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Math Help - [SOLVED] Trignometry and Radians Question

  1. #1
    Member looi76's Avatar
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    [SOLVED] Trignometry and Radians Question

    Question:


    In the diagram, AB is an arc of a circle, centre O and radius r \ cm, and angle AOB = \theta \ radians. The point X lies on OB and AX is perpendicular OB.

    (i) Show that the area, A\ cm^2, of the shaded region AXB is given by
    A = \frac{1}{2}r^2(\theta - \sin\theta\cos\theta)
    (ii) In the case where r = 12 and \theta = \frac{1}{6}\pi, find the perimeter of the shaded region AXB, leaving your answer in terms of \sqrt{3} and \pi.


    Attempt:

    (i) Area of shaded region = Area of sector of circle - Area of triangle
    Area of sector of circle = \frac{1}{2}r^2\theta

    Area of Triangle = \frac{1}{2} \times Lenght \times Breath


    Length \rightarrow \sin\theta = \frac{AX}{r}

    Length \rightarrow AX = \sin\theta \times r

    Length \rightarrow AX = \sin\theta{r}


    Breath \rightarrow \cos\theta = \frac{OX}{r}

    Breath \rightarrow OX = \cos\theta \times r

    Breath \rightarrow OX = \cos\theta{r}


    Area of Triangle  = \frac{1}{2} \times \sin\theta{r} \times \cos\theta{r}

    Area of Triangle  = \frac{1}{2}r^2\sin\theta\cos\theta

    Area of shaded region = Area of sector of circle - Area of triangle

    A \ = \  \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta\cos\theta

    A \ = \ \frac{1}{2}r^2(\theta - \sin\theta\cos\theta)

    are my steps for showing the area of the shaded region correct?



    (ii) Perimeter of Shaded Region = Arc lenght of circle AB + AX + XB

    Arc length of circle = r\theta

    Arc length of circle = 12 \times \frac{1}{6}\pi

    Length of AX = \sin{\frac{1}{6}\pi} \times 12

    How can I get the length of AX in terms of \pi? I know that \sin{30^o} = \frac{1}{2} but if I used this would the \pi vanish?
    Last edited by looi76; May 4th 2008 at 01:21 PM.
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  2. #2
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    Quote Originally Posted by looi76 View Post
    Question:


    In the diagram, AB is an arc of a circle, centre O and radius r \ cm, and angle AOB = \theta \ radians. The point X lies on OB and AX is perpendicular OB.

    (i) Show that the area, A\ cm^2, of the shaded region AXB is given by
    A = \frac{1}{2}r^2(\theta - \sin\theta\cos\theta)
    (ii) In the case where r = 12 and \theta = \frac{1}{6}\pi, find the perimeter of the shaded region AXB, leaving your answer in terms of \sqrt{3} and \pi.

    Attempt:

    (i) Area of shaded region = Area of sector of circle - Area of triangle
    Area of sector of circle = \frac{1}{2}r^2\theta

    Area of circle = \frac{1}{2} \times Lenght \times Breath


    Length \rightarrow \sin\theta = \frac{AX}{r}

    Length \rightarrow AX = \sin\theta \times r

    Length \rightarrow AX = \sin\theta{r}


    Breath \rightarrow \cos\theta = \frac{OX}{r}

    Breath \rightarrow OX = \cos\theta \times r

    Breath \rightarrow OX = \cos\theta{r}


    Area of circle  = \frac{1}{2} \times \sin\theta{r} \times \cos\theta{r}

    Area of circle  = \frac{1}{2}r^2\sin\theta\cos\theta

    Area of shaded region = Area of sector of circle - Area of triangle

    A \ = \ \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta\cos\theta

    A \ = \ \frac{1}{2}r^2(\theta - \sin\theta\cos\theta)

    are my steps for showing the area of the shaded region correct?



    (ii) Perimeter of Shaded Region = Arc lenght of circle AB + AX + XB

    Arc length of circle = r\theta

    Arc length of circle = 12 \times \frac{1}{6}\pi

    Length of AX = \sin{\frac{1}{6}\pi} \times 12

    How can I get the length of AX in terms of \pi? I know that \sin{30^o} = \frac{1}{2} but if I used this would the \pi vanish?
    For part 1: Aside from calling your triangle a circle a few times, the work for the formula is correct.

    For part 2: The length of AX is not going to be in terms of pi. It is simply r \sin{(\theta)}, which, if \theta = \frac{\pi}{6}, is equal to 6.
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  3. #3
    Moo
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    Hi again !

    Quote Originally Posted by looi76 View Post
    are my steps for showing the area of the shaded region correct?
    Well, I find it correct and well-explained

    Quote Originally Posted by looi76 View Post
    (ii) Perimeter of Shaded Region = Arc lenght of circle AB + AX + XB

    Arc length of circle = r\theta

    Arc length of circle = 12 \times \frac{1}{6}\pi

    Length of AX = \sin{\frac{1}{6}\pi} \times 12

    How can I get the length of AX in terms of \pi? I know that \sin{30^o} = \frac{1}{2} but if I used this would the \pi vanish?[/SIZE][/FONT]
    There is still \pi due to the arc length of circle, isn't it ? You don't necessarily have \pi in AX.

    And what about XB ? ( \sqrt{3} will be used here)
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  4. #4
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    Do you know the formula \frac{1}{2}ab\sin C for the area of a triangle.

    It makes this question much simpler.
    One side of the triangle is r, another is r\cos \theta...
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  5. #5
    Member looi76's Avatar
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    Quote Originally Posted by a tutor View Post
    Do you know the formula \frac{1}{2}ab\sin C for the area of a triangle.

    It makes this question much simpler.
    One side of the triangle is r, another is r\cos \theta...
    Thanks for reminding me. HOW DID I FORGET THAT?
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