How to calculate the point(x2,y2) with the point (x1,y1) and the angle

Again i came back with a question.

1.I have a line drawing from a point (x2,y2) for which i do no the end point(Say point unknown as in the fig),but i knew the line length and the angle from the vector (x2,y2). Can any one help me how to cal开发者_StackOverflow中文版culate the unknown points.

Thanks, Lokesh.

This looks like a job for ... Trigonometry-man.

Consider the following diagram:

          /|
         /B|
        /  |
       /   |
      /    |
   c /     | a
    /      |
   /       |
  /       _|
 /A      |C|
*----------+
      b

You know the angle B, it's 45^o. You also know C is 90^o because that's the "right angle" bit of the right angle triangle.

And, because the angles inside a triangle add up to 180^o, angle A must also be 45^o.

You also know the length of the hypotenuse c. With trigonometry^(a), you can get the lengths of the other two sides with:

a = c sin A
b = c cos B

Then simply add those to your starting point (making sure you get the signs right) and you have your ending point.

For example, let's say your hypotenuse was 1.414213562, a number I just just picked at random off the top of my head :-)

The length a is c sin A or 1.414213562 * sin 45, which is 1.414213562 * 0.707106781 or 1.

Wow, what were the chances of that? :-)

^(a) My children reminded me of the "soh cah toa" (pronounced so car toe ah) rule, where "opposite" and "adjacent" sides are location from the angle X:

soh : sin X = opposite / hypotenuse
cah : cos X = adjacent / hypotenuse
toa : tan X = opposite / adjacent

          /|
         / |
        /  |
       /   |
      /    |
 hyp /     | opp
    /      |
   /       |
  /       _|
 /X      | |
*----------+
     adj