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Copyright © 2005 Pearson Education, Inc. Chapter 2 Acute Angles and Right Triangles Copyright © 2005 Pearson Education, Inc. 2.1 Trigonometric Functions of Acute Angles Copyright © 2005 Pearson Education, Inc. Right-triangle Based Definitions of Trigonometric Functions For any acute angle A in standard position. y side opposite sin A r hypotenuse x side adjacent cos A r hypotenuse y side opposite tan A x side adjacent x side adjacent cot A . y side opposite Copyright © 2005 Pearson Education, Inc. r hypotenuse csc A y side opposite r hypotenuse sec A x side adjacent Slide 2-4 Example: Finding Trig Functions of Acute Angles Find the values of sin A, cos A, and tan A in the right triangle shown. side opposite 20 sin A hypotenuse 52 side adjacent 48 cos A hypotenuse 52 side opposite 20 tan A side adjacent 48 Copyright © 2005 Pearson Education, Inc. 48 A C 20 52 B Slide 2-5 Cofunction Identities For any acute angle A, sin A = cos(90 A) csc A = sec(90 A) tan A = cot(90 A) cos A = sin(90 A) sec A = csc(90 A) cot A = tan(90 A) Copyright © 2005 Pearson Education, Inc. Slide 2-6 Example: Write Functions in Terms of Cofunctions Write each function in terms of its cofunction. a) cos 38 b) sec 78 sec 78 = csc (90 78) = csc 12 cos 38 = sin (90 38) = sin 52 Copyright © 2005 Pearson Education, Inc. Slide 2-7 Example: Solving Equations Find one solution for the equation cot(4 8 ) tan(2 4 ) . o o Assume all angles are acute angles. cot(4 8o) tan(2 4o) (4 8o) (2 4o) 90o 6 12o 90o 6 78o 13o Copyright © 2005 Pearson Education, Inc. Slide 2-8 Example: Comparing Function Values Tell whether the statement is true or false. sin 31 > sin 29 In the interval from 0 to 90, as the angle increases, so does the sine of the angle, which makes sin 31 > sin 29 a true statement. Copyright © 2005 Pearson Education, Inc. Slide 2-9 Special Triangles 30-60-90 Triangle Copyright © 2005 Pearson Education, Inc. 45-45-90 Triangle Slide 2-10 Function Values of Special Angles sin 30 1 2 3 2 3 3 3 2 3 3 2 45 2 2 2 2 1 1 2 2 60 3 2 1 2 3 3 3 2 2 3 3 Copyright © 2005 Pearson Education, Inc. cos tan cot sec csc Slide 2-11 2.2 Trigonometric Functions of Non-Acute Angles Copyright © 2005 Pearson Education, Inc. Answers to homework 1. 2. 3. 4. 5. 6. 7. 8. Sin 21/29 45/53 n/p k/z C H B G Copyright © 2005 Pearson Education, Inc. cos 20/29 28/53 m/p y/z tan 21/20 45/28 n/m k/y E 10. A 9. Slide 2-13 Homework answers 11. Sin Cos Tan Csc Sec cot 12/13 5/13 12/5 13/12 13/5 5/12 12 5/3 13 14 3/5 6/7 7/12 Copyright © 2005 Pearson Education, Inc. 7/6 12/7 Slide 2-14 Reference Angles A reference angle for an angle is the positive acute angle made by the terminal side of angle and the x-axis. Copyright © 2005 Pearson Education, Inc. Slide 2-15 Example: Find the reference angle for each angle. a) 218 Positive acute angle made by the terminal side of the angle and the xaxis is 218 180 = 38. Copyright © 2005 Pearson Education, Inc. 1387 Divide 1387 by 360 to get a quotient of about 3.9. Begin by subtracting 360 three times. 1387 – 3(360) = 307. The reference angle for 307 is 360 – 307 = 53 Slide 2-16 Example: Finding Trigonometric Function Values of a Quadrant Angle Find the values of the trigonometric functions for 210. Reference angle: 210 – 180 = 30 Choose point P on the terminal side of the angle so the distance from the origin to P is 2. Copyright © 2005 Pearson Education, Inc. Slide 2-17 Example: Finding Trigonometric Function Values of a Quadrant Angle continued The coordinates of P are x = 3 y = 1 r=2 1 sin 210 2 3 cos 210 2 csc 210 2 sec 210 Copyright © 2005 Pearson Education, Inc. 3, 1 2 3 3 3 tan 210 3 cot 210 3 Slide 2-18 Finding Trigonometric Function Values for Any Nonquadrantal Angle Step 1 Step 2 Step 3 Step 4 If > 360, or if < 0, then find a coterminal angle by adding or subtracting 360 as many times as needed to get an angle greater than 0 but less than 360. Find the reference angle '. Find the trigonometric function values for reference angle '. Determine the correct signs for the values found in Step 3. (Use the table of signs in section 5.2, if necessary.) This gives the values of the trigonometric functions for angle . Copyright © 2005 Pearson Education, Inc. Slide 2-19 Example: Finding Trig Function Values Using Reference Angles Find the exact value of each expression. cos (240) Since an angle of 240 is coterminal with and angle of 240 + 360 = 120, the reference angles is 180 120 = 60, as shown. Copyright © 2005 Pearson Education, Inc. cos(240 ) cos120 1 cos 60 2 Slide 2-20 Homework Pg 59-60 # 1-6, 10-13 Copyright © 2005 Pearson Education, Inc. Slide 2-21 Example: Evaluating an Expression with Function Values of Special Angles Evaluate cos 120 + 2 sin2 60 tan2 30. Since 1 3 3 cos 120 , sin 60 , and tan 30 , 2 2 3 cos 120 + 2 sin2 60 tan2 30 = 2 3 3 1 + 2 2 2 3 1 3 3 2 2 4 9 2 3 Copyright © 2005 Pearson Education, Inc. 2 Slide 2-22 Copyright © 2005 Pearson Education, Inc. Slide 2-23 Copyright © 2005 Pearson Education, Inc. Slide 2-24 Example: Using Coterminal Angles Evaluate each function by first expressing the function in terms of an angle between 0 and 360. cos 780 cos 780 = cos (780 2(360) = cos 60 = 1 2 Copyright © 2005 Pearson Education, Inc. Slide 2-25 2.3 Finding Trigonometric Function Values Using a Calculator Copyright © 2005 Pearson Education, Inc. Function Values Using a Calculator Calculators are capable of finding trigonometric function values. When evaluating trigonometric functions of angles given in degrees, remember that the calculator must be set in degree mode. Remember that most calculator values of trigonometric functions are approximations. Copyright © 2005 Pearson Education, Inc. Slide 2-27 Example: Finding Function Values with a Calculator a) sin 38 24 Convert 38 24 to decimal degrees. 24 38.4 60 sin 38 24 sin 38.4 .6211477 38 24 38 Copyright © 2005 Pearson Education, Inc. b) cot 68.4832 Use the identity 1 cot . tan cot 68.4832 .3942492 Slide 2-28 Angle Measures Using a Calculator Graphing calculators have three inverse functions. If x is an appropriate number, then sin 1 x,cos 1 x, or tan -1 x gives the measure of an angle whose sine, cosine, or tangent is x. Copyright © 2005 Pearson Education, Inc. Slide 2-29 Example: Using Inverse Trigonometric Functions to Find Angles Use a calculator to find an angle in the interval [0 ,90 ] that satisfies each condition. sin .8535508 Using the degree mode and the inverse sine function, we find that an angle having sine value .8535508 is 58.6 . We write the result as sin 1 .8535508 58.6 Copyright © 2005 Pearson Education, Inc. Slide 2-30 Example: Using Inverse Trigonometric Functions to Find Angles continued sec 2.486879 1 . Find the Use the identity cos sec reciprocal of 2.48679 to get cos .4021104. Now find using the inverse cosine function. The result is 66.289824 Copyright © 2005 Pearson Education, Inc. Slide 2-31 Page 64 # 22-29 22. 23. 24. 25. 26. 27. 28. 29. 57.99717206 55.84549629 81.166807334 16.16664145 30.50274845 38.49157974 46.17358205 68.6732406 Copyright © 2005 Pearson Education, Inc. Slide 2-32 2.4 Solving Right Triangles Copyright © 2005 Pearson Education, Inc. Significant Digits for Angles A significant digit is a digit obtained by actual measurement. Your answer is no more accurate then the least accurate number in your calculation. Number of Significant Digits Angle Measure to Nearest: 2 Degree 3 Ten minutes, or nearest tenth of a degree 4 Minute, or nearest hundredth of a degree 5 Tenth of a minute, or nearest thousandth of a degree Copyright © 2005 Pearson Education, Inc. Slide 2-34 Example: Solving a Right Triangle, Given an Angle and a Side Solve right triangle ABC, if A = 42 30' and c = 18.4. B = 90 42 30' B B = 47 30' c = 18.4 a sin A c a 18.4 a 18.4sin 42o30' a 18.4(.675590207) sin 42o30' a 12.43 Copyright © 2005 Pearson Education, Inc. 4230' C A b cos A c b cos 42 30' 18.4 b 18.4cos 42o30' b 13.57 o Slide 2-35 Example: Solving a Right Triangle Given Two Sides Solve right triangle ABC if a = 11.47 cm and c = 27.82 cm. side opposite sin A hypotenuse 11.47 .412293314 27.82 sin 1 A 24.35 B = 90 24.35 B = 65.65 Copyright © 2005 Pearson Education, Inc. B a = 11.47 c = 27.82 C A b2 c 2 a 2 b 2 27.822 11.47 2 b 25.35 Slide 2-36 Solve the right triangle A= 28.00o, and c = 17.4 ft Copyright © 2005 Pearson Education, Inc. Slide 2-37 Homework Page 73 # 9-14 Copyright © 2005 Pearson Education, Inc. Slide 2-38 Definitions Angle of Elevation: from point X to point Y (above X) is the acute angle formed by ray XY and a horizontal ray with endpoint X. Copyright © 2005 Pearson Education, Inc. Angle of Depression: from point X to point Y (below) is the acute angle formed by ray XY and a horizontal ray with endpoint X. Slide 2-39 Solving an Applied Trigonometry Problem Step 1 Step 2 Step 3 Draw a sketch, and label it with the given information. Label the quantity to be found with a variable. Use the sketch to write an equation relating the given quantities to the variable. Solve the equation, and check that your answer makes sense. Copyright © 2005 Pearson Education, Inc. Slide 2-40 Example: Application The length of the shadow of a tree 22.02 m tall is 28.34 m. Find the angle of elevation of the sun. Draw a sketch. 22.02 tan B 28.34 1 22.02 B tan 37.85 28.34 22.02 m B 28.34 m The angle of elevation of the sun is 37.85. Copyright © 2005 Pearson Education, Inc. Slide 2-41 2.5 Further Applications of Right Triangles Copyright © 2005 Pearson Education, Inc. Bearing Other applications of right triangles involve bearing, an important idea in navigation. Copyright © 2005 Pearson Education, Inc. Slide 2-43 Example An airplane leave the airport flying at a bearing of N 32E for 200 miles and lands. How far east of its starting point is the plane? e e sin 32 200 e 200 sin 32o e 106 o 200 32º The airplane is approximately 106 miles east of its starting point. Copyright © 2005 Pearson Education, Inc. Slide 2-44 Example: Using Trigonometry to Measure a Distance A method that surveyors use to determine a small distance d between two points P and Q is called the subtense bar method. The subtense bar with length b is centered at Q and situated perpendicular to the line of sight between P and Q. Angle is measured, then the distance d can be determined. a) Find d when = 1 23'12" and b = 2.0000 cm b) Angle usually cannot be measured more accurately than to the nearest 1 second. How much change would there be in the value of d if were measured 1 second larger? Copyright © 2005 Pearson Education, Inc. Slide 2-45 Example: Using Trigonometry to Measure a Distance continued cot 2 d b 2 b d cot 2 2 Let b = 2, change to decimal degrees. 2 1.386944 d cot 82.6176 cm 2 2 1 23'12" 1.386667 2 1.386667 d cot 82.6341 cm 2 2 Copyright © 2005 Pearson Education, Inc. b) Since is 1 second larger, use 1.386944. The difference is .0170 cm. Slide 2-46 Example: Solving a Problem Involving Angles of Elevation Sean wants to know the height of a Ferris wheel. From a given point on the ground, he finds the angle of elevation to the top of the Ferris wheel is 42.3 . He then moves back 75 ft. From the second point, the angle of elevation to the top of the Ferris wheel is 25.4 . Find the height of the Ferris wheel. Copyright © 2005 Pearson Education, Inc. Slide 2-47 Example: Solving a Problem Involving Angles of Elevation continued The figure shows two unknowns: x and h. Since nothing is given about the length of the hypotenuse, of either triangle, use a ratio that does not involve the hypotenuse, (the tangent ratio). In triangle ABC, In triangle BCD, h C 42.3 x 25.4 A 75 ft D h tan 42.3 or h x tan 42.3 . x tan 25.4 Copyright © 2005 Pearson Education, Inc. B h or h (75 x) tan 25.4 . 75 x Slide 2-48 Example: Solving a Problem Involving Angles of Elevation continued Since each expression equals h, the expressions must be equal to each other. x tan 42.3 (75 x) tan 25.4 Solve for x. x tan 42.3 75 tan 25.4 x tan 25.4 Distributive Property x tan 42.3 x tan 25.4 75 tan 25.4 Get x-terms on one side. x(tan 42.3 tan 25.4 ) 75 tan 25.4 75 tan 25.4 x tan 42.3 tan 25.4 Copyright © 2005 Pearson Education, Inc. Factor out x. Divide by the coefficient of x. Slide 2-49 Example: Solving a Problem Involving Angles of Elevation continued We saw above that h x tan 42.3 . Substituting for x. 75 tan 25.4 h tan 42.3 . tan 42.3 tan 25.4 tan 42.3 = .9099299 and tan 25.4 = .4748349. So, tan 42.3 - tan 25.4 = .9099299 - .4748349 = .435095 and 75 .4748349 .435095 .9099299 74.47796 The height of the Ferris wheel is approximately 74.48 ft. Copyright © 2005 Pearson Education, Inc. Slide 2-50