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Math Help - Help with Special Triangle Problem

  1. #1
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    Help with Special Triangle Problem

    Would appreciate any help on the following problem, Thank You!
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  2. #2
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    given AB = 2

    BC is the hypotenuse of the 45-45-90 triangle ...

    BC = (AB)\sqrt{2} = 2\sqrt{2}

    BC is also the longer leg of a 30-60-90 triangle ... BC is the short leg.

    BD = \frac{2\sqrt{2}}{\sqrt{3}} = \frac{2\sqrt{6}}{3}


    from this point, you should be able to finish up.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    given AB = 2

    BC is the hypotenuse of the 45-45-90 triangle ...

    BC = (AB)\sqrt{2} = 2\sqrt{2}

    BC is also the longer leg of a 30-60-90 triangle ... BC is the short leg.

    BD = \frac{2\sqrt{2}}{\sqrt{3}} = \frac{2\sqrt{6}}{3}


    from this point, you should be able to finish up.
    Thanks!

    So if you can just confirm for me:

    CD = \frac{2\sqrt{42}}{3} (Used Pythagorean Theorem)
    and
    x = \frac{2\sqrt{7}}{\sqrt{3}} (Used Pythagorean Theorem again)

    Is this correct and in simplest form?
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  4. #4
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    bump.... Anybody able to confirm if my previous answer was correct?
    Thanks!
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