trianglestrigonometry
I simply have a triangle.. say abc . Given coordinates of a & b. and the length of
ac and bc..
I can calculate the length between ab via the distance square rule.
$D = \sqrt{(X_2 – X_1)^2 + (Y_2 – Y_1)^2} $
I have tried to use the distance rule to compute the third vertex, but it has a lot of steps..
I wonder if there is another method.. that yields the two possible solutions for the vertex
You have two points $A=(a,b)$ and $B=(c,d)$ and want a point $P$ at given distances from $A$ and $B$, say $l$ and $m$. Then $|PA|^2=l^2$ and $|PB|^2=m^2$ that is $$(x-a)^2+(y-b)^2=l^2\qquad\qquad(1)$$ and $$(x-c)^2+(y-d)^2=m^2.\qquad\qquad(2)$$ Subtracting (2) from (1) gives a linear equation. Use this to eliminate one variable from (1). This yields a quadratic equation in the other variable. Solving this will give the two possible positions for $P$.
Here is a step by step approach.
$1.$ As we have the coordinates of $B$ and $C$, we can compute a unit normal vector to the line segment $BC$. This is how yo may do this
$$\begin{align} {\bf{L}}&= {\bf{x}}_B-{\bf{x}}_C\\ &=(x_B-x_C){\bf{i}}+(y_B-y_C){\bf{j}} \\ \\ {\bf{N}}&= {\bf{k}} \times {\bf{L}}\\ &=-(y_B-y_C){\bf{i}}+(x_B-x_C){\bf{j}} \\ \\ {\bf{N}} \cdot {\bf{L}} &= 0 \\ \\ {\bf{n}} &= \frac{\bf{N}}{\|\bf{N}\|} \\ &= \frac{{\bf{k}} \times {\bf{L}}}{\|\bf{L}\|} \\ &= \frac{{\bf{k}} \times ({\bf{x}}_B-{\bf{x}}_C)}{\|{\bf{x}}_B-{\bf{x}}_C\|}\\ &=\frac{-(y_B-y_C){\bf{i}}+(x_B-x_C){\bf{j}}}{\sqrt{(x_B-x_C)^2+(y_B-y_C)^2}} \end{align}$$
$2.$ We can observe from the figure below that
$$d = ({\bf{x}}_B-{\bf{x}}_A)\cdot{\bf{n}}$$
so we have found the length of $AD$.
$3.$ The coordinates of $D$ will be
$$\begin{align} {\bf{x}}_D &= {\bf{x}}_A + d{\bf{n}} \\ &= {\bf{x}}_A + [({\bf{x}}_B-{\bf{x}}_A)\cdot{\bf{n}}]{\bf{n}} \\ \\ &= {\bf{x}}_A + \left[({\bf{x}}_B-{\bf{x}}_A)\cdot\frac{{\bf{k}} \times ({\bf{x}}_B-{\bf{x}}_C)}{\|{\bf{x}}_B-{\bf{x}}_C\|}\right]\frac{{\bf{k}} \times ({\bf{x}}_B-{\bf{x}}_C)}{\|{\bf{x}}_B-{\bf{x}}_C\|} \\ \end{align}$$
and finally
$$\boxed{{\bf{x}}_D= {\bf{x}}_A + \frac{{\bf{k}} \cdot \left[ ({\bf{x}}_B-{\bf{x}}_C) \times ({\bf{x}}_B-{\bf{x}}_A) \right] \ }{\|{\bf{x}}_B-{\bf{x}}_C\|^2} {\bf{k}} \times ({\bf{x}}_B-{\bf{x}}_C)}$$
you can write the above vector equation as the following two scalar ones
$$\boxed{ \begin{align} x_D &= x_A - \frac{\begin{vmatrix} x_B-x_C & y_B-y_C \\ x_B-x_A & y_B-y_A \\ \end{vmatrix}}{(x_B-x_C)^2+(y_B-y_C)^2} (y_B-y_C) \\ \\ y_D &= y_A + \frac{\begin{vmatrix} x_B-x_C & y_B-y_C \\ x_B-x_A & y_B-y_A \end{vmatrix}}{(x_B-x_C)^2+(y_B-y_C)^2} (x_B-x_C) \end{align} }$$
Also, here is a MAPLE file that can help the reader of this post for the computation of final formulas mentioned here and the verification of the example mentioned in the question.
Best Answer
Let $A = (x_A, y_A)$ and $B = (x_B,y_B)$ the known vertices of your triangle. Let's call $d_{AB}$, $d_{BC}$ and $d_{CA}$ the lengths of each side.
$$A' = (0, 0), B' = (x_B-x_A, y_B-y_A ) = (x_B', y_B').$$
$$A'' = (0,0), B'' = (d_{AB}, 0).$$
Anyway, the value of the rotation angle is important for the next steps. In particular it is $$\theta = \arctan2\left(y_B-y_A,x_B-x_A\right),$$
where $\arctan2(\cdot, \cdot)$ is defined in details here.
$$x_C'' = \frac{d_{AB}^2+d_{AC}^2-d_{BC}^2}{2d_{AB}},$$
and
$$y_C'' = \pm\frac{\sqrt{(d_{AB}+d_{AC}+d_{BC})(d_{AB}+d_{AC}-d_{BC})(d_{AB}-d_{AC}+d_{BC})(-d_{AB}+d_{AC}+d_{BC})}}{2d_{AB}}.$$