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  • Exercise 2.1 Degrees and Radians

    18

    Chapter 2 Trigonometric Functions

    EXERCISE 2.1 Degrees and Radians

    4. Since 45° corresponds to a radian measure of π/4 rad, we have:

    90° = 2 · 45° corresponds to 2 · π/4 or π/2 rad. 315° = 7 · 45° corresponds to 7 · π/4 or 7π/4 rad. 135° = 3 · 45° corresponds to 3 · π/4 or 3π/4 rad. 360° = 8 · 45° corresponds to 8 · π/4 or 2π rad. 180° = 4 · 45° corresponds to 4 · π/4 or π rad. 225° = 5 · 45° corresponds to 5 · π/4 or 5π/4 rad. 270° = 6 · 45° corresponds to 6 · π/4 or 3π/2 rad.

    6. The angle of degree measure 20 is smaller, since

    !

    1

    2 radian corresponds to a degree measure of

    !

    180

    2"

    which is approximately 28.6°.

    8.

    Coterminal angles have measures 135° + 360° = 495° and 135° – 360° = –225°

    10.

    Coterminal angles have measures 60° + 360° = 420° and 60° – 360° = –300°

    12.

    Coterminal angles have measures –140° + 360° = 220° and –140° – 360° = –500°

    14.

    Coterminal angles have measures

    !

    "

    3 + 2π =

    !

    7"

    3 rad and

    !

    "

    3 – 2π = –

    !

    5"

    3 rad

  • Chapter 2 Trigonometric Functions

    19

    16.

    Coterminal angles have measures

    !

    7"

    4 + 2π =

    !

    "

    4 rad and

    !

    7"

    4 – 2π = –

    !

    15"

    4 rad

    18.

    Coterminal angles have measures

    !

    6"

    5 + 2π =

    !

    16"

    5 rad and

    !

    6"

    5 – 2π = –

    !

    4"

    5 rad

    22. 1° corresponds to

    !

    "

    180 rad; 5° corresponds to 5 ·

    !

    "

    180 =

    !

    "

    36 rad ≈ 0.08727 rad

    24. 190° corresponds to 190 ·

    !

    "

    180 rad =

    !

    19"

    18 rad ≈ 3.316 rad

    26. 108° corresponds to 108 ·

    !

    "

    180 rad =

    !

    3"

    5 rad ≈ 1.885 rad

    28. 1 rad corresponds to

    !

    180

    "

    #

    $ %

    &

    ' ( °; 3 rad corresponds to

    !

    3 " 180

    #

    $

    % &

    '

    ( ) ° =

    !

    540

    "

    #

    $ %

    &

    ' ( ° ≈ 171.9°

    30. 1.4 rad corresponds to

    !

    (1.4)(180)

    "

    #

    $ %

    &

    ' ( ° =

    !

    252

    "

    #

    $ %

    &

    ' ( ° ≈ 80.21°

    32.

    !

    2"

    9 rad corresponds to

    !

    2"

    9 # 180

    "

    $

    % &

    '

    ( ) ° = 40° (Exact)

    34. Set calculator in degree mode. (A) (0.750) rad is 42.972° (B) (1.5) rad is 85.944° (C) (3.80) rad is 217.724° (D) (–7.21) rad is –413.103°

    36. Set calculator in radian mode. (A) 35° is 0.611 rad (B) 187° is 3.264 rad (C) 437° is 7.627 rad (D) –175.3° is –3.060 rad

    38. Since θ = s

    r , we have

    (A) θ =

    !

    15cm

    40 cm = 0.375 rad; 21.5° (B) θ =

    !

    65cm

    40 cm = 1.625 rad; 93.1°

    (C) θ =

    !

    123.5cm

    40 cm = 3.0875 rad; 176.9° (D) θ =

    !

    210 cm

    40 cm = 5.25 rad; 300.8°

  • Exercise 2.1 Degrees and Radians

    20

    40. (A) s = rθ (B) s = rθ

    = (5)(0.228) = 1.14 in = (5)

    !

    37° " #

    180°

    $

    % &

    '

    ( ) = 3.23 in

    (C) s = rθ (D) s = rθ

    = (5)(1.537) = 7.685 in = (5)

    !

    175° " #

    180°

    $

    % &

    '

    ( ) = 15.27 in

    42. Since θr =

    !

    "

    180 θd , if both sides of this equation are cut in half, then

    !

    1

    2 θr =

    !

    "

    180 ·

    !

    1

    2 θd

    Thus if θd is cut in half, θr is also cut in half.

    44. Since θr =

    !

    s

    r , and if s (the numerator) is doubled while r is held constant, then θr will be doubled.

    46. (A) A =

    !

    1

    2 r2θ (B) A =

    !

    1

    2 r2θ

    =

    !

    1

    2 (10.5)2 (0.150) ≈ 8.27 ft2 =

    !

    1

    2 (10.5)2

    !

    15.0° " #

    180°

    $

    % &

    '

    ( ) ≈ 14.43 ft2

    (C) A =

    !

    1

    2 r2θ (D) A =

    !

    1

    2 r2θ

    =

    !

    1

    2 (10.5)2 (1.74) ≈ 95.92 ft2 =

    !

    1

    2 (10.5)2

    !

    105° " #

    180°

    $

    % &

    '

    ( ) ≈ 101.02 ft2

    48. Since θ =

    !

    s

    r , θ in radian measure, θ =

    !

    s

    1 = s.

    50.

    !

    9"

    4 = 2π +

    !

    "

    4 . So

    !

    9"

    4 is coterminal with 2π +

    !

    "

    4 – 2π =

    !

    "

    4 . Since

    !

    "

    4 is between 0 and

    !

    "

    2 , its

    terminal side lies in quadrant I.

    52. 696° is coterminal with (696 – 360)° = 336°. Since 336° is between 270° and 360°, its terminal side lies in quadrant IV.

    54. –

    !

    11"

    3 is coterminal with –

    !

    11"

    3 + 2π = –

    !

    5"

    3 . Since –

    !

    5"

    3 is between –

    !

    3"

    2 and –2π, its terminal side

    lies in quadrant I.

    56. –672° is coterminal with (–672 + 360)° = –312°. Since –312° is between –270° and –360°, its terminal side lies in quadrant I.

    58. –20 rad is approximately –1146° which is coterminal with –1146° + 360° = –786° and with –786° + 360° = –426° and with –426° + 360° = –66°. Since –66° is between 0° and –90°, its terminal side lies in quadrant IV.

    60. 116.9853° = (116.9853) ·

    !

    "

    180 rad = 2.0418 rad

    62. 2.562 rad = (2.562) ·

    !

    180

    " = 146.7918°

  • Chapter 2 Trigonometric Functions

    21

    64. 261°15'45" =

    !

    261+ 15

    60 +

    45

    3,600

    "

    # $

    %

    & ' ° = 261.2625° = (261.2625) ·

    !

    "

    180 = 4.560 rad

    66.

    !

    43"

    11 rad =

    !

    43"

    11 ·

    !

    180

    " = 703.6364°

    68. Since θ =

    !

    s

    r , we have

    !

    "

    2 =

    !

    s

    2r =

    !

    5

    2(4) = 0.625 rad

    70. Since θ =

    !

    180

    " # s

    r

    $

    % &

    '

    ( ) °, we have

    !

    "

    2 =

    !

    180

    2" # s

    r

    $

    % &

    '

    ( ) ° =

    !

    180

    2" # 34

    15

    $

    % &

    '

    ( ) ° = 65°

    72. s = rθ, where r = 52 km and θ = 2(27°) = 54°, s = (52)

    !

    (54°) " #

    180°

    $

    % &

    '

    ( ) ≈ 49 km

    74. θ = 2(0.25) = 0.5

    r =

    !

    s

    " =

    !

    26

    0.5 = 52 cm

    76. At 4:30, the minute hand has moved

    !

    1

    2 of a circumference from its position at the top of the clock.

    The hour hand has moved 4

    !

    1

    2 twelfths of a circumference from the same position. Therefore, they

    form an angle of

    !

    1

    2 C –

    !

    4 1

    2

    12 C radians,

    where C = a total circumference or 2π radians. Thus, the desired angle is

    !

    1

    2 (2π) –

    !

    4 1

    2

    12 (2π) = π –

    !

    9"

    12 =

    !

    3"

    12 =

    !

    "

    4 rad ≈ 0.79 rad

    78. We are to find a central angle θ, in a circle of radius 22 cm, when the arc length is 9.5 cm.

    s =

    !

    "

    180 rθ ; θ =

    !

    180s

    "r =

    !

    180(9.5)

    "(22) ≈ 25°

    80. Since s = rθ we have

    r =

    !

    s

    " =

    !

    18

    21° " # 180°

    ≈ 49 in

    82. Since s = rθ , we have

    s = (102 cm)

    !

    2"

    3 ≈ 214 cm

    84. Since s = rθ, we have

    diameter = s = (5.000 km)

    !

    2.44° " #

    180°

    $

    % &

    '

    ( ) ≈ 213 m

    86. Since s =

    !

    "

    180 rθ, we have

    width of field ≈ s =

    !

    "

    180 (865 ft)(2.5) ≈ 38 ft

  • Exercise 2.1 Degrees and Radians

    22

    88. ! = 5