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2.4 Solving Right Triangles Significant Digits ▪ Solving Triangles ▪ Angles of Elevation or Depression Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-1 2.4 Example 1 Solving a Right Triangle Given an Angle and a Side (page 69) Solve right triangle ABC, if B = 28°40′ and a = 25.3 cm. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-2 2.4 Example 1 Solving a Right Triangle Given an Angle and a Side (cont.) Three significant digits Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-3 2.4 Example 1 Solving a Right Triangle Given an Angle and a Side (cont.) Three significant digits Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-4 2.4 Example 1 Solving a Right Triangle Given an Angle and a Side (cont.) Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-5 2.4 Example 2 Solving a Right Triangle Given Two Sides (page 70) Solve right triangle ABC, if a = 44.25 cm and b = 55.87 cm. Use the Pythagorean theorem to find c: Four significant digits Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-6 2.4 Example 2 Solving a Right Triangle Given Two Sides (cont.) Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-7 2.4 Example 2 Solving a Right Triangle Given Two Sides (cont.) Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-8 2.4 Extra Example Finding a Length When the Angle of Elevation is Known The angle of depression from the top of a tree to a point on the ground 15.5 m from the base of the tree is 60.4°. Find the height of the tree. The measure of equals the measure of the angle of depression because the two angles are alternate interior angles, so Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-9 2.4 Extra Example Finding a Length When the Angle of Elevation is Known (cont.) Three significant digits The tree is about 27.3 m tall. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-10 2.4 Example 3 Finding a Length When the Angle of Elevation is Known (page 71) The length of a shadow of a flagpole 55.20 ft tall is 27.65 ft. Find the angle of elevation of the sun. Four significant digits The angle of elevation of the sun is 63.39°. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-11 2.5 Further Applications of Right Triangles Bearing ▪ Further Applications Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-12 2.5 Example 4 Solving a Problem Involving Angles of Elevation (page 85) – similar to Homework #28 Marla needs to find the height of a building. From a given point on the ground, she finds that the angle of elevation to the top of the building is 74.2°. She then walks back 35 feet. From the second point, the angle of elevation to the top of the building is 51.8°. Find the height of the building. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-13 2.5 Example 4 Solving a Problem Involving Angles of Elevation (cont.) There are two unknowns, the distance from the base of the building, x, and the height of the building, h. In triangle ABC In triangle BCD Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-14 2.5 Example 4 Solving a Problem Involving Angles of Elevation (cont.) Set the two expressions for h equal and solve for x. Since h = x tan 74.2°, substitute the expression for x to find h. The building is about 69 feet tall. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 2-15 2.5 Examples for homework • See section 2.5 #15 and #25 • Help with sketch on #26 Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-16 7 Applications of Trigonometry and Vectors 7.1 Oblique Triangles and the Law of Sines (Unit 2) 7.2 The Ambiguous Case of the Law of Sines (Unit 2) 7.3 The Law of Cosines (Unit 2) 7.4 Vectors, Operations, and the Dot Product (Unit 4) 7.5 Applications of Vectors (Unit 4) Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-17 7.1 Oblique Triangles and the Law of Sines Congruency and Oblique Triangles ▪ Derivation of the Law of Sines ▪ Solving SAA and ASA Triangles (Case 1) ▪ Area of a Triangle Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-18 7.1 Example 1 Using the Law of Sines to Solve a Triangle (AAS) Solve triangle ABC if A = 28.8°, C = 102.6°, and c = 25.3 in. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-19 7.1 Example 1 Using the Law of Sines to Solve a Triangle (AAS) (cont.) Use the Law of Sines to find the lengths of the missing sides. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-20 7.1 Example 2 Using the Law of Sines in an Application (ASA) Jerry wishes to measure the distance across the Big Muddy River. He determines that C = 117.2°, A = 28.8°, and b = 75.6 ft. Find the distance a across the river. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-21 7.1 Example 2 Using the Law of Sines in an Application (ASA) (cont.) Use the Law of Sines to find the length of side a. The distance across the river is about 65.1 ft. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-22 7.1 Example 3 Using the Law of Sines in an Application (ASA) The bearing of a lighthouse from a ship was found to be N 52° W. After the ship sailed 5.8 km due south, the new bearing was N 23° W. Find the distance between the ship and the lighthouse at each location. Let x = the distance to the lighthouse at bearing N 52° W and y = the distance to the lighthouse at bearing N 23° W. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-23 7.1 Example 3 Using the Law of Sines in an Application (ASA) (cont.) The lighthouse is located at Z, and the ship is first located at Y and then at X. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-24 7.1 Example 3 Using the Law of Sines in an Application (ASA) (cont.) The distance between the ship and the first location is about 4.7 km. The distance between the ship and the second location is about 9.4 km. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-25 7.1 Example 4 Finding the Area of a Triangle (SAS) Find the area of triangle DEF in the figure. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-26 7.1 Example 5 Finding the Area of a Triangle (ASA) Find the area of triangle ABC if B = 58°10′, a = 32.5 cm, and C = 73°30′. We must find AC (side b) or AB (side c) in order to find the area of the triangle. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-27 7.1 Example 5 Finding the Area of a Triangle (ASA) (cont.) Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-28 7.1 Example 5 Finding the Area of a Triangle (ASA) (cont.) 48°20′ 37.0 cm Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-29 7.2 The Ambiguous Case of the Law of Sines Description of the Ambiguous Case ▪ Solving SSA Triangles (Case 2) ▪ Analyzing Data for Possible Number of Triangles Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-30 7.2 Example 1 Solving the Ambiguous Case (No Such Triangle) Solve triangle ABC if a = 17.9 cm, c = 13.2 cm, and C = 75°30′. Use the Law of Sines to find A. Since sin A > 1 is impossible, no such triangle exists. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-31 7.2 Example 1 Solving the Ambiguous Case (No Such Triangle) (cont.) An attempt to sketch the triangle leads to this figure. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-32 7.2 Example 2 Solving the Ambiguous Case (Two Triangles) Solve triangle ABC if A = 61.4°, a = 35.5 cm, and b = 39.2 cm. Use the Law of Sines to find B. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-33 7.2 Example 2 Solving the Ambiguous Case (Two Triangles) (cont.) There are two angles between 0° and 180° such that sin B ≈ .9695: so is a valid possibility. Solve separately for triangles Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-34 7.2 Example 2 Solving the Ambiguous Case (Two Triangles) (cont.) Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-35 7.2 Example 2 Solving the Ambiguous Case (Two Triangles) (cont.) Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-36 7.2 Example 3 Solving the Ambiguous Case (One Triangle) Solve triangle ABC if B = 68.7°, b = 25.4 in., and a = 19.6 in. Use the Law of Sines to find A. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-37 7.2 Example 3 Solving the Ambiguous Case (One Triangle) (cont.) There are two angles between 0° and 180° such that sin A ≈ .7189: is not a valid possibility. There is only one triangle Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-38 7.2 Example 3 Solving the Ambiguous Case (One Triangle) (cont.) Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-39 7.2 Example 4 Analyzing Data Involving an Obtuse Angle Without using the law of sines, explain why no triangle exists satisfying B = 93°, b = 42 cm, and c = 48 cm. Because B is an obtuse angle, it must be the largest angle of the triangle. Thus, b must be the longest side of the triangle. We are given that c > b, so no such triangle exists. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-40 7.3 The Law of Cosines Derivation of the Law of Cosines ▪ Solving SAS and SSS Triangles (Cases 3 and 4) ▪ Heron’s Formula for the Area of a Triangle Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-41 7.3 Example 1 Using the Law of Cosines in an Application (SAS) Two boats leave a harbor at the same time, traveling on courses that make an angle of 82°20′ between them. When the slower boat has traveled 62.5 km, the faster one has traveled 79.4 km. At that time, what is the distance between the boats? The harbor is at C. The slower boat is at A, and the faster boat is at B. We are seeking the length of AB. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-42 7.3 Example 1 Using the Law of Cosines in an Application (SAS) Use the law of cosines because we know the lengths of two sides of the triangle and the measure of the included angle. The two boats are about 94.3 km apart. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-43 7.3 Example 2 Using the Law of Cosines to Solve a Triangle (SAS) Solve triangle ABC if B = 73.5°, a = 28.2 ft, and c = 46.7 ft. Use the law of cosines to find b because we know the lengths of two sides of the triangle and the measure of the included angle. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-44 7.3 Example 2 Using the Law of Cosines to Solve a Triangle (SAS) (cont.) Now use the law of sines to find the measure of another angle. 47.2 ft Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-45 7.3 Example 3 Using the Law of Cosines to Solve a Triangle (SSS) Solve triangle ABC if a = 25.4 cm, b = 42.8 cm, and c = 59.3 cm. Use the law of cosines to find the measure of the largest angle, C. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-46 7.3 Example 3 Using the Law of Cosines to Solve a Triangle (SSS) (cont.) Use either the law of sines or the law of cosines to find the measure of angle B. 118.6° Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-47 7.3 Example 3 Using the Law of Cosines to Solve a Triangle (SSS) (cont.) Alternative: 118.6° Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-48 7.3 Example 4 Designing a Roof Truss (SSS) Find the measure of angle C in the figure. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-49 4.5 Harmonic Motion Simple Harmonic Motion ▪ Damped Oscillatory Motion 4-50 4.5 Example 1 Modeling the Motion of a Spring Suppose that an object is attached to a coiled spring. It is pulled down a distance of 16 cm from its equilibrium position and then released. The time for one complete oscillation is 6 seconds. 4-51 4.5 Example 1(a) Modeling the Motion of a Spring (a) Give an equation that models the position of the object at time t. When the object is released at t = 0, the object is at distance −16 cm from equilibrium. Since the time needed to complete one oscillation is 6 sec, P = 6, and 4-52 4.5 Example 1(b, c) Modeling the Motion of a Spring (b) Determine the position at t = 1.5 seconds. At t = 1.5 seconds, the object is at the equilibrium position. (c) Find the frequency. The frequency is the reciprocal of the period. 4-53 4.5 Example 2 Analyzing Harmonic Motion Suppose that an object oscillates according to the model s(t) = 2.5 sin 5t, where t is in seconds and s(t) is in meters. Analyze the motion. The motion is harmonic because the model is of the form a = 2.5, so the object oscillates 2.5 meters in either direction from the starting point. and 4-54