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I have 3 triangles joined together with common legs between them. For triangle ABC (the blue one in the diagram below) we know all its angles and the length of all it's sides. For the other two triangles, BCO and ACO (the red ones in the diagram below) we known one side and one angle for each. I want to calculate the unknown sides and angles of triangles BCO and ACO.

How would I go about doing this?

enter image description here

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  • $\begingroup$ Something is missing. you can move point $O$ around so there are multiple solutions. $\endgroup$
    – Vasili
    Apr 28, 2022 at 16:25
  • $\begingroup$ For the sake of constraining this in a coordinate system you could assume that O is located at the origin (0,0). Further we can say that the rotation about O is fixed by known angles for sides BO, CO, AO to the vertical axes. $\endgroup$
    – math-anon54
    Apr 28, 2022 at 16:37
  • $\begingroup$ my point is, there will not be a unique solution based on the information $\endgroup$
    – Vasili
    Apr 28, 2022 at 17:19

2 Answers 2

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From the midpoint $M$ of $AC$ construct the perpendicular bisector of $AC$, and take on it (outside triangle $ABC$) a point $P$ such that $\angle CPM=\angle COA$.

From the midpoint $N$ of $BC$ construct the perpendicular bisector of $BC$, and take on it (outside triangle $ABC$) a point $Q$ such that $\angle CQN=\angle COB$.

Point $O$ is the second intersection of the circle centred at $P$ through $C$, with the circle centred at $Q$ through $C$.

enter image description here

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  • $\begingroup$ Very interesting, how would you use this to arrive at the unknown sides of triangle ACO and ACO? Also does this hold true if angles BOC and COA are not equal? $\endgroup$
    – math-anon54
    Apr 28, 2022 at 22:20
  • $\begingroup$ @math-anon54 Radii $PC$, $QC$ can be easily computed, as well as $\angle PCQ$. From them one can find $PQ$ and the distance $CH$ of $C$ from $PQ$, which is one half of $CO$. Finally you can also find $\angle PCO$ and $\angle QCO$. This works for any value of $\angle AOC$ and $\angle BOC$. $\endgroup$
    – Intelligenti pauca
    Apr 29, 2022 at 9:18
  • $\begingroup$ @math-anon54 By the way, this construction shows that the solution is indeed unique. And you don't need to solve a sixth-degree equation. $\endgroup$
    – Intelligenti pauca
    Apr 29, 2022 at 9:21
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enter image description here

You have the situation depicted in the figure above. All the known information is drawn in black. Let $x = OA, y = OB, z = OC$ and let $a = BC , b = AC, c= AB$. Further, let $\theta = \angle AOC, \phi = \angle BOC $

Then from the law of cosines,

$ b^2 = x^2 + z^2 - 2 x z \cos(\theta) $

$ a^2 = z^2 + y^2 - 2 y z \cos(\phi) $

$ c^2 = x^2 + y^2 - 2 x y \cos(\theta + \phi) $

These are three equations in $x,y,z$ and can be solved numerically by an iterative method such as the Newton-Raphson multivariate method. Once $x,y,z$ are found, then everything else follows.

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  • $\begingroup$ This is perfect and exactly what I was looking for. I should be able to further constrain the problem for my particular use case. For example I know that my result must be a positive real integer since I'm dealing with actual lengths. $\endgroup$
    – math-anon54
    Apr 28, 2022 at 21:17
  • $\begingroup$ Hi @GrabaCoffee, this is unrelated to your answer, hence I would be deleting this comment soonest. Please could you help out with this question math.stackexchange.com/q/4435979/585488 $\endgroup$
    – linker
    Apr 29, 2022 at 8:09
  • $\begingroup$ @LiNKeR I took a look at your question, but don't know the answer. $\endgroup$
    – Hosam H
    Apr 29, 2022 at 9:23

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