Solution manual for trigonometry 4th edition by young

Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-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-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-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-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-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-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-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-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-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-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-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-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-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-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-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-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-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-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-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-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Section 1.2 24 First, note that 35m = 3,500cm Now, observe that by similarity, c b = , so that f e 3,500cm b = Solving for b yields 14cm 10cm (14cm)b = 35, 000cm b= 25 By similarity, b e = , so that c f 35,000 c m 14 cm = 2,500cm = 25m in e = Solving for e yields in in 2 ( in.) e = ( in.)( in.) e= 26 By similarity, d a = , so that e b ( 16 ( in.)( in.) ( in.) mm = mm mm ) a = ( e= 16 = 12 25 in a Solving for a yields mm mm )( mm ) ( 716 mm )( mm ) ( mm ) = 35 16 mm 27 Note that 14 14 = 14.25 Now, consider the following two diagrams Let y = height of the tree (in feet) Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain 14.25 ft y = ft 1.5 ft (1.5 ft.) y = 57 ft.2 57 ft.2 y= = 38 ft 1.5 ft So, the tree is 38 feet tall 21 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young 28 Note that = 0.75 Now, consider the following two diagrams Let y = height of the flag pole (in feet) Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain y 15 ft = ft 0.75 ft (0.75 ft.) y = 30 ft.2 30 ft.2 y= = 40 ft 0.75 ft So, the flag pole is 40 feet tall 29 Consider the following two diagrams 48 ft Let y = height of the lighthouse (in feet) Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain ft 1.2 ft = y 48 ft (1.2 ft.) y = 240 ft.2 240 ft.2 = 200 ft y= 1.2 ft So, the lighthouse is 200 feet tall 22 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Section 1.2 30 Consider the following two diagrams Let y = length of the son’s shadow (in feet) Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain ft ft ft.2 = so that (6 ft.) y = ft.2 and hence, y = = ft ft y ft So, the son’s shadow is 23 ft long Since ft = 12 in , this is equivalent to in 31 First, make certain to convert all quantities involved to a common unit In order to avoid using decimals, use the smallest unit – in this particular problem, use cm Now, consider the following two diagrams: Let y = height of the lighthouse (in cm) Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain 200cm 5cm = 300cm y (5cm) y = 60, 000 cm2 60, 000 cm2 y= = 12, 000cm = 120m 5cm So, the lighthouse is 120 m tall 23 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young 32 Let H = the height of the pyramid (in meters) (Important! Don’t confuse this with the slant height.) Using the fact that the height is the length of the segment that extends from the apex of the pyramid down to the center of the square base, we have the following diagram Also, we have Then, using similarity (which applies since sunlight rays act like parallel lines – see Text Example 3), we obtain H (115 + 16)m = 1m 0.9m 0.9 m H 131 m = ( ) m H = 131 0.9 m = 145.56m ≈ 146m So, the pyramid is approximately 146 m tall 33 Let x = length of the planned island (in inches) (Note that 2′4′′ = 28in ) Using similarity, we obtain x 28 in = 11 in 16 in 11 28 ( 16 ) in = ( 14 in.) x x= 11 28 ( 16 ) in.2 in = 77 in = ft in So, the planned island is 6ft 5in long 24 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Section 1.2 34 Let x = width of the pantry (in inches) Using similarity, we obtain 28 in x = in 16 in 28 ( 16 ) in = ( 14 in.) x x= 28 ( 167 ) in.2 in = 49 in = ft in So, the pantry is 4ft 1in long 35 Consider the following two triangles: The triangles are similar by AAA, so that 36 Consider the following diagram: = 3.5 x Solving for x yields x = 2.8ft Since this is a 30 − 60 − 90 triangle and x is the longer side (being opposite the larger of the two acute angles), we know that x = 15 ≈ 26 ft 60 37 The two triangles are similar by AAA So, we have 1.6 in x = ⇒ ( 2.2 in.) x = (1.6 in.)(1.2 in.) ⇒ x ≈ 0.87 in 2.2 in 1.2 in 38 If a right triangle is isosceles, then it must be a 45 − 45 − 90 triangle Let x = length of a leg Then, the hypotenuse is x = , so that x = in 25 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young 39 The information provided is: f d = cm, dt = 2.5cm, d s = 80cm We must determine the value of f s Using similar triangles, we have the following proportion: d s d s + dt 80 2.5 + 80 = ⇒ = fs fd fs 320 Solving for f s yields: 82.5 f s = 320 ⇒ f s = ≈ 3.87878 82.5 Thus, f s ≈ 7.8cm 40 The information provided is: f d = cm, so that f d = cm dt = 3.5cm f s = 3.8cm, so that f s = 1.9cm We must determine the value of d s Using similar triangles, we have the following proportion: d s d s + dt d s d s + 3.5 = ⇒ = fs fd 1.9 Solving for d s yields: 2d s = 1.9d s + 6.65 ⇒ 0.1d s = 6.65 ⇒ d s = 66.5cm 41 The information provided is: f d = cm, so that f d = cm d s = 60cm f s = 3.75cm, so that f s = 1.875cm We must determine the value of dt Using similar triangles, we have the following proportion: d s d s + dt 60 + dt 60 = ⇒ = fs fd 1.875 Solving for dt yields: 120 = 112.5 + 1.875dt ⇒ 7.5 = 1.875dt ⇒ dt = 4cm 26 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Section 1.2 42 The information provided is: f s = 4.15 cm, dt = 4.5cm, d s = 60cm We must determine the value of f d Using similar triangles, we have the following proportion: d s d s + dt 60 60 + 4.5 = ⇒ = 2.075 fs fd fd 133.8375 Solving for f d yields: 60 f d = 133.8375 ⇒ f d = ≈ 2.2306 60 Thus, f d ≈ 4.5cm 43 The ratio is set up incorrectly It should be A = B D E 44 In order to find F, one needs C instead of B In such case, A = C D F 45 True Two similar triangles must have equal corresponding angles, by definition 46 True If two triangles are congruent, then the ratio of corresponding sides equals 1, for all three pairs of sides, and all corresponding angles are equal Hence, they must be similar (Conversely, two similar triangles need not be congruent – see Problem 36.) 47 False The third angle in such case would have measure 180 − ( 82 + 67 ) = 31 48 True Consider two equilateral triangles, one whose sides have length x and a second whose sides have length y Observe then that the ratio of corresponding sides is always xy (or y x , depending on which you refer to as Triangle and Triangle 2) Since all angles of an equilateral triangle have measure 60 , we conclude that they must be similar However, if x = and y = , then the triangles are NOT congruent 49 False Such angles are congruent, and hence unless they are both 90 , they need not be supplementary 50 True 51 False You need the corresponding angles to be the same to ensure similarity of two isosceles triangles 52 False All we can say here is that corresponding sides of similar triangles are in proportion 27 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young 53 We label the given diagram as follows: Observe that ACB = DCE (vertical angles) Since we are given that CBA = CDE , the third pair of angles must have the same measures Hence, ABC is similar to CDE The corresponding sides between these two triangles are identified as follows: AB corresponds to ED, AC corresponds to EC, BC corresponds to DC Using the ratio for similar triangles, we have AB = AC , so that = , so that x = ED EC x 54 Consider the following diagram: 18 Observe that ABE and ACD are similar triangles (since we are given that ABE = ACD and the triangles share A Corresponding sides of these triangles are identified as follows: AB corresponds to AC, AE corresponds to AD, BE corresponds to CD Using the ratio for similar triangles, we know that AB AE BE = = , so that AC AD CD y = = = + x y + 18 20 (1) We now use different equalities from (1) to find x and y Find x: = so that 16 = + x and hence, x = 12 4+ x y Find y: = so that y = y + 18 and hence, y = y + 18 28 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Section 1.2 55 Triangles and are similar because they both have a 90 angle, and the angles formed at the horizontal have the same measure since they are vertical angles Similarly, Triangles and are similar 56 a) Consider the following diagram: H0 D0 − f f Hi The two right triangles above are similar since they both have a 90 , and the measures of the angles opposite the sides with lengths H and H i are equal because they are vertical H D −f angles Hence, using the similar triangles ratio yields = Hi f b) Consider the following diagram: H0 Di − f f Again, from similarity if follows that Hi H0 f = H i Di − f 29 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young 1 1 + = D0 Di f Proof Observe that from Problem 56, we obtain the following identities: ⎡H ⎤ ⎛H ⎞ f ⎢ + 1⎥ = f ⎜ ⎟ + f = D0 (from (a)) ⎣ Hi ⎦ ⎝ Hi ⎠ 57 Claim (Lens Law): ⎡H ⎤ ⎛H ⎞ f ⎢ i + 1⎥ = f ⎜ i ⎟ + f = Di (from (b)) ⎣ H0 ⎦ ⎝ H0 ⎠ Hence, observe that 1 Di + D0 + = D0 Di D0 Di ⎡H ⎤ ⎡H ⎤ f ⎢ + 1⎥ + f ⎢ i + 1⎥ ⎣ Hi ⎦ ⎣ H0 ⎦ = ⎧⎪ ⎡ H ⎤ ⎫⎪ ⎧⎪ ⎡ H i ⎤ ⎫⎪ + 1⎥ ⎬ ⋅ ⎨ f ⎢ + 1⎥ ⎬ ⎨f ⎢ ⎦ ⎭⎪ ⎩⎪ ⎣ H ⎦ ⎭⎪ ⎩⎪ ⎣ H i = = ⎡H H ⎤ f ⎢ i + + 2⎥ ⎣ H0 Hi ⎦ ⎡ ⎤ ⎢ ⎥ H H H H f ⎢ ⋅ i + + i + 1⎥ Hi H0 ⎢ Hi H0 ⎥ ⎢⎣ =1 ⎥⎦ ⎡ Hi H0 ⎤ ⎢ H + H + 2⎥ i ⎣ ⎦ f ⎡ H0 Hi ⎤ ⎢2 + H + H ⎥ i 0⎦ ⎣ = , f as desired ☻ 30 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Section 1.2 58 We begin with several observations: 1.) m ( ∠ABE ) = 90 since BF is parallel to CG, and ∠ABE ≈ ∠GCD since they are corresponding angles 2.) m ( ∠AEB ) = m ( ∠CGA ) = m ( ∠FEG ) = 60 3.) m ( ∠CGD ) = m ( ∠CDG ) = 45 since ΔCDG is an isosceles right triangle 4.) BC and FG have the same length Hence, ΔABE ∼ ΔACG ∼ ΔGFE 59 We begin with several observations: 1.) m ( ∠ABE ) = 90 since BF is parallel to CG, and ∠ABE ≈ ∠GCD since they are corresponding angles 2.) m ( ∠AEB ) = m ( ∠CGA ) = m ( ∠FEG ) = 60 3.) m ( ∠CGD ) = m ( ∠CDG ) = 45 since ΔCDG is an isosceles right triangle 4.) BC and FG have the same length Hence, ΔABE ∼ ΔACG ∼ ΔGFE Now, using this information, consider the following triangles: Since ΔACG is a 30 − 60 − 90 triangle, we know that m( AC ) = ⋅ m(CG ), and so m(CG ) = that 3 Hence, m( AG ) = ⋅ m(CG ) = 103 Next, since ΔACG ∼ ΔGFE , we know 103 5 33 = ⇒ y = and = ⇒ x= 3 y x So, m( EF ) = cm, m( EG ) = cm 60 Since m(CG ) = m( DG ) = 3 ( ) cm = 3 cm and ΔDCG is 45 − 45 − 90 , we know that cm 31 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Section 1.3 Solutions -8 sin θ = = cos θ = = 10 10 1 = = sin θ tan θ = = 3 csc θ = Note that by the Pythagorean Theorem, the hypotenuse has length So, = 5 Note that by the Pythagorean Theorem, the hypotenuse has length So, 1 = = sec θ = cos θ cos θ = 11 tan θ = = 13 sin θ = 5 34 = 34 34 1 = = cos θ 1 cot θ = = = tan θ Note that by the Pythagorean Theorem, the hypotenuse has length So, sec θ = 2 = 5 10 Note that by the Pythagorean Theorem, the hypotenuse has length So, sin θ = csc θ = = sin θ = 2 1 = tan θ 14 Using the Pythagorean Theorem, we know that the leg adjacent to angle θ has 12 cot θ = length So, cos θ = 15 Using the Pythagorean Theorem, we know that the leg adjacent to angle θ has length So, tan θ = 17 Using the Pythagorean Theorem, we know that the leg adjacent to angle θ has length So, sec θ = = cos θ 34 ( ) 19 sin ( 60 ) = cos ( 90 − 60 ) = cos 30 16 csc θ = = sin θ 3 34 = 34 34 34 18 Using the Pythagorean Theorem, we know that the leg adjacent to angle θ has = length So, cot θ = tan θ ( ) 20 sin ( 45 ) = cos ( 90 − 45 ) = cos 45 32 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Section 1.3 ( 21 cos ( x ) = sin 90 − x ) ( ) 23 csc ( 30 ) = sec ( 90 − 30 ) = sec 60 24 25 26 sin ( x + y ) = cos ( 90 − ( x + y ) ) ( = cos 90 − x − y 27 ( ) ( ( ) cot ( 45 − x ) = tan 90 − ( 45 − x ) ( = tan 45 + x ) ( sin ( 60 − x ) = cos 90 − ( 60 − x ) ( = cos 30 + x cos ( 20 + A ) = sin 90 − ( 20 + A ) = sin 70 − A 29 ) ( sec ( B ) = csc ( 90 − B ) 22 cot ( A ) = tan 90 − A ) 28 ) cos ( A + B ) = sin ( 90 − ( A + B ) ) ( = sin 90 − A − B ) 31 From the given information, we obtain the following triangle: 30 ) ( ) sec ( 30 − θ ) = csc 90 − ( 30 − θ ) ( = csc 60 + θ ) ) 32 Consider the following triangle: θ θ Hence, a round trip on the ATV corresponds to twice the length of the hypotenuse, which is × ( miles ) = 10 miles opp y = = , so y = x adj x Since we are given that ( x + y ) = 200 Observe that tan θ = (since a round trip is 200 yards), we see that x + y = 100 and hence, x = y = 50 As such, by the Pythagorean Theorem, the hypotenuse has length 50 So, the round trip of the ATV is 100 yards ≈ 141 yards 33 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at Chapter https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young 33 Consider the triangle: Applying the Pythagorean theorem yields 52 + 122 = z ⇒ z = 13 Thus, sin θ = 135 θ 34 The scenario can be described by the following two triangles: Bob Neighbor θ θ Applying the Pythagorean theorem yields z = 13 and w = 193 Hence, we have: Neighbor: cos θ = 12193 Bob: cos θ = 12 13 The value of cos θ is larger for Bob’s roof The steeper the roof for the same horizontal distance, the longer the hypotenuse and hence denominator used in computing cos θ 34 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young Section 1.3 Since the hypotenuse has length by the Pythagorean theorem, this is a 30 − 60 − 90 triangle Hence, since θ is opposite the longer leg, it must be 60 35 If tan θ = , we have the following diagram: θ 36 Since θ = 60 we have the following triangle: Observe that cos 30 = 37 Consider the following triangle: Observe that ft W ⇒ W= cos30 ≈ 3.46 ft sin B = 53 35 Full file at https://TestbankDirect.eu/Solution-Manual-for-Trigonometry-4th-Edition-by-Young ... https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 4th- Edition- by- Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 4th- Edition- by- Young. .. https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 4th- Edition- by- Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 4th- Edition- by- Young. .. https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 4th- Edition- by- Young Solution Manual for Trigonometry 4th Edition by Young Full file at https://TestbankDirect.eu /Solution- Manual- for- Trigonometry- 4th- Edition- by- Young