se. Think out this efore reding further. Exmple 11..3. Ssquth Islnd lies off the ost of Ippizuti Lke. Two sightings, tken 5 miles prt, re mde to the islnd. The ngle etween the shore nd the islnd t the first oservtion point is 30 nd t the seond point the ngle is 45. Assuming stright ostline, find the distne from the seond oservtion 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 prolem elow with the first oservtion point leled 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 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 = 180 45 = 135. We n now find = 180 30 γ = 180 30 135 = 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 leled s C, we need to find the perpendiulr distne from the islnd to the ost. 11 10 Rememer, we hve lredy rgued tht tringle exists in this se! 11 Do you see why C must lie to the right of Q?
11. The Lw of Sines 903 Let x denote the distne from the seond oservtion 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 ) 9.66 6.83 miles. Hene, the islnd is pproximtely 6.83 miles from the ost. To find the distne from Q to C, we note tht = 180 90 45 = 45 so y 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 oservtion point. Ssquth Islnd Ssquth Islnd 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 y tringle. Its proof uses the sme ses nd digrms s the proof of the Lw of Sines nd is left s n exerise. Theorem 11.4. Suppose (, ), (, ) nd (γ, ) re the ngle-side opposite pirs of tringle. Then the re A enlosed y the tringle is given y A = 1 sin() = 1 sin() = 1 sin(γ) Exmple 11..4. Find the re of the tringle in Exmple 11.. numer 1. Solution. From our work in Exmple 11.. numer 1, we hve ll three ngles nd ll three sides to work with. However, to minimize propgted error, we hoose A = 1 sin() from Theorem 11.4 euse it uses the most piees of given informtion. We re ( given ) = 7 nd = 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 otined using the other formuls in Theorem 11.4. 1 Or y Theorem 10.4 gin...
904 Applitions of Trigonometry 11..1 Exerises In Exerises 1-0, solve for the remining side(s) nd ngle(s) if possile. As in the text, (, ), (, ) nd (γ, ) re ngle-side opposite pirs. 1. = 13, = 17, = 5. = 73., = 54.1, = 117 3. = 95, = 85, = 33.33 4. = 95, = 6, = 33.33 5. = 117, = 35, = 4 6. = 117, = 45, = 4 7. = 68.7, = 88, = 9 8. = 4, = 17, = 3.5 9. = 68.7, = 70, = 90 10. = 30, = 7, = 14 11. = 4, = 39, = 3.5 1. γ = 53, = 53, = 8.01 13. = 6, = 57, = 100 14. γ = 74.6, = 3, = 3.05 15. = 10, = 16.75, = 13 16. = 10, = 16.75, = 18 17. = 10, γ = 35, = 16.75 18. = 9.13, γ = 83.95, = 314.15 19. γ = 10, = 61, = 4 0. = 50, = 5, = 1.5 1. Find the re of the tringles given in Exerises 1, 1 nd 0 ove. (Another Clssi Applition: Grde of Rod) The grde of rod is muh like the pith of roof (See Exmple 10.6.6) in tht it expresses the rtio of rise/run. In the se of rod, this rtio is lwys positive euse 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 prolem 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 out 4 ngle with the horizontl. (It will not e extly 4, ut it s pretty lose.) 3. Wht grde is given y 9.65 ngle mde y the rod nd the horizontl? 13 13 I hve friends who live in Pifi, CA nd their rod is tully this steep. It s not nie rod to drive.
11. The Lw of Sines 905 4. Along long, stright streth of mountin rod with 7% grde, you see tll tree stnding perfetly plum 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 erings. Simply put, ering is the diretion you re heding ording to ompss. The lssi nomenlture for erings, however, is not given s n ngle in stndrd position, so we must first understnd the nottion. A ering 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 ering whih is rotted lokwise 40 from due north. If we imgine stnding t the origin in the Crtesin Plne, this ering 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 euse we strted out pointing due south (long θ = 70 ) nd rotted lokwise 50 k to 0. Counter-lokwise rottions would e found in the erings 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 erings re drwn in the plne elow. 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 erings in the fshion desried ove, ut 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 ove.) 5. Find the ngle θ in stndrd position with 0 θ < 360 whih orresponds to eh of the erings given elow. () due west () S83 E () N5.5 E (d) due south 14 The word plum here mens tht the tree is perpendiulr to the horizontl.
906 Applitions of Trigonometry (e) N31.5 W (f) S7 41 1 W 15 (g) N45 E (h) S45 W 6. The Colonel spots mpfire t of ering N4 E from his urrent position. Srge, who is positioned 3000 feet due est of the Colonel, rekons the ering to the fire to e 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 Chupr Trilhed efore she relizes tht she hsn t reset her pedometer. From the Chupr Trilhed she hikes for 5 miles long ering of N53 W whih rings her to the Muffin Ridge Oservtory. From there, she knows ering of S65 E will tke her stright k to Ssquth Point. How fr will she hve to wlk to get from the Muffin Ridge Oservtory to Ssquh Point? Wht is the distne etween Ssquth Point nd the Chupr Trilhed? 8. The ptin of the SS Bigfoot sees signl flre t ering of N15 E from her urrent lotion. From his position, the ptin of the HMS Ssquth finds the signl flre to e t ering of N75 W. If the SS Bigfoot is 5 miles from the HMS Ssquth nd the ering 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 ering of N10 E nd rdios Jeff, who is t ering of N50 E from Crl s position. From Jeff s position, the nest is t ering 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 ering to lodge from her urrent position is S40 W. She proeeds to hike miles t ering of S0 E t whih point she determines the ering 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 ering of N70 E. A seond tower, whih is 50 miles from the first t ering of S80 E from the first tower, determines the ering to the ship to e N5 W. How fr is the ot 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 ndoned streth of desert runwy. An hour into their investigtion, Skippy spies UFO hovering over spot on the runwy diretly etween him nd Slly. He reords the ngle of inlintion from the ground to the rft to e 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 10.1.1 in Setion 10.1 for review of the DMS system.
11. The Lw of Sines 907 33. The ngle of depression from n oserver in n prtment omplex to grgoyle on the uilding next door is 55. From point five stories elow the originl oserver, the ngle of inlintion to the grgoyle is 0. Find the distne from eh oserver to the grgoyle nd the distne from the grgoyle to the prtment omplex. Round your nswers to the nerest foot. (Use the rule of thum tht one story of uilding 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 e 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 etween them re given. (Sid nother wy, explin why the Lw of Sines nnot e used in the SSS nd SAS ses.) 37. Given = 30 nd = 10, hoose four different vlues for so tht () the informtion yields no tringle () the informtion yields extly one right tringle () the informtion yields two distint tringles (d) the informtion yields extly one otuse tringle Explin why you nnot hoose in suh wy s to hve = 30, = 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 11.4. Why do those formuls yield squre units when four quntities re eing multiplied together?
908 Applitions of Trigonometry 11.. Answers 1. = 13 = 17 γ = 150 = 5 6.50 11.11. = 73. = 54.1 γ = 5.7 = 117 99.00 97. 3. Informtion does not produe tringle 4. = 95 = 6 γ = 3 = 33.33 9.54 13.07 5. Informtion does not produe tringle 6. = 117 56.3 γ 6.7 = 45 = 4 5.89 7. = 68.7 76.9 γ 34.4 = 88 = 9 53.36 8. = 4 67.66 γ 70.34 = 17 = 3.5 3.93 = 68.7 103.1 γ 8. = 88 = 9 13.47 = 4 11.34 γ 5.66 = 17 = 3.5 11.00 9. Informtion does not produe tringle 10. = 30 = 90 γ = 60 = 7 = 14 = 7 3 11. = 4 3.78 γ 114. = 39 = 3.5 53.15 1. = 53 = 74 γ = 53 = 8.01 33.71 = 8.01 13. = 6 169.43 γ 4.57 = 57 = 100 43.45 14. 78.59 6.81 γ = 74.6 = 3.05 1.40 = 3 = 6 10.57 γ 163.43 = 57 = 100 155.51 101.41 3.99 γ = 74.6 = 3.05 0.17 = 3 15. 8.61 = 10 γ 49.39 8.0 = 16.75 = 13 16. Informtion does not produe tringle 17. = 43 = 10 γ = 35 11.68 = 16.75 9.8 18. = 66.9 = 9.13 γ = 83.95 593.69 = 314.15 641.75 19. Informtion does not produe tringle 0. = 50.5 γ 107.48 = 5 = 1.5 31.13 1. The re of the tringle from Exerise 1 is out 8.1 squre units. The re of the tringle from Exerise 1 is out 377.1 squre units. The re of the tringle from Exerise 0 is out 149 squre units.. rtn ( ) 7 100 0.699 rdins, whih is equivlent to 4.004 3. Aout 17% 4. Aout 53 feet
11. The Lw of Sines 909 5. () θ = 180 () θ = 353 () θ = 84.5 (d) θ = 70 (e) θ = 11.5 (f) θ = 197 18 48 (g) θ = 45 (h) θ = 5 6. The Colonel is out 3193 feet from the mpfire. Srge is out 55 feet to the mpfire. 7. The distne from the Muffin Ridge Oservtory to Ssquh Point is out 7.1 miles. The distne from Ssquth Point to the Chupr Trilhed is out.46 miles. 8. The SS Bigfoot is out 4.1 miles from the flre. The HMS Ssquth is out.9 miles from the flre. 9. Jeff is out 371 feet from the nest. 30. She is out 3.0 miles from the lodge 31. The ot is out 5.1 miles from the seond tower. 3. The UFO is hovering out 9539 feet ove the ground. 33. The grgoyle is out 44 feet from the oserver on the upper floor. The grgoyle is out 7 feet from the oserver on the lower floor. The grgoyle is out 5 feet from the other uilding.