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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- Dec 10th 2008, 03:03 PM #1

- Joined
- Nov 2008
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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)=?

- Dec 10th 2008, 09:51 PM #2
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(?)

- Dec 10th 2008, 09:57 PM #3

- Joined
- Nov 2008
- Posts
- 50

- Dec 12th 2008, 04:08 AM #4
Since $\displaystyle |\overline{AB}| = 5$ and $\displaystyle \overline{AB}$ is placed on the diagonal of the third quadrant the coordinates of B and $\displaystyle \overline{AB}$ form an isosceles right triangle.

Use Pythagorean theorem to calculate the coordinates:

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

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

I'm still confused about the angle of 200° ... (?)