Solution manual for analytic trigonometry with applications 11th edition by barnett

Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Chapter Chapter Right Triangle Ratios EXERCISE 1.1 Angles, Degrees, and Arcs Since one complete revolution has measure 360°, Similarly, 1 revolution has measure (360°) = 90° 4 11 (360°) = 120°, (360°) = 315°, (360°) = 330° 12 € € Since one complete revolution has measure 360°, the fraction of a counterclockwise revolution that 30 225 240 € form a 30° angle € is will = €Similarly, = , = 360 12 360 360 Since 88° angle is between 0° and 90°, this is an acute angle € None of these € € € € € 10 Since 175° is between 90° and 180°, this is an obtuse angle 12 It is a right angle 14 A minute of angle measure is 16 ⎛ 43 30 ⎞ ° 43° 30° + Since 43' = and 30" = , then 71°43'30" = ⎜ 71+ ≈ 71.725° to three decimal € € ⎟⎠ 60 3,600 60 3,600 ⎝ places 18 ⎛ ⎞ ° 3° 1° Since € 3' = 60 and€1" = 3,600 , then 9°3'1"€= ⎜⎝ + 60 + 3,600 ⎟⎠ ≈ 9.050° to three decimal places 20 1 of one degree; a second of angle measure is of one minute 60 60 ⎛ 11 25 ⎞ ° 11° 25° + Since 11' = and 25" = , then 267°11'25" = ⎜ 267 + ⎟ ≈ 267.190° to three € 60 3,600 ⎠ 60 3,600 € ⎝ € decimal places 22 € 35.425° = 35°(0.425 € × 60)' = 35°25.5' = 35°25'(0.5 × 60)" = 35°25'30" 24 52.927° = 52°(0.927 × 60)' = 52°55.62' = 52°55'(0.62 × 60)" ≈ 52°55'37" 26 235.253° = 235°(0.253 × 60)' = 235°15.18' = 235°15'(0.18 × 60)" ≈ 235°15'11" € Right Triangle Ratios Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Exercise 1.1 28 There are two methods a Convert the first one to decimal hour form and compare with the second one 43 24 + ≈ 3.723 to three decimal places 3+ 60 3,600 Since 3.723 < 3.732, Runner A is faster b € 30 32 34 36 € Angles, Degrees, and Arcs Convert the second one to HMS form and compare with the first one 3.732 = hr + (0.732 × 60) € = hr + 43.92 = hr + 43 + (0.92 × 60) sec ≈ hr, 43 min, 55 sec This is bigger than hr, 43 min, 24 sec, and hence Runner A is faster To compare α and β, we convert α to decimal form Since 51' = ⎛ 51 54 ⎞ ° + 32°51'54" = ⎜ 32 + ⎟ = 32.865° Thus α = β 60 3,600 ⎝ ⎠ 40° € 20° € To compare α and β, we convert β to decimal form Since 40' = and 20" = , then 60 3,600 ⎛ 40 20 ⎞ ° € + 80°40'20" = ⎜80 + ⎟ ≈ 80.672° Thus α < β 60 3,600 ⎠ ⎝ € € 18° 32° We convert β to decimal form Since 18' = and 32" = , then 242°18'32" = 60 3,600 ⎛ € 18 ⎞ 32 ° + ⎜ 242 + ⎟ ≈ 242.309° Thus α > β 60 3,600 ⎠ ⎝ € € 105°53'22" + 26°38'55" → DMS 132°32'17" 38 180° – 121°51'22" → DMS 58°8'38" 40 The circumference of a circle with radius r is 2πr If 2πr = 6, then estimate of r to the nearest whole number would be The value of r to two decimal places is r = ≈ 0.95 2π 42 The arc length of a semicircle that has diameter 10 is 44 Since € € 1 C= (2πr) = (10π) = 5π An estimate 2 of 5π to the nearest whole number is 16 The length of€the semicircle of radius (or diameter 10) to two decimal places is 15.71 s θ = , then C 360° 12 θ = 108 360° ⎛ 12 ⎞ θ = ⎜ € ⎟ 360° = 40° ⎝ 108 ⎠ € € Since € s = θ€ , then 46 C 360° s 72° = 740 360° 72 s = (740) = 148 mi € € 360 € € € € 51° 54° and 54" = , then 60 3,600 Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Chapter 48 Right Triangle Ratios s θ = and C = 2πr, then C 360° s θ = 2πr 360° 38,000 cm 45.3° = € 2πr 360° Since € € € 50 € € 52 € r = (38,000 cm)(360) ≈ 48,000 cm (to the nearest 1,000 cm) (2π)(45.3) € ⎛ 16 34 ⎞ ° s θ + = and θ = 24°16'34" = ⎜ 24 + ⎟ ≈ 24.276°, then 60 3,600 C 360° ⎝ ⎠ € 14.23 m 24.276° ≈ C 360° (14.23 m)(360) € C ≈ € ≈ 211.0 m (to one decimal place) 24.276 Since € A A θ 24.6° , then = 360° 360° π(7.38 ft) πr € 24.6 A = (π)(7.38 ft)2 ≈ 11.7 ft2 (to one decimal place) 360 € € € € 347 in θ θ A 54 Since = , then = 2 360° 360° π(32.4€ in) πr 347 θ = · 360° ≈ 37.9° π(32.4) € € € € Since = 56 Note that the central angles corresponding € hence to the arcs AB and CD are 72° and ∠A = 36°, ∠D = 36° Since the sum of the angles in a triangle is 180°, the angle ∠x should be 180° - (36° + 36°) = 108° 58 Since s θ s θ = and C = 2πr, then = C 360° 2πr 360° s 12.1 mm θ = · 360° = · 360° ≈ 131.8° 2πr 2(π)(5.26 mm) € € € € s θ s θ 60 Since = and C = 2πr, then = C 360° 2πr 360° € s € 360° 360 11.8 mm r = · = · ≈ 5.73 mm 2π θ 117.9 2(π) € € € € € € € € Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Exercise 1.1 62 Since Angles, Degrees, and Arcs s θ = and C = 2πr, then C 360° s θ = 2πr 360° ⎛ 10 ⎞ θ = 40°40' – 33°30' = 7°10' = ⎜ + ⎟ ° ⎝ 60 ⎠ + 10 € € € € θ 60 ≈ 495 mi s = 2πr · = 2(π)(3,960 mi) · 360 360° € s θ s θ 64 Since = and C = 2πr, then = C 360° 2πr€ 360° € ⎛ 50 ⎞ θ = 42°50' – 36°0' = 6°50' = ⎜ + ⎟ ° ⎝ 60 ⎠ + 50 € € € € θ 60 ≈ 472 mi s = 2πr · = 2(π)(3,960 mi) · 360 360° € 66 To find the length of s in nautical miles, since nautical mile is the length of 1' on the circle shown in the diagram, we need only to find how many € minutes are in the angle € θ Since θ = 40°40' – 33°30' = 7°10' = (7 × 60 + 10)' = 430' Therefore, s = 430 nautical miles 68 To find the length of s in nautical miles, since nautical mile is the length of 1' on the circle shown in the diagram, we need only to find how many minutes are in the angle θ Since θ = 42°50' – 36°0' = 6°50' = (6 × 60 + 50)' = 410' Therefore, s = 410 nautical miles 70 72 € € The arc length of a circular sector is very close to the chord length if the central angle of the sector is small and the radius of the sector is large, which is the case in this problem minute of arc is π in ≈ 1.05 in at 100 yd; the diameter of the quarter is slightly less than in s θ = and C = 2πr, then C 360° s θ = 2πr 360° s 864,000 · 360˚ = · 360° ≈ 0.028° € θ = 2πr 2(π)(1,780,000,000) Since € 74 € s θ Since = and C = 2πr, then C 360° € s θ€ = 2πr 360° s 360° 360° 52 r = · = · ≈ 40,000 miles € € 2π θ 2(π) 0.075° € € € € € € Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Chapter EXERCISE 1.2 Right Triangle Ratios Similar Triangles The measures of the second acute angle in each right triangle are the same, since the sum of the measures of the two acute angles in any right triangle is 90° Thus if A + B = 90° and A' + B' = 90° and A = A', then clearly B = B' (10)(0.33) = 3.3 and according to Appendix A.3, the correct rounded value is Since b c 24 (24)(1) = by Euclid's Theorem, then = , c' = =8 bʹ′ cʹ′ cʹ′ a b 51 b (51)(8) Since = by Euclid's Theorem, then = ,b= = 24 a ʹ′ b ʹ′ 17 17 € € € € € a b 640, 000 b (640, 000)(0.75) 10 Since = by Euclid's Theorem, then = ,b= = 32,000 15 0.75 15 € € aʹ′ bʹ′ € € € 12 Yes, since the two triangles will then be congruent (the angle-side-angle of one triangle is equal to € the€angle-side-angle of the other) € € € 14 Since the triangles are similar, the sides are proportional and we can write a 32 cm c 32 cm = = 0.47 1.0 1.1 1.0 (0.47)(32 cm) (1.1)(32 cm) ≈ 15 cm c = ≈ 35 cm a = 1.0 1.0 € 16 € the triangles are similar, the sides are € proportional € Since and we can write b 63.19 cm c 63.19 cm = € 1.0 = € 0.47 1.1 0.47 63.19 cm (1.1)(63.19 cm) b = ≈ 130 cm c = ≈ 150 cm 0.47 0.47 € the triangles are similar, the sides are € proportional € 18.€ Since and we can write 13 a 1.037 × 10 m c 1.037 × 1013 m = = € € 0.47 1.0 1.1 1.0 (0.47)(1.037 × 1013 m) a = ≈ 4.9 × 1012 m c = (1.1)(1.037 × 1013 m) ≈ 1.1 × 1013 m 1.0 € € € € 20 Since the triangles are similar, the sides are proportional and we can write a 2.86 × 10 –8 cm b 2.86 × 10 –8 cm = = € 0.47 1.1 1.0 1.1 (0.47)(2.86 × 10 –8 cm) 2.86 × 10 –8 cm ≈ 1.2 × 10–8 cm b = ≈ 2.6 × 10–8 cm a = 1.1 1.1 € € € € € € Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Exercise 1.2 22 Similar Triangles We make a scale drawing of the triangle, choosing b' to be 3.00 in, ∠A' = 50°, ∠C' = 90° Now measure c' (approximately 4.67 in) and set up a proportion Thus, c 36 ft = 4.67 in 3.00 in 4.67 c ≈ (36 ft) ≈ 56 ft 3.00 Drawing not to scale € € In the drawing, we note that triangles LBT and LNM are similar MN = 8.5 ft.€ 1 NB = (length of court) = (78 ft) = 39 ft TB = ft BL is to be found Let 2 BL TB BL = x Then, = M NL MN T x 8.5 ft = € NL = NB + BL Thus, € ft 39 + x 8.5 N x B 39 ft 8.5x = 3(39 + x) = 117 + 3x € € 5.5x = 117 39 + x ft 117 Drawing not to scale x =€ ≈ 21.27 ft € 5.5 AB AC 30 Since the triangles ABC and DEC in the figure are similar, we can write = Then, DE CD AB 25 ft =€ 5.75 ft 2.25 ft 5.75 € € AB = (25 ft) ≈ 64 ft 2.25 28 €32 € drawing, we note that triangles In the ABC and ABC are similar So, € Bʹ′C ʹ′ Aʹ′Bʹ′ BC = BC x 42 = or x = ≈ ft € € 34 In the drawing, we note that triangles ABC and ABC are similar So, ABʹ′ Bʹ′C ʹ′ = € € € AB BC x = x + 25 12 or € € 12x = 5x + 125 7x = 125 € € Aʹ′Bʹ′ x = ≈ 18 ft BC € L Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Chapter 36 € 38 Let us make a scale drawing of the figure in the text as follows: pick any convenient length, say in, for A'C' Copy the 28° angle CAB and 90° angle ACB using a protractor Now, measure B'C' (approximately 1.06 in) and set up a proportion Thus, x 4.0 × 10 m = 1.06 in in 1.06 x = (4.0 × 103 m) ≈ 2.1 × 103 m Right Triangle Ratios A' in C' Measure (approx 1.06 in) 28° B' (A)€u(m) = 10 or u(mm) = 10,000 and 1 hence + = , v 50 10,000 € 1 199 = – = v 50 10,000 10,000 and € € € 10, 000 v= ≈ 50.25 mm 199 € € € € Continuing in this manner we obtain u(mm) €v(mm) 10,000 50.2 50.25 20,000 50.13 30,000 50.08 40,000 50.06 50,000 50.05 60,000 50.04 (B) v approaches 50 mm 40 From the graph on the right, we have AREA = (Area of the rectangle ABDE) + (Area of the triangle ⎛ ⎞ BCD) = (15w) + ⎜ (4)(w) ⎟ = 15w + 2w = 17w Since the total ⎝ ⎠ area is given to be 340 square feet, then 17w = 340 or 340 w= = 20 ft 17 € C ft B θ D θ 15 ft A E w B 42 € Triangles ABC and DBE are similar, and hence x 12 = = 8+ x+4 y Note that € A Trigonometric Ratios and Right Triangles € of protractor € 2.€ With the help and ruler € with θ€= 20° and c = 10 cm, we arrive at b ≈ 3.4 and α = 10 − (3.4) ≈ 9.4 from which we 3.4 9.4 have sin θ ≈ = 0.34 and cos θ = = 0.94 10 10 € € D θ 8.0 θ E 12 y (12) 144 x 12 €8 = € implies y = = = 18, and = implies 8x + 32 = 12x or x = 12 8 12 x+4 y EXERCISE 1.3 € x € 4.0 C Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Exercise 1.3 Trigonometric Ratios and Right Triangles With the help of protractor and ruler with θ = 40° and c = 10 cm, we arrive at b ≈ 6.4 and α = 10 − (6.4) ≈ 7.7 from which we 6.4 7.7 have sin θ ≈ = 0.64 and cos θ = = 0.77 10 10 € With the help of protractor and ruler with θ = 60° and c = 10 cm, we€arrive at α ≈ 5.0 and b € = 10 − 52 ≈ 8.7 from which we have sin θ ≈ 8.7 5.0 = 0.87 and cos θ = = 0.50 10 10 € With the help of protractor and ruler with θ = 80° and c = 10 cm, € € we arrive at α ≈ 1.7 and b = 10 − (1.7) ≈ 9.8 from which we 9.8 1.7 have sin θ ≈ = 0.98 and cos θ = = 0.17 10 10 € € € With the help of protractor and ruler with θ = 18° and α = 10 cm, 3.2 we arrive at b ≈ 3.2 for which tan θ ≈ = 0.32 10 10 € With the help of protractor and ruler with θ = 27° and α = 10 cm, we arrive at b ≈ 5.1 for which 5.1 tan θ ≈ = 0.51 10 12 14 With the help of protractor and ruler with θ = 36° and € α = 10 cm, we arrive at b ≈ 7.3 for which 7.3 tan θ ≈ = 0.73 10 € 16 With the help of protractor and ruler with θ = 54° and α = 10 cm, we arrive at b ≈ 13.8 for which 13.8 tan θ ≈ = 1.38 10 18 € Set calculator in degree mode and use sin key sin 75.6° = 0.969 c Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Chapter 20 Right Triangle Ratios Set calculator in degree mode, convert to decimal degrees and use tan key ⎛ 25 ⎞ ° 88°25ʹ′ = ⎜88 + ⎟ = (88.41666666…)° ⎝ 60 ⎠ tan 88°25ʹ′ = tan(88.41666666…)° ≈ 36.2 Use the reciprocal relation sec θ = Set calculator in degree mode and use cos key, then take € cos θ reciprocal sec 15.1° = ≈ 1.04 cos15.1° € and use cos key 24 Set calculator in degree mode cos 44.8° = 0.710 € calculator in degree mode, use tan key, then take reciprocal 26 Set ≈ 2.32 cot 23.3° = tan 23.3° 22 28 Set calculator in degree mode, convert to decimal degrees, sin θ and € use sin key, then take reciprocal ⎛ 12 ⎞ ° 2°12ʹ′ = ⎜ + ⎟ = 2.2° ⎝ 60 ⎠ € csc 2°12ʹ′ = csc 2.2° = ≈ 26.0 sin 2.2° Use the reciprocal relation csc θ = 30 €If tan θ = 1, then θ = tan–1 = 45° 34 32 If tan θ = 2.25, then θ = tan–1 2.25 = 66.04° € θ = arc cos 0.2557 = 75.18° ≈ 75°10' 36 θ = cos–1(0.0125) = 89.284° ≈ 89°17' 38 The triangle is uniquely determined The other angle can be found by subtracting the given angle from 90° The other two sides can be found by using a trigonometric ratio that involves either of the two acute angles and a given side 40 The triangle is not uniquely determined There are infinitely many triangles with the same hypotenuse Select side a as any number between and c and construct a triangle with side a and the given hypotenuse 42 Solve for the complementary angle: 90° – θ = 90° – 62°10' = 27°50' ⎛ 10 ⎞ ° Solve for b: Since θ = 62°10' = ⎜ 62 + ⎟ = (62.1666…)° and c = 33.0 cm, we look for a ⎝ 60 ⎠ trigonometric ratio that involves θ and c (the known quantities) and b (the unknown quantity) We choose the sine b € sin θ = c b = c sin θ = (33.0 cm)(sin 62.1666…°) ≈ 29.2 cm Solve for a: We choose the cosine to find a Thus, a cos θ = c a = c cos θ = (33.0 cm)(cos 62.1666…°) ≈ 15.4 cm € Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Exercise 1.3 44 46 48 Trigonometric Ratios and Right Triangles Solve for the complementary angle: 90° – θ = 90° – 32.4° = 57.6° Solve for b: Since θ = 32.4° and a = 42.3 m, we look for a trigonometric ratio that involves θ and a (the known quantities) and b (the unknown quantity) We choose the tangent b tan θ = a b = a tan θ = (42.3 m) tan 32.4° = 26.8 m Solve for c: We choose the cosine to find c Thus, a cos θ = c a 42.3 m c = = = 50.1 m cos θ cos 32.4° € Solve for the complementary angle: 90° – θ = 90° – 44.5° = 45.5° Solve Since θ = 44.5° and a = 2.30 × 106 m, we look for a trigonometric ratio that involves θ € for b: € and a (the known quantities) and b (the unknown quantity) We choose the tangent b tan θ = a b = a tan θ = (2.30 × 106 m) tan(44.5°) = 2.26 × 106 m Solve for c: We choose the cosine to find c Thus, a cos θ = c a 2.30 × 10 in c = = = 3.22 × 106 m cos θ cos 44.5° € b 22.0 km Solve € for θ: € tan θ = a = 46.2 km = 0.4762 θ = tan–1 0.4762 = 25.46° = 25° + (0.46 × 60)' ≈ 25°30' (to the nearest 10') Solve for the complementary angle: 90° – θ = 90° – 25°30' = 64°30' b € Solve for c: We will use the sine Thus, sin θ = c c = 50 52 € a 134 m Solve for θ: cos θ = = = 0.7363 € € c 182 m –1 θ = cos 0.7363 = 42.6° Solve for the complementary angle: 90° – θ = 90° – 42.6° = 47.4° € € b Solve for b: We will use the tangent Thus, tan θ = a b = a tan θ = (134) tan (42.6°) ≈ 123 m € An error was made in computing β, which should be β = 90° – 32°10' = 57°50' So a = 27.8 cos(57°50') = 14.8003, as the top two lines on the graphing calculator screen indicated 10 b 22.0 km = ≈ 51.2 km sin θ sin 25.46° Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Chapter 54 Right Triangle Ratios (A) If θ = 34°, (sin θ)2 + (cos θ)2 = (sin 34°)2 + (cos 34°)2 = (0.55919…)2 + (0.82903…)2 = (B) If θ = 37.281°, (sin θ)2 + (cos θ)2 = (sin 37.281°)2 + (cos 37.281°)2 = (0.60572…)2 + (0.79567…)2 (C) If θ = 87°23'41", (sin θ)2 + (cos θ)2 = (sin 87°23'41")2 + (cos 87°23'41")2 = (0.99896…)2 + (0.04545.…)2 = 56 (A) If θ = 17°, tan θ – cot(90° – θ) = tan 17° – cot(90° – 17°) = tan 17° – cot(73°) = 0.3057 – 0.3057 = (B) If θ = 27.143°, tan θ – cot(90° – θ) = tan 27.143° – cot(90° – 27.143°) = tan 27.143° – cot 62.857° = 0.51267 – 0.51267 = (C) If θ = 14°12'33", tan θ – cot(90° – θ) = tan(14°12'33") – cot(90° – 14°12'33") = tan(14°12'33") – cot(75°47'27") = 0.2532 – 0.2532 = 58 Solve for the complementary angle: 90° – θ = 90° – 35°44' = 54°16' b Solve for b: We choose the tangent to find b Thus, tan θ = a b = a tan θ = (6.482 m)(tan 35°44') = 4.664 m € a Solve for c: We choose the cosine to find c Thus, cos θ = c a 6.482 m c = = = 7.985 m cos θ (cos 35°44ʹ′) 60 Solve for θ: cos θ = 62 Solve for θ: tan θ = 64 (A) In right triangle OAD, cos θ = € a 123.4 ft = = 0.7533; θ = cos–1 0.7533 = 41.12° = 41°7' c 163.8 ft € € Solve for the complementary angle: 90° – θ = 90° – 41°7' = 48°53' b Solve for b: We choose to use the tangent to find b Thus, tan θ = a € € b = a tan θ = (123.4 ft)(tan 41°7') = 107.7 ft € b 5.207 mm = = 0.6484; θ = tan–1 0.6484 = 32.96° a 8.030 mm Solve for the complementary angle: 90° – θ = 90° – 32.96° = 57.04° b Solve for c: We will use the sine Thus, sin θ = c € € b 5.207 mm = = 9.570 mm c = sin θ (sin 32.96°) Adj€ = Hyp € Adj (B) In right triangle ODE, cot θ = = Opp € € (∠OED = θ) € € € OA OA = = OA OD € DE DE = = DE OD € 11 Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Exercise 1.3 Trigonometric Ratios and Right Triangles (C) In right triangle OCD, cos θ = Thus OC = 66 OD Adj = = OC Hyp OC = sec θ cos θ € € € (A) As θ approaches 90°, OA approaches O and cos θ = OA approaches O € (B) As θ approaches 90°, cot θ = DE approaches O (C) As θ approaches 90°, cos θ approaches O (see part (A)) and hence sec θ = increases cos θ without bound 68 (A) As θ approaches O ° , AD approaches O and hence sin θ = AD approaches O € (B) As θ approaches O ° , tan θ = DC approaches O (C) As θ approaches O ° , csc θ = OE increases without bound 70 E AREA = (Area of the square) + (Area of the triangle BEC) ⎛ ⎞ = (x2) + ⎜ (7)(24) ⎟ ⎝ ⎠ 24 m 7m B C x From the right triangle BEC we have x2 = (7)2 + (24)2 = 625, € therefore the total area is 625 + 84 = 709 m2 x A D 72 Total area = (Area of ∆ACB) + (Area of rectangle BCDE) + (Area of ∆EDF) ⎛ ⎞ ⎛ ⎞ = ⎜ yz ⎟ + (7z) + ⎜ xz ⎟ ⎝ ⎠ ⎝ ⎠ We need to find x, y and z From Triangle ABC we have y2 + z2 = 25 and from triangle EDF we have x2 + z2 = 121 From these two equations we arrive at x2 – y2 = 121 – 25 = 96 or (x – y)(x + y) = 96 We know that x + y + = 19 or cm B E € € x + y = 12 and hence 12(x– y) = 96 or x – y = 11 cm cm z ⎧ x + y = 12 Thus, ⎨ implies 2x = 20 or x = 10 and ⎩ x − y = A y C D 19 cm consequently y = and z = 25− y = 25 − = 21 So, the total area of the polygon is ⎛ ⎛ ⎞ ⎞ € ⎜ (2)( 21) ⎟ + (7 21 ) + ⎜ (10)( 21) ⎟ = 21 + 21 + 21 = 13 21 cm2 ⎝ ⎝ ⎠ € € € ⎠ € € € € € 12 € € x F Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Chapter Right Triangle Ratios 74 The area inscribed in a circle of radius is equal to ⋅ (area of triangle OAB) The right triangle O × B has angle ∠O = 30° and hence Ox = cos 30° = So, the area of the triangle ⎛ 3 ⎞ OAB is ⎜⎜1× and consequently the area of the circle ⎟ = ⎝ ⎟⎠ € ⎛ ⎞ 3 2.598 inside the hexagon is ⎜⎜ ⎟⎟ = ≈ 2.598 which is = 0.827 π ⎝ ⎠ € € € 82.7% of the area of the circle of radius (area of the circle of or radius r is πr2) € € € EXERCISE 1.4 Right Triangle Applications From the drawing, we note that 14.5 sin θ = 18 or ⎛ ⎛ 14.5 ⎞⎞ ⎛ 180° ⎞ θ ° = ⎜sin−1⎜ ⎟⎟ ⎜ ⎟ = 54° ⎝ 18 ⎠⎠ ⎝ π ⎠ ⎝ € Let α be the angle € of elevation of the sun Then € 14 tan(90˚ – α) = 21 ⎛ 14 ⎞ or 90˚ – α = tan–1 ⎜ ⎟ ≈ 34˚ ⎝ 21 ⎠ or α € ≈ 56˚ From the figure on the right we have h = 400 € sin 28.5˚ ≈ 191 ft 400 ft h 28.5˚ From the drawing, we note that x tan 64° 30ʹ′ = 38.4 or x = 38.4 tan(64.5° ) = 80.5 m € 13 Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Exercise 1.4 10 Right Triangle Applications From the triangle ABC, we note that 3, 000 x= ≈ 5,000 ft cos 53° € 12 From the triangle ABC, we note that x = tan 84° 21 or x = 21 tan° ≈ 200 m € 14 18°20' 70.0 m x 16 tan 28° = = y θ Opp Adj 70.0 m y 70.0 m y = = 223 m sin18°20ʹ′ (It is clear that θ = 18°20'.) € sin 18°20' = € A x 4.0 × 103 m 28° x 4.0 × 10 m € x = (4.0 × 103 m)(tan 28°) = 2.1 × 103 m 18 20 22 B We first sketch a figure and label the known parts € h tan 23.4° = 1.00 km h = (1.00 km)(tan 23.4°) = 433 m We first sketch a figure and label the known parts € 3, 600 ft Opp tan θ = = = 0.387 9, 300 ft Adj θ = tan–1 0.387 ≈ 21° (A) In triangle ABC, ∠θ is complementary to 82°, € € thus θ = 8° BC = x = roof overhang AC = 19 ft x Opp tan θ = = 19 ft Adj x = (19 ft)(tan 8°) = 2.7 ft € h 23.4° = α b = 1.00 km 3,600 ft θ 9,300 ft C θ' 19 ft A' B θ 82° 35° A € 14 C Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Chapter Right Triangle Ratios (B) In triangle A'BC, θ' = 35° A'C = y = how far down shadow will reach y Opp tan 35° = = 2.7 ft Adj y = (2.7 ft)(tan 35°) = 1.9 ft 24 26 28 We note: by the symmetry of the cone, ATC is an isosceles right triangle,€hence € ∠TAC = ∠TCA = 45° Since the mast is perpendicular to the deck, TBA and TBC are also right triangles, and since each has a 45° angle, these are also isosceles right triangles Then it follows that TB = AB and TB = BC Since the length AC = 100 feet, the top of the lightning rod should be TB = 50 feet above the water (A) We note that since ∠TSC = 90° – α€, hence ∠C = α r Adj €Thus, in triangle CST, cos α = Hyp€ = r + h (B) (r + h)cos α = r r cos α + h cos α = r h cos α = r – r cos α € = r(1 – cos α) r(1€– cos α ) h = cos α 3.75 km sin 33° = x 3.75 km x = = 6.9 km sin 33° € 18' C y € 8.0' 68° A x B S h α T r r C Earth We note that since ABC is an isosceles triangle, 1 € ∠FCA = ∠BCA = (2θ) = θ, and also, 1 AF = AB = (7.5 km) = 3.75 km 2 € € We are to find AC, the radius of the circle € C D (3,960 mi)(1 – cos 24°14' ) = 383 mi (cos 24°14' ) € In right triangle AFC, AF € sin€∠FCA = AC 3.75 km sin θ = x B A Label the required sides AD and AB Then AD = AC + CD = AC + 18 In right triangle ABC, BC BC sin A = AB tan A = AC 8.0 feet 8.0 feet sin 68° = tan 68° = x y 8.0 feet 8.0 feet x = = 8.6 feet y = = 3.2 feet sin 68° tan 68° (C) h = 30 T 15 B 2θ C θ P F x } 12 A B A Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Exercise 1.4 32 34 Right Triangle Applications In the figure, r denotes the radius of the parallel of latitude, R the radius of the r Adj earth, i.e., the radius of the equator Clearly, cos θ = = R Hyp Since L = 2πr and E = 2πR, we have r = R cos θ 2πr = 2πR cos θ € € L = E cos θ = (40,100 km)(cos 66°33') = 1.60 × 104 km R Equator (A) The lifeguard runs a distance d – y = d – c cot θ and then swims a distance x = c csc θ Thus the distance D covered by the lifeguard from the tower to the distressed swimmer is D = (d – c cot θ) + c csc θ (B) From Problem 33C, d = 380 m, c = 76 m, and θ = 51° Thus, for these values D = ((380 m) – (76 m)(cot 51°)) + (76 m)(csc 51°) = 416 m (C) 227.96 sec The shortest distance requires the lifeguard to travel entirely through the water, and his rate in the water is about of that of his rate on the beach From the table in Problem 33(D), T has a minimum value of 116.66 sec for θ = 70°, so his best bet is to run on the beach first, then swim € 36 r θ θ Distressed swimmer x P L Lifeguard tower d c θ y B Water Beach (A) We note that the pipeline consists of ocean section TP, and shore section PW = 10 mi – SP mi Let x = TP and y = SP, then L = x + (10 mi – y) In right triangle SPT, cos θ = , x y tan θ = and hence x = (4 mi) sec θ, y = (4 mi) tan θ mi Thus, L = (4 mi) sec θ + ((10 mi) – (4 mi) tan θ), € or L = sec θ + (10 – tan θ) € (B) For θ = 35°, L = sec 35° + (10 – tan 35°) ≈ 4.88 + 7.20 = 12.08 mi (C) The length of the shortest pipeline is mi TP = + 10 = 10.77 mi entirely in the water The cost will be C = (10.77)(40,000) = $430,800 Island T θ x y P W S Shore 10 mi €The shortest pipeline is more costly, since it is entirely in the water, and the pipeline in the water costs twice as much as on land From the table in Problem 35(D), C has the minimum value of $338,600 when θ = 30°, so minimum cost requires the pipeline to be laid both on land and water 16 Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Chapter 38 From the drawing, we note that the triangle ABC gives us y = tan 18° x + 45 The triangle DBC gives us y = tan 23° or y = x tan 23° x € Substituting y = x tan 23° in the first equation, we obtain x tan 23° = tan 18˚ or x + 45 x tan 23˚ = x tan 18˚ + 45 tan 18˚ € Solving for x, we have: x= 40 € 45 tan18° ≈ 150 m tan 23° − tan18° € (A) From the right triangle ABC we have d+x cot α = and from the right triangle DBC, h x we have cot β = Thus h d + h cot β cot α = € h h cot α = d + h cot β € h(cot α – cot β) = d d € h = cot α – cot β B h α A (B) From Problem 39, α = 25°, β = 42°, and d = 1.0 km Thus 1.0 € km h= = 0.97 km cot 25° – cot 42° 42 € From the drawing, we note that the triangle ABC gives us € x = tan 42° or x = 65 tan 42° ≈ 59 ft 65 The triangle ABD gives us x+y 59 + y = tan 55° or = tan 55° 65 65 Solving for y, we obtain y = 65 tan 55° – 59 ≈ 34 ft € € 17 Right Triangle Ratios d D β x C Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Exercise 1.4 44 46 Right Triangle Applications (A) From the right triangle ADB, B AD cot α = and from the right triangle BDC, h h DC cot β = Thus AD = h cot α and DC = h cot β, α β h A and hence d = AD + DC = h(cot α + cot β), D € d d h= cot α + cot β € (B) From Problem 43, α = 90° – 51.4° = 38.6°, β = 90° – 43.2° = 46.8°, and d = 847 ft Thus, 847 ft h= ≈ 386 ft € cot 38.6° + cot 46.8° We are given t = 1.50 sec, v = 12.4 ft/sec, θ = 15.0° Thus, € g= 48 v 12.4 ft/sec = ≈ 31.9 ft/sec2 (sin θ )t (sin15.0°)(1.50sec) We first sketch a figure and label the known parts From geometry we €know that€each angle of an equilateral triangle has measure 60° 5.0 cm sin 60° = x 5.0 cm x = ≈ 5.8 cm sin 60° x 5.0 cm 60° € 50 360° The angle α is = 22.5˚ € 16 x = tan α or 14 x = 14 tan 22.5˚ € Perimeter = 16x = (16)(14 tan 22.5˚) ≈ 93 yd x 22.5˚ € 52 We note that since AB is a side of a nine-sided regular polygon, ∠BCA = (360°) = 40° Since ABC is an isosceles triangle, 1 ∠FCA = ∠BCA = (40˚) = 20°, 2 and€also, AC = 8.32 cm Let€x = CF = radius € of a circle inscribed in the polygon x cos 20° = 8.32 cm x = (8.32 cm)(cos 20°) = 7.82 cm € 18 x α 14 B C 20° F 8.32 cm A C Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Chapter 54 AD = AO + OD = x + r In the right triangle ADC, cot 20° = Right Triangle Ratios x+r 0.89 A 20° or x + r = 0.89 cot 20° In the right triangle AEO, € r = r B From the figure on the right we observe that the distance from earth to sun (denoted by x) is 250, 000 250, 000 250,000 x = = = cos θ 0.0029 cos(89°50ʹ′) = 86,206,897 ≈ 90,000,000 miles (rounded to the nearest 10,000,000 miles) € € 19 C 0.89 cot 20° ≈ 0.632 cm 1+ csc 20° 360° Each central angle is = 60˚, so all the sides of the € triangle ABC are equal This implies that x = 2(20.0) = 40.0 cm From the right triangle Aʹ′Bʹ′C ʹ′ , we have € ⎛ y ⎞2 ⎜ ⎟ = (20)2 – (10)2 = 300 ⎝ ⎠ € or y2 = 1,200 or y ≈ 35 cm € D 0.89 € 58 E r O r + r = 0.89 cot 20° sin 20° r(1 + € csc 20°) = 0.89 cot 20° € 56 x € r r sin 20° = or x = Thus, x sin 20° B 60˚ x A B' A' 20.0 60˚ 60˚ C y C' .. .Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Exercise 1.1 28 There are two methods a Convert the first one to decimal hour form and compare with the... € 4.0 C Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett Exercise 1.3 Trigonometric Ratios and Right Triangles With the help of protractor and ruler with θ... Solve for a: We choose the cosine to find a Thus, a cos θ = c a = c cos θ = (33.0 cm)(cos 62.1666…°) ≈ 15.4 cm € Solution Manual for Analytic Trigonometry with Applications 11th Edition by Barnett