Triangulation and the Law of Sines A triangle has three sides and - PowerPoint PPT Presentation

Triangulation and the Law of Sines A triangle has three sides and three angles; their values provide six pieces of information about the triangle. In ordinary geometry, most of the time three pieces of information are Elementary Functions sufficient to give us the other three pieces of information. Part 5, Trigonometry In order to more easily discuss the angles and sides of a triangle, we will Lecture 5.3a, The Laws of Sines and Triangulation label the angles by capital letters (such as A, B, C ) and label the sides by small letters (such as a, b, c .) Dr. Ken W. Smith We will assume that a side labeled with a small letter is the side opposite Sam Houston State University the angle with the same, but capitalized, letter. For example, the side a is opposite the angle A . 2013 We will also use these letters for the values or magnitudes of these sides and angles, writing a = 3 feet or A = 14 ◦ . Smith (SHSU) Elementary Functions 2013 1 / 26 Smith (SHSU) Elementary Functions 2013 2 / 26 Three Bits of a Triangle AA and AAA Suppose we have three bits of information about a triangle. Can we recover all the information? AAA If the 3 bits of information are the magnitudes of the 3 angles , and if we If we know the values of two angles then since the angles sum to π have no information about lengths of sides, then the answer is No. ( = 180 ◦ ), we really know all 3 angles. Given a triangle with angles A, B, C (in ordinary Euclidean geometry) we can always expand (or contract) the triangle to a similar triangle which So AAA is really no better than AA ! has the same angles but whose sides have all been stretched (or shrunk) We need to know at least one of the sides. by some constant factor. This “Angle-Angle-Angle” information ( AAA ) is not enough information to describe the entire triangle. However, we will discover that in almost every other situation, we can recover the entire triangle. Smith (SHSU) Elementary Functions 2013 3 / 26 Smith (SHSU) Elementary Functions 2013 4 / 26

ASA and AAS The Law of Sines and ASA Suppose we know two angles and one side. If the known side is between The law of sines says that give a triangle △ ABC , with lengths of sides the two angles we abbreviate this ASA . a, b and c then sin A = sin B = sin C If we know two angles and one side, but the known side is not between the (1) a b c two angles, then we are in the Angle-Angle-Side ( AAS ) situation and we can also recover the remaining bits of the triangle. Why is this true? We do this using the Law of Sines. Smith (SHSU) Elementary Functions 2013 5 / 26 Smith (SHSU) Elementary Functions 2013 6 / 26 The Law of Sines The Law of Sines Start with a triangle with vertices A, B, C sin A = h b and sin B = h a . Draw a perpendicular from the vertex C onto the side AB and let D Solve for h : h = b sin A and h = a sin B. represent the right angle formed there. Since both b sin A and a sin B are equal to h then If h is the height of the triangle then b sin A = a sin B. sin A = h b and sin B = h a. Dividing both sides by ab we have sin A = sin B b . a A similar argument tells us that sin A = sin C a c Smith (SHSU) Elementary Functions 2013 7 / 26 Smith (SHSU) Elementary Functions 2013 8 / 26

The Law of Sines Two Angles and the Law of Sines If we know two angles of a triangle, then since the three angles add to 180 ◦ = π then we can figure out the third angle. As long as we know one more bit of information, the length of a side, then the law of sines gives us sin A = sin B = sin C the length of all sides. (2) In our abbreviated notation for the known information, this includes the a b c cases ASA and AAS , that is, the cases in which we know two angles and an included side or two angles and a side not between them. These cases are equivalent since, essentially, we know all three angles once we know two. Smith (SHSU) Elementary Functions 2013 9 / 26 Smith (SHSU) Elementary Functions 2013 10 / 26 Some Worked Problems. Some Worked Problems. Use the Law of Sines to solve the following triangles. Solve the triangle A = 60 ◦ , B = 70 ◦ , c = 10 feet A = 62 ◦ , B = 74 ◦ , c = 14 feet Solution. If A = 60 ◦ , B = 70 ◦ , c = 10 feet then C = 50 ◦ since the sum of Solution. If A = 62 ◦ , B = 74 ◦ , c = 14 feet then C = 54 ◦ . Then, by the the angles of a triangle is 180 ◦ . law of sines, By the law of sines, sin 54 ◦ = sin 62 ◦ = sin 75 ◦ sin 60 ◦ = sin 70 ◦ = sin 50 ◦ 14 a b . 10 so a b So a = 10 sin 60 ◦ = 11 . 31 feet and b = 10 sin 70 ◦ a = 14 sin 62 ◦ ≈ 15 . 28 feet and b = 14 sin 75 ◦ = 12 . 27 feet ≈ 16 . 72 feet. sin 50 ◦ sin 50 ◦ sin 54 ◦ sin 54 ◦ Smith (SHSU) Elementary Functions 2013 11 / 26 Smith (SHSU) Elementary Functions 2013 12 / 26

Law of Sines Elementary Functions Part 5, Trigonometry In the next presentation, we will continue to look at the Law of Sines, Lecture 5.3b, The Laws of Sines and Triangulation investigating the SSA case. Dr. Ken W. Smith (End) Sam Houston State University 2013 Smith (SHSU) Elementary Functions 2013 13 / 26 Smith (SHSU) Elementary Functions 2013 14 / 26 The law of sines and SSA Some Worked Problems Solve the following triangle. (Find all sides and all angles.) A = 30 ◦ , a = 6 feet , b = 8 feet If we know only one angle of a triangle but two sides, sometimes the Law of Sines is sufficient. Solution. To solve A = 30 ◦ , a = 6 feet , b = 8 feet apply the law of sines: To use the laws of sines, we need one of the sides to be opposite the = sin B = sin C sin 30 ◦ c 6 8 known angle. This forces In this case, our information is often abbreviated SSA – we know two sin B = 8 sin 30 ◦ = 2 sides and then an angle not between them. 6 3 3 ) ≈ 41 . 81 ◦ or B = 180 − 41 . 81 ◦ = 138 . 19 ◦ . so B is either sin − 1 ( 2 We do have to be a bit careful here. Just because we know the sine of an angle does not mean we know the value of the angle! If B = 41 . 81 ◦ then C = 108 . 19 ◦ . If B = 138 . 19 ◦ then C = 11 . 81 ◦ . By the law of sines c = 6 sin C sin 30 ◦ = 12 sin C. There could be two angles with the same sine! If C = 108 . 19 ◦ then c = 12 sin 108 . 19 ◦ = 11 . 40 But if C = 11 . 81 ◦ then c = 12 sin 11 . 81 ◦ = 2 . 46 Smith (SHSU) Elementary Functions 2013 15 / 26 Smith (SHSU) Elementary Functions 2013 16 / 26

Some Worked Problems Some Worked Problems Solve the triangle A = 60 ◦ , a = 10 feet , c = 12 feet Solutions. If A = 60 ◦ , a = 10 feet , c = 12 feet then by the law of sines We solved the triangle A = 30 ◦ , a = 6 feet , b = 8 feet: sin 60 ◦ = sin C ◦ 10 12 There are two solutions, two triangles: So √ sin C = 12 sin 60 ◦ = 3 3 A B C a b c ≈ 1 . 039 > 1 . 10 5 30 ◦ 41 . 81 ◦ 108 . 19 ◦ 6 8 11 . 41 30 ◦ 138 . 19 ◦ 11 . 81 ◦ 6 8 2 . 46 There is no angle with sine greater than 1 so there is no solution; This triangle cannot exist. Smith (SHSU) Elementary Functions 2013 17 / 26 Smith (SHSU) Elementary Functions 2013 18 / 26 Some Worked Problems Some Worked Problems Solve the triangle a = 10 , b = 6 , B = 20 ◦ . Solution. Use the Law of Sines to see that A could be either A = 34 . 75 ◦ or A = 145 . 25 ◦ , Solve the triangle b = 12 , c = 10 , C = 60 ◦ . Then C = 125 . 25 ◦ or C = 14 . 75 ◦ . √ Solution. By the Law of Sines, sin B = 12 10(sin 60 ◦ ) = 3 3 ≈ 1 . 039 . 5 Finally use the Law of Sines to finish off the problem. Since the sine of B cannot be greater than one, There are two answers. Both must be listed. no such triangle is possible. a = 10 , b = 6 , c = 14 . 33 , A = 34 . 75 ◦ , B = 20 ◦ , C = 125 . 25 ◦ (The length of b is not large enough to “reach” the side a .) and a = 10 , b = 6 , c = 4 . 47 , A = 145 . 25 ◦ , B = 20 ◦ , C = 14 . 75 ◦ . Smith (SHSU) Elementary Functions 2013 19 / 26 Smith (SHSU) Elementary Functions 2013 20 / 26

Some Worked Problems Triangulation Solve the triangle b = 12 , c = 10 , C = 45 ◦ . Solution. By the Law of Sines either B ≈ 58 . 06 ◦ or B ≈ 121 . 95 ◦ , Then A ≈ 76 . 95 ◦ or A ≈ 13 . 05 ◦ . One can often measure distance to an object by working out a baseline and then measuring the angles formed by lines from the distant object to the ends of the baseline. Finally use the Law of Sines to finish off the problem. This technique is called ”triangulation”. There are two answers. Both must be listed. This is a classic case of ASA and is perfect for the law of sines. a = 13 . 78 , b = 12 , c = 10 , A = 76 . 95 ◦ , B = 58 . 45 ◦ , C = 45 ◦ , and a = 3 . 19 , b = 12 , c = 10 , A = 13 . 05 ◦ , B = 121 . 95 ◦ , C = 45 ◦ Smith (SHSU) Elementary Functions 2013 21 / 26 Smith (SHSU) Elementary Functions 2013 22 / 26 Triangulation Triangulation Suppose that a ship is out in the harbor. Standing at the east end of the Draw an east-west baseline representing the 200 yards and then draw lines dock, a compass reveals that the ship is at a bearing of 30 ◦ east of north. from the ends of the baseline to the ship. We have a triangle in which the But if one walks 200 yards west of the dock, the ship is now at a bearing vertex on the west side of the baseline has an (interior) angle of of 40 ◦ east of north. 90 ◦ − 40 ◦ = 50 ◦ . The vertex on the east side of the baseline has an angle of 90 ◦ + 30 ◦ = 120 ◦ . Call the length of the baseline c = 200 , and so we How far is the ship from the dock? have angles A = 50 ◦ and B = 120 ◦ . Smith (SHSU) Elementary Functions 2013 23 / 26 Smith (SHSU) Elementary Functions 2013 24 / 26