Results 1 to 3 of 3

Thread: Arcs/Arc Length

  1. #1
    Junior Member
    Joined
    Sep 2009
    From
    Russia
    Posts
    47

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    Thanks
    1
    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 $\displaystyle \theta$ is measured in radians, then the formulae for arc length, $\displaystyle s$, and area of sector, $\displaystyle A$, are:
    $\displaystyle s =r\theta$

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

    So:

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

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

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

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

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

    Grandad
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    From
    Russia
    Posts
    47
    Thank you so much, I appreciate the help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Arcs
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Jul 1st 2010, 05:24 AM
  2. Angles and Arcs
    Posted in the Geometry Forum
    Replies: 8
    Last Post: Jul 15th 2009, 04:50 AM
  3. Arcs and Chords
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 6th 2008, 10:00 PM
  4. Clock and arcs
    Posted in the Geometry Forum
    Replies: 1
    Last Post: Apr 30th 2008, 04:54 PM
  5. length of arcs
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: Oct 17th 2007, 03:49 AM

Search Tags


/mathhelpforum @mathhelpforum