Find all triangles with a fixed base and opposite angleI 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-axisthen 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 tastenow 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 degI then need to extend to arbitrary bases and opposite angles, but thought starting with a nice simple one might be a good stepping stone.

Question

Find all triangles with a fixed base and opposite angleI 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-axisthen 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 tastenow 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 degI then need to extend to arbitrary bases and opposite angles, but thought starting with a nice simple one might be a good stepping stone.

kilinumad · Accepted Answer

Step 1A 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        =  kwhere 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  ⁡  BStep 2Similarly,   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.

Find all triangles with a fixed base and opposite angle

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