Solution manual for trigonometry 8th edition by mckeague

Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Chapter The Six Trigonometric Functions 1.1 Angles, Degrees, and Special Triangles EVEN SOLUTIONS 10 12 14 16 18 20 22 Two angles with a sum of 90° are called complementary angles, and when the sum is 180° they are called supplementary angles In a right triangle, the longest side opposite the right angle is called the hypotenuse, and the other two sides are called legs In a 30°–60°–90° triangle, the hypotenuse is always twice the shortest side, and the side opposite the 60° angle is always times the shortest side a ii b iv c i d iii 70° is an acute angle The complement is 70° – 50° = 20°, and the supplement is 180° – 70° = 110° 90° is neither an acute or obtuse angle (it is a right angle) The complement is 90° – 90° = 0°, and the supplement is 180° – 90° = 90° 150° is an obtuse angle The complement is 90° – 150° = –60°, and the supplement is 180° – 150° = 30° y is neither an acute or obtuse angle (its measure is unknown) The complement is 90° – y, and the supplement is 180° – y Adding the angles in triangle BDC: B + 90° + 50° = 180° B +140° = 180° B = 40° Adding the angles in triangle ADC: α + α + 90° = 180° 2α + 90° = 180° 2α = 90° α = 45° Since A = 50°, α = 180° − 50° − 90° = 40° Since α + β = 75° : 40° + β = 75° β = 35° Since β = 35° , B = 180° − 35° − 90° = 55° 24 No The posts must be perpendicular to the ground for α and β to be complementary 26 28 Since α and β are complementary, α = 90° −15° = 75° Since revolution = 360°, which corresponds to 24 hours, then in hours the hour earth will turn ( 360°) = ( 360°) = 45° 24 Let x represent the measure of each angle Since the angles must add to 180°: x + x + x = 180° 30 3x = 180° x = 60° Chapter Page Problem Set 1.1 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 32 Using the Pythagorean Theorem: 2 + b = 10 + b = 100 b = 96 34 b = 96 = Using the Pythagorean Theorem: 22 + b2 = 62 + b = 36 b = 32 36 b = 32 = Using the Pythagorean Theorem: a + 10 = 26 a + 100 = 676 a = 576 38 a = 24 Using the Pythagorean Theorem: 42 + 42 = x2 16 +16 = x x = 32 40 x = 32 = Using the Pythagorean Theorem: 12 + x = 2 + x2 = x2 = 42 x= Using the Pythagorean Theorem: x + = ( 2x −1) x +16 = 4x − 4x +1 = 3x − 4x −15 3x − 4x −15 = (3x + 5)(x − 3) = Since x represents the side of a triangle, it cannot be a negative number So the only solution is x = First find AC using the Pythagorean Theorem: x = 3,− 44 (AC)2 + = 132 (AC)2 + 25 = 169 (AC)2 = 144 AC = 12 Chapter Page Problem Set 1.1 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Thus DC = AC − AD = 12 − = Now find BD using the Pythagorean Theorem: + = (BD)2 64 + 25 = (BD)2 (BD)2 = 89 46 BD = 89 Let AB = x Since AC = AB + BC = x + , using the Pythagorean Theorem: 12 + = ( x + ) 144 + 25 = ( x + ) ( x + 5)2 = 169 x + = ± 169 = ±13 48 x = −18,8 Since AB must be positive, AB = Let x represent the distance across the pond Using the Pythagorean Theorem: 25 + 60 = x 625 + 3600 = x x = 4225 50 52 54 56 x = 65 The distance across the pond is 65 yards If the shortest side is 4, the side opposite the 60° angle is i = , and the longest side is i = 5 If the longest side is 5, the shortest side is i = , and the side opposite the 60° angle is i = 2 2 3 i = , and the longest side is i = If the side opposite the 60° angle is 3, the shortest side is 3 Since 20 ft is now opposite the 60° angle, the length of the shortest side is 20 20 i = The length of the 3 20 40 = ≈ 23.09 ft 3 To solve this problem we need to find the widths of the base and sides of the tent Since ft is the side opposite the 3 i = ft and the longest side is ft The width of the base and 60° angle at the end, the shortest side is 3 escalator is the longest side, which is i 58 sides are therefore ft If l represents the length of the tent, the sides and floor of the tent have a total area of 3i i l = 6l ft, and the ends (which are triangles) have a total area of i i i = ft2 Thus the total area of material needed is 90: 6l + = 90 (l +1) = 90 l +1 = 90 i =5 3 l = −1 ≈ 7.7 The length of the tent should be approximately 7.7 feet 60 Since the shorter sides are each Chapter 1 , the longer side is 2= 2 Page Problem Set 1.1 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu =6 62 Since the longest side is , the shorter sides are 64 Since the longest side is 12, the shorter sides are 66 Since the height is 1,000 ft, which represents the shorter side, the longer side is 1000 ft Thus, if the bullet is 68 1000 ft ≈ sec 2828 ft/sec Since d = and c is the hypotenuse of a 45°-45°-90° triangle, c = Since b is the shorter side of a 30°-60°-90° 4 i = Since a is the hypotenuse of a 30°-60°-90° triangle, a = i triangle, b = = 3 3 70 a Since GD = cm and ΔGCD is a 45°–45°–90° triangle, then each side has a length of b Since GD = cm and from part a and BD = 12 i =6 2 traveling at 2,828 ft/sec, the total time is 5 i = cm 2 5 cm, using the Pythagorean Theorem: (GD)2 + (BD)2 = (GB)2 ! $2 & = (GB)2 " % ( )2 + # 25 = (GB)2 75 (GB)2 = 75 5 GB = = i = cm 2 2 The measure of ∠GDH is 45°, since ΔGHD is a 45°–45°–90° triangle a Since ΔODB is a right triangle, using the Pythagorean Theorem: 25 + 72 74 (OD)2 + (DB)2 = (OB)2 12 + 2 = (OB)2 (OB)2 = OB = b Since OB = OE, OE = c Since CE = CO + OE, CE = + CE + This is called the golden ratio = EF Using the Pythagorean Theorem: 22 + b2 = 52 d 76 Finding the ratio: + b = 25 b = 21 78 b = 21 The correct answer is c Since the leg is 24 feet in a 45°–45°–90° triangle, the length is 24 feet The correct answer is b Chapter Page Problem Set 1.1 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu ODD SOLUTIONS 11 13 180! counterclockwise, clockwise hypotenuse, sum, legs equal, ! 10 is an acute angle The complement of 10! is 80! because 10! + 80! = 90! The supplement of 10! is 170! because 10! +170! = 180! 45! is an acute angle The complement of 45! is 45! because 45! + 45! = 90! The supplement of 45! is 135! because 45! +135! = 180! 120! is an obtuse angle ! The complement of 120! is −30! because 120! + (−30 ) = 90! 15 The supplement of 120! is 60! because 120! + 60! = 180! We can’t tell if x ! is acute or obtuse (or neither) The complement of x ! is 90! − x ! because x ! + ( 90! − x ! ) = 90! The supplement of x ! is 180! − x ! because x ! + (180! − x ! ) = 180! 17 The sum of the angles of a triangle is 180! α = 180! − (∠A + ∠D ) = 180! − ( 30! + 90! ) ! 19 Substitute given values ! = 180 −120 = 60! α = 180! − (∠A + ∠D ) Simplify The sum of the angles of a triangle is 180! = 180! − ( 2α + 90! ) 21 Substitute the given values = 90! − 2α 3α = 90! α = 30! ∠A = 180! − (α + β + ∠B) Simplify right side Add α to both sides Divide both sides by The sum of the angles of a triangle is 180! = 180! − (100! + 30! ) 23 Substitute given values = 180! −130! Simplify = 50! Angles α and β are complementary because α + β + 90! = 180! β α + β = 90! 25 α + β = 90! α = 90 − β ! 27 29 β α and β are complementary ! Subtract β from both sides ! = 90 − 52 Substitute given value = 38! Simplify One complete revolution equals 360! Therefore, it rotates 360! in seconds and 90! in second Let α = the degree measure of each base angle Then α + α + 40! = 180! 2α + 40! = 180! 2α = 140! α α = 70! Each base angle of this isosceles triangle is 70! Chapter Page Problem Set 1.1 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 31 33 c2 = a2 + b2 Pythagorean Theorem = + 32 Substitute given values = 16 + Simplify = 25 Therefore, c = ±5 The only solution is c = 5, because we cannot use c = −5 a2 + b2 = c2 Pythagorean Theorem 2 Subtract a from both sides b = c −a = 17 − Substitute given values = 289 − 64 Simplify = 225 Therefore, b = ±15 The only solution is b = 15, because we cannot use b = −15 35 a2 + b2 = c2 Pythagorean Theorem Subtract b from both sides a2 = c2 − b2 = 132 −12 Substitute given values = 169 −144 Simplify = 25 Therefore, a = ±5 Our only solution is a = 5, because we cannot use a = −5 37 ( x + 52 = ) Pythagorean Theorem x + 25 = 50 Simplify x = 25 Subtract 25 from both sides Therefore, x = ±5 Our only solution is x = because we cannot use x = −5 Note: This must be a 45! − 45! − 90! triangle 39 ( ) x = ( ) + Pythagorean Theorem = +12 Simplify = 16 Therefore x = ±4 Our only solution is x = , because we cannot use x = −4 Note: This must be a 30! − 60! − 90! triangle 41 ( 10 ) = x + ( x + 2) Pythagorean Theorem 10 = x + x + 4x + Simplify 10 = 2x + 4x + Combine like terms = 2x + 4x − = x + 2x − Subtract 10 from both sides Divide both sides by = ( x + 3) ( x −1) Factor Set each factor equal to zero x + = or x −1 = x =1 x = −3 Therefore, x = because x = −3 is not possible 43 = (CD ) + ( BC ) 2 ( BD ) = (CD ) + ( ) Pythagorean Theorem Substitute given values 25 = (CD ) +16 = (CD ) Simplify Subtract 16 from both sides CD = or CD = −3 Take square root of both sides CD = Eliminate negative solution Therefore, AC = + = AC = AD + DC ( AB) 2 = ( AC ) + ( BC ) =5 +4 Chapter 2 Pythagorean Theorem Substitute given values Page Problem Set 1.1 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 45 = 25 +16 = 41 AB = 41 or AB = − 41 Take square root of both sides AB = 41 Eliminate negative solution 2 ( AB + BC ) = (CD ) + ( AD ) (4 + r ) = r + 82 47 53 Pythagorean Theorem Substitute given values 16 + 8r + r = r + 64 Simplify 16 + 8r = 64 Subtract r from both sides 8r = 48 Subtract 16 from both sides r=6 Divide both sides by This is an isosceles triangle Therefore, the altitude must bisect the base 51 2 x = (18 ) + (13.5 ) 49 Simplify Pythagorean Theorem = 324 +182.5 Simplify = 506.25 x = 22.5 or x = −22.5 Take square root of both sides x = 22.5 ft Eliminate negative solution The shortest side is The longest side is twice the shortest side Therefore, it is The side opposite the 60! angle is or The longest side is which is twice the shortest side Therefore, the shortest side is The side opposite the 60! angle is Let t = the shortest side, 2t = the longest side, and t = the side opposite 60! Therefore, t = Side opposite 60! is 6 t= Divide both sides by 3 = =2 3 ( Rationalize the denominator ) Since t = 3, 2t = 2 = 55 57 The shortest side is and the longest side is The shortest side is 20 feet The longest side is twice the shortest side Therefore, x = ( 20 ) x = 40 feet The tent is made up of congruent rectangles and congruent triangles First we’ll find the sides of the 30! − 60! − 90! triangle The side opposite 60! is ft Let t = the shortest side t 3=4 t= t= 20 ft 30o 4 ⋅ = 3 The shortest side is Chapter x !4 3$ &= The hypotenuse is # 3 " % Page Problem Set 1.1 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu !4 3$ &= Also the base of the triangular sides are # " % Note: This is an equilateral triangle Area of rectangles = length ⋅ width !8 3$ & Area of rectangles = # " % = 16 ft2 Area of triangles = ⋅ base ⋅ height 1!8 3$ = # &(4) 2" % 16 ft Area of tent = rectangles + triangles ! 16 $ = 16 + # & " % = ( ) 61 32 3 ≈ 101.6 ft2 hypotenuse = ⋅ Hypotenuse is t Simplify = hypotenuse = t t is the shorter side 63 Substitute given value =t = t Divide both sides by hypotenuse = t t is the shorter side = 48 + 59 4=t =t 65 Substitute given value Divide both sides by Rationalize denominator by multiplying numerator and t =2 We are looking for the hypotenuse of a 45! − 45! − 90! triangle where the shorter sides are 1000 feet hypotenuse = 1000 Hypotenuse is t ≈ 1414 ft Rounded to the nearest foot The bullet travels 1414 feet Chapter Page Problem Set 1.1 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 67 First, we will find the leg of the 45! − 45! − 90! triangle where the hypotenuse is 3: Hypotenuse = d 3= d 3 = d or d = 2 Divide both sides by and rationalize the denominator Next, we will find the shorter side of the 30! − 60! − 90! triangle where the side opposite the 60! angle is 3: b 3=3 b = the side opposite 60! = b= Divide both sides by and rationalize the denominator Last, we will find the hypotenuse of the 30! − 60! − 90! triangle: a = (b ) = Hypotenuse = twice the shorter side 69 a hypotenuse = t t is the edge of the cube Substitute given value =1 Simplify = Therefore, CH = inches b (CF ) 2 = (CH ) + ( FH ) = ( 2) + (1) Pythagorean Theorem Substitute given values = +1 =3 CF = ± CF = inches 71 Simplify Take easy square root of both sides Eliminate the negative solution a hypotenuse = x x is the edge of the cube The length of the diagonal of any face of the cube will be x b (CF ) ( 2 = (CH ) + ( FH ) ) = x + ( x) = 2x + x = 3x 75 77 Chapter 2 Pythagorean Theorem Substitute given values Simplify or CF = 3x CF = − 3x This is impossible =x The length of any diagonal that passes through the center of the cube will be x The complement of 61! is 90! − 61! = 29! The supplement of 61! is 180! − 61! = 119! The answer is d The longest side is The shortest side is one-half the longest side Therefore it is The side opposite the 60! angle is 3 The answer is c Page Problem Set 1.1 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 1.2 The Rectangular Coordinate System EVEN SOLUTIONS The unit circle has center (0, 0) and a radius of The notation θ ∈QIII means that θ is in standard position and its terminal side lies in quadrant three Coterminal angles are two angles in standard position having the same terminal side We can find a coterminal angle by adding or subtracting any multiple of 360° The formula for a circle with center (h, k) and radius r is ( x − h ) + ( y − k ) = r 10 The ordered pair ( 4,2 ) lies in the first quadrant 12 The ordered pair −1,− lies in the third quadrant 14 Graphing the line: 16 Graphing the line: 18 All the points have positive y-coordinates in quadrants I and II x The ratio is also negative in quadrant IV, since x is positive and y is negative in that quadrant y Graphing the parabola: 24 Graphing the parabola: 20 22 Chapter Page 10 ( ) Problem Set 1.2 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 86 88 90 For any integer k, –60° + 360°k will be coterminal with –60° Other answers are possible For any integer k, 180° + 360°k will be coterminal with 180° Other answers are possible Drawing 60° in standard position and labeling the point ( 2, b ) : 92 Since is the shorter side of the triangle (opposite the 30° angle), the value of b must be b = Drawing an angle whose terminal side contains the point ( 2, −3) : 94 Using the distance formula: r = (2 − 0)2 + (−3 − 0)2 = 2 + (−3)2 = + = 13 Plotting the points: Note that a = 3, b = 4, and c = (−3 − 0)2 + (−2 − 2)2 = (−3)2 + (−4)2 = + 16 = 25 = Since a + b = c , by the Pythagorean Theorem these points form a right triangle Chapter Page 16 Problem Set 1.2 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 96 Pascal’s triangle appears as: 1 1 1 1 10 10 15 20 15 These numbers are the coefficients of the expansion of (a + b)n , where n represents the line number of the triangle The first few expansions are: (a + b)2 = a + 2ab + b (a + b)3 = a + 3a b + 3ab + b (a + b)4 = a + 4a 3b + 6a b + 4ab + b (a + b)5 = a + 5a b + 10a 3b + 10a b + 5ab + b (a + b)6 = a + 6a b + 15a b + 20a 3b + 15a b + 6ab + b This pattern continues for all natural numbers n 98 Using the distance formula: d = (4 + 2)2 + (5 − 8)2 = + ( −3) = 36 + = 45 = The correct answer is b 100 A coterminal angle is 160° – 360° = –200° The correct answer is d ODD SOLUTIONS quadrants, I, IV, counterclockwise (0, 0), quadrantal d= Since x is positive and y is negative, the point ( 2,−4 ) must lie in quadrant IV 11 Since x is negative and y is positive, the point − 3,1 must lie in quadrant II 13 If we let x = 0, the equation y = x becomes: y = This gives us the point ( 0,0 ) ( ( x2 − x1 ) + ( y2 − y1 ) ) If we let x = 2, the equation y = x becomes: y = This gives us the point ( 2,2 ) Graphing the points (0, 0) and (2, 2) and then drawing a line through them, we have the graph of y = x 15 17 19 21 1 x becomes: y = ( ) = 2 This gives us (0, 0) as one solution to y = x 1 If we let x = 4, the equation y = x becomes: y = ( ) = 2 This gives us (4, 2) as a second solution to y = x Graphing the points (0, 0) and (4, 2) and then drawing a line through them, we have the graph of y= x Quadrants II and III lie to the left of the y-axis Therefore, all points in these two quadrants have negative x-coordinates x In quadrant III, x and y are always negative Therefore the ratio will always be positive y If we let x = 0, the equation y = The vertex of this parabola is at ( 0,−4 ) Chapter Page 17 Problem Set 1.2 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu If we let x = −2, the equation y = x − becomes y = (−2 ) − =4−4 =0 This gives us (−2,0 ) as a point on the curve If we let x = −1, the equation y = x − becomes y = (−1) − = 1− = −3 This gives us (−1,−3) as a point on the curve Using the symmetry of a parabola, the points ( 2,0 ) and (1,−3) will also be points on the curve Graphing the points (−2,0 ) , (−1,−3) , ( 0,−4 ) , (1,−3) , and ( 2,0 ) , and then drawing a smooth curve through them, we have the graph of the parabola y = x − 23 The vertex of this parabola is (−2, ) If we let x = −1, the equation y = ( x + ) + becomes y = (−1+ ) + = 12 + = This gives us (−1,5 ) as a point on the curve If we let x = 0, the equation y = ( x + ) + becomes y = (0 + 2) + = 22 + = This gives us (0, 8) as a point on the curve Using the symmetry of a parabola, the points (−3,5 ) and (−4,8 ) will also be points on the curve Graphing the points ( 0,8 ) , (−1,5 ) , (−2, ) , (−3,5 ) and (−4,8 ) and then drawing a smooth curve through them, we have the graph of the parabola y = ( x + ) + 29 The cannonball is on the ground (y = 0) when x = and when x = 160 The x-coordinate of the vertex (the maximum) will be at (160 ) or 80, and the y-coordinate will be 60 We can now sketch the parabola through the points (0, 0), (80, 60), and (160, 0) The equation will be in the form y = a ( x − 80 ) + 60 We will use the point (160, 0) to find a Let x = 160 and y = 0: y = a ( x − 80 ) + 60 = a (160 − 80 ) + 60 −60 = a ( 80 ) −60 = 6400a a=− 320 Therefore, the equation is y = − 31 r= ( x2 − x1 ) + ( y2 − y1 ) = ( 3− 6) + ( − 3) = (−3) + 42 2 ( x − 80) + 60 320 Distance formula Substitute given values Simplify = +16 = 25 = Chapter Page 18 Problem Set 1.2 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 33 ( x2 − x1 ) + ( y2 − y1 ) r= 2 = ( − 5) + (12 − 0) = (−5) Distance formula Substitute given values +12 Simplify = 25 +144 = 169 = 13 35 ( x2 − x1 ) + ( y2 − y1 ) r= Distance formula 2 = "#−1− (−10 )$% + (−2 − ) = + (−7 ) = 81+ 49 = 130 37 ( x2 − x1 ) + ( y2 − y1 ) r= ( 3− 0) + (−4 − 0) = = ( 3) + (−4 ) Substitute given values Simplify Distance formula Substitute given values Simplify = +16 = 25 = 39 r= ( x2 − x1 ) + ( y2 − y1 ) 13 = ( x −1) + ( − 5) 13 = ( x −1) + (−3) 13 = ( x −1) 2 2 Distance formula Substitute given values Simplify +9 13 = ( x −1) + Square both sides 41 Subtract from both sides = ( x −1) Use the square root method to solve ±2 = x −1 x −1 = or x −1 = −2 x = or x = −1 First, we convert 1.2 miles to feet: 1.2 mi = 1.2(5,280) ft = 6,336 ft Pythagorean Theorem c2 = a2 + b2 43 45 Substitute given values = ( 2,640 ) + ( 6, 336 ) Simplify = 6,969,600 + 40,144,896 = 47,114, 496 Take square root of both sides c = 6,864 ft (or 1.3 mi) 60 The x-axis goes from home plate to first base, a distance of 60 feet Therefore, home plate is (0, 0) and first base is (60, 0) The y-axis goes from home plate to third base Therefore, third base is (0, 60) Second base will be at (60, 60) We substitute x = and y = −1 into the equation of the unit circle and we get: x + y = + (−1) = +1 = Chapter 60 It checks Page 19 Problem Set 1.2 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 47 49 57 59 and y = into the equation of the unit circle and we get: 2 2 !1$ ! 3$ x + y2 = # & + # & "2% " % ! $ ! 3$ = # &+# & = It checks "4% "4% The center of this circle is (0, 0) and the radius is From the graph of x + y = 25, we can see that (0, 5) and (5, 0) are the points at which the line x + y = will intersect the circle Solving this system of equations by substitution,, we get: x2 + x2 = x + y2 = We substitute x = 2x = y=x x2 = x =± =± 2 2 2 If x = − ,y= , y=− 2 2 ! 2$ " 2% , ,− & and $− ' The solution is # & " 2 % # If x = 61 63 65 67 69 71 73 75 77 The complement of a 45! angle is 90! − 45! = 45! The complement of a 60! angle is 90! − 60! = 30! The supplement of a 120! angle is 180! −120! = 60! The supplement of a 90! angle is 180! − 90! = 90! An angle coterminal with an angle of −135! is −135! + 360! = 225! An angle coterminal with an angle of −210! is −210! + 360! = 150! This angle in standard position lies in quadrant IV One revolution in a positive direction gives us: 300! + 360! = 660! One revolution in a negative direction gives us: 300! − 360! = −60! This angle in standard position lies in quadrant III One revolution in a positive direction gives us: −150! + 360! = 210! One revolution in a negative direction gives us: −150! − 360! = −510! a If we draw 135! in standard position, we see that the terminal side is along the line y = −x Since the terminal side lies in the second quadrant, x is negative and y is positive Some of ( ) points on the terminal side are (−1,1) , (−3, 3) , and − , b To find the distance from (0, 0) to (−3, 3) , we use the distance formula: r= = (−3− 0) + ( 3− 0) (−3) + ( 3) 2 = 9+9 = 18 = c One revolution in a negative direction gives us: 135! − 360! = −225! Chapter Page 20 Problem Set 1.2 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 79 a If we draw 225! in standard position, we see that the terminal side is along the line y = x Since the terminal side lies in the third quadrant, x and y are both negative Some of ( ) points on the terminal side are (−1,−1) , (−3,−3) , and − ,− b To find the distance from (0, 0) to (−3,−3) , we use the distance formula: r= = 81 83 (−3− 0) + (−3− 0) (−3) + (−3) 2 = + = 18 = c One revolution in a negative direction gives us: 225! − 360! = −135! a If we draw 90! in standard position, we see that the terminal side is the positive y-axis Some of the points on the terminal side are ( 0,1) , ( 0,2 ) , and ( 0, 3) b The distance between (0, 0) and (0, 3) is units c One revolution in a negative direction gives us: 90! − 360! = −270! a If we draw −45! in standard position, we see that the terminal side is along the line y = −x Since the terminal side lies in the fourth quadrant, x is positive and y is negative Some of points on the terminal side are (1,−1) , ( 3,−3) , and ( ) ,− b To find the distance from (0, 0) to ( 3,−3) , we use the distance formula: r= = 85 87 89 ( 3− 0) + (−3− 0) ( 3) + (−3) = 9+9 = 18 = c One revolution in a positive direction gives us: −45! + 360! = 315! For any integer k, 30! + 360! k will be coterminal with 30! For any integer k, −135! + 360! k will be coterminal with −135! The side opposite 30! is The side opposite 60! is or Therefore, the point is r= 91 = 93 ( ( 3− 0) + (−2 − 0) ( 3) + (−2) ) 3,1 2 = + = 13 We will find the lengths of the three sides: From (0, 0) to (5, 0) is units From (5, 0) to (5, 12) is 12 units From (0 0) to (5, 12), we use the distance formula: r= ( − 0) + (12 − 0) = +12 = 25 +144 = 169 = 13 If this is a right triangle, then c = a + b We check: 132 = +12 169 = 25 +144 169 = 169 It checks Chapter Page 21 Problem Set 1.2 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 97 If a point ( x, y ) lies on the unit circle, then x + y = Testing the given points, we find that (a) is true: 99 ! 3$ ! $ 16 & = + = =1 # & +# " % " % 16 16 16 To draw 140! in standard position, place the vertex at the origin and draw the terminal side 140! counterclockwise from the positive x-axis The answer is a 1.3 Definition 1: Trigonometric Functions EVEN SOLUTIONS In quadrant I, all of the six trigonometric functions are positive In each of the other three quadrants, only two of the six functions are positive The tangent and cotangent functions not depend on the value of r Begin by finding the distance r from the origin to (−4,−3) : r = (−4 − 0)2 + (−3− 0)2 = 16 + = 25 = Now applying the definitions for the six trigonometric functions using the values x = –4, y = –3, and r = 5: r y cscθ = = − sin θ = = − y r x r cosθ = = − secθ = = − r x y −3 x −4 tan θ = = = cot θ = = = x −4 y −3 Begin by finding r = (12)2 + (−5)2 = 144 + 25 = 169 = 13 Now applying the definitions for the six trigonometric functions using the values x = 12, y = –5, and r = 13: y r 13 sin θ = = − csc θ = = − r 13 y x 12 r 13 cos θ = = sec θ = = r 13 x 12 y x 12 tan θ = = − cot θ = = − x 12 y 10 Begin by finding r = + = 36 + 36 = 72 = Now applying the definitions for the six trigonometric functions using the values x = , y = , and r = : sin θ = 6 = = 2 = = 2 y tan θ = = = x cosθ = Chapter cscθ = r = = y r = = x x cot θ = = = y secθ = Page 22 Problem Set 1.3 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 12 (−2)2 + ( Begin by finding r = ) = + = = Now applying the definitions for the six trigonometric functions using the values x = –2, y = , and r = 3: y = r x cosθ = = − r y tan θ = = − x r 3 = = y 5 r secθ = = − x x 2 cot θ = = − =− y 5 sin θ = cscθ = 14 Begin by finding r = + = 16 = Now applying the definitions for the six trigonometric functions using the values x = 4, y = 0, and r = 4: y r sin θ = = = cscθ = = , which is undefined r y x r cosθ = = = secθ = = = r x y x tan θ = = = cot θ = = , which is undefined x y 16 Begin by finding r = m + n Now applying the definitions for the six trigonometric functions using the values x = m , y = n , and r = m + n : 18 sin θ = y = r n cos θ = x = r tan θ = y n = x m m2 + n2 m m2 + n2 csc θ = r = y m2 + n2 n sec θ = r = x m2 + n2 m cot θ = x m = y n Begin by finding r = (−6)2 + (−2)2 = 36 + = 40 = 10 Now applying the definitions for sin θ , cos θ , and tan θ using x = –6, y = –2, and r = 10 : sin θ = 20 10 y −2 =− = 10 r 10 cos θ = −6 10 =− 10 10 tan θ = y −2 = = x −6 Begin by finding r = (3)2 + (−8)2 = + 64 = 73 Now applying the definitions for sin θ , cos θ , and tan θ using x = 3, y = –8, and r = 73 : sin θ = 22 73 y =− =− 73 r 73 cos θ = x 3 73 = = r 73 73 tan θ = y =− x Begin by finding r: r = (3.63)2 + (6.25)2 = 13.1769 + 39.0625 = 52.2394 ≈ 7.2277 Now applying the definitions for sin θ and cos θ using x = 3.63, y = 6.25, and r = 7.2277: sin θ = Chapter 6.25 y = ≈ 0.865 r 7.2277 cosθ = x 3.63 = ≈ 0.502 r 7.2277 Page 23 Problem Set 1.3 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 24 Drawing 225° in standard position: A point on the terminal side is ( −1, −1) Now find r = (−1)2 + (−1)2 = + = Appling the definitions for sin 225°, cos 225°, and tan 225°: 26 y =− =− r Drawing 180° in standard position: 28 A point on the terminal side is ( −1, ) with r = Appling the definitions for sin 180°, cos 180°, and tan 180°: y x −1 y sin180° = = = cos180° = = = −1 tan180° = = =0 r r x −1 Drawing –90° in standard position: sin 225° = Chapter cos 225° = x =− =− r 2 Page 24 tan 225° = y −1 = =1 x −1 Problem Set 1.3 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 30 A point on the terminal side is ( 0, −1) with r = Appling the definitions for sin (–90°), cos (–90°), and tan (–90°): y −1 x y −1 sin(−90°) = = = −1 cos(−90°) = = = tan(−90°) = = , which is undefined r r x Drawing –135° in standard position: A point on the terminal side is ( −1, −1) Now find r = (−1)2 + (−1)2 = + = Appling the definitions for sin (–135°), cos (–135°), and tan (–135°): x y −1 y =− cos(−135°) = = − =− tan(−135°) = = =1 =− r x −1 r 2 This statement is true 34 This statement is false r Since r ≥ x, ≥ Thus sec θ ≥ , so there is no angle θ such that sec θ = x y Since y ≤ r, ≤ Thus sin θ ≤ for any angle θ r The value of cos θ tends toward as θ increases from 0° to 90° As θ increases from 0° to 90°, the value of sec θ tends towards θ could terminate in quadrant II or quadrant IV θ could terminate in quadrant III or quadrant IV θ could terminate in quadrant II or quadrant IV θ could terminate in quadrant I or quadrant II If sin θ is positive, y > 0, and if cos θ is negative, x < 0, so the terminal side of θ must lie in quadrant II cscθ and cot θ are both positive in quadrant I, and cscθ and cot θ are both negative in quadrant IV Thus cscθ and cot θ have the same sign in quadrants I and IV 24 If cos θ = , we can choose x = 24 and r = 25 To find y, use x + y = r : 25 sin(−135°) = 32 36 38 40 42 44 46 48 50 52 54 56 24 + y = 25 576 + y = 625 y = 49 y = ±7 Since θ terminates in quadrant IV, y < and thus y = –7 Using x = 24, y = –7, and r = 25 in the definitions: y r 25 r 25 sin θ = = − sec θ = = csc θ = = − r 25 x 24 y y x 24 tan θ = = − cot θ = = − x 24 y Chapter Page 25 Problem Set 1.3 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 58 If sin θ = − 20 , we can choose y = –20 and r = 29 To find y, use x + y = r : 29 x + (−20)2 = 29 x + 400 = 841 x = 441 x = ±21 Since θ terminates in quadrant III, x < and thus x = –21 Using x = –21, y = –20, and r = 29 in the definitions: x 21 r 29 r 29 cos θ = = − csc θ = = − sec θ = = − r 29 y 20 x 21 y 20 x 21 tan θ = = cot θ = = x 21 y 20 60 If sin θ = , we can choose y = and r = To find x, use x + y = r : x2 + ( 2) = 22 x2 + = x2 = x =± Since θ terminates in quadrant II, x < and thus x = − Using x = − , y = , and r = in the definitions: sin θ = y = r cscθ = r = = y secθ = r = =− x − y x − = = −1 cot θ = = = −1 x − y If tan θ = −3 and θ terminates in quadrant IV, x > and y < 0, so choose y = –3 and x = Finding r using tan θ = 62 x + y2 = r : 12 + (−3)2 = r 1+ = r r = 10 r = 10 Using x = 1, y = –3, and r = 10 in the definitions: 64 sin θ = 10 y −3 = =− 10 r 10 cscθ = r 10 =− y cosθ = x 10 = = r 10 10 secθ = r = 10 x If sec θ = cot θ = x =− y 13 , we can choose x = and r = 13 To find y, use x + y = r : 5 + y = 132 + y = 169 y = 144 y = ±12 Chapter Page 26 Problem Set 1.3 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 66 Since sin θ < , y < and thus y = –12 Using x = 5, y = –12, and r = 13 in the definitions: y 12 y 12 r 13 sin θ = = − tan θ = = − csc θ = = − r 13 x y 12 x x cos θ = = cot θ = = − r 13 y 12 If cot θ = − and sin θ > , y > 0, so choose y = and x = –1 Finding r using x + y = r : (−1)2 + = r 1+16 = r r = 17 r = 17 Using x = –1, y = 4, and r = 17 in the definitions: 68 70 72 74 sin θ = 4 17 y = = 17 r 17 cscθ = r 17 = y cosθ = x 17 =− =− r 17 17 secθ = r = − 17 x If cot θ = tan θ = x = −4 y m , we can choose x = m and y = n, so r = m + n Using the definitions: n sin θ = y = r cos θ = x = r n m2 + n2 m csc θ = r = y m2 + n2 n sec θ = r = x m2 + n2 m tan θ = y n = x m m2 + n2 a If tan θ = , the angel is θ = 45° b If tan θ = −1 , the angel is θ = 135° A point on y = x in quadrant III is (−2,−1) , so choose x = –2 and y = –1 Thus r = (−2)2 + (−1)2 = +1 = Using the definitions: x 2 y =− cosθ = = − =− sin θ = = − r r 5 A point on y = −3x in quadrant IV is (1, −3) , so choose x = and y = –3 Thus r = 12 + (−3)2 = + = 10 Using the definitions: sin θ = Chapter 10 y =− =− 10 r 10 tan θ = y −3 = = −3 x Page 27 Problem Set 1.3 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 76 78 Drawing 45° and –45° in standard position: Let ( x, y ) represent a point on the terminal side of 45° Then ( x, −y ) will represent a point on the terminal side −y y of –45° Therefore: sin(−45°) = = − = − sin 45° r r If sin θ = − , we can choose y = –6 and r = 10 To find x, use x + y = r : x + (−6)2 = 10 x + 36 = 100 x = 64 x = ±8 Note that both values of x are possible, since the points ( 8,−6 ) and (−8,−6 ) both satisfy the condition sin θ = − 80 82 The value of cos θ tends toward as θ increases from 0° to 90° The correct answer is a If sin θ = , we can choose y = and r = To find x, use x + y = r : x + = 52 x + 16 = 25 x2 = x = ±3 Since θ terminates in quadrant II, x < and thus x = –3 Thus tan θ = y = − The correct answer is a x ODD SOLUTIONS terminal, distance, origin tangent and secant, cotangent and cosecant y sin θ = = ( x, y) = ( 3, ) r x x = and y = cosθ = = r y 2 r = +4 tan θ = = x x = y r secθ = = x r cscθ = = y cot θ = = +16 = 25 = Chapter Page 28 Problem Set 1.3 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu y 12 = r 13 x −5 cosθ = = =− 13 r 13 y 12 12 tan θ = = =− x −5 ( x, y) = (−5,12) sin θ = x = −5 and y = 12 r= (−5) +12 x −5 = =− 12 y 12 r 13 13 secθ = = =− x −5 r 13 cscθ = = y 12 cot θ = = 25 +144 = 169 = 13 ( x, y) = (−1,−2) sin θ = y −2 = =− r 5 cot θ = x −1 = = y −2 x = −1 and y = −2 cosθ = x −1 5 = =− r 5 secθ = r = =− x −1 tan θ = y −2 = =2 x −1 cscθ = r 5 = =− y −2 3,−1 sin θ = y −1 = =− r cot θ = x =− =− y x = and y = −1 cosθ = x = r secθ = 2 r = = x tan θ = y −1 = =− x cscθ = r = = −2 y −1 r= (−1) + (−2) = 1+ = 11 ( x, y) = ( r= ) ( 3) + (−1) = 3+1 = =2 13 ( x, y) = ( 0,−5) x = and y = −5 r = + (−5 ) y −5 = = −1 r x cosθ = = = r y −5 is undefined tan θ = = x x = =0 y −5 r secθ = = is undefined x r = −1 cscθ = = y −5 y −12a = =− r 15a x −9a =− cosθ = = r 15a y −12a = tan θ = = x −9a x −9a = = y −12a r 15a =− secθ = = x −9a r 15a =− cscθ = = y −12a sin θ = cot θ = = 25 = 15 ( x, y) = (−9a,−12a) sin θ = x = −9a and y = −12a r= (−9a) + (−12a) cot θ = = 81a +144a = 225a = 15a 17 19 (4, 3) lies on the terminal side of θ Therefore, x = 4, y = 3, and r = + 32 = 16 + = 25 = sin θ = y = r (−7,6) lies on the terminal side of θ Therefore, x = −7, y = 6, and r = sin θ = 6 85 y = = 85 r 85 Chapter cosθ = x = r cosθ = tan θ = 85 x −7 = =− 85 r 85 Page 29 y = x (−7) tan θ = + = 49 + 36 = 85 y = =− x −7 Problem Set 1.3 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu 21 y 7.02 = = 0.6 r 11.7 x 9.36 cosθ = = = 0.8 r 11.7 ( x, y) = ( 9.36, 7.02) sin θ = x = 9.36 and y = 7.02 r= ( 9.36) + ( 7.02) = 87.6096 + 49.2804 = 136.89 = 11.7 23 The terminal side of 135! in standard position lies on the line y = −x A point on this line in quadrant II is (−1,1) x = −1, y = and r = (−1) +12 = 1+1 = 25 y x −1 2 cosθ = = = = =− r r 2 2 ! A point on the terminal side of an angle of 90 is ( 0,1) 27 x = 0, y = and r = +12 = = y x y sin θ = = = cosθ = = = tan θ = = (undefined) r r x The terminal side of −45! in standard position lies on the line y = −x A point on this line sin θ = tan θ = y = = −1 x −1 in quadrant IV is (1,−1) x = , y = −1 , and r = 12 + (−1) = 1+1 = 2 y −1 x = =− cosθ = = = r r 2 A point on the terminal side of an angle of 0! is (1, 0) sin θ = 29 31 33 tan θ = y −1 = = −1 x x = , y = , and r = 12 + = = y x y sin θ = = = cosθ = = = tan θ = = = r r x We have chosen P and Q on the terminal side of 35! and 45! , so that the x-coordinate is the same for both Because 35! is less than 45! , we can see that r1 < r2 x x We also know that cos 35! = and cos 45! = As the denominator of a fraction r1 r2 its value decreases Therefore, cos 35! > cos 45! and the statement is false We have chosen P and Q on the terminal side of 60! and 75! , so that the x-coordinate is the same for both Because 60! is less than 75! , we can see that r1 < r2 r1 r and sec 75! = x x Since r1 < r2 , sec 60! < sec 75! and the statement is true r2 r1 increases, 35o o 45 x We also know that sec 60! = r2 r1 60o o 45 x Chapter Page 30 Problem Set 1.3 © 2017 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-8th-Edition-by-McKeagu ... https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 8th- Edition- by- McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 8th- Edition- by- McKeagu... https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 8th- Edition- by- McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 8th- Edition- by- McKeagu... https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 8th- Edition- by- McKeagu Solution Manual for Trigonometry 8th Edition by McKeague Full file at https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 8th- Edition- by- McKeagu