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Law of Sines MATH 160, Precalculus J. Robert Buchanan Department of Mathematics Fall 2011 J. Robert Buchanan Law of Sines Objectives In this lesson we will learn to: use the Law of Sines to solve oblique triangles (AAS, ASA, and SSA), find the areas of oblique triangles, use the the Law of Sines to model and solve real-world problems. J. Robert Buchanan Law of Sines Oblique Triangles Oblique triangles are triangles having no right angles. C a b c A J. Robert Buchanan B Law of Sines Solving an Oblique Triangle Solving a triangle involves determining to the lengths of its three sides and the measures of its three angles. J. Robert Buchanan Law of Sines Solving an Oblique Triangle Solving a triangle involves determining to the lengths of its three sides and the measures of its three angles. To solve a triangle we must know the length of at least one side and any two other measures of the triangle: 1 Any side and two angles (AAS or ASA). 2 Two sides and the angle opposite one of them (SSA). 3 Three sides (SSS). 4 Two sides and their included angle. J. Robert Buchanan Law of Sines Solving an Oblique Triangle Solving a triangle involves determining to the lengths of its three sides and the measures of its three angles. To solve a triangle we must know the length of at least one side and any two other measures of the triangle: 1 Any side and two angles (AAS or ASA). 2 Two sides and the angle opposite one of them (SSA). 3 Three sides (SSS). 4 Two sides and their included angle. Today we will explore the first two cases. J. Robert Buchanan Law of Sines Law of Sines Law of Sines If ABC is a triangle with sides a, b, and c, then b c a = = sin A sin B sin C sin A sin B sin C = = . a b c or C C b a h h A c B Acute triangle J. Robert Buchanan a b A c Obtuse triangle Law of Sines B Example Solve the following triangle (not drawn to scale). C a b 40 ° 35 ° A c =10 J. Robert Buchanan B Law of Sines Example Solve the following triangle (not drawn to scale). C a b 40 ° 35 ° A c =10 B This is the ASA case. C = 180◦ − 35◦ − 40◦ = 105◦ c 10 sin 35◦ a = sin A = ≈ 5.94 sin C sin 105◦ c 10 sin 40◦ b = sin B = ≈ 6.65 sin C sin 105◦ J. Robert Buchanan Law of Sines Example A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 12◦ with the horizontal. The flagpole’s shadow is 16 meters long and points directly up the slope. The angle of elevation from the tip of the shadow to the sun is 20◦ . Find the height of the flagpole. J. Robert Buchanan Law of Sines Example A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 12◦ with the horizontal. The flagpole’s shadow is 16 meters long and points directly up the slope. The angle of elevation from the tip of the shadow to the sun is 20◦ . Find the height of the flagpole. This is the ASA situation. Let A = 78◦ , B = 32◦ , and c = 16 meters. J. Robert Buchanan Law of Sines Example A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 12◦ with the horizontal. The flagpole’s shadow is 16 meters long and points directly up the slope. The angle of elevation from the tip of the shadow to the sun is 20◦ . Find the height of the flagpole. This is the ASA situation. Let A = 78◦ , B = 32◦ , and c = 16 meters. C = 180◦ − 78◦ − 32◦ = 70circ c 16 sin 70◦ b = sin B = ≈ 9.02 sin C sin 32◦ J. Robert Buchanan Law of Sines meters SSA: Ambiguous Case If we know two side lengths and the measure of the angle opposite one of the sides, three possibles outcomes may present themselves when solving the oblique triangle. No such triangle exists. One such triangle exists. Two distinct triangles exist. J. Robert Buchanan Law of Sines Example Use the Law of Sines to solve the triangle, if there is a solution. If two solutions exist, find them both. A = 110◦ , a = 125, J. Robert Buchanan Law of Sines b = 200. Example Use the Law of Sines to solve the triangle, if there is a solution. If two solutions exist, find them both. A = 110◦ , sin A a a = 125, b = 200. sin B b b sin A sin B = a 200 sin 110◦ = 125 = 1.50351 > 1 = No solution. J. Robert Buchanan Law of Sines Example Use the Law of Sines to solve the triangle, if there is a solution. If two solutions exist, find them both. A = 76◦ , a = 34, J. Robert Buchanan Law of Sines b = 21. Example Use the Law of Sines to solve the triangle, if there is a solution. If two solutions exist, find them both. A = 76◦ , sin A a a = 34, b = 21. sin B b b sin A sin B = a 21 sin 76◦ = 34 ≈ 0.5993 = B ≈ 36.82◦ C ≈ 67.18◦ a 34 sin 67.18◦ c = sin C ≈ ≈ 32.30 sin A sin 76◦ J. Robert Buchanan Law of Sines Example Use the Law of Sines to solve the triangle, if there is a solution. If two solutions exist, find them both. A = 25◦ , a = 9.5, J. Robert Buchanan Law of Sines b = 22. Example Use the Law of Sines to solve the triangle, if there is a solution. If two solutions exist, find them both. A = 25◦ , a = 9.5, sin A a sin B b = 22. sin B b b sin A = a 22 sin 25◦ = 9.5 ≈ 0.978695 = B ≈ 78.15◦ or B ≈ 101.85◦ C ≈ 76.85◦ or C ≈ 53.15◦ 9.5 sin 76.85◦ 9.5 sin 53.15◦ c≈ ≈ 21.89 or c ≈ ≈ 17.99 sin 25◦ sin 25◦ J. Robert Buchanan Law of Sines Area of an Oblique Triangle Area The area of any triangle is one-half the product of the lengths of two sides times the sine of their included angle. Area = 1 1 1 bc sin A = ac sin B = ab sin C 2 2 2 C C b a h h A c B Acute triangle J. Robert Buchanan a b A c Obtuse triangle Law of Sines B Example Find the area of the triangle having the following angle and sides. B = 130◦ , a = 62, c = 20. J. Robert Buchanan Law of Sines Example Find the area of the triangle having the following angle and sides. B = 130◦ , a = 62, c = 20. Area = 1 1 ac sin B = (62)(20) sin 130◦ = 474.948 2 2 J. Robert Buchanan Law of Sines Application The circular arc of a railroad curve has a chord of length 3000 feet corresponding to a central angle of 40◦ . Find the radius r of the circular arc. Find the length s of the circular arc. J. Robert Buchanan Law of Sines Application The circular arc of a railroad curve has a chord of length 3000 feet corresponding to a central angle of 40◦ . Find the radius r of the circular arc. r sin 70◦ = r = 3000 sin 40◦ 3000 sin 70◦ ≈ 4385.71 sin 40◦ Find the length s of the circular arc. J. Robert Buchanan Law of Sines feet Application The circular arc of a railroad curve has a chord of length 3000 feet corresponding to a central angle of 40◦ . Find the radius r of the circular arc. r sin 70◦ = r = 3000 sin 40◦ 3000 sin 70◦ ≈ 4385.71 sin 40◦ feet Find the length s of the circular arc. The central angle of 40◦ is equivalent to 2π/9 radians 2π s = r θ = 4385.71 = 3061.8 feet 9 J. Robert Buchanan Law of Sines Homework Read Section 5.6. Exercises: 1, 5, 9, 13, . . . , 49, 53 J. Robert Buchanan Law of Sines