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7.1 The Sines Law 1. Determining if the Law of Sines Can Be Used to Solve an Oblique Triangle 2. Using the Law of Sines to Solve the SAA Case or ASA Case 3. Using the Law of Sines to Solve the SSA Case 4. Using the Law of Sines to Solve Applied 5. Problems Involving Oblique Triangles H.Melikian/1200 Dr .Hayk Melikyan/ Departmen of Mathematics and CS/ [email protected] 1 The Law of Sines If A, B, and C are the measures of the angles of any triangle and if a, b, and c are the lengths of the sides opposite the corresponding angles, then a b c sin A sin B sin C or sin A sin B sin C a b c H.Melikian/1200 2 The Law of Sines The following information is needed to use the Law of Sines. 1. The measure of an angle must be known. 2. The length of the side opposite the known angle must be known. 3. At least one more side or one more angle must be known. H.Melikian/1200 3 Determining of the Law of Sines Can Be Used to Solve an Oblique Triangle Decide whether or not the Law of Sines can be used to solve each triangle. A a. b. c. A 47° A 99° b b = 40 c = 54 c b = 15.4 c = 14 37° 50° B C B C a a C a = 28 B No; We are not given the measure of any angle. H.Melikian/1200 Yes; we are given an angle and the measure of its opposite side and an additional angle. Yes; we are given an angle and the measure of its opposite side and an additional angle. 4 Text Example (ASA) Solve triangle ABC if A 50º, C 33.5º, and b 76. Solution We begin by drawing a picture of triangle ABC and labeling it with the given information. The figure shows the triangle that we must solve. We begin by finding B. C A + B + C 180º 33.5º b 76 a 50º + B + 33.5º 180º 83.5º + B 180º 50º A c H.Melikian/1200 B B 96.5º The sum of the measurements of a triangle’s interior angles is 180º. A = 50º and C = 33.5º. Add. Subtract 83.5º from both sides. 5 Text Example cont. Solve triangle ABC if A 50º, C 33.5º, and b 76. Keep in mind that we must be given one of the three ratios to apply the Law of Sines. In this example, we are given that b 76 and we found that B 96.5º. Thus, we use the ratio b/sin B, or 76/sin96.5º, to find the other two sides. Use the Law of Sines to find a and c. Solution C Find a: Find c: This is the known ratio. 33.5º b 76 a 50º A c H.Melikian/1200 B a b sin A sin B a 76 sin 50 sin 96.5 76sin 50 a 59 sin 96.5 c b sin C sin B c 76 sin 33.5 sin 96.5 76sin 33.5 c 42 sin 96.5 The solution is B 96.5º, a 59, and c 42. 6 Solving a SAA Triangle Using the Law of Sines Solve the given oblique triangle. Round lengths to one decimal place. Angles A 99° b c = 14 50° C B Sides A = 99 a=? B = ?? b=? C = 50 c = 14 a A + B + C 180 99 + B + 50 180 149 + B 180 B 31 H.Melikian/1200 7 Solving a SAA Triangle Using the Law of Sines Solve the given oblique triangle. Round lengths to one decimal place. Find a: Find b: A c = 14 99° b 50° C B a Angles Sides A = 99 a=? B = 31 b=? C = 50 c = 14 H.Melikian/1200 a c sin A sin C a 14 sin 99 sin 50 14sin 99 a sin 50 a 18.1 b c sin B sin C b 14 sin 31 sin 50 14sin 31 b sin 50 b 9.4 8 Solving a ASA Triangle Using the Law of Sines Solve oblique triangle ABC if B = 42, A = 57, and c = 18.6 cm. C 180 57 42 81 A c = 18.6 57° b 42° B a C Angles Sides A = 57 a=? B = 42 b=? C= c = 18.6 81 H.Melikian/1200 a c sin A sin C a 18.6 sin 57 sin81 18.6sin 57 a sin81 a 15.8 b c sin B sin C b 18.6 sin 42 sin81 18.6sin 42 a sin81 a 12.6 9 The Ambiguous Case (SSA) a is less than h and not long enough to form a triangle. a b h = b sin A A a = h and is just the right length to form a right triangle. b a h = b sin A A No Triangle One Right Triangle a is greater than h and a is less than b. Two distinct triangles are formed. A b a a h = b sin A a is greater than h and a is greater than b. One triangle is formed. b a A Two Triangles H.Melikian/1200 One Triangle 10 H.Melikian/1200 11 Solving a SSA Triangle Using the Law of Sines ( Quiz No) Solve the triangle; if possible. Angles Sides A=? a = 20 B=? b= C = 50 c = 10 sin A sin C a c sin A sin 50 20 10 20sin 50 sin A 10 sin A 1.5 Since there is no angle A for which sin A > 1, there can be no triangle with the given measurements. H.Melikian/1200 12 Solving a SSA Triangle Using the Law of Sines (One Triangle) Solve the triangle; if possible. Angles Sides A = 40 a = 30 B=? b = 20 C=? c=? Two possible angles: sin A sin B a b sin 40 sin B 30 20 20sin 40 sin A 30 sin B 0.4285 25.4 or 180 25.4 154.6 154.6 is too large when combined with the given angle. H.Melikian/1200 13 Solving a SSA Triangle Using the Law of Sines (One Triangle)-cont Solve the triangle; if possible. Angles Sides A = 40 a = 30 B = 25.4 b = 20 C = 114.6 c=? H.Melikian/1200 C 180 40 25.4 114.6 sin A sin C a c sin 40 sin114.6 30 c 30sin114.6 c sin 40 c 42.3 14 Solving a SSA Triangle Using the Law of Sines (Two Triangles) Solve the triangle; if possible. Angles Sides A = 35 a = 60 B=? b = 80 C=? c=? sin A sin B a b sin 35 sin B 60 80 80sin 35 sin B 60 sin B 0.7648 Two possible angles: 49.9 or 180 49.9 130.1 There are two possible triangles. H.Melikian/1200 15 Solving a SSA Triangle Using the Law of Sines (Two Triangles) Solve the triangle; if possible. Angles Sides Angles Sides A = 35 a = 60 A = 35 a = 60 B = 49.9 b = 80 B = 130.1 b = 80 C= c = 104.2 C = 14.9 95.1 C 180 35 49.9 95.1 or c= sin A sin C a c sin 35 sin 95.1 60 c 60sin 95.1 c sin 35 c 104.2 C 180 35 130.1 14.9 H.Melikian/1200 16 Solving a SSA Triangle Using the Law of Sines (Two Triangles) Solve the triangle; if possible. Angles Sides Angles Sides A = 35 a = 60 A = 35 a = 60 B = 49.9 b = 80 B = 130.1 b = 80 C= 95.1 H.Melikian/1200 c = 104.2 C = 14.9 c= 26.9 sin A sin C a c sin 35 sin14.9 60 c 60sin14.9 c sin 35 c 26.9 17 Solving an Applied Problem Martin wants to measure the distance across a river. He has made a sketch. Find the distance across the river, a. C 180 23 126 31 23° 348 feet 31 a H.Melikian/1200 126° a b sin A sin B a 348 sin 23 sin 31 348sin 23 a sin 31 a 264 ft 18 Determining the Distance a Ship is from Port A ship set sail from port at a bearing of N 53E and sailed 63 km to point B. The ship then turned and sailed an additional 69 km to point C. Determine the distance from port to point C if the ship’s final bearing is N 74E. Draw a diagram. Find angle A. A 74 53 21 H.Melikian/1200 19 Determining the Distance a Ship is from Port-cont A ship set sail from port at a bearing of N 53E and sailed 63 km to point B. The ship then turned and sailed an additional 69 km to point C. Determine the distance from port to point C if the ship’s final bearing is N 74E. Angles Sides A = 21 a = 69 B= b= C= c = 63 Two possible angles: 19.1 or 180 19.1 160.9 H.Melikian/1200 sin C sin A c a sin C sin 21 63 69 63sin 21 sin A 69 sin A 0.3272 160.9 will not work 20 Determining the Distance a Ship is from Port-cont A ship set sail from port at a bearing of N 53E and sailed 63 km to point B. The ship then turned and sailed an additional 69 km to point C. Determine the distance from port to point C if the ship’s final bearing is N 74E. Angles Sides A = 21 a = 69 B = 139.9 b= C = 19.1 c = 63 b a sin B sin A b 69 sin139.9 sin 21 69sin139.9 b sin 21 b 124.0 The ship is 124 miles from port to point C. H.Melikian/1200 21