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Finding third cordinates of a triangle

I have a right angle triangle and I know two cordinates and the length of all thres sides. I need to find the third cordiante in VB. e.g. I have (0,0) and (3,0) and lengths as 3 (for base), 4 (for perp), 5 for (hypoteneous). How to find third cordinate in VB?

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aqk139
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aqk139
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1 Solution
 
TheRedGuyCommented:
Sound like homeworkj to me! Look up Pythagoras Theorem!
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cquinnCommented:
for a 3,4,5 triangle the coordinates would be:

0,0
3,0
0,4


 0,4
 | \
 |  \
 |___\
0,0   3,0

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

If you take the time to read the original question, He/She states EXPLICITLY that they are working with RIGHT TRIANGLES.
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YourBuddyTooCommented:
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)
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YourBuddyTooCommented:
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).
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aqk139Author Commented:
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.
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Arthur_WoodCommented:
are you ALWAYS dfealing with right triangles?
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YourBuddyTooCommented:
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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andyclapCommented:
>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.

Unfortunately you cannot have one. There are two possible answers to this problem: Think about a mirror image of the triangle, flipped along the side defined by the two points you have - you cannot know which of the two answers is correct.

So, how would you like a general function which takes two coordinates and three lengths, and returns two sets of two coordinates represenging the two possible answers? Or do you have any extra information which would help you determine which of the two possible positions for the third point you want.
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NetminderCommented:
There has been no activity in this question in quite some time, and it looks like it has been abandoned. As part of our ongoing mission to clean up the topic areas, a Moderator will finalize this question within the next seven (7) days. At that time, either I or one of the other Moderators will force/accept the comment of andyclap.

DO NOT ACCEPT THIS COMMENT AS AN ANSWER. If you have further comments on this question or the recommendation, please leave them here.

aqk139,

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NetminderCommented:
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