Chapter 5. Triangles and Vectors

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1 Chpter 5. Tringles nd Vectors 5.3 The Lw of Sines Lerning Objectives Understnd how both forms of the Lw of Sines re obtined. Apply the Lw of Sines when you know two ngles nd non-included side nd if you know two ngles nd the included side. Use the Lw of Sines in rel-world nd pplied problems. We hve lerned bout the Lw of Cosines, which is generliztion of the Pythgoren Theorem for non-right tringles. We know tht we cn use the Lw of Cosines when: 1. We know two sides of tringle nd the included ngle (SAS) or 2. We know ll three sides of the tringle (SSS) But, wht hppens if the tringle we re working with doesn t fit either of those scenrios? Here we introduce the Lw of Sines. The Lw of Sines is sttement bout the reltionship between the sides nd the ngles in ny tringle. While the Lw of Sines will yield one correct nswer in mny situtions, there re times when it is mbiguous, mening tht it cn produce more thn one nswer. We will explore the mbiguity of the Lw of Sines in the next section. We cn use the Lw of Sines when: 1. We know two ngles nd non-included side (AAS) or 2. We know two ngles nd the included side (ASA) Deriving the Lw of Sines ABC contins ltitude CE, which extends from C nd intersects AB. We will refer to the length of ltitude CE s x. We know tht sina= x b nd sinb= x, by the definition of sine. If we cross-multiply both equtions nd substitute, we will hve the Lw of Sines. 331

2 5.3. The Lw of Sines b(sina)=x nd (sinb)=x ց sina b(sina)=(sinb) = sinb b or ւ sina = b sinb Extending these rtios to ngle C nd side c, we rrive t both forms of the Lw of Sines: Form 1 : (sines over sides) Form 2 : (sides over sines) sina = sinb b = sinc c sina = b sinb = c sinc AAS (Angle-Angle-Side) One cse where we cn to use the Lw of Sines is when we know two of the ngles in tringle nd non-included side (AAS). Exmple 1: Using GMN, G=42, N = 73 nd g=12. Find n. Since we know two ngles nd one non-included side(g), we cn find the other non-included side(n). sin73 = sin42 n 12 nsin42 = 12sin73 n= 12sin73 sin42 n Exmple 2: Continuing on from Exmple 1, find M nd m. 332

3 Chpter 5. Tringles nd Vectors Solution: M is simply = 65. To find side m, you cn now use either the Lw of Sines or Lw of Cosines. Considering tht the Lw of Sines is bit simpler nd new, let s use it. It does not mtter which side nd opposite ngle you use in the rtio with M nd m. Option 1: G nd g Option 2: N nd n sin65 m = sin42 12 msin42 = 12sin65 m= 12sin65 sin42 m sin65 m = sin msin73 = 17.15sin65 m= 17.15sin65 sin73 m Exmple 3: A business group wnts to build golf course on plot of lnd tht ws once frm. The deed to the lnd is old nd informtion bout the lnd is incomplete. If AB is 5382 feet, BC is 3862 feet, AEB is 101, BDC is 74, EAB is 41 nd DCB is 32, wht re the lengths of the sides of ech tringulr piece of lnd? Wht is the totl re of the lnd? Solution: Before we cn figure out the re of the lnd, we need to figure out the length of ech side. In tringle ABE, we know two ngles nd non-included side. This is the AAS cse. First, we will find the third ngle in tringle ABE by using the Tringle Sum Theorem. Then, we cn use the Lw of Sines to find both AE nd EB. ABE = 180 (41+101)=38 sin = sin38 sin 101 AE 5382 = sin41 EB AE(sin 101) = 5382(sin 38) EB(sin 101) = 5382(sin 41) AE = 5382(sin38) EB= 5382(sin41) sin 101 sin 101 AE = f eet EB f eet 333

4 5.3. The Lw of Sines Next, we need to find the missing side lengths in tringle DCB. In this tringle, we gin know two ngles nd non-included side (AAS), which mens we cn use the Lw of Sines. First, let s find DBC= 180 (74+32)=74. Since both BDC nd DBC mesure 74, tringle DCB is n isosceles tringle. This mens tht since BC is 3862 feet, DC is lso 3862 feet. All we hve left to find now is DB. sin = sin32 DB DB(sin 74) = 3862(sin 32) DB= 3862(sin32) sin74 DB f eet Finlly, we need to clculte the re of ech tringle nd then dd the two res together to get the totl re. From the lst section, we lerned two re formuls, K = 1 2 bcsina nd Heron s Formul. In this cse, since we hve enough informtion to use either formul, we will use K = 1 2 bcsina since it is less computtionlly intense. First, we will find the re of tringle ABE. Tringle ABE: K = 1 2 (3375.5)(5382)sin41 K = 5,959,292.8 ft 2 Tringle DBC: K = 1 2 (3862)(3862)sin32 K = 3,951,884.6 ft 2 The totl re is 5,959, ,951,884.6=9,911,177.4 ft 2. ASA (Angle-Side-Angle) The second cse where we use the Lw of Sines is when we know two ngles in tringle nd the included side (ASA). For instnce, in T RI: 334

5 Chpter 5. Tringles nd Vectors T, R, nd i re known T, I,nd r re known R, I, nd t re known In this cse, the Lw of Sines llows us to find either of the non-included sides. Exmple 4: (Use the picture bove) In T RI, T = 83, R=24, nd i=18.5. Find the mesure of t. Solution: Since we know two ngles nd the included side, we cn find either of the non-included sides using the Lw of Sines. Since we lredy know two of the ngles in the tringle, we cn find the third ngle using the fct tht the sum of ll of the ngles in tringle must equl 180. I = 180 (83+24) I = I = 73 Now tht we know I = 73, we cn use the Lw of Sines to find t. sin = sin83 t t(sin73)=18.5(sin83) t = 18.5(sin83) sin73 t 19.2 Notice how we wit until the lst step to input the vlues into the clcultor. This is so our nswer is s ccurte s possible. Exmple 5: In order to void lrge nd dngerous snowstorm on flight from Chicgo to Bufflo, pilot John strts out 27 off of the norml flight pth. After flying 412 miles in this direction, he turns the plne towrd Bufflo. The ngle formed by the first flight course nd the second flight course is 88. For the pilot, two issues re pressing: 1. Wht is the totl distnce of the modified flight pth? 2. How much further did he trvel thn if he hd styed on course? Solution, Prt 1: In order to find the totl distnce of the modified flight pth, we need to know side x. To find side x, we will need to use the Lw of Sines. Since we know two ngles nd the included side, this is n ASA cse. Remember tht in the ASA cse, we need to first find the third ngle in the tringle. 335

6 5.3. The Lw of Sines MissingAngle=180 (27+88)=65 The sum of ngles in tringle is 180 sin = sin27 x Lw of Sines x(sin 65) = 412(sin 27) Cross multiply x= 412(sin27) sin65 x miles Divide by sin 65 The totl distnce of the modified flight pth is =618.4 miles. Solution, Prt 2: To find how much frther John hd to trvel, we need to know the distnce of the originl flight pth, y. We cn use the Lw of Sines gin to find y. sin = sin88 y y(sin 65) = 412(sin 88) y= 412(sin88) sin65 y miles Lw of Sines Cross multiply Divide by sin 65 John hd to trvel =164.1 miles frther. Solving Tringles The Lw of Sines cn be pplied in mny wys. Below re some exmples of the different wys nd situtions to which we my pply the Lw of Sines. In mny wys, the Lw of Sines is much esier to use thn the Lw of Cosines since there is much less computtion involved. Exmple 6: In the figure below, C=22,BC=12,DC=14.3, BDA=65, nd ABD=11. Find AB. Solution: In order to find AB, we need to know one side in ABD. In BCD, we know two sides nd n ngle, which mens we cn use the Lw of Cosines to find BD. In this cse, we will refer to side BD s c. c 2 = (12)(14.3)cos22 c c 5.5 Lw of Cosines 336

7 Chpter 5. Tringles nd Vectors Now tht we know BD 5.5, we cn use the Lw of Sines to find AB. In this cse, we will refer to AB s x. A=180 (11+65)=104 Tringle Sum Theorem sin 104 = sin x x= 5.5sin65 sin 104 x 5.14 Lw of Sines Cross multiply nd divide by sin104 Exmple 7: A group of forest rngers re hiking through Denli Ntionl Prk towrds Mt. McKinley, the tllest mountin in North Americ. From their cmpsite, they cn see Mt. McKinley, nd the ngle of elevtion from their cmpsite to the summit is 21. They know tht the slope of mountin forms 127 ngle with ground nd tht the verticl height of Mt. McKinley is 20,320 feet. How fr wy is their cmpsite from the bse of the mountin? If they cn hike 2.9 miles in n hour, how long will it tke them to get the bse? Solution: As you cn see from the figure bove, we hve two tringles to del with here: right tringle ( MON) nd non-right tringle ( MOU). In order to find the distnce from the cmpsite to the bse of the mountin, y, we first need to find one side of our non-right tringle, MOU. If we look t M in MNO, we cn see tht side ON is our opposite side nd side x is our hypotenuse. Remember tht the sine function is opposite/hypotenuse. Therefore we cn find side x using the sine function. sin21 = x xsin21 = x= sin21 x Now tht we know side x, we know two ngles nd the non-included side in MOU. We cn use the Lw of Sines to solve for side y. First, MOU = = 32 by the Tringle Sum Theorem. sin = sin32 y ysin127 = sin32 y= sin32 sin127 x or 7.1 miles If they cn hike 2.9 miles per hour, then they will hike the 7.1 miles in 2.45 hours, or 2 hours nd 27 minutes. 337

8 5.3. The Lw of Sines Points to Consider Are there ny situtions where we might not be ble to use the Lw of Sines or the Lw of Cosines? Considering wht you lredy know bout the sine function, is it possible for two ngles to hve the sme sine? How might this ffect using the Lw of Sines to solve for n ngle? By using both the Lw of Sines nd the Lw of Cosines, it is possible to solve ny tringle we re given? Review Questions 1. In the tble below, you re given figure nd informtion known bout tht figure. Decide if ech sitution represents the AAS cse or the ASA cse. TABLE 5.4: Given Figure Cse. b=16,a=11.7,c=23.8 b. e=214.9,d=39.7,e = 41.3 c. G=22,I = 18,H = 140 d. k=6.3,j = 16.2,L=40.3 e. M = 31,O=9,m=15 f. Q=127,R=21.8,r= Even though ASA nd AAS tringles represent two different cses of the Lw of Sines, wht do they both hve in common?

9 Chpter 5. Tringles nd Vectors 3. Using the figures nd the given informtion from the tble bove, find the following if possible:. side b. side d c. side i d. side l e. side o f. side q 4. In GHI, I = 21.3, H = 62.1, nd i=108. Find g nd h. 5. Use the Lw of Sines to show tht b = sina sinb is true. 6. Use the Lw of Sines, the Lw of Cosines, nd trigonometry functions to solve for x.. b. 7. In order to void storm, pilot strts out 11 off pth. After he hs flown 218 miles, he turns the plne towrd his destintion. The ngle formed between his first pth nd his second pth is 105. If the plne trveled t n verge speed of 495 miles per hour, how much longer did the modified flight tke? 8. A delivery truck driver hs three stops to mke before she must return to the wrehouse to pick up more pckges. The wrehouse, Stop A, nd Stop B re ll on First Street. Stop A is on the corner of First Street nd Route 52, which intersect t 41 ngle. Stop B is on the corner of First Street nd Min Street, which intersect t 103 ngle. Stop C is t the intersection of Min Street nd Route 52. The driver knows tht Stop A nd Stop B re 12.3 miles prt nd tht the wrehouse is 1.1 miles from Stop A. If she must be bck to the wrehouse by 10:00.m., trvels t speed of 45 MPH, nd tkes 2 minutes to deliver ech pckge, t wht time must she leve? Review Answers ASA 2. AAS 3. neither 4. ASA 5. AAS 6. AAS 2. Student nswers will vry but they should notice tht in both cses you know or cn find n ngle nd the side cross from it. 339

10 5.3. The Lw of Sines sin = sin ,=5.6 sin = sin39.7 d,d = not enough informtion sin l = sin ,l = 4.9 sin9 5. o = sin31 15,o=4.6 sin q = sin ,q= G= = 96.6 sin96.6 g = sin21.3,g= sin62.1 h = sin21.3,h= sina = sinb b (sin B) = b(sin A) b = sina sinb Lw of Sines Cross multiply Divide by b(sinb) 1. tn54 = h 7.15 h=9.8,cos67 = 9.8 x x= The ngle we re finding is the one t the fr left side of the tringle = cosA A=43.4, sin43.4 x = sin x= First we need to find the other two sides in the tringle. sin = sin11 x = sin105 y,x=46.3,y=234.3, where y is the length of the originl fight pln. The modified flight pln is = Dividing both by 495 mi/hr, we get 32 min (modified) nd 28.4 min (originl). Therefore, the modified flight pln is 3.6 minutes longer. 6. First, we need to find the distnce between Stop B (B) nd Stop C (C). sin = sin41 B = sin103 C B=13.7,C = The totl length of her route is =48.6 miles. Dividing this by 45 mi/hr, we get tht it will tke her 1.08 hours, or 64.8 minutes, of ctul driving time. In ddition to the driving time, it will tke her 6 minutes (three stops t 2 minutes per stop) to deliver the three pckges, for totl roundtrip time of 70.8 minutes. Subtrcting this 70.8 minutes from 10:00 m, she will need to leve by 8:49 m. 340

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