ready

[alexbg]

ewqe

[hamsterman]

well, if I recall correctly,

c2 = a2 + b2 - 2ab*cos(a^b)

If you need the smallest angle, make c to be the shortest side.
If you need more than that, ask specific questions.

[alexbg]

deleted

[ne555]

Use a convention.

Point p[3];
Line l[3];
double angles[3]; //the angle between l[i] and l[(i+1)%3] 

Please just post in one place:
http://www.cplusplus.com/forum/general/31179/
http://www.cplusplus.com/forum/beginner/31182/

[alexbg]

I am sorry about the posts in two places, but i decided that i posted a comment on a wrong page.


Mmm sorry, i dont understand that :

Point p[3];
Line l[3];
double angles[3]; //the angle between l[i] and l[(i+1)%3]

I wrote my first program - hello world, 2 weeks ago :(
Can u please explain it?

[hamsterman]

step by step:
1. get three points pt1, pt2 and pt3.
2. get the length of each line. store it in variables a, b and c.
3. make sure that c is the shortest line. if c > a, swap a, c. if c > b swap b, c.
4. calculate cos(a^b) from the formula I gave you.

[alexbg]

deleted

[hamsterman]

How should i store the program to get the length of the lines

see http://en.wikipedia.org/wiki/Pythagorean_theorem
Though most likely there already is a function for doing that in whatever library you're using.

decide which angle is the smallest?

the smallest angle is in front of (don't know the right word) the shortest line.

[alexbg]

deleted

[hamsterman]

you wrote:

what i have to write so the program automaticly will decide which is the smallest angle?

I wrote:

3. make sure that c is the shortest line. if c > a, swap a, c. if c > b swap b, c.

Line a = Line (p1, p2); That doesn't really do much.. I'll write a function for Pythagoras theorem:

float GetLength(Point a, Point b){
   return sqrt((a.get_x()-b.get_x())*(a.get_x()-b.get_x()) + (a.get_y()-b.get_y())*(a.get_y()-b.get_y()));
   //don;t forget to #include <cmath>
}

[alexbg]

ohh, my brain will explode..I really dont know how to write the program...

[alexbg]

Thank u so much about the answers, but if anyone can tell me what exactly i should do, will help me a lot...

[alexbg]

Please help me !!!!!!