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Math Help - lengths of sides in a right angled triangle

  1. #1
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    lengths of sides in a right angled triangle

    "A right angled triangle with hypotenuse of length 6 contains the angle π/4. What are the exact lengths of the remaining two sides?"

    my working leads me no where, any suggestions? thanks, any help will be appreciated.
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  2. #2
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    Quote Originally Posted by [nuthing] View Post
    "A right angled triangle with hypotenuse of length 6 contains the angle π/4. What are the exact lengths of the remaining two sides?"
    If a right triangle contains an angle of \frac{\pi}{4} then it is isosceles. So solve this equation: 2s^2=36~.
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  3. #3
    Zap
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    I'm sorry, I'll will learn the latex editor as soon as I'm done here ... frown.. although Plato is is beyond brillient I wondered if I could offer a different and heavier handed approach. I wondered if your teacher wants you to resolve this with sin and cos functions.. if that is where you are in your book. If this is the case:

    hypotenuse = 6
    angle = pi/4 is a 45 degree angle
    side adjacent is the leg on the x axis
    side opposite is the leg perpendicular to the x axis

    soooo, to get the side adjacent you use: cos (pi/4) = side adjacent / hypontenuse (this is the identity for a cosine)
    rearrange here to get hypotenuse * cos (pi/4) = side adjacent
    cos (pi/4) is sqrt(2)/2 and hyp = 6 So: 6*sqrt(2)/2=3*sqrt(2)

    for the side opposite same deal, but you use the formula for sin: sin (pi/4)= side opposite / hyp (this is the identity for a sine)
    giving: hyp * sin(pi/4)=side opposite
    Now sin (pi/4) also equals sqrt(2)/2
    so you will get the same answer: 6*sqrt(2)/2=3*sqrt(2)

    they ask for exact lengths so you would leave the answers in the square root form.
    in the end both legs then are 3*sqrt(2)

    once again sorry for not using the editor.
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