Find all triangles with a fixed base and opposite angle

Bobby Mitchell 2022-08-12 Answered
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 / = 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.
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Answers (1)

kilinumad
Answered 2022-08-13 Author has 21 answers
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 ( 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: a sin A = b sin B = 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 sin B
Step 2
Similarly, c = k sin ( 180 ( B + A ) ) = k 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 = ( x 3 x 2 ) 2 + ( y 3 y 2 ) 2 and c = ( 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 sin 2 ( B + A )
( x 3 x 2 ) 2 + ( y 3 y 2 ) 2 = k 2 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.
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