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Thread: find lengths of triangle's 2 sides

  1. #1
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    find lengths of triangle's 2 sides

    hi

    The perimeter of triangle ABC = 15 cm. Given that AB = 7 cm and angle BAC = 60 degreez, find the lengths of AC and BC.

    Anyone could give me some hint how to go with this ? Because all i know is that b+c=8 ..
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  2. #2
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    Re: find lengths of triangle's 2 sides

    Quote Originally Posted by AbYz View Post
    hi

    The perimeter of triangle ABC = 15 cm. Given that AB = 7 cm and angle BAC = 60 degreez, find the lengths of AC and BC.
    Anyone could give me some hint how to go with this ? Because all i know is that b+c=8 ..
    The rule of cosines tells us that
    $(8-b)^2=b^2+49-7b\cos(60^o)$
    Thanks from topsquark, AbYz and sakonpure6
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  3. #3
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    Re: find lengths of triangle's 2 sides

    Quote Originally Posted by Plato View Post
    The rule of cosines tells us that
    $(8-b)^2=b^2+49-\color{red}{2} \cdot 7b\cos(60^o)$
    ...

    method 2 ... use Pythagoras on the larger right triangle

    Thanks from AbYz
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    Re: find lengths of triangle's 2 sides

    Quote Originally Posted by AbYz View Post
    Anyone could give me some hint how to go with this ? Because all i know is that b+c=8 ..
    You didn't know that COS(60) = 1/2 ?
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  5. #5
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    Re: find lengths of triangle's 2 sides

    Quote Originally Posted by DenisB View Post
    You didn't know that COS(60) = 1/2 ?
    somehow i ruled out using sine/cosine rules at the beginning..
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