| Triangle Math <- Easy one? | |
| cbx | OK I am trying to find the formula to find the length of 2 sides of a 2D triangle.
I already know what each of the 3 inner angles of the triangle are as well as the length of one side. So does anyone know how to calculate the length of the other 2 sides of the triangle? The formula must work on ANY type of triangle weather it be equilateral, isosceles, scalene or right-angled etc.
This is what I have. I have 2 points in 2D space. I know the X/Y location of each point. That is how I know what the distance of one side of the triangle is. Now each point is facing in a direction and because I know what direction each of the 2 points is facing I can calculate what each of the inner anges of the triangle are. Again that is how I know what the distance of one side of the triangle is. What I an trying to find is the formula to determin at what X/Y position the third point in the triangle is locate at. Now I can find this position if I can find the length of the 2nd or 3rd side of the triangle.
See the attached pic for what I am talking about...
Image Insert: [img]http://www.vbgamer.com/msgboard/uploaded/cbx/2004621171949_Triangle NTK.gif[/img] Even if I could find the length of one other side would be helpfull. Thanx. |
| masterbooda | Are you trying to find the total length for the triangle? Because otherwise, couldn't you use the same formula to calculate every line of the triangle? Just input the points of that side... or is this not what you are asking? This code will give you the distance between two points which is the same thing as length... [code]Public Function FindDistance(X1 as Single, Y1 as Single, X2 as Single, Y2 as Single) as Double FindDistance = Sqr((X2 - X1) * (X2 - X1) + (Y2 - Y1) * (Y2 - Y1)) End Function[/code] DaBooda out...[:D] |
| Sr. Guapo | hehe... Law of Sines (and I thought I'd never have to use it after trig...). Okay, basically givin triangle ABC(excuse the awful ASCII art [:)]): A |\ |.\ |..\ B--C where the side opposite the angles are a,b, and c (respectively): Sin(A)/a = Sin(B)/b or in you case: a = b*Sin(A)/Sin(B) where the letters can be changed around. That will allow you to find the remaining sides. It looks like it may be a little computationally expensive, so use sparingly... |
| game_maker | Hi OK ,,, I will assume that you have 4 points ( or 2 points and 2 vectors) I don't exactly know the information you have [:)] . anyway the idea would be the same ... [img]http://axdteam.com/temp/2.jpg[/img] notice that p3 is any point lies on Line1,p4 is any point lies on Line2 [code] Ya = m1 * Xa + b1 ' P1 & P3 satisfied (lies on it) Yb = m2 * Xb + b2 ' P2 & P4 satisfied (lies on it) b1 = Ya - m1 * Xa use p1 or p3 b1 = y1 - m1 * x1 b2 = y2 - m2 * x2 m1 = (y3-y1)/(x3-x1) , x3 # x1 m2 = (y4-y2)/(x4-x2) , x4 # x2 _____________ Ya = m1 * Xa + b1 Yb = m2 * Xb + b2 to find intersection point we let Ya = Yb = y , Xa = Xb = x Ya = Yb m1 * x + b1 = m2 * x + b2 m1 * x - m2 * x = b2 - b1 x (m1 - m2) = b2 - b1 x = (b2 - b1)/(m1 - m2) notice : m1 never = m2 otherwise the are || i.e. never intersect hence , mx1 = 1 /(x3-x1) mx2 = 1 /(x4-x2) m1 = (y3-y1) * mx1 m2 = (y4-y2) * mx2 b1 = y1 - m1 * x1 b2 = y2 - m2 * x2 x = (b2 - b1)/(m1 - m2) y = m1 * x + b1 or y = m2 * x + b2 [/code] [code] Option Explicit Private Type Vector2D X As Single Y As Single End Type Private Sub Form_Load() Dim P1 As Vector2D, P2 As Vector2D, P3 As Vector2D, P4 As Vector2D P1 = SetV2(3000, 4000) P3 = SetV2(1000, 1000) P2 = SetV2(3000, 1000) P4 = SetV2(1000, 3000) Me.AutoRedraw = True Line (P1.X, P1.Y)-(P3.X, P3.Y) Line (P2.X, P2.Y)-(P4.X, P4.Y) Me.Circle (P1.X, P1.Y), 100: Print " 1" Me.Circle (P2.X, P2.Y), 100: Print " 2" Me.Circle (P3.X, P3.Y), 100: Print " 3" Me.Circle (P4.X, P4.Y), 100: Print " 4" Dim IP As Vector2D Dim M1 As Single, M2 As Single Dim Mx1 As Single, Mx2 As Single Dim B1 As Single, B2 As Single Mx1 = IIf(P3.X = P1.X, 0, 1 / (P3.X - P1.X)) Mx2 = IIf(P4.X = P2.X, 0, 1 / (P4.X - P2.X)) M1 = (P3.Y - P1.Y) * Mx1 M2 = (P4.Y - P2.Y) * Mx2 B1 = P1.Y - M1 * P1.X B2 = P2.Y - M2 * P2.X IP.X = (B2 - B1) / (M1 - M2) IP.Y = M1 * IP.X + B1 Me.ForeColor = RGB(255, 0, 0): Me.DrawWidth = 3 Circle (IP.X, IP.Y), 100: Print " IP" End Sub Private Function SetV2(X As Single, Y As Single) As Vector2D SetV2.X = X SetV2.Y = Y End Function [/code] regards [:)] |
| game_maker | If you didn't understand what I was doing: At first, we have 2 unknowns (x & y) this means you have to find 2 equation at least to solve it. To find the point of intersection first we find 2 lines passing through that point (satisfy x & y) then we equalize them and hence we get x & y To find an equation of a line we need to find (2 points) lies on it or one point and the slope If you have a vector or ray and point then: a = <x1, y1> p = (x2, y2) m = y1/x1 b = y2 - m * x2 That's how to find an equation of a line by knowing a point & vector (direction) [we get the slope from the vector , point] If you know 2 point's then read my prev post [:)] |