1/14 MTH 112: Elementry Functions Section 8.1: Lw of Sines Lern out olique tringles. Derive the Lw os Sines. Solve tringles. Solve the miguous cse.
8.1:Lw of Sines. 2/14 Solving olique tringles Solving Olique tringles S (ngle side ngle) or S (ngle ngle side), depending on whether the given side is etween the ngles. SS (side side ngle). Two sides nd n ngle opposite one of the sides re given. SSS (side side side) SS (side ngle side). Two sides nd the ngle etween re given. Note: First two cses re solved using the lw of sines. Previously we solve right tringles, now we solve olique tringles (olique tringle hve no right ngles).
8.1:Lw of Sines. 3/14 Stndrd leling C γ α c β B
8.1:Lw of Sines. 4/14 Lw of sines ny tringle with stndrd leling stisfies: Lw of sines or equivlently, sin(α) sin(α) = = sin(β) sin(β) = = sin(γ) c c sin(γ) γ C α c β B
8.1:Lw of Sines. 5/14 Exmple (S) If β = 85,γ = 40, nd = 26, solve tringle BC. C First, find α= 180 40 85 = 55 Next, side cn e found using the lw of sines. 40 sin(α) = sin(β) 26 B 85 α c 21.38
8.1:Lw of Sines. 6/14 Exmple (S) continues C Finding sided c: 40 21.38 26 B 85 55 c c 16.78
8.1:Lw of Sines. 7/14 Solving the miguous cse (SS) Given two sides nd n ngle opposite one side, the following steps cn e used to determine whether there re zero, one, or two tringles tht stisfy the conditions. For simplicity, ssume tht,, nd α re given. C γ =? α c =? β =? B
8.1:Lw of Sines. 8/14 Use the lw of sines to find sin(β), where β is the unknown ngle opposite one of the given sides. Cse: 1. No Solutions: If sin(β) > 1, there re no possile tringles. Cse 2. One Solution: If sin(β) = 1, then β = 90 nd γ = 90 α. Use the lw of sines or the Pythgoren theorem to find the unknown side c. There is one right tringle. Cse 3: If sin(β) = k, where 0 < k < 1, clculte sin 1 (k) = β 1 nd β 2 = 180 β 1. Note tht β 1 nd β 2 re the two possile solutions to sin(β) = k. One Solution: If α + β 2 180, there is one possile tringle determined y β 1. Let γ = 180 α β 1. Use the lw of sines to find c. Two solutions: If α + β 2 < 180, there re two possile tringles. Let γ 1 = 180 α β 1 nd γ 2 = 180 α β 2. Use the lw of sines to find c 1 nd c 2.
8.1:Lw of Sines. 9/14 Exmple Exmple 1 Let α = 62, = 6, nd = 10. If possile, solve the tringle. We egin y using the lw of sines to find sin(β). sin(β) = sin(α) sin(β) 1.47 > 1
8.1:Lw of Sines. 10/14 Exmple continues Let α = 62, = 6, nd = 10. If possile, solve the tringle. = 10 = 6 62 Since the sine function is never greter thn 1, there re no solutions. No such tringle exists.
8.1:Lw of Sines. 11/14 Exmple Exmple 2 Let α = 62, = 9, nd = 10. If possile, solve the tringle. We egin y using the lw of sines to find sin(β). sin(β) = sin(α) sin(β) 0.9811
8.1:Lw of Sines. 12/14 Exmple continued β 1 78.83 nd γ 1 = 180 62 78.83 = 39.17. C β 2 = 180 β 1 101.17 nd γ 2 = 180 62 101.17 = 16.83. C = 10 = 9 = 10 = 9 62 78.83 B We cn find c 1 using the lw of sines. 62 101.17 B We cn find c 2 using the lw of sines. c 1 6.44 c 2 2.95
8.1:Lw of Sines. 13/14 Exmple Exmple 3 Let α = 62, = 16, nd = 10. If possile, solve the tringle. We egin y using the lw of sines to find sin(β). sin(β) = sin(α) sin(β) 0.5518
8.1:Lw of Sines. 14/14 Exmple continued We hve α = 62, = 16, nd = 10. β 1 33.49 nd γ 1 = 180 62 33.49 = 84.51. β 2 = 180 β 1 146.51 nd γ 2 = 180 62 146.51 = 28.51. = 10 = 16 62 33.49 We cn find c using the lw of sines. = 10 62 146.51 = 16 c 18.04