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Math Help - Finding The Angle In Radians

  1. #1
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    Finding The Angle In Radians

    1. In a circle with radius 12 cm, an arc of length 20 cm subtends a central angle of theta. Determine the measure of theta in radians.

    Answer: 1.67 (Rounded to two decimals)

    My work:
    (theta in radians) x Radius = arclength
    (theta in radians) x 12 = 20
    theta in radians = 20/12
    Unsure how to continue...

    2. Solve: 7tan(x) = -3 0 <= x <= 2pi

    Answer: 2.74, 5.88
    (Rounded to two decimals)

    My work:
    tan(x) = -3/7

    3. For the function f(x) = 3sin(bx) + d, where b and d are positive constants, determine an expression for the smallest positive value of x that produces the maximum value of f(x)


    Answer: pi/2b


    How and Why?
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  2. #2
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    Quote Originally Posted by AlphaRock View Post
    1. In a circle with radius 12 cm, an arc of length 20 cm subtends a central angle of theta. Determine the measure of theta in radians.

    Answer: 1.67 (Rounded to two decimals)

    My work:
    (theta in radians) x Radius = arclength
    (theta in radians) x 12 = 20
    theta in radians = 20/12
    Unsure how to continue...
    \frac{20}{12} = \frac{5}{3} \approx 1.67

    2. Solve: 7tan(x) = -3 0 <= x <= 2pi

    Answer: 2.74, 5.88
    (Rounded to two decimals)

    My work:
    tan(x) = -3/7
    x = \tan^{-1}\left(-\frac{3}{7}\right)
    x \approx -0.40

    You need answers in 0 \le x \le 2\pi, but when you take the inverse tangent, you get an answer in -\pi/2 \le x \le \pi/2 (because the latter is the range of the inverse tangent). Since our answer is negative, add \pi to it to get x \approx 2.74. Then add \pi to it again to get x \approx 5.88. Don't add \pi a third time, because you'll then get an answer greater than 2\pi.


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  3. #3
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    Quote Originally Posted by AlphaRock View Post
    3. For the function f(x) = 3sin(bx) + d, where b and d are positive constants, determine an expression for the smallest positive value of x that produces the maximum value of f(x)

    Answer: pi/2b


    How and Why?
    If you look at the basic sine function
    y = \sin x,
    the maximum value (for y) is 1, so the smallest positive value of x that gives the maximum value is
    x = \frac{\pi}{2}.

    Now, looking at your function
    f(x) = 3\sin (bx) + d,
    since there is no reflection on the x-axis (ie. no negative in front of the 3), and since b and d are positive, the smallest positive value of x that gives the maximum value is when what's inside the sine is \frac{\pi}{2}, or
    bx = \frac{\pi}{2}.
    Solve for x to get
    x = \frac{\pi}{2b}.


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