I'm attempting to program something and I'm reaching the limits of what I can remember about basic geometry (its been like 15 years since I've taken trig). I feel like I have figured out a large chunk of the problem but I'm struggling with the final triangle equations. I've tried googling but I'm having trouble coming down to the solution because google's info on isoceles triangles aren't described in cartesian form. Here is what I got.
Imagine a circle centered on the point (135, 135). The circle has a radius of 125. This circle is divided into 24 pie slices. The first slice starts at 0 degrees off the x axis and goes to 15 degrees and the slices continue all the way around the circle filling up all 360 degrees.
This part of the question isn't particularly important, but hopefully adds some context of what I am doing.
Now its the triangle part that is confusing me. I want to generate an isosceles triangle with its vertex at a point along the circle. I know the angle of the vertex (and therefore all the angles of the triangle) and I know the lengths of all the sides of the triangle. And I also know how to generate the X and Y coordinates of the vertex of the triangle using trig as shown below.
So the X coordinate of the point on the circle is 135 + (125 * cosθ)
The Y coordinate of the point on the circle is 135 - (125 * sinθ)
The lengths of the sides of the triangle are 10, 10, & 12
The vertex angle of the isosceles triangle is the arccos(1 - 144/200)
Based on my googling, if I set the angle I'm inputting to 90 degrees or pi/2 radians, i can determine the location of the all the points using the idea that lengths of multiples of 5, 5, & 6 make a triangle that works. Therefore at 90 degrees, the vertices of the triangle would be at A(135,10), B(129,2), and C(141,2). But I don't know how to produce a formula that produces B's & C's X and Y coordinates. I hope that makes sense. Any help is appreciated.
I do not provide a closed form for the coordinates but some formulae that you can use.
The distance formula can be the first step here.
$$10=\sqrt{(Ax-Bx)^2+(Ay-By)^2}$$ $$10=\sqrt{(Ax-Cx)^2+(Ay-Cy)^2}$$ $$12 = \sqrt{(Bx-Cx)^2 + (By - Cy)^2}$$ Secondly, the angle between the two lines $|AC|$ and $|AB|$ you have given to be $a$ such that $cos \ \ a= 1-\frac{144}{200}$ Now I define
$$m_1 = \frac{Ay - By}{Ax-Bx} \\ m_2 = \frac{Ay-Cy}{Ax-Cx}$$ then I also have the condition:
$$ | \frac{m_1 - m_2}{1+m_1m_2} | = tan(cos^{-1}(1 - \frac{144}{200})) $$ where $|x|$ is the absolute value of $x$.
Now unfortunately I do not have access to any computer resource that can let me use these equations to find a closed form for $Bx$, $By$, $Cx$, and $Cy$. Maybe you can try to use these equations and put them to a CAS to try to get closed expressions for the same?
After determining $A$, you can click here and put the coordinates of $A$ with the scheme ($Ax = X$, $Ay = Y$, $Bx = x$, $By = y$, $Cx = p$, $Cy = q$) to find triangles that pass through $A$ satisfying the given constraints.