Reputation: 102
I want to draw a triangle with only the 3 sides length. (In C# Winforms)
Example: S1(3), S2(4), S3(5) with SN(x) as the length of each side.
I've tried to do this and the result is not a right triangle but it looks like it.
For the first side i just draw it like a line. and after i triy to find the angle with Pythagore and the same for the second one, but i realised that if i enter (5,4,3) it's something else.I'm just try to understand how can i find coordinates of a triangle with only the length of the sides.
Point a = new Point(0, 0);
Point b = new Point(s1, 0);
double y = (Math.Pow(s1, 2) + Math.Pow(s3, 2) - Math.Pow(s2, 2)) / (2 * s1);
double x = Math.Sqrt(Math.Pow(s3, 2) - Math.Pow(y, 2));
Point c = new Point((int)x, (int)y);
e.Graphics.DrawLine(Pens.Black, a, b);
e.Graphics.DrawLine(Pens.Black, b, c);
e.Graphics.DrawLine(Pens.Black, c, a);
That's the result:
Can someone help me? because I think I don't understand how can I do this.
Upvotes: 1
Views: 1460
Reputation: 25972
This is more a math problem. At point A you have the sides s1, s3 with opposing side s2. The cosine formula then gives
2*s1*s3*cos(alpha) = s1^2+s3^2-s2^2.
Now the cosine is the projection of the angle to the horizontal axis, so you should have
x = s3*cos(alpha) = (s1^2+s3^2-s2^2)/(2*s1)
and correspondingly
y = sqrt(s3^2-x^2).
For the test side lengths 3,4,5 this would give
x = (3^2 + (5^2-4^2))/(2*3) = 3
y = sqrt(5^2-3^2) = 4
producing the points for the rectangular triangle.
Upvotes: 5