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Math Help - Finding areas of sectors using radians

  1. #1
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    Finding areas of sectors using radians

    There is a question on the area of a sector of a circle on which I am stuck and was wondering whether anyone could give me a pointer.


    I have attached an image of the diagram to which the question refers...


    The diagram shows an arc of ABC of a circle centre o and radius 10cm.
    Angle AOB = \theta radians and angle AOC is a right angle.


    a. Write down, in terms of theta, the area of sector AOB.


    I can do this easily enough:  area = \frac{1}{2} r^2 * \theta


    area =  50 \theta


    b. Show that the area of the shaded segment is  25(\pi -2\theta - 2cos \theta)cm^2


    This is where I get stuck.


    I can find the area of the sector OBC using the formula and get 50(\frac{\pi }{2} - \theta)$ = $ 25\pi - 50 \theta


    But I can't see how to get the area of the triangle.


    Having said that, multiplying out  25(\pi - 2\theta - 2 cos \theta) gives:

     25\pi - 50\theta -50 cos\theta , so the area of the triangle must be  50 cos \theta .


    I just can't see how to get at this area for the triangle, though. Any advice would be greatly appreciated!
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    Last edited by pturo1; February 15th 2013 at 11:59 AM.
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  2. #2
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    Re: Finding areas of sectors using radians

    Quote Originally Posted by pturo1 View Post
    The diagram shows an arc of ABC of a circle centre o and radius 10cm. Angle AOB = $\theta$ radians and angle AOC is a right angle.
    a. Write down, in terms of theta, the area of sector AOB.

    I can do this easily enough:  area = \frac{1}{2} r^2 * \theta
    area =  50 \theta

    b. Show that the area of the shaded segment is  25(\pi -2\theta - 2cos \theta)cm^2

    You can use this webpage.
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  3. #3
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    Re: Finding areas of sectors using radians

    That's a isosceles triangle having two sides of length 10 and one angle of measure \theta. If you drop a perpendicular, you divide it into two right triangles having hypotenuse 10 and angle \theta/2. So the two legs of the right triangle have length 10cos(\theta) and  10sin(\theta). Since the legs are the "base" and "altitude" of a right triangle, the area of each is (1/2)(10 cos(\theta/2))(10 sin(\theta/2) and the area of the entire triangle is 100 sin(\theta/2)cos(\theta/2).
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  4. #4
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    Re: Finding areas of sectors using radians

    Thank you for your replies. Plato, thanks for the link; it looks interesting but involves cotangents which I have not yet got to in my mathematical education.

    Thanks for your explanation Halls of Ivy. I can follow your working but there is just one thing I don't understand (I am probably failing to see something obvious):

    How do you know one of the angles in the isocoles triangle is  \theta? The only conclusion I could draw was that the angle would be \frac{\pi}{2} - \theta radians.

    Thanks again for your help.
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