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Thread: [SOLVED] Trignometry and Radians Question

  1. #1
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    [SOLVED] Trignometry and Radians Question

    Question:


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

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


    Attempt:

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

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


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

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

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


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

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

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


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

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

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

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

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

    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 $\displaystyle = r\theta$

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

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

    How can I get the length of $\displaystyle AX$ in terms of $\displaystyle \pi$? I know that $\displaystyle \sin{30^o} = \frac{1}{2}$ but if I used this would the $\displaystyle \pi$ vanish?
    Last edited by looi76; May 4th 2008 at 12:21 PM.
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  2. #2
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    Quote Originally Posted by looi76 View Post
    Question:


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

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

    Attempt:

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

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


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

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

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


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

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

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


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

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

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

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

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

    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 $\displaystyle = r\theta$

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

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

    How can I get the length of $\displaystyle AX$ in terms of $\displaystyle \pi$? I know that $\displaystyle \sin{30^o} = \frac{1}{2}$ but if I used this would the $\displaystyle \pi$ 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 $\displaystyle r \sin{(\theta)}$, which, if $\displaystyle \theta = \frac{\pi}{6}$, 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 $\displaystyle = r\theta$

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

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

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

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

    It makes this question much simpler.
    One side of the triangle is r, another is $\displaystyle r\cos \theta$...
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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 $\displaystyle \frac{1}{2}ab\sin C$ for the area of a triangle.

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