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Trigonometric Functions of Acute Angles
2.1 Trigonometric Functions of Acute Angles Right-Triangle-Based Definitions of the Trigonometric Functions ▪ Cofunctions ▪ Trigonometric Function Values of Special Angles 2.1 2.2 2.3 2.4 2.5 Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.1 Example 1 Finding Trigonometric Function Values of An Acute Angle (page 51)
Find the sine, cosine, and tangent values for angles D and E in the figure. Soh Cah Toa
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Explanation of Co-Functions
Sin θ = y/r Cos (90˚ – θ) = y/r 90˚ - θ r Sin θ = Cos (90˚ – θ) y Tan θ = y/x Cot (90˚ – θ) = y/x θ x Tan θ = Cot (90˚ – θ) Sec θ = r/x Csc (90˚ – θ) = r/x Copyright © 2008 Pearson Addison-Wesley. All rights reserved. Sec θ = Csc (90˚ – θ)
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2.1 Example 2 Writing Functions in Terms of Cofunctions (page 51)
Write each function in terms of its cofunction. (a) sin 9° = cos (90° – 9°) = cos 81° (b) cot 76° = tan (90° – 76°) = tan 14° (c) csc 45° = sec (90° – 45°) = sec 45° Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.1 Example 3(a&b) Solving Equations Using the Cofunction Identities (page 52)
Find one solution for the equation. Assume all angles involved are acute angles. Since cotangent and tangent are cofunctions, the equation is true if the sum of the angles is 90º. (a) (b) Since secant and cosecant are cofunctions, the equation is true if the sum of the angles is 90º.
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2.1 Example 5 Finding Trigonometric Values for 30° (page 54)
Find the six trigonometric function values for a 30° angle. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.1 Summary Soh Cah Toa Special Right Triangles 30:60:90 1: :2 45:45:90 1:1: Co-Functions Sin θ = Cos (90˚ – θ) Tan θ = Cot (90˚ – θ) Sec θ = Csc (90˚ – θ) Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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Trigonometric Functions of Non-Acute Angles
2.2 Trigonometric Functions of Non-Acute Angles Reference Angles ▪ Special Angles as Reference Angles ▪ Finding Angle Measures with Special Angles Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.2 Example 1(a) Finding Reference Angles (page 59)
Find the reference angle for 294°. 294 ° lies in quadrant IV. The reference angle is 360° – 294° = 66°. Find the reference angle for 883°. 883° is coterminal with 163°. The reference angle is 180° – 163° = 17°. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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Find the values of the six trigonometric functions for 135°.
2.2 Example 2 Finding Trigonometric Functions of a Quadrant II Angle (page 60) Find the values of the six trigonometric functions for 135°. The reference angle for 135° is 45°. 45˚ reference angle, use 1:1: for the sides
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Find the exact value of sin(–150°).
2.2 Example 3(a) Finding Trigonometric Function Values Using Reference Angles (page 61) Find the exact value of sin(–150°). An angle of –150° is coterminal with an angle of –150° + 360° = 210°. The reference angle is 210° – 180° = 30°. Since an angle of –150° lies in quadrant III, its sine is negative. sin(-150˚) = – sin30˚ = -1/2 Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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Find the exact value of cot(780°).
2.2 Example 3(b) Finding Trigonometric Function Values Using Reference Angles (page 61) Find the exact value of cot(780°). An angle of 780° is coterminal with an angle of 780° – 2 ∙ 360° = 60°. The reference angle is 60°. Since an angle of 780° lies in quadrant I, its cotangent is positive. cot780˚= cot60˚= Reciprocal of tan60˚ = Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.2 Example 4 Evaluating an Expression with Function Values of Special Angles (page 62)
1 1 45˚ 45˚ -1 1 -1 45˚ -1 Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.2 Example 5 Using Coterminal Angles to Find Function Values (page 62)
Evaluate each function by first expressing the function in terms of an angle between 0° and 360°. (a) -1 45˚ -1 (b) 2 1 30˚ Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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Find all values of θ, if θ is in the interval [0°, 360°) and sin θ =
2.2 Example 6 Finding Angle Measures Given an Interval and a Function Value (page 63) Find all values of θ, if θ is in the interval [0°, 360°) and sin θ = Since sin θ is negative, θ must lie in quadrants III or IV. To find the reference angle, ask yourself the sin of what angle is ? Then use the and right triangle to find the reference angle of 60˚. 1 √2 1 The angle in quadrant III is 60° + 180° = 240°. The angle in quadrant IV is 360° – 60° = 300°. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.2 Summary Coterminal Angles: ±360˚
Reference Angles: the angle measure from the vector to the x-axis Special Right Triangles: Soh Cah Toa 30:60: : :2 45:45: :1: Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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Finding Trigonometric Functions Using a Calculator
2.3 Finding Trigonometric Functions Using a Calculator Finding Function Values Using a Calculator ▪ Finding Angle Measures Using a Calculator Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.3 Example 1 Finding Function Values with a Calculator (page 67)
Approximate the value of each expression. = 1/sin ˚ = 1/cos(-287˚) Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.3 Example 2 Using Inverse Trigonometric Functions to Find Angles (page 67)
Use a calculator to find an angle θ in the interval [0°, 90°] that satisfies each condition.
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2.3 Example 3 Finding Grade Resistance (page 68)
The force F in pounds when an automobile travels uphill or downhill on a highway is called grade resistance and is modeled by the equation , where θ is the grade and W is the weight of the automobile. If the automobile is moving uphill, then θ > 0°; if it is moving downhill, then θ < 0°. (a) Calculate F to the nearest 10 pounds for a 5500-lb car traveling an uphill grade with θ = 3.9°. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.3 Example 3 Finding Grade Resistance (cont.)
(b) Calculate F to the nearest 10 pounds for a 2800-lb car traveling a downhill grade with θ = –4.8°. (c) A 2400-lb car traveling uphill has a grade resistance of 288 pounds. What is the angle of the grade? Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.3 Summary Using a calculator to find: function values angle values
Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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Solving Right Triangles
2.4 Solving Right Triangles Significant Digits ▪ Solving Triangles ▪ Angles of Elevation or Depression Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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Solve right triangle ABC, if B = 28°40′ and a = 25.3 cm.
2.4 Example 1 Solving a Right Triangle Given an Angle and a Side (page 75) Solve right triangle ABC, if B = 28°40′ and a = 25.3 cm. Soh Cah Toa c ≈ 28.8 cm b ≈ 13.8 cm Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.4 Example 2 Solving a Right Triangle Given Two Sides (page 76)
Solve right triangle ABC, if a = cm and b = cm. Soh Cah Toa Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.4 Example 3 Finding a Length When the Angle of Elevation is Known (page 77)
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 angle B equals the measure of the angle of depression because the two angles are alternate interior angles, so measure of angle B = 60.4˚ 60.4˚ Soh Cah Toa The tree is about 27.3 m tall. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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The angle of elevation of the sun is 63.39°.
2.4 Example 4 Finding a Length When the Angle of Elevation is Known (page 78) The length of a shadow of a flagpole ft tall is ft. Find the angle of elevation of the sun. Soh Cah Toa The angle of elevation of the sun is 63.39°. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.4 Summary Solving for a right triangle: means to find all missing lengths and angle measures Angle of elevation is from the horizontal upward Angle of depression is form the horizontal downward #4 – directions tricky for the angle answer (round to the nearest 10 minutes) So how would you round 61⁰56′48.129? Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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Further Applications of Right Triangles
2.5 Further Applications of Right Triangles Bearing ▪ Further Applications Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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The distance from B to C is about 5.2 km.
2.5 Example 1 Solving Problem Involving Bearing (First Method) (page 83) Radar stations A and B are on an east-west line, 8.6 km apart. Station A detects a plane at C, on a bearing of 53°. Stations B simultaneously detects the same plane, on a bearing of 323°. Find the distance from B to C. C 90⁰ a 53⁰ 37⁰ 53⁰ cos 53˚ 323⁰ A 8.6km B The distance from B to C is about 5.2 km. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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The distance from B to C is about 352.29 miles.
2.5 Example 2 Solving Problem Involving Bearing (Second Method) (page 84) The bearing from A to C is N 64° W. The bearing from A to B is S 82° W. The bearing from B to C is N 26° E. A plane flying at 350 mph take 1.8 hours to go from A to B. Find the distance from B to C. C 90° a 64⁰ 26⁰ 34° A ≈ 56° B 82⁰ 630 mi 98° The distance from B to C is about miles. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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The building is about 69.45 ft
2.5 Example 4 Solving a Problem Involving Angles of Elevation (page 85) 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. 22.4⁰ h x 105.8⁰ 51.8⁰ 74.2⁰ 35 ft The building is about ft Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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The building is about 69.45 feet tall.
2.5 Example 4 Solving a Problem Involving Angles of Elevation (page 85) 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. Triangle ABC Triangle DBC Since h = x tan 74.2°, substitute the expression for x to find h. ≈ 69.45 The building is about feet tall. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.5 Example 4 Solving a Problem Involving Angles of Elevation (cont.)
Graphing Calculator Solution Superimpose coordinate axes on the figure with D at the origin. The coordinates of A are (35, 0). Equation of line DB is Equation of line AB is Graph y1 and y2, then find the point of intersection. The y-coordinate gives the height, h. The building is about 69 feet tall. Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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2.5 Summary Further Application of Right Triangles
Boats and Planes: start north and then the angle is measured clockwise Plane math problems: start on the positive x-axis and then the angle is measured counterclockwise Angle of elevation is from the horizontal upward Angle of depression is form the horizontal downward Copyright © 2008 Pearson Addison-Wesley. All rights reserved.
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