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Math Help - sine and cosine rule help

  1. #1
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    sine and cosine rule help

    I dont really understand what question 'a' is asking, can anyone explain it to me?

    Thanks
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  2. #2
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    <QPR is just the angle created by the segment PQ and PR(the lower left angle of the triangle), and they want you to find the cosine of that angle

    They tell you that sin(<QPR)= \frac{2}{3} so using this, and the identity sin^2(x)+cos^2(x)=1 you can get the answer

    \left(\frac{2}{3}\right)^2+cos^2(x)=1

    \frac{4}{9}+cos^2(x)=1

    cos^2(x)=1-\frac{4}{9}=\frac{5}{9} and take the sqrt of both sides to get

    cos(x)=\pm\frac{\sqrt{5}}{3} and if you go by the picture, the angle is acute, so the angle must lie in the first quadrant, so we can ignore the negative value
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  3. #3
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    Quote Originally Posted by Tweety View Post
    I dont really understand what question 'a' is asking, can anyone explain it to me?

    Thanks
    You are asked to calculate the \cos(\angle(QPR))

    1. You know already \sin(\angle(QPR))=\dfrac23

    2. You know that (\sin(\alpha))^2+(\cos(\alpha))^2=1

    3. Therefore:

    (\cos(\angle(QPR)))^2+(\sin(\angle(QPR)))^2=1

    (\cos(\angle(QPR)))^2+\dfrac49=1

    (\cos(\angle(QPR)))^2=\dfrac59

    \cos(\angle(QPR))=\dfrac13 \sqrt{5}
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  4. #4
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    Thanks, didn't realise I had to use that identity;

    Can you also help me with part 'b', to find the side QR I would have to use te cosine rule (?), do I use the answer from part 'a'?

    so I find the inverse cosine of  \frac{\sqrt{5}}{3} = 41.8.

    Is this the angle I use?
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  5. #5
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    Quote Originally Posted by Tweety View Post
    Thanks, didn't realise I had to use that identity;

    Can you also help me with part 'b', to find the side QR I would have to use te cosine rule (?), do I use the answer from part 'a'?

    so I find the inverse cosine of  \frac{\sqrt{5}}{3} = 41.8.

    Is this the angle I use?
    That isn't necessary here. As you've stated you have to use the cosine rule:

    QR=\sqrt{7^2+(3\sqrt{5})^2-2 \cdot 7 \cdot 3\sqrt{5} \cdot \cos(\angle(QPR))}

    QR=\sqrt{49+45-42 \cdot \sqrt{5} \cdot \dfrac13 \sqrt{5}}

    QR=\sqrt{49+45- 70 } = \sqrt{24}=2\sqrt{6}
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