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Math Help - problem using sine and cosine rule

  1. #1
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    problem using sine and cosine rule

    I have attached the image of the problem to look at.


    a). Apply the Sine rule to triangle QRS to determine the length of QS

    b). Apply the cosine rule to triangle QPS to determine the length of QP [IMG]file:///C:/Documents%20and%20Settings/hunter/Desktop/ICA1%20-%20SPECIMEN-5.jpg[/IMG]
    Attached Thumbnails Attached Thumbnails problem using sine and cosine rule-ica1-specimen-5.jpg  
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  2. #2
    Senior Member Stroodle's Avatar
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    Hi there.

    Do you know what the sine and cosine rules are? Because if you do it's simply just a matter of plugging the values into the equations.

    The sine rule is \frac{a}{sin A}=\frac{b}{sin B}=\frac{c}{sinC}

    So if you take a to be the side QS then A would be 30^{\circ} , b would be 15, and B would be 135^{\circ} (as 180-30-15=135)

    After solving this you'll have the side length you need to apply the relevant cosine rule to obtain the length of QP.

    Let us know if you have trouble with this.
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  3. #3
    MHF Contributor
    Grandad's Avatar
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    Hello davyboy123
    Quote Originally Posted by davyboy123 View Post
    I have attached the image of the problem to look at.


    a). Apply the Sine rule to triangle QRS to determine the length of QS

    b). Apply the cosine rule to triangle QPS to determine the length of QP [IMG]file:///C:/Documents%20and%20Settings/hunter/Desktop/ICA1%20-%20SPECIMEN-5.jpg[/IMG]
    I'm not quite sure where your problem is here. \angle QSR =180-(15+30)=135^o. So in \triangle QSR:
    \frac{QS}{\sin 30^o}= \frac{15}{\sin 135^o}

    ...etc
    And \angle QSP =180-135= 45^o. So in \triangle QSP:
    QP^2 = QS^2+PS^2-2.QS.PS.\cos45^o

    ... etc
    Can you complete it now?

    Grandad
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