Solution manual for trigonometry 3rd edition by young

Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young CHAPTER Section 1.1 Solutions -1 x x Solve for x: = Solve for x: = 360 360 360 = x, so that x = 180 x Solve for x: − = 360 360 = x, so that x = 90 x Solve for x: − = 360 360 = −3x, so that x = −120 (Note: The angle has a negative measure since it is a clockwise rotation.) x Solve for x: = 360 720 = 2(360 ) = −3x, so that x = −240 (Note: The angle has a negative measure since it is a clockwise rotation.) x Solve for x: = 12 360 1800 = 5(360 ) = x, so that x = 300 x Solve for x: − = 360 2520 = 7(360 ) = 12 x, so that x = 210 x Solve for x: − = 360 1440 = 4(360 ) = −5 x, so that x = −288 (Note: The angle has a negative measure since it is a clockwise rotation.) 1800 = 5(360 ) = −9 x, so that x = −200 (Note: The angle has a negative measure since it is a clockwise rotation.) 10 a) complement: 90 − 18 = 72 a) complement: 90 − 39 = 51 b) supplement: 180 − 18 = 162 11 b) supplement: 180 − 39 = 141 12 a) complement: 90 − 42 = 48 a) complement: 90 − 57 = 33 b) supplement: 180 − 42 = 138 13 b) supplement: 180 − 57 = 123 14 a) complement: 90 − 89 = a) complement: 90 − 75 = 15 b) supplement: 180 − 89 = 91 b) supplement: 180 − 75 = 105 15 Since the angles with measures ( 4x ) and ( 6x ) are assumed to be complementary, we know that ( x ) + ( x ) = 90 Simplifying this yields (10 x ) = 90 , so that x = So, the two angles have measures 36 and 54 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young 16 Since the angles with measures ( 3x ) and (15x ) are assumed to be supplementary, we know that ( 3x ) + (15 x ) = 180 Simplifying this yields (18 x ) = 180 , so that x = 10 So, the two angles have measures 30 and 150 17 Since the angles with measures ( 8x ) and ( 4x ) are assumed to be supplementary, we know that ( x ) + ( x ) = 180 Simplifying this yields (12 x ) = 180 , so that x = 15 So, the two angles have measures 60 and 120 18 Since the angles with measures ( 3x + 15 ) and (10 x + 10 ) are assumed to be complementary, we know that ( 3x + 15 ) + (10 x + 10 ) = 90 Simplifying this yields (13x + 25) = 90 , so that (13x ) = 65 and thus, x = So, the two angles have measures 30 and 60 19 Since α + β + γ = 180 , we know that 20 Since α + β + γ = 180 , we know that 117 + 33 + γ = 180 and so, γ = 30 110 + 45 + γ = 180 and so, γ = 25 = 150 = 155 21 Since α + β + γ = 180 , we know that ( 4β ) + β + ( β ) = 180 and so, β = 30 22 Since α + β + γ = 180 , we know that ( 3β ) + β + ( β ) = 180 and so, β = 36 = 6β = 5β Thus, α = β = 120 and γ = β = 30 ( Thus, α = 3β = 108 and γ = β = 36 ) ( 23 α = 180 − 53.3 + 23.6 = 103.1 ) 24 β = 180 − 105.6 + 13.2 = 61.2 25 Since this is a right triangle, we know from the Pythagorean Theorem that a + b = c Using the given information, this becomes 42 + 32 = c , which simplifies to c = 25 , so we conclude that c = 26 Since this is a right triangle, we know from the Pythagorean Theorem that a + b = c Using the given information, this becomes 32 + 32 = c , which simplifies to c = 18 , so we conclude that c = 18 = 27 Since this is a right triangle, we know from the Pythagorean Theorem that a + b = c Using the given information, this becomes 62 + b = 102 , which simplifies to 36 + b = 100 and then to, b = 64 , so we conclude that b = Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Section 1.1 28 Since this is a right triangle, we know from the Pythagorean Theorem that a + b = c Using the given information, this becomes a + = 122 , which simplifies to a = 95 , so we conclude that a = 95 29 Since this is a right triangle, we know from the Pythagorean Theorem that a + b = c Using the given information, this becomes 82 + 52 = c , which simplifies to c = 89 , so we conclude that c = 89 30 Since this is a right triangle, we know from the Pythagorean Theorem that a + b = c Using the given information, this becomes 62 + 52 = c , which simplifies to c = 61 , so we conclude that c = 61 31 Since this is a right triangle, we know from the Pythagorean Theorem that a + b = c Using the given information, this becomes + b = 112 , which simplifies to b = 72 , so we conclude that b = 72 = 32 Since this is a right triangle, we know from the Pythagorean Theorem that a + b = c Using the given information, this becomes a + 52 = 92 , which simplifies to a = 56 , so we conclude that a = 56 = 14 33 Since this is a right triangle, we know from the Pythagorean Theorem that a + b = c Using the given information, this becomes a + ( 7) = 52 , which simplifies to a = 18 , so we conclude that a = 18 = 34 Since this is a right triangle, we know from the Pythagorean Theorem that a + b = c Using the given information, this becomes 52 + b = 102 , which simplifies to b = 75 , so we conclude that b = 75 = 35 If x = 10 in., then the hypotenuse of 36 If x = m, then the hypotenuse of this triangle has length ≈ 11.31 m this triangle has length 10 ≈ 14.14 in 37 Let x be the length of a leg in the given 45 − 45 − 90 triangle If the hypotenuse of this triangle has length 2 cm, then x = 2, so that x = Hence, the length of each of the two legs is cm 38 Let x be the length of a leg in the given 45 − 45 − 90 triangle If the hypotenuse 10 10 of this triangle has length 10 ft., then x = 10, so that x = = = 2 Hence, the length of each of the two legs is ft Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young 39 The hypotenuse has length in = 8in ( 40 Since ) x = 6m ⇒ x = 2 = 2m , each leg has length m 41 Since the lengths of the two legs of the given 30 − 60 − 90 triangle are x and x , the shorter leg must have length x Hence, using the given information, we know that x = m Thus, the two legs have lengths m and ≈ 8.66 m, and the hypotenuse has length 10 m 42 Since the lengths of the two legs of the given 30 − 60 − 90 triangle are x and x , the shorter leg must have length x Hence, using the given information, we know that x = ft Thus, the two legs have lengths ft and ≈ 15.59 ft., and the hypotenuse has length 18 ft 43 The length of the longer leg of the given triangle is x = 12 yards So, 12 12 = = As such, the length of the shorter leg is ≈ 6.93 yards, and 3 the hypotenuse has length ≈ 13.9 yards x= 44 The length of the longer leg of the given triangle is 3x = n units So, n n n units, and the hypotenuse = As such, the length of the shorter leg is 3 2n units has length 45 The length of the hypotenuse is x = 10 inches So, x = Thus, the length of the shorter leg is inches, and the length of the longer leg is ≈ 8.66 inches 46 The length of the hypotenuse is x = cm So, x = Thus, the length of the shorter leg is cm, and the length of the longer leg is ≈ 6.93 cm x= Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Section 1.1 47 For simplicity, we assume that the minute hand is on the 12 Let α = measure of the desired angle, as indicated in the diagram below Since the measure of the angle formed using two rays emanating from the center of the clock out toward consecutive hours is always ( 360 ) = 30 , it immediately follows that 12 α = ⋅ ( −30 ) = −120 (Negative since measured clockwise.) α 48 For simplicity, we assume that the minute hand is on the Let α = measure of the desired angle, as indicated in the diagram below Since the measure of the angle formed using two rays emanating from the center of the clock out toward consecutive hours is always ( 360 ) = 30 , it immediately follows that 12 α = ⋅ ( −30 ) = −150 (Negative since measured clockwise.) α Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young 49 The key to solving this problem is setting up the correct proportion Let x = the measure of the desired angle From the given information, we know that since complete revolution corresponds to 360 , we obtain the following proportion: 360 x = 30 minutes 12 minutes Solving for x then yields ⎛ ⎞ 360 x = 12 minutes ⎜ ⎟ = 144 ⎝ 30 minutes ⎠ ( ) 50 The key to solving this problem is setting up the correct proportion Let x = the measure of the desired angle From the given information, we know that since complete revolution corresponds to 360 , we obtain the following proportion: 360 x = 30 minutes minutes Solving for x then yields ⎛ ⎞ 360 x = minutes ⎜ ⎟ = 60 30 minutes ⎝ ⎠ ( ) 51 We know that complete revolution corresponds to 360 Let x = time (in minutes) it takes to make complete revolution about the circle Then, we have the following proportion: 270 360 = x 45 minutes Solving for x then yields 270 x = 360 ( 45 minutes ) x= 360 ( 45 minutes ) = 60 minutes 270 So, it takes one hour to make one complete revolution 52 We know that complete revolution corresponds to 360 Let x = time (in minutes) it takes to make complete revolution about the circle Then, we have the following proportion: 72 360 = x minutes Solving for x then yields 72 x = 360 ( minutes ) x= 360 ( minutes ) = 45 minutes 72 So, it takes 45 minutes to make one complete revolution Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Section 1.1 53 Let d = distance (in feet) the dog runs along the hypotenuse Then, from the Pythagorean Theorem, we know that 302 + 802 = d 7,300 = d 85 ≈ 7,300 = d So, d ≈ 85 feet 54 Let d = distance (in feet) the dog runs along the hypotenuse Then, from the Pythagorean Theorem, we know that 252 + 1002 = d 10, 625 = d 103 ≈ 10, 625 = d So, d ≈ 103 feet 55 Consider the following triangle T 45 45 Since T is a 45 − 45 − 90 triangle, the two legs (i.e., the sides opposite the angles with measure 45 ) have the same length Call this length x Since the hypotenuse of such a 100 100 = = 50 triangle has measure 2x , we have that x = 100 , so that x = 2 So, since lights are to be over both legs and the hypotenuse, the couple should buy 50 + 50 + 100 = 100 + 100 ≈ 241 feet of Christmas lights Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young 56 Consider the following triangle T 45 45 Since T is a 45 − 45 − 90 triangle, the two legs (i.e., the sides opposite the angles with measure 45 ) have the same length Call this length x Since the hypotenuse of such a 60 60 = = 30 triangle has measure 2x , we have that x = 60 , so that x = 2 So, since lights are to be over both legs and the hypotenuse, the couple should buy 30 + 30 + 60 = 60 + 60 ≈ 145 feet of Christmas lights Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Section 1.1 57 Consider the following diagram: 30 30 60 60 The dashed line segment AD represents the TREE and the vertices of the triangle ABC represent STAKES Also, note that the two right triangles ADB and ADC are congruent (using the Side-Angle-Side Postulate from Euclidean geometry) Let x = distance between the base of the tree and one staked rope (measured in feet) For definiteness, consider the right triangle ADC Since it is a 30 − 60 − 90 triangle, the side opposite the 30 -angle (namely DC) is the shorter leg, which has length x feet Then, we know that the hypotenuse must have length 2x Thus, by the Pythagorean Theorem, it follows that: x + 17 = (2 x) x + 289 = x 289 = 3x 289 = x2 289 9.8 ≈ =x So, the ropes should be staked approximately 9.8 feet from the base of the tree 58 Using the computations from Problem 57, we observe that since the length of the 289 , it follows that the length of each of the two ropes hypotenuse is 2x, and x = 289 ≈ 19.6299 feet Thus, one should have × 19.6299 ≈ 39.3 feet of rope in order to have such stakes support the tree should be Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young 59 Consider the following diagram: 60 60 30 30 The dashed line segment AD represents the TREE and the vertices of the triangle ABC represent STAKES Also, note that the two right triangles ADB and ADC are congruent (using the Side-Angle-Side Postulate from Euclidean geometry) Let x = distance between the base of the tree and one staked rope (measured in feet) For definiteness, consider the right triangle ADC Since it is a 30 − 60 − 90 triangle, the side opposite the 30 -angle (namely AD) is the shorter leg, which has length 10 feet Then, we know that the hypotenuse must have length 2(10) = 20 feet Thus, by the Pythagorean Theorem, it follows that: x + 102 = 202 x + 100 = 400 x = 300 x = 300 ≈ 17.3 feet So, the ropes should be staked approximately 17.3 feet from the base of the tree 60 Using the computations from Problem 59, we observe that since the length of the hypotenuse is 20 feet, it follows that the length of each of rope tied from tree to the stake in this manner should be 20 feet in length Hence, for four stakes, one should have × 20 ≈ 80 feet of rope 10 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young 31 Let x = width of the built-in refrigerator (in inches) Using similarity, we obtain in in ⎛ ⎞ ⎛ ft ⎞ = so that x = ⎜ in ⎟ ⎜ ⎟ = ft = 48 in ft x ⎝ ⎠ ⎝ in ⎠ So, the built-in refrigerator is 48 inches wide 32 Let x = width of the pantry (in inches) Using similarity, we obtain in in ⎛ ⎞ ⎛ ft ⎞ 15 ⎛ 15 ⎞ = so that x = ⎜ in ⎟ ⎜ ft = ⎜ ⎟ (12 in.) = 45 in ⎟= ft x ⎝ ⎠ ⎝ in ⎠ ⎝ 4⎠ So, the pantry is 45in long 33 Consider the following diagram: 34 Using the diagram and value of h in Exercise 33, we see that 3 13 = 13 13 35 Using the diagram and information from Exercise 33, we see that sin θ = 13 = cos θ 36 Using Exercise 34, we see that sec θ = θ From the Pythagorean Theorem, we see that h = 32 + 22 = 13 Thus, cos θ = 2 13 = 13 13 ( ) 39 cos ( 90 − 30 ) = cos 60 13 = sin θ 37 First, note that using the diagram and value of h in Exercise 33, we see that sin θ = 13 As such, we have sin θ = 13 = tan θ = cos θ 13 38 Using Exercise 37, we see that cot θ = = tan θ csc θ = ( 40 cos ( A ) = sin 90 − A ) 76 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Chapter Review 41 ( ) tan ( 45 ) = cot ( 90 − 45 ) = cot 45 43 ( sin ( 30 − x ) = cos 90 − ( 30 − x ) ) ( = cos x + 60 45 ( csc ( 45 − x ) = sec 90 − ( 45 − x ) ( = sec 45 + x 47 b 53 tan 30 = 48 a ) ) = ) = 3 sin 60 = 12 = cos 60 1 = = 57 csc 45 = sin 45 55 tan 60 = 1 2 = = = cos 30 3 1 =1 = 61 sec 60 = cos 60 59 sec 30 = 63 cot 45 = 1 = =1 tan 45 65 0.6691 69 1.5399 ( ) csc ( 60 ) = sec ( 90 − 60 ) = sec 30 44 ( cos ( 55 + A ) = sin 90 − ( 55 + A ) ) ( = sin 35 − A 49 b sin 30 = cos 30 42 66 0.5446 70 1.0446 46 ( sec ( 60 − θ ) = csc 90 − ( 60 − θ ) ) ( = csc θ + 30 50 a 51 c sin 45 54 tan 45 = = cos 45 56 csc 30 = ) ) 52 c 2 2 =1 1 = = sin 30 1 2 = = = sin 60 3 1 = = 60 sec 45 = cos 45 58 csc 60 = 62 cot 30 = 1 = = tan 30 3 1 = = tan 60 3 67 0.9548 68 0.4706 71 1.5477 72 2.7837 64 cot 60 = ⎛ 17 ⎞ 73 First, note that 17′ = ⎜ ⎟ ≈ 0.28 ⎝ 60 ⎠ ⎛ 15 ⎞ 74 First, note that 15′ = ⎜ ⎟ = 0.25 ⎝ 60 ⎠ So, 39 17′ ≈ 39.28 So, 68 15′ = 68.25 77 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young 75 First, note that 76 First, note that ⎛ 25 ⎞ 25′′ = ⎜ ⎟ ≈ 0.007 ⎝ 3600 ⎠ ⎛ 15 ⎞ 15′′ = ⎜ ⎟ ≈ 0.004 ⎝ 3600 ⎠ ⎛ 30 ⎞ 30′ = ⎜ ⎟ = 0.50 ⎝ 60 ⎠ ⎛ 45 ⎞ 45′ = ⎜ ⎟ = 0.75 ⎝ 60 ⎠ So, 29 30′25′′ = 29.507 77 Observe that 42.25 = 42 + 0.25 So, 25 45′15′′ = 25.754 78 Observe that 60.45 = 60 + 0.45 ⎛ 60′ ⎞ = 42 + ( 0.25 ) ⎜ ⎟ ⎝1 ⎠ = 42 + 15′ ⎛ 60′ ⎞ = 60 + ( 0.45 ) ⎜ ⎟ ⎝1 ⎠ = 60 + 27′ So, 42.25 = 42 15′ So, 60.45 = 60 27′ 79 Observe that 30.175 = 30 + 0.175 80 Observe that 25.258 = 25 + 0.258 ⎛ 60′ ⎞ = 30 + ( 0.175 ) ⎜ ⎟ ⎝1 ⎠ = 30 + 10.5′ = 30 + 10′ + 0.5′ ⎛ 60′′ ⎞ = 30 + 10′ + ( 0.5′ ) ⎜ ⎟ ⎝ 1′ ⎠ = 30 + 10′ + 30′′ ⎛ 60′ ⎞ = 25 + ( 0.258 ) ⎜ ⎟ ⎝1 ⎠ = 25 + 15.48′ = 25 + 15′ + 0.48′ ⎛ 60′′ ⎞ = 25 + 15′ + ( 0.48′ ) ⎜ ⎟ ⎝ 1′ ⎠ = 25 + 15′ + 28.8′′ So, 30.175 = 30 10′ 30′′ So, 25.258 ≈ 25 15′ 29′′ 81 First, observe that 82 First, observe that ⎛ 15 ⎞ 37 15′ = 37 + ⎜ ⎟ = 37.25 ⎝ 60 ⎠ So, sin ( 37.25 ) ≈ 0.6053 ⎛ 35 ⎞ 42 35′ = 42 + ⎜ ⎟ = 42.583 ⎝ 60 ⎠ So, cos ( 42.583 ) ≈ 0.7363 83 First, observe that 84 First, observe that ⎛ 48 ⎞ 61 48′ = 61 + ⎜ ⎟ = 61.80 ⎝ 60 ⎠ So, cos ( 61.80 ) ≈ 0.4726 ⎛ 17 ⎞ 20 17′ = 20 + ⎜ ⎟ = 20.283 ⎝ 60 ⎠ So, sin ( 20.283 ) ≈ 0.3467 78 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Chapter Review 85 Using ni sin (θi ) = nr sin (θ r ) , observe 86 Using ni sin (θ i ) = nr sin (θ r ) , observe that 1.00sin ( 60 ) = nr sin ( 35.26 that nr = 1.00sin ( 60 ) 1.00sin ( 60 ) ≈ 1.50 sin ( 35.26 ) 87 Consider this triangle: ) = n sin ( 36.09 ) 1.00sin ( 60 ) n = ≈ 1.47 sin ( 36.09 ) r r 88 Consider this triangle: 50 25 Since cos ( 25 ) = adj a = , we have hyp 15 in a = (15 in.) cos ( 25 ) ≈ 13.59 in ≈ 14 in Since sin ( 50 ) = opp a = , we have hyp 27 ft a = ( 27 ft.) sin ( 50 ) ≈ 20.68 ft ≈ 21 ft 79 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young 89 Consider this triangle: 90 Consider this triangle: 33.5 Since tan ( 33.5 ) = opp a = , we adj 21.9 mi 47.45 Since sin ( 47.45 ) = opp 19.22 cm = , we c hyp have a = ( 21.9 mi.) tan ( 33.5 ) ≈ 14.5 mi have 91 Consider this triangle: 92 Consider this triangle: c= 37.75 19.22 cm ≈ 26.09 cm sin ( 47.45 ) β First, observe that ⎛1 ⎞ ⎟ ≈ 37.75 ⎝ 60′ ⎠ β = 37 45′ = 37 + 45′ ⎜ Since tan ( 37.75 ) = opp = adj First, observe that ⎛1 ⎞ ⎟ ≈ 75.16667 ′ 60 ⎝ ⎠ β = 75 10′ = 75 + 10′ ⎜ b , we 120.0 yd opp 96.5 km , = hyp c 96.5 km we have c = ≈ 99.8 km sin ( 75.16667 ) Since sin ( 75.16667 ) = have b = (120.0 yd.) tan ( 37.75 ) ≈ 92.91 yd 80 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Chapter Review 93 Consider this triangle: 94 Consider this triangle: α 30 65 β First, observe that β = 180 − ( 90 + 30 ) = 60 opp a = , we have Since sin ( 30 ) = hyp 21 ft a = ( 21 ft.) ( sin 30 ) ≈ 11 ft adj b = , Similarly, since cos ( 30 ) = hyp 21 ft we have b = ( 21 ft.) ( cos 30 ) ≈ 18 ft First, observe that α = 180 − ( 90 + 65 ) = 25 Since sin ( 65 ) = opp = hyp b , we have 8.5 mm b = ( 8.5 mm ) ( sin 65 ) ≈ 7.7 mm Similarly, since adj a cos ( 65 ) = = , we have hyp 8.5 mm a = ( 8.5 mm ) ( cos 65 ) ≈ 3.6 mm 81 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young 95 Consider this triangle: 96 Consider this triangle: 48.5 30.25 β β First, observe that β = 180 − ( 90 + 48.5 ) = 41.5 Since sin ( 48.5 ) = First, observe that β = 180 − ( 90 + 30 15′ ) = 59 45′ opp 215 mi = , we c hyp ⎛1 ⎞ = 59 + 45′ ⎜ ⎟ = 59.75 ′ 60 ⎝ ⎠ ⎛1 ⎞ α = 30 15′ = 30 + 15′ ⎜ ⎟ = 30.25 ⎝ 60′ ⎠ have c= 215 mi ≈ 287 mi sin ( 48.5 ) Similarly, since opp 215 mi tan ( 48.5 ) = = , we have b adj 215 mi b= ≈ 190 mi tan ( 48.5 ) Since cos ( 30.25 ) = adj 2,154 ft = , we c hyp have c= 2,154 ft ≈ 2, 494 ft cos ( 30.25 ) Similarly, since opp a tan ( 30.25 ) = = , we have adj 2,154 ft a = ( 2,154 ft.) tan ( 30.25 ) ≈ 1, 256 ft 82 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Chapter Review 97 Consider this triangle: 98 Consider this triangle: α α β 30.5 ft , we obtain 45.7 ft ⎛ 30.5 ⎞ α = tan −1 ⎜ ⎟ ≈ 33.7 ⎝ 45.7 ⎠ So, β ≈ 180 − ( 90 + 33.7 ) ≈ 56.3 β 11,798 km , we obtain 32,525 km First, since tan α = First, since sin α = Next, using the Pythagorean Theorem, we obtain 2 c = ( 30.5 ft.) + ( 45.7 ft.) = 3, 018.74 ft.2 ⎛ 11,798 ⎞ ⎟ ≈ 21.268 ⎝ 32,525 ⎠ 11,798 km Similarly, since cos β = , we 32,525 km obtain ⎛ 11,798 ⎞ β = cos −1 ⎜ ⎟ ≈ 68.732 ⎝ 32,525 ⎠ Next, using the Pythagorean Theorem, we obtain 2 b + (11, 798 km ) = ( 32,525 km ) and so, c ≈ 54.9 ft α = sin −1 ⎜ b = 918, 682,821 km b ≈ 30,310 km 83 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young 99 Consider this triangle: 100 Consider this triangle: 30 30 100 ft , so that H a , so that Observe that sin 30 = 150 ft a = ( sin 30 ) (150 ft.) ≈ 75 ft Observe that sin 30 = So, the altitude should be about 75 ft 101 Consider this diagram: So, the hose should be about 200 ft long 102 Consider this diagram: H= 100 ft ≈ 200 ft sin 30 9mi 10mi θ 3mi θ 15 Observe that tan (15 + θ ) = 20 5mi 10 mi , so that Observe that tan 20 + θ = mi , so that ( ) mi mi ⎛ 10 ⎞ 15 + θ = tan −1 ⎜ ⎟ ⎝ 3⎠ ⎛ 10 ⎞ θ = tan −1 ⎜ ⎟ − 15 ≈ 58 ⎝ 3⎠ So, the bearing would be N 58 E ⎛9⎞ 20 + θ = tan −1 ⎜ ⎟ ⎝5⎠ ⎛9⎞ θ = tan −1 ⎜ ⎟ − 20 ≈ 41 ⎝5⎠ So, the bearing would be N 41 W 84 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Chapter Review 103 Since the lengths of the two legs of the given 30 − 60 − 90 triangle are x and x , the shorter leg must have length x Hence, using the given information, we know that x = 41.32 ft Thus, the two legs have lengths 41.32 ft and 41.32 ≈ 71.57 ft., and the hypotenuse has length 82.64 ft 104 Since the lengths of the two legs of the given 30 − 60 − 90 triangle are x and x , the shorter leg must have length x Hence, using the given information, we know that x = 87.65 cm Thus, the two legs have lengths 87.65 cm and 87.65 ≈ 50.60 cm, and the hypotenuse has length 101.20 cm 1 ≈ ≈ 1.02041 sin 78.4 0.980 b The keystroke sequence 78.4, sin, 1/x yields 1.02085 1 ≈ ≈ 1.43885 106 a tan 34.8 ≈ 0.695 and so, tan 34.8 0.695 b The keystroke sequence 34.8, tan, 1/x yields 1.43881 107 2.612 108 0.125 109 -1 110 Error – dividing by zero 105 a sin 78.4 ≈ 0.980 and so, Chapter Practice Test Solutions -1 Let x = measure of smallest angle in the triangle Then, (1) implies that 5x = measure of the largest angle in the triangle, (2) implies that 3x = measure of the third angle in the triangle So, x + x + x = 180 so that x = 180 and thus, x = 20 Thus, the three angles are 20 , 60 , 100 Consider the following triangle: 60 The angle opposite the 30 angle in a 30 − 60 − 90 triangle is the smallest of the three angles So, the other leg must have length cm and the hypotenuse must have length 10 cm 30 85 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Consider the following two diagrams: θ θ Let H = height of the Grand Canyon (in feet) Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain ⎛ ft ⎞ ft H = so that H = 600 ft ( ) ⎜ ⎟ = 6, 000 ft 600 ft ft ⎝ ft ⎠ So, the Grand Canyon is 6,000 ft tall Consider the following triangle: θ From the Pythagorean Theorem, we see that c = 32 + 12 = 10 a sin θ = 10 = 10 10 3 10 = 10 10 sin θ 10 = c tan θ = = cos θ 10 b cos θ = 10 = cos θ 10 e csc θ = = = 10 sin θ 1 f cot θ = =3 tan θ d sec θ = 86 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Chapter Practice Test 10 (by Exercise 4b) 10 b sec ( 90 − θ ) = csc θ = 10 (by Exercise 4e) a sin ( 90 − θ ) = cos θ = (by Exercise 4c) In the following computations, we use sin θ 1 tan θ = , cot θ = , sec θ = , csc θ = cos θ tan θ cos θ sin θ c cot ( 90 − θ ) = tan θ = θ 30 sin θ 45 2 60 sec ( 42.8 ) = cos ( 42.8 cos θ 2 2 ) ≈ 1.3629 First, note that tan θ 3 cot θ sec θ 3 csc θ 2 3 3 The first value ( cos θ = ) is an exact value of the cosine function, while the second ( cos θ = 0.66667 ) would serve as an approximation to the first 10 Consider the following triangle: ⎛ 20 ⎞ 20′′ = ⎜ ⎟ ≈ 0.00556 ⎝ 3600 ⎠ ⎛ 45 ⎞ 45′ = ⎜ ⎟ = 0.75 ⎝ 60 ⎠ 20 So, 33 45′20′′ ≈ 33.756 Since tan ( 20 ) = b= opp 10 cm = , we have b adj 10 cm ≈ 27 cm tan ( 20 ) 87 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young 11 Consider this triangle: 9.2 km , we obtain 23 km ⎛ 9.2 ⎞ β = cos −1 ⎜ ⎟ ≈ 66 ⎝ 23 ⎠ Since cos β = β 12 Consider this triangle: 13 Consider this triangle: α 50 β Since cos 50 = 12 cmm , we see that mm c = 12 ≈ 19 mm cos50 13.3 , we obtain Since sin β = 14.0 13.3 β = sin −1 ( 14.0 ) ≈ 71.8 14 Consider this triangle: 15 Consider this triangle: α α First, we convert α to DD: 33 10′ = 33 + 10′ 601 ′ ≈ 33.17 Since sin ( 33.17 ) = ( ) a 47m Since tan α = 3.45 6.78 , we obtain α = tan −1 ( 3.45 6.78 ) ≈ 27.0 , we see that a = ( 47m ) sin ( 33.17 ) ≈ 26m 88 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Chapter Practice Test 16 Consider the following diagram: Let α = measure of the desired angle between the hour hand and minute hand Since the measure of the angle formed using two rays emanating from the center of the clock out toward consecutive hours is always ( 360 ) = 30 , it immediately follows 12 that α = 30 17 Using ni sin (θi ) = nr sin (θ r ) , observe that 1.00sin ( 25 ) = nr sin (16 nr = 18 Consider the following diagram: 20 20 ) 1.00sin ( 25 sin (16 ) ) ≈ 1.53 r , we have 150 ft r = (150 ft.) ( tan 20 ) ≈ 54.6 ft Since tan 20 = So, the diameter of the circle is approximately 109 ft 89 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young Solution Manual for Trigonometry 3rd Edition by Young θ Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young ( ) = 44 + 16.20′ = 44 + 16′ + ( 0.20 )′ ( ) = 44 16′12′′ 22 10′23′′ = 22 + 10′ ( ) + 23′′ ( ) ≈ 22.1731 19 44.27 = 44 + ( 0.27 ) 20 60′ 1 60′ 21 Since csc 75 = sin175 = 22 sin 60 + 60′′ 1′ 3600′′ 6+ cos 45 = + cot 30 2 , we have cos15 = sin 75 = = 2 6+ 3 3+ + ⋅ = + = 2 6 23 12 40′ + 55 49′ = 67 89′ = 67 + (60 + 29)′ = 67 + 60′ + 29′ = 68 29′ =1 24 82 27′ − 35 39′ = 81 87′ − 35 39′ = ( 87 − 35 ) + ( 87 − 39 )′ = 46 48′ 25 Consider the following triangle: We have tan 72 = h 200 yards , so that h = ( 200 yards ) tan 72 ≈ 620 yards 72 90 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-3rd-Edition-by-Young ... https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 3rd- Edition- by- Young Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 3rd- Edition- by- Young. .. https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 3rd- Edition- by- Young Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 3rd- Edition- by- Young. .. https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 3rd- Edition- by- Young Solution Manual for Trigonometry 3rd Edition by Young Full file at https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 3rd- Edition- by- Young