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

Convert the first one to decimal hour form and compare with the second one.. Convert the second one to HMS form and compare with the first one.. To compare α and β, we convert α to decim

Trang 1

1

EXERCISE 1.1 Angles, Degrees, and Arcs

2 Since one complete revolution has measure 360°,

€

1

4 revolution has measure

€

1

4 (360°) = 90° Similarly,

€

1

3(360°) = 120°,

€

7

8(360°) = 315°,

€

11

12(360°) = 330°

4 Since one complete revolution has measure 360°, the fraction of a counterclockwise revolution that will form a 30° angle is

€

30

360 =

€

1

12 Similarly,

€

225

360 =

€

5

8,

€

240

360 =

€

2

3

6 Since 88° angle is between 0° and 90°, this is an acute angle

8 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

€

1

60 of one degree; a second of angle measure is

€

1

60 of one minute

16 Since 43' =

€

43°

60 and 30" =

€

30°

3, 600, then 71°43'30" =

€

71+ 43

60+ 30

3, 600

⎛

⎝

⎠

⎟° ≈ 71.725° to three decimal places

18 Since 3' =

€

3°

60 and 1" =

€

1°

3, 600, then 9°3'1" =

€

9+ 3

60+ 1

3, 600

⎛

⎝

⎠

⎟° ≈ 9.050° to three decimal places

20 Since 11' =

€

11°

60 and 25" =

€

25°

3, 600, then 267°11'25" =

€

267+11

60+ 25

3, 600

⎛

⎝

⎠

⎟° ≈ 267.190° to three

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"

Trang 2

28 There are two methods

a Convert the first one to decimal hour form and compare with the second one

3 +

€

43

60 +

€

24

3, 600 ≈ 3.723 to three decimal places

Since 3.723 < 3.732, Runner A is faster

b Convert the second one to HMS form and compare with the first one

3.732 = 3 hr + (0.732 × 60) min

= 3 hr + 43.92 min

= 3 hr + 43 min + (0.92 × 60) sec

≈ 3 hr, 43 min, 55 sec

This is bigger than 3 hr, 43 min, 24 sec, and hence Runner A is faster

30 To compare α and β, we convert α to decimal form Since 51' =

€

51°

60 and 54" =

€

54°

3, 600, then 32°51'54" =

€

32+51

60+ 54

3, 600

⎛

⎝

⎠

⎟° = 32.865° Thus α = β

32 To compare α and β, we convert β to decimal form Since 40' =

€

40°

60 and 20" =

€

20°

3, 600, then 80°40'20" =

€

80+40

60+ 20

3, 600

⎛

⎝

⎠

⎟° ≈ 80.672° Thus α < β

34 We convert β to decimal form Since 18' =

€

18°

60 and 32" =

€

32°

3, 600, then 242°18'32" =

€

242+18

60+ 32

3, 600

⎛

⎝

⎠

⎟° ≈ 242.309° Thus α > β

36 105°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 2 πr If 2πr = 6, then estimate of r to the nearest whole number would be 1 The value of r to two decimal places is r =

€

6 2π ≈ 0.95

42 The arc length of a semicircle that has diameter 10 is

€

1

2C =

€

1

2 (2πr) =

€

1

2(10π) = 5π An estimate

of 5π to the nearest whole number is 16 The length of the semicircle of radius 5 (or diameter 10) to

two decimal places is 15.71

44 Since

€

s

C =

€

θ

€

s

C =

€

θ

360°, then

€

12

108 =

€

θ

€

s

740 =

€

72°

360°

θ =

€

12 108

⎛

⎝

⎜ ⎞

⎠

€

72

360(740) = 148 mi

Exercise 1.1 Angles, Degrees, and Arcs

.

Trang 3

3

48 Since

€

s

C =

€

θ

360° and C = 2πr, then

€

s

2πr =

€

θ

360°

€

38, 000 cm 2πr =

€

45.3°

360°

r =

€

(38, 000 cm)(360) (2π)(45.3) ≈ 48,000 cm (to the nearest 1,000 cm)

50 Since

€

s

C =

€

θ

360° and θ = 24°16'34" =

€

24+16

60+ 34

3, 600

⎛

⎝

⎠

⎟° ≈ 24.276°, then

€

14.23 m

C ≈

€

24.276°

360°

C ≈

€

(14.23 m)(360) 24.276 ≈ 211.0 m (to one decimal place)

52 Since

€

A

πr2

=

€

θ

360°, then

€

A

π(7.38 ft)2

=

€

24.6°

360°

A =

€

24.6

360(π)(7.38 ft)2 ≈ 11.7 ft2 (to one decimal place)

54 Since

€

A

πr2 =

€

θ

360°, then

€

347 in2 π(32.4 in)2 =

€

θ

360°

θ =

€

347 π(32.4)2 · 360° ≈ 37.9°

58 Since

€

s

C =

€

θ

€

s

2πr =

€

θ

360°

θ =

€

s

2πr · 360° =

€

12.1 mm 2(π)(5.26 mm) · 360° ≈ 131.8°

60 Since

€

s

C =

€

θ

€

s

2πr =

€

θ

360°

r =

€

s

2π ·

€

360°

θ =

€

11.8 mm 2(π) ·

€

360 117.9 ≈ 5.73 mm

56 Note that the central angles corresponding

to the arcs AB and CD are 72° and hence

∠A = 36°, ∠D = 36° Since the sum of the

angles in a triangle is 180°, the angle ∠x should be 180° - (36° + 36°) = 108°

Trang 4

62 Since

€

s

C =

€

θ

€

s

2πr =

€

θ

360°

θ = 40°40' – 33°30' = 7°10' =

€

7+10

60

⎛

⎝

⎜ ⎞

⎠

⎟°

s = 2πr ·

€

θ

360° = 2(π)(3,960 mi) ·

€

7+10 60

360 ≈ 495 mi

64 Since

€

s

C =

€

θ

€

s

2πr =

€

θ

360°

θ = 42°50' – 36°0' = 6°50' =

€

6+50

60

⎛

⎝

⎜ ⎞

⎠

⎟°

s = 2πr ·

€

θ

360° = 2(π)(3,960 mi) ·

€

6+50 60

360 ≈ 472 mi

66 To find the length of s in nautical miles, since 1 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 1 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 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 1 minute of arc is

€

π

3 in ≈ 1.05 in at 100 yd; the diameter of the quarter is slightly less than 1 in

72 Since

€

s

C =

€

θ

360° and C = 2 πr, then

€

s

2πr =

€

θ

360°

θ =

€

s

2πr · 360˚ =

€

864, 000 2(π)(1,780,000,000) · 360° ≈ 0.028°

74 Since

€

s

C =

€

θ

€

s

2πr =

€

θ

360°

r =

€

s

2π ·

€

360°

θ =

€

52 2(π) ·

€

360°

0.075° ≈ 40,000 miles

Exercise 1.1 Angles, Degrees, and Arcs

.

Trang 5

5

EXERCISE 1.2 Similar Triangles

2 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'

4 (10)(0.33) = 3.3 and according to Appendix A.3, the correct rounded value is 3

6 Since

€

b

ʹ′

b =

€

c

ʹ′

c by Euclid's Theorem, then

€

3

1 =

€

24 ʹ′

c , c' =

€

(24)(1)

3 = 8

8 Since

€

a

ʹ′

a =

€

b

ʹ′

b by Euclid's Theorem, then

€

51

17 =

€

b

8, b =

€

(51)(8)

17 = 24

10 Since

€

a

ʹ′

a =

€

b

ʹ′

b by Euclid's Theorem, then

€

640, 000

15 =

€

b

0.75, b =

€

(640, 000)(0.75)

15 = 32,000

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

0.47 =

€

32 cm 1.0

€

c

1.1 =

€

32 cm 1.0

a =

€

(0.47)(32 cm)

€

(1.1)(32 cm) 1.0 ≈ 35 cm

€

b

1.0 =

€

63.19 cm 0.47

€

c

1.1 =

€

63.19 cm 0.47

b =

€

63.19 cm

€

(1.1)(63.19 cm) 0.47 ≈ 150 cm

€

a

0.47 =

€

1.037× 1013m 1.0

€

c

1.1 =

€

1.037× 1013m 1.0

a =

€

(0.47)(1.037× 1013m)

1.0 ≈ 4.9 × 1012 m c = (1.1)(1.037 × 1013 m) ≈ 1.1 × 1013 m

€

a

0.47 =

€

2.86× 10–8cm 1.1

€

b

1.0 =

€

2.86× 10–8cm 1.1

a =

€

(0.47)(2.86× 10–8cm)

1.1 ≈ 1.2 × 10–8 cm b =

€

2.86× 10–8cm 1.1 ≈ 2.6 × 10–8 cm

Trang 6

22 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

4.67 in =

€

36 ft 3.00 in

c ≈

€

4.67 3.00(36 ft) ≈ 56 ft

Drawing not to scale

28 In the drawing, we note that triangles LBT and LNM are similar

MN = 8.5 ft

NB =

€

1

2(length of court) =

€

1

2(78 ft) = 39 ft TB = 3 ft BL is to be found Let

BL = x Then,

€

BL

NL =

€

TB

MN

NL = NB + BL Thus,

€

x

39+x =

€

3 8.5

8.5x = 3(39 + x) = 117 + 3x

5.5x = 117

x =

€

117 5.5 ≈ 21.27 ft

3 ft 8.5 ft

M

N

B

39 ft

39 + x ft

T

Drawing not to scale

30 Since the triangles ABC and DEC in the figure are similar, we can write

€

AB

DE =

€

AC

CD Then,

€

AB

5.75 ft =

€

25 ft 2.25 ft

AB =

€

5.75 2.25(25 ft) ≈ 64 ft

32 In the drawing, we note that triangles ABC and ABC are similar So,

€

ʹ′

B C ʹ′

BC =

€

ʹ′

A B ʹ′

BC

€

x

6 =

€

7

8 or x =

€

42

8 ≈ 5 ft

34 In the drawing, we note that triangles ABC and ABC are similar So,

€

A B ʹ′

AB =

€

ʹ′

B C ʹ′

BC

€

x

x+25 =

€

5 12

or

12x = 5x + 125 7x = 125

x =

€

ʹ′

A B ʹ′

BC ≈ 18 ft

Exercise 1.2 Similar Triangles

.

Trang 7

7

36 Let us make a scale drawing of the figure in the text as

follows: pick any convenient length, say 2 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

1.06 in =

€

4.0× 103m

2 in

x =

€

1.06

2 (4.0 × 103 m) ≈ 2.1 × 103 m

C'

B'

Measure (approx 1.06 in) 28°

38 (A) u(m) = 10 or u(mm) = 10,000 and

hence

€

1

10, 000 +

€

1

v =

€

1

50,

€

1

v =

€

1

50 –

€

1

10, 000 =

€

199

10, 000 and

v =

€

10, 000

199 ≈ 50.25 mm Continuing in this manner we obtain

u(mm) 10,000 20,000 30,000 40,000 50,000 60,000

v(mm) 50.25 50.2 50.13 50.08 50.06 50.05 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) +

€

1

2(4)(w)

⎛

⎝

⎠

⎟ = 15w + 2w = 17w Since the total area is given to be 340 square feet, then 17w = 340 or

w =

€

340

17 = 20 ft

A

15 ft

4 ft

E

C

w

42 Triangles ABC and DBE are similar, and hence

€

8

8+4 =

€

x

x+4 =

€

12

y

E

B x D

y

8 0

4.0

θ θ

€

8

12 =

€

12

y implies y =

€

(12)2

8 =

€

144

8 = 18, and

€

8

12 =

€

x

x+4 implies 8x + 32 = 12x or x = 8

EXERCISE 1.3 Trigonometric Ratios and Right Triangles

2 With the help of protractor and ruler with θ = 20° and c = 10 cm,

we arrive at b ≈ 3.4 and α =

€

102− (3.4)2 ≈ 9.4 from which we have sin θ ≈

€

3.4

10 = 0.34 and cos θ =

€

9.4

10 = 0.94

Trang 8

we arrive at b ≈ 6.4 and α =

€

6.4

10 = 0.64 and cos θ =

€

7.7

10 = 0.77

we arrive at α ≈ 5.0 and b =

€

102− 52 ≈ 8.7 from which we have sin θ ≈

€

8.7

10 = 0.87 and cos θ =

€

5.0

10 = 0.50

we arrive at α ≈ 1.7 and b =

€

9.8

10 = 0.98 and cos θ =

€

1.7

10 = 0.17

10 With the help of protractor and ruler with θ = 18° and α = 10 cm,

we arrive at b ≈ 3.2 for which tan θ ≈

€

3.2

10 = 0.32

12 With the help of protractor and ruler with θ = 27° and

α = 10 cm, we arrive at b ≈ 5.1 for which

tan θ ≈

€

5.1

10 = 0.51

tan θ ≈

€

7.3

10 = 0.73

tan θ ≈

€

13.8

10 = 1.38

18 Set calculator in degree mode and use sin key

sin 75.6° = 0.969

c

Exercise 1.3 Trigonometric Ratios and Right Triangles

.

Trang 9

9

20 Set calculator in degree mode, convert to decimal degrees and use tan key

88°25ʹ′ =

€

88+25

60

⎛

⎝

⎠

⎟° = (88.41666666…)°

tan 88°25ʹ′ = tan(88.41666666…)° ≈ 36.2

22 Use the reciprocal relation sec θ =

€

1 cosθ Set calculator in degree mode and use cos key, then take reciprocal

sec 15.1° =

€

1 cos15.1° ≈ 1.04

24 Set calculator in degree mode and use cos key

cos 44.8° = 0.710

26 Set calculator in degree mode, use tan key, then take reciprocal

cot 23.3° =

€

1 tan 23.3° ≈ 2.32

28 Use the reciprocal relation csc θ =

€

1 sinθ Set calculator in degree mode, convert to decimal degrees, and use sin key, then take reciprocal

2°12ʹ′ =

€

2+12

60

⎛

⎝

⎜ ⎞

⎠

⎟° = 2.2°

csc 2°12ʹ′ = csc 2.2° =

€

1 sin 2.2° ≈ 26.0

30 If tan θ = 1, then θ = tan–1 1 = 45° 32 If tan θ = 2.25, then θ = tan–1 2.25 = 66.04°

34 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 0 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'

Solve for b: Since θ = 62°10' =

€

62+10

60

⎛

⎝

⎠

⎟° = (62.1666…)° and c = 33.0 cm, we look for a

trigonometric ratio that involves θ and c (the known quantities) and b (the unknown quantity) We

choose the sine

sin θ = b 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,

cos θ =

€

a c

a = c cos θ = (33.0 cm)(cos 62.1666…°) ≈ 15.4 cm

Trang 10

44 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

tan θ = b 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,

cos θ =

€

a c

c =

€

a

cosθ =

€

42.3 m cos 32.4° = 50.1 m

46 Solve for the complementary angle: 90° – θ = 90° – 44.5° = 45.5°

Solve for b: Since θ = 44.5° and a = 2.30 × 106 m, we look for a trigonometric ratio that involves θ

and a (the known quantities) and b (the unknown quantity) We choose the tangent

tan θ = b 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,

cos θ =

€

a c

c =

€

a

cosθ =

€

2.30× 106in cos 44.5° = 3.22 × 106 m

48 Solve for θ: tan θ = b a =

€

22.0 km 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'

Solve for c: We will use the sine Thus, sin θ =

€

b c

c =

€

b

sinθ =

€

22.0 km sin 25.46° ≈ 51.2 km

50 Solve for θ: cos θ =

€

a

c =

€

134 m

182 m = 0.7363

θ = cos–1 0.7363 = 42.6°

Solve for the complementary angle: 90° – θ = 90° – 42.6° = 47.4°

Solve for b: We will use the tangent Thus, tan θ =

€

b a

b = a tan θ = (134) tan (42.6°) ≈ 123 m

52 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

Exercise 1.3 Trigonometric Ratios and Right Triangles

.

Định dạng
Số trang	19
Dung lượng	794,82 KB