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Math Help - Arcs/Arc Length

  1. #1
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    Exclamation Arcs/Arc Length

    Hey guys, another question for ya:

    0 = theda

    Suppose that 0 = 150 degrees

    a. Find the length of the circular arc corresponding to 0(theda) on a circle of radius 5.
    (For this question I got, 13.09, so could someone check to see if it's correct?)

    b. Find the area of the slice corresponding to 0(theda) on a circle of radius 5

    c. Find sin 0 and cos 0 (again 0 = theda)

    d. Find the angle that the radial line segment of the unit circle ending at (sin 0, cos 0) makes with the positive horizontal axis.

    I would GREATLY appreciate it if you guys could also provide the answers just for proofing and rechecking.
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  2. #2
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    Hello goliath
    Quote Originally Posted by goliath View Post
    Hey guys, another question for ya:

    0 = theda

    Suppose that 0 = 150 degrees

    a. Find the length of the circular arc corresponding to 0(theda) on a circle of radius 5.
    (For this question I got, 13.09, so could someone check to see if it's correct?)

    b. Find the area of the slice corresponding to 0(theda) on a circle of radius 5

    c. Find sin 0 and cos 0 (again 0 = theda)

    d. Find the angle that the radial line segment of the unit circle ending at (sin 0, cos 0) makes with the positive horizontal axis.

    I would GREATLY appreciate it if you guys could also provide the answers just for proofing and rechecking.
    If \theta is measured in radians, then the formulae for arc length, s, and area of sector, A, are:
    s =r\theta

    A = \tfrac12r^2\theta
    In this question \theta = 150^o=\frac{5\pi}{6} radians and r = 5.

    So:

    (a) Arc length s = r\theta=5\times \frac{5\pi}{6}\approx 13.09\text{ cm}. So you are right!

    (b) Area of sector A = \tfrac12r^2\theta=\frac{5^2\cdot5\pi}{2\cdot6}\app  rox32.7\text{ cm}^2

    (c) \sin\left(\frac{5\pi}{6}\right)=\sin\left(\pi - \frac{5\pi}{6}\right)=\sin\left(\frac{\pi}{6}\righ  t)=\frac12

    \cos\left(\frac{5\pi}{6}\right)=-\cos\left(\pi - \frac{5\pi}{6}\right)=-\cos\left(\frac{\pi}{6}\right)=-\frac{\sqrt3}{2}

    (d) We want the point whose coordinates are \left(\frac12,-\frac{\sqrt3}{2}\right). Sketch a diagram, then, and you'll see that the point is in the fourth quadrant, with the radial line making an angle 60^o below the x-axis. So the angle we want is -60^o (or 300^o), measuring angles positive anticlockwise.

    Grandad
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  3. #3
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    Thank you so much, I appreciate the help!
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