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?
2 Answers
$\begingroup$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 $C$.
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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