Find all triangles with a fixed base and opposite angle

Question

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 / A = 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.

apologies if my formatting is poor - first post on math stack exchange - i am more a stack overflow sort of guy..

Answer 1

The locus of the vertex is a part circle, due to the theorem " angles in the same segment are equal"

Answer 2

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 $(x_1,y_1)$ and $(x_2,y_2)$. For a fixed $A$, we need to find the mathematical equation of the locus of third point, $(x_3,y_3)$.

Using sine rule, we have: $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k $$ where $k$ is easily found since we know $a$ and $A$.

In the triangle, angle $C = 180 - (B+A)$. We also see that: $$ b = k \cdot \sin B $$ Similarly, $$ c = k \cdot \sin (180-(B+A)) = k \cdot \sin (B+A) $$ 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: $b= \sqrt{(x_3-x_2)^2+(y_3-y_2)^2}$ and $c= \sqrt{(x_3-x_1)^2+(y_3-y_1)^2}$.

The locus of $(x_3,y_3)$ is given by simultaneously solving: $$ (x_3-x_1)^2+(y_3-y_1)^2 = k^2 \cdot \sin ^2 (B+A) $$ $$ (x_3-x_2)^2+(y_3-y_2)^2 = k^2 \cdot \sin ^2 (B) $$ when $B$ varies from $0$ to $(180-A)$ degrees. The locus will be the arc of the circle. The base $a$ serves as a chord.

Answer 3

Here is the fleshing out of my geometric approach, inspired by David Quinn's input utilising the 'angles for the same segment are equal' theorem and the 'central angle theorem'

for case where chord is horizontal (parallel to x-axis)

(and any chord can be made to be parallel by rotating the circle appropriately)

$$ a = \frac{x_2-x_1}{2} $$ $$ r = \frac{a}{\sin(\theta)} $$ $$ r = \frac{\left(\frac{x_2-x_1}{2}\right)}{\sin(\theta)} $$ $$ b = \frac{a}{\tan(\theta)} $$ $$ b = \frac{\left(\frac{x_2-x_1}{2}\right)}{\tan(\theta)} $$ therefore if the origin, $o$ has cartesian coordinates $[o_x,o_y]$ $$ o_x = x_1 + \frac{x_2-x_1}{2} $$ $$ o_x = \frac{x_1+x2}{2} $$ $$ o_y = y_1 + \frac{\left(\frac{x_1+x2}{2}\right)}{\tan(\theta)} $$ so we know the origin, $[o_x,o_y]$ and radius, $r$, of the circle prescribing the locus of all solutions

so if we want to convert back from polar coordinates to cartesian coordinates we use the following:

$$ x = r\cos(\phi) $$ $$ y = r\sin(\phi) $$ $\phi$ is the angle as we travel round the circle, $r$ the radius

this assumes the centre of the circle is at the origin so that has to be corrected, e.g. $$ x = r\cos(\theta) + \frac{x_1+x2}{2} $$ $$ y = r\sin(\phi) + y_1 + \frac{\left(\frac{x_1+x2}{2}\right)}{\tan(\theta)} $$

Answer 4

Vertices of the triangles sought form two circle's arcs.

See Inscribed angle — Theorem on Wikipedia.