# Math Help - 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)=?

2. Originally Posted by tukilala

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)=?

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(?)

3. AB goes diagonally downward into the direction of the third "quadrant"

4. Originally Posted by tukilala
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° ... (?)