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Math Help - help me please............

  1. #1
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    help me please............




    there is a triangel ABC
    A(0,0)
    AB=5 BC=4 CA=3
    CB is going with the angel of 200 degrees!
    i dont know the angels of the triangel
    i just know that CB is having a shape of 200 degrees
    any way, i need to find the poing B of the triangel
    B(x,y)=?
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  2. #2
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    Quote Originally Posted by tukilala View Post



    there is a triangel ABC
    A(0,0)
    AB=5 BC=4 CA=3
    CB is going with the angel of 200 degrees!
    i dont know the angels of the triangel
    i just know that CB is having a shape of 200 degrees
    any way, i need to find the poing B of the triangel
    B(x,y)=?
    I don't understand your question:

    1. If AB=5 BC=4 CA=3 are the measures of the lengths of the sides then ABC is a right triangle with AB as hypotenuse.

    2. It's this statemant: CB is going with the angel of 200 degrees which I don't understand. To determine an angle you need:
    - three points
    - two straight lines
    - an arc
    - two curves
    - ....

    Please tell us exactly where this angle of 200 is to be found. Maybe you can provide us with a sketch(?)
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  3. #3
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    AB goes diagonally downward into the direction of the third "quadrant"
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  4. #4
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    Quote Originally Posted by tukilala View Post
    AB goes diagonally downward into the direction of the third "quadrant"
    Since |\overline{AB}| = 5 and \overline{AB} is placed on the diagonal of the third quadrant the coordinates of B and \overline{AB} form an isosceles right triangle.

    Use Pythagorean theorem to calculate the coordinates:

    x_B^2+y_B^2=25~\wedge~|x_B| = |y_B|~\implies~|x_B|= |y_B| =\frac52\sqrt{2}

    Since B has to be in the third quadrant both coordinates must be negative. Therefore the point B has the coordinates B\left( -\frac52\sqrt{2}\ ,\ -\frac52\sqrt{2} \right)

    I'm still confused about the angle of 200 ... (?)
    Attached Thumbnails Attached Thumbnails help me please............-koord_vonb.png  
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