Report

8 Applications of Trigonometry Copyright © 2009 Pearson Addison-Wesley 8.1-1 8 Applications of Trigonometry 8.1 The Law of Sines 8.2 The Law of Cosines 8.3 Vectors, Operations, and the Dot Product 8.4 Applications of Vectors 8.5 Trigonometric (Polar) Form of Complex Numbers; Products and Quotients 8.4 De Moivre’s Theorem; Powers and Roots of Complex Numbers 8.5 Polar Equations and Graphs 8.6 Parametric Equations, Graphs, and Applications Copyright © 2009 Pearson Addison-Wesley 8.1-2 8.1 The Law of Sines Congruency and Oblique Triangles ▪ Derivation of the Law of Sines ▪ Using the Law of Sines ▪ Ambiguous Case ▪ Area of a Triangle Copyright © 2009 Pearson Addison-Wesley 1.1-3 8.1-3 Congruence Axioms Side-Angle-Side (SAS) If two sides and the included angle of one triangle are equal, respectively, to two sides and the included angle of a second triangle, then the triangles are congruent. Angle-Side-Angle (ASA) If two angles and the included side of one triangle are equal, respectively, to two angles and the included side of a second triangle, then the triangles are congruent. Copyright © 2009 Pearson Addison-Wesley 1.1-4 8.1-4 Congruence Axioms Side-Side-Side (SSS) If three sides of one triangle are equal, respectively, to three sides of a second triangle, then the triangles are congruent. Copyright © 2009 Pearson Addison-Wesley 1.1-5 8.1-5 Oblique Triangles Oblique triangle A triangle that is not a right triangle The measures of the three sides and the three angles of a triangle can be found if at least one side and any other two measures are known. Copyright © 2009 Pearson Addison-Wesley 8.1-6 Data Required for Solving Oblique Triangles Case 1 One side and two angles are known (SAA or ASA). Case 2 Two sides and one angle not included between the two sides are known (SSA). This case may lead to more than one triangle. Case 3 Two sides and the angle included between the two sides are known (SAS). Case 4 Three sides are known (SSS). Copyright © 2009 Pearson Addison-Wesley 1.1-7 8.1-7 Note If three angles of a triangle are known, unique side lengths cannot be found because AAA assures only similarity, not congruence. Copyright © 2009 Pearson Addison-Wesley 1.1-8 8.1-8 Derivation of the Law of Sines Start with an oblique triangle, either acute or obtuse. Let h be the length of the perpendicular from vertex B to side AC (or its extension). Then c is the hypotenuse of right triangle ABD, and a is the hypotenuse of right triangle BDC. Copyright © 2009 Pearson Addison-Wesley 8.1-9 Derivation of the Law of Sines In triangle ADB, = = Since h = c sin A and h = a sin C, Copyright © 2009 Pearson Addison-Wesley 8.1-10 Derivation of the Law of Sines In triangle BDC, Similarly, it can be shown that and Copyright © 2009 Pearson Addison-Wesley 8.1-11 Law of Sines In any triangle ABC, with sides a, b, and c, Copyright © 2009 Pearson Addison-Wesley 1.1-12 8.1-12 Example 1 USING THE LAW OF SINES TO SOLVE A TRIANGLE (SAA) Solve triangle ABC if A = 32.0°, C = 81.8°, and a = 42.9 cm. Law of sines Copyright © 2009 Pearson Addison-Wesley 1.1-13 8.1-13 Example 1 USING THE LAW OF SINES TO SOLVE A TRIANGLE (SAA) (continued) A + B + C = 180° C = 180° – A – B C = 180° – 32.0° – 81.8° = 66.2° Use the Law of Sines to find c. Copyright © 2009 Pearson Addison-Wesley 1.1-14 8.1-14 Example 2 USING THE LAW OF SINES IN AN APPLICATION (ASA) Kurt wishes to measure the distance across the Big Muddy River. He determines that C = 112.90°, A = 31.10°, and b = 347.6 ft. Find the distance a across the river. First find the measure of angle B. B = 180° – A – C = 180° – 31.10° – 112.90° = 36.00° Copyright © 2009 Pearson Addison-Wesley 1.1-15 8.1-15 Example 2 USING THE LAW OF SINES IN AN APPLICATION (ASA) (continued) Now use the Law of Sines to find the length of side a. The distance across the river is about 305.5 feet. Copyright © 2009 Pearson Addison-Wesley 1.1-16 8.1-16 Example 3 USING THE LAW OF SINES IN AN APPLICATION (ASA) Two ranger stations are on an east-west line 110 mi apart. A forest fire is located on a bearing N 42° E from the western station at A and a bearing of N 15° E from the eastern station at B. How far is the fire from the western station? First, find the measures of the angles in the triangle. Copyright © 2009 Pearson Addison-Wesley 1.1-17 8.1-17 Example 3 USING THE LAW OF SINES IN AN APPLICATION (ASA) (continued) Now use the Law of Sines to find b. The fire is about 234 miles from the western station. Copyright © 2009 Pearson Addison-Wesley 1.1-18 8.1-18 Description of the Ambiguous Case If the lengths of two sides and the angle opposite one of them are given (Case 2, SSA), then zero, one, or two such triangles may exist. Copyright © 2009 Pearson Addison-Wesley 8.1-19 If A is acute, there are four possible outcomes. Copyright © 2009 Pearson Addison-Wesley 1.1-20 8.1-20 If A is obtuse, there are two possible outcomes. Copyright © 2009 Pearson Addison-Wesley 1.1-21 8.1-21 Applying the Law of Sines 1. For any angle θ of a triangle, 0 < sin θ ≤ 1. If sin θ = 1, then θ = 90° and the triangle is a right triangle. 2. sin θ = sin(180° – θ) (Supplementary angles have the same sine value.) 3. The smallest angle is opposite the shortest side, the largest angle is opposite the longest side, and the middle-value angle is opposite the intermediate side (assuming the triangle has sides that are all of different lengths). Copyright © 2009 Pearson Addison-Wesley 1.1-22 8.1-22 Example 4 SOLVING THE AMBIGUOUS CASE (NO SUCH TRIANGLE) Solve triangle ABC if B = 55°40′, b = 8.94 m, and a = 25.1 m. Law of sines (alternative form) Since sin A > 1 is impossible, no such triangle exists. Copyright © 2009 Pearson Addison-Wesley 1.1-23 8.1-23 Example 4 SOLVING THE AMBIGUOUS CASE (NO SUCH TRIANGLE) (continued) An attempt to sketch the triangle leads to this figure. Copyright © 2009 Pearson Addison-Wesley 1.1-24 8.1-24 Note In the ambiguous case, we are given two sides and an angle opposite one of the sides (SSA). Copyright © 2009 Pearson Addison-Wesley 1.1-25 8.1-25 Example 5 SOLVING THE AMBIGUOUS CASE (TWO TRIANGLES) Solve triangle ABC if A = 55.3°, a = 22.8 ft, and b = 24.9 ft. There are two angles between 0° and 180° such that sin B ≈ .897867: Copyright © 2009 Pearson Addison-Wesley 1.1-26 8.1-26 Example 5 SOLVING THE AMBIGUOUS CASE (TWO TRIANGLES) (continued) Solve separately for triangles Copyright © 2009 Pearson Addison-Wesley 1.1-27 8.1-27 Example 5 Copyright © 2009 Pearson Addison-Wesley SOLVING THE AMBIGUOUS CASE (TWO TRIANGLES) (continued) 1.1-28 8.1-28 Example 5 Copyright © 2009 Pearson Addison-Wesley SOLVING THE AMBIGUOUS CASE (TWO TRIANGLES) (continued) 1.1-29 8.1-29 Number of Triangles Satisfying the Ambiguous Case (SSA) Let sides a and b and angle A be given in triangle ABC. (The law of sines can be used to calculate the value of sin B.) 1. If applying the law of sines results in an equation having sin B > 1, then no triangle satisfies the given conditions. 2. If sin B = 1, then one triangle satisfies the given conditions and B = 90°. Copyright © 2009 Pearson Addison-Wesley 1.1-30 8.1-30 Number of Triangles Satisfying the Ambiguous Case (SSA) 3. If 0 < sin B < 1, then either one or two triangles satisfy the given conditions. (a) If sin B = k, then let B1 = sin–1 k and use B1 for B in the first triangle. (b) Let B2 = 180° – B1. If A + B2 < 180°, then a second triangle exists. In this case, use B2 for B in the second triangle. Copyright © 2009 Pearson Addison-Wesley 1.1-31 8.1-31 SOLVING THE AMBIGUOUS CASE (ONE TRIANGLE) Example 6 Solve triangle ABC given A = 43.5°, a = 10.7 in., and c = 7.2 in. There is another angle C with sine value .46319186: C = 180° – 27.6° = 152.4° Copyright © 2009 Pearson Addison-Wesley 1.1-32 8.1-32 Example 6 SOLVING THE AMBIGUOUS CASE (ONE TRIANGLE) (continued) Since c < a, C must be less than A. So C = 152.4° is not possible. B = 180° – 27.6° – 43.5° = 108.9° Copyright © 2009 Pearson Addison-Wesley 1.1-33 8.1-33 Example 7 ANALYZING DATA INVOLVING AN OBTUSE ANGLE Without using the law of sines, explain why A = 104°, a = 26.8 m, and b = 31.3 m cannot be valid for a triangle ABC. Because A is an obtuse angle, it must be the largest angle of the triangle. Thus, a must be the longest side of the triangle. We are given that b > a, so no such triangle exists. Copyright © 2009 Pearson Addison-Wesley 1.1-34 8.1-34 Area of a Triangle (SAS) In any triangle ABC, the area A is given by the following formulas: Copyright © 2009 Pearson Addison-Wesley 1.1-35 8.1-35 Note If the included angle measures 90°, its sine is 1, and the formula becomes the familiar Copyright © 2009 Pearson Addison-Wesley 1.1-36 8.1-36 Example 8 FINDING THE AREA OF A TRIANGLE (SAS) Find the area of triangle ABC. Copyright © 2009 Pearson Addison-Wesley 1.1-37 8.1-37 Example 9 FINDING THE AREA OF A TRIANGLE (ASA) Find the area of triangle ABC if A = 24°40′, b = 27.3 cm, and C = 52°40′. Draw a diagram. Before the area formula can be used, we must find either a or c. B = 180° – 24°40′ – 52°40′ = 102°40′ Copyright © 2009 Pearson Addison-Wesley 1.1-38 8.1-38 Example 9 FINDING THE AREA OF A TRIANGLE (ASA) (continued) Now find the area. Copyright © 2009 Pearson Addison-Wesley 1.1-39 8.1-39 Caution Whenever possible, use given values in solving triangles or finding areas rather than values obtained in intermediate steps to avoid possible rounding errors. Copyright © 2009 Pearson Addison-Wesley 1.1-40 8.1-40