# Find all triangles with a fixed base and opposite angle

Find all triangles with a fixed base and opposite angle
I have a situation where I know the cartesian coordinates of the 2 vertices of a triangle that form its base, hence I know the length of the base and this is fixed.
I also know the angle opposite the base and this is also fixed.
Now what I want to do is figure out how to compute all possible positions for the third vertex.
My maths is rusty, I reverted to drawing lots of pictures and with the help of some tracing paper I believe that the set of all possible vertices that satisfies the fixed base and opposite angle prescribes a circle or possibly some sort of ellipse, my drawings are too rough to discern which.
I started with a simple case of an equilateral triangle, with a base length of two, i.e. the 3rd vertex is directly above the x origin, 0, base runs from -1 to 1 along the x-axis
then i started drawing other triangles that had that same base, -1 to 1 and the same opposite angle of 60 degrees or pi/3 depending on your taste
now i need to take it to the next step and compute the x and y coordinates for all possible positions of that opposite vertex.
struggling with the maths, do i use the sin rule, i.e. sin a / $=\mathrm{sin}b/B$ and so on, or do I need to break it down into right angle triangles and then just use something along the lines of ${a}^{2}+{b}^{2}={c}^{2}$
ultimately, I intend to plot the line that represents all the possible vertex positions but i have to figure out the mathematical relationship between that and the facts, namely,
base is fixed running from (-1,0) to (1,0) angle opposite the base is 60 deg
I then need to extend to arbitrary bases and opposite angles, but thought starting with a nice simple one might be a good stepping stone.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Step 1
A general solution to the problem:
Let the triangle have sides a,b,c and angles opposite to these sides as A,B,C respectively. We fix a between points $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$. For a fixed A, we need to find the mathematical equation of the locus of third point, $\left({x}_{3},{y}_{3}\right)$.
Using sine rule, we have: $\frac{a}{\mathrm{sin}A}=\frac{b}{\mathrm{sin}B}=\frac{c}{\mathrm{sin}C}=k$
where k is easily found since we know a and A.
In the triangle, angle $C=180-\left(B+A\right)$. We also see that:
$b=k\cdot \mathrm{sin}B$
Step 2
Similarly, $c=k\cdot \mathrm{sin}\left(180-\left(B+A\right)\right)=k\cdot \mathrm{sin}\left(B+A\right)$.
For different values of B, we will have corresponding values of b and c. Knowing a,b,c fixes the triangle.
From basic geometry, we have:
$b=\sqrt{\left({x}_{3}-{x}_{2}{\right)}^{2}+\left({y}_{3}-{y}_{2}{\right)}^{2}}$ and $c=\sqrt{\left({x}_{3}-{x}_{1}{\right)}^{2}+\left({y}_{3}-{y}_{1}{\right)}^{2}}$.
The locus of $\left({x}_{3},{y}_{3}\right)$ is given by simultaneously solving:
$\left({x}_{3}-{x}_{1}{\right)}^{2}+\left({y}_{3}-{y}_{1}{\right)}^{2}={k}^{2}\cdot {\mathrm{sin}}^{2}\left(B+A\right)$
$\left({x}_{3}-{x}_{2}{\right)}^{2}+\left({y}_{3}-{y}_{2}{\right)}^{2}={k}^{2}\cdot {\mathrm{sin}}^{2}\left(B\right)$
when B varies from 0 to $\left(180-A\right)$ degrees. The locus will be the arc of the circle. The base a serves as a chord.