3rd point of triangle, given 2 lengths, included angle

[barhoooooom]

Hi...

I am having a hard time!

I have a triangle that is made from three points: A, B and C.
The length of the side AB is known.
It is also known that AB = BC.
The angle A(B)C is known (lets call it theta).
The coordinates of the points A and B are known.

I need two formulas to calculate the the x and y coordinates of the third point C, something like:

Cx = Ax + cos(theta) + .... etc..
Cy = Ay * sine(theta) * whatever ....

If I have the values of the knowns , then I know how to calculate the answer by hand. The problem is in representing the answer in the form of two formulas. I am not interested in an answer; I am interested in a formula and that is what I am not able to do.

I have tried solving the two equations:
distance(A,C) = sqrt((Ax-Cx)^2 + (Ay-Cy)^2)
distance(B,C) = sqrt((Bx-Cx)^2 + (By-Cy)^2)
Since these are two equations with two unknowns: Cx and Cy.
But as I start working it out the solution starts getting too long and no longer within my control. This is because I don't have the values of the knowns and therefore I cannot simplify.

The same happens when I try to use instead of the distance equations the equations of the lines AC and BC.

My background in math is not good, but I have a feeling that it might be useful to look at the problem from other perspectives like:

- ABC is a sector in a circle. Where B is the center (known), AB is the radius (known) and the angle between A(B)C is theta (known). What is required is to find the third point C in this circle sector.

- Given a line segment AB, where point A and point B are known. What will point A be if we fix B and rotate the segment theta degrees?

As I said, my background in Math is not good, so I don't known if I am complicating matters by thinking this way or not.

One more thing that I was able to figure out is that C will always have two solutions unless theta is 180 degrees.

Any help will be appreciated.

Many thanks in advance

Ibrahim

 

[Deleted member 4993]

Re: third point of a triangle

Let

\(\displaystyle coordinates \, of \, A \, be \, (x_1,y_1)\)

\(\displaystyle coordinates \, of \, B \, be \, (x_2,y_2)\)

\(\displaystyle coordinates \, of \, C \, be \, (x_3,y_3)\)

Then slope of line AB

\(\displaystyle \tan(\alpha) \, = \, \frac{y_2 \, - \, y_1}{x_2 \, - \, x_1}\)

calculate \(\displaystyle \alpha\) from above.

let the slope of the line BC be:

\(\displaystyle \tan(\beta) \, = \, \frac{y_2 \, - \, y_3}{x_2 \, - \, x_3}\)

and

\(\displaystyle \beta \, = \, \alpha \, + \, \theta\)

so

\(\displaystyle \tan(\alpha \, + \, \theta) \, = \, \frac{y_2 \, - \, y_3}{x_2 \, - \, x_3}\)

above is your first equation.

Then the equality of distance will give you the second equation.

You will have multiple solutions - but only one of those will make physical sense.

 

[barhoooooom]

Re: third point of a triangle

Hey!

Thanks a lot for the quick and clear reply.
Your equation 1 simplified many things for me.

Thanks again for your time.

 

[Denis]

Re: third point of a triangle

I have a triangle that is made from three points: A, B and C.
The length of the side AB is known.
It is also known that AB = BC.
The angle A(B)C is known (lets call it theta).
The coordinates of the points A and B are known.

Why are you not using this easy calculation:
Let a = AB = BC (isosceles triangle)
Let g = angle ABC
Then: AC = aSIN(g/2)

 

[Deleted member 4993]

Re: third point of a triangle

Then: AC = aSIN(g/2)

That gives him the length of line AC - however he needed location of point C (the x & y co-ordinates).

 

[Denis]

Re: third point of a triangle

That gives him the length of line AC - however he needed location of point C (the x & y co-ordinates).

Right. But we now have a triangle with given sides; with equal sides = a and base = b:
(x1 - x3)^2 + (y1 - y3)^2 = b^2
(x2 - x3)^2 + (y2 - y3)^2 = a^2
No need for slopes and.....