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Math Help - Problem with IB past paper, Triangle and sin

  1. #1
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    Problem with IB past paper, NOT SOLVED

    Hello everyone, I do Math SL for the IB, and I don't fully understand the following question that I've attached as an image to this post. I would appreciate any kind of help.

    I know that sin of th. is the opposite / hypotenuse, and the hypotenuse is 3, but why is the opposite sqrt. of 5?

    also question b I don't full understand.
    Attached Thumbnails Attached Thumbnails Problem with IB past paper, Triangle and sin-picture-36.png  
    Last edited by Leitzge; March 15th 2010 at 06:57 AM. Reason: had to change title
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  2. #2
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    Quote Originally Posted by Leitzge View Post
    Hello everyone, I do Math SL for the IB, and I don't fully understand the following question that I've attached as an image to this post. I would appreciate any kind of help.

    I know that sin of th. is the opposite / hypotenuse, and the hypotenuse is 3, but why is the opposite sqrt. of 5?

    also question b I don't full understand.
    Dear Leitzge,

    If you apply the Phytogoras theorem to the traingle,

    AB^{2}=AC^{2}+CB^{2}

    CB^{2}=AB^{2}-AC^{2}

    BC=\sqrt{AB^{2}-AC^{2}}=\sqrt{3^{2}-2^{2}}=\sqrt{9-4}=\sqrt{5}

    Therefore, sin\theta=\frac{\sqrt{5}}{3}

    Similarly you could calculate cos\theta then use,

    sin2\theta=2\sin{\theta}\cos{\theta}

    Hope this will help you.
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  3. #3
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    Yeah thanks that's helpful!
    Could you explain (b) as well?
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  4. #4
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    Dear Leitzge,

    I have told you how to get the answer. All you got to do is calculate sin\theta~and~ cos\theta by looking at the traingle. Then you have to substitute those values in the given equation, sin2\theta=2\sin{\theta}\cos{\theta}

    Hope this will help you.
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