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Math Help - Trig Functions (Angles subtended?)

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    Trig Functions (Angles subtended?)

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    Last edited by MathIsPower; May 30th 2007 at 02:30 PM.
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    Quote Originally Posted by MathIsPower View Post
    Hey everyone I really need help with this question for my homework and I'm having a hard time with it.


    Q- The angle subtended at the center by the arc PQ is 3Pi/8 radians. R is the midpoint of arc PQ. Find the shaded area.

    http://olympic.txc.net.au/AK/help.JPG (Diagram)

    Any help will be much appreciated please Thanks in advance!
    Hello,

    the shaded area can be calculated by:

    area of sector - area of right triangle

    The angle(ROP) = \frac{3 \pi}{16}

    Thus the area of the sector is: \frac{A_{sector}}{\pi r^2}=\frac{\frac{3 \pi}{16}}{2\pi} \ \Longleftrightarrow \ A_{sector}=\frac{3}{32} \pi r^2

    The area of the right triangle is A_{\text{right triangle}}=\frac{1}{2} \cdot {base} \cdot {height}

    If r is the length of the radius (here r = 4 cm) then the length of the base: b=r \cdot \cos \left(\frac{3 \pi}{16} \right) and

    the length of the height is: h=r \cdot \sin \left(\frac{3 \pi}{16} \right)

    Therefore the shaded area is:

    A_{\text{shaded area}}=\frac{3}{32} \pi r^2 - \frac{1}{2} \cdot r \cdot \cos \left(\frac{3 \pi}{16} \right) \cdot r \cdot \sin \left(\frac{3 \pi}{16} \right)

    Plug in the values you know to calculate the area.

    (For confirmation only: I've got A = 1.017 cm²)
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    Thanks Heaps!
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