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Determine the lenght of arc and the area of the sector subtended by an

  1. Jan 6, 2012 #1
    Determine the lenght of arc and the area of the sector subtended by an angle of 60° in circle of radius 3 m

    Ok First change 60to radian measure. 60 x pi / 180° = 2pi / 6

    Then.. I used the formula s = rθ 3(2pi/6) = 6pi/6 = pi <--- Is this right??

    And how can I find the area which formula can I use?
     
  2. jcsd
  3. Jan 6, 2012 #2
    Re: Trigonometric

    You could just have kept the angle in degrees and worked out the arc length by using,

    [itex]\ c=pi*d[/itex]

    and then multiplied that answer by,

    [itex]\frac{60}{360}[/itex]
     
  4. Jan 6, 2012 #3
    Re: Trigonometric

    Ok but my answer is right?
     
  5. Jan 6, 2012 #4
    Re: Trigonometric

    Yes your answer is correct.

    That might give you a clue how to find the area of the sector. Think of the fraction of the total circles area you are looking for?
     
  6. Jan 6, 2012 #5
    Re: Trigonometric

    A = 60 / 360 x 2 ∏ 3 ??
     
  7. Jan 6, 2012 #6
    Re: Trigonometric

    Nearly. You're along the right lines but your calculation for the total area is wrong, [itex]\ a = pi * r^{2}[/itex]

    So for the sector area your equation should be,

    [itex] Area of Sector = \pi r ^ {2} * \frac{Angle of Sector°}{360°}[/itex]
     
    Last edited: Jan 6, 2012
  8. Jan 6, 2012 #7
    Re: Trigonometric

    Ok so A = 9 pi x 60 / 360 = 540 pi / 360 = 3pi / 2
     
  9. Jan 6, 2012 #8
    Re: Trigonometric

    Perfect.
     
  10. Jan 6, 2012 #9
    Re: Trigonometric

    Yay!! Hey do you know how to determine cos or sin without calculator?? like cos 75° any hint?
     
  11. Jan 6, 2012 #10
    Re: Trigonometric

    I found this on another website.

    For sin(x)

    [itex] x - \frac{x^ {3}}{3!} + \frac{x^ {5}}{5!} -\frac{x^ {7}}{7!} + ...[/itex]

    For cos(x)

    [itex] 1 - \frac{x^ {2}}{2!} + \frac{x^ {4}}{4!} -\frac{x^ {6}}{6!} +...[/itex]

    From a bit of quick testing here these series seem to converge to the right value fairly quickly.

    Not sure if that really helps you much

    AL
     
  12. Jan 6, 2012 #11
    Re: Trigonometric

    Also I forgot to say that you need to be using radian values for those 2 series to work.

    edit.

    I just realized you would still need a basic calculator to do that.

    Without a calculator of any sort your options are really either getting a trig table sheet or learning some of the common ones like 0, 30, 45, 60, 90 etc
     
    Last edited: Jan 6, 2012
  13. Jan 6, 2012 #12

    Mark44

    Staff: Mentor

    Re: Trigonometric

    These are the Maclaurin series for the sine and cosine functions. Maclaurin series are special cases of Taylor series.
     
  14. Jan 6, 2012 #13
    Re: Trigonometric

    Thanks Mark, I haven't really studied series in school in much detail yet so thanks for telling me what those where.

    Thanks
    AL
     
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