Finding third cordinates of a triangle

Sound like homeworkj to me! Look up Pythagoras Theorem!

for a 3,4,5 triangle the coordinates would be:

0,0
3,0
0,4


 0,4
 | \
 |  \
 |___\
0,0   3,0

You would need to use that standard Trig functions:

Sin(angle) = Side opposite/hypotenuse
Cos(Angle) = side adjacent/hypotenuse

you can also use the Pythogorean theorem:

   H^2 = A^2 + B^2

However, this IS NOT a VB question, but rather a basic question about Trigonometry and Algebra.

Pythogrean's theorem only works on right triangles.  The question at hand is not necessarily limited to right triangles.

YourBuddyToo:

If you take the time to read the original question, He/She states EXPLICITLY that they are working with RIGHT TRIANGLES.

The solution looks something like this:

Let A be the left side of the triangle, B be the base of the trinangle, C the right Side.  ThetaA the angle between side A and side B, ThetaB the angle between side B and side C.

First, the A + C must be > B (or it won't form a triangle, the sides are too short.

Next

B = A Cos(ThetaA) + C Cos(ThetaB)
A Sin(ThetaA) = C Sin(ThetaB)

You now have two equations w/ two unknowns (ThetaA & ThetaB).  Solve for ThetaA & ThetaB

But wait, we are not done yet, we need to calculate the coordinates for that point.  If we assume the base is horizontal (if not, we need to do some more trig), we can calculate the offsets from the intersection of sides A & B by:

DeltaX = A Cos(ThetaA)
DeltaY = +/- A Sin(ThetaA)  (two solutions, one where the point is above line B, one where it is below line B)

Thanks Arthur, leave it to an engineer to make a question harder than it needs to be.  For right triangles, the answer is easier.

The Y cord is the length of A,  X cord is either 0 (the x position of the intersection of A & B) or B (the intersection of B & C).

I think u all r confuse due to my example. I need a general function that give me the third cordinate of a trinangle by giving that function all three length of triangle and two cordinates. The triangle my be at any angle with x-axis or y-axis.


Clear.

are you ALWAYS dfealing with right triangles?

And if so, is the 90? angle on the left or the right side of the base.  Note that the equaitons I gave above work for any triangle (including right triangles).  It did not include any angle for the base itself (which appears to be part of the quesiton now).

For a base that could be at an angle,

ThetaBase = Atn((BaseYRight - BaseYLeft) / (BaseXRight - BaseXLeft)

XCoord = A Cos(ThetaA) + B Cos(ThetaBase) + BaseXLeft
YCoord = +/- A Sin(ThetaA) + B Sin(ThetaBase) + BaseYLeft

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