geometrytrianglestrigonometry
I have a triangle with 3 Points. The positions of two points are available. And also i know the distances between the three Points. How is it possible to calculate the position of the third point?
There are many ways to find your desired point. Here is one way. Since you calculated a length as an approximation rather than exactly, I'll also use approximations for irrational values.
First, let's find the distance $d$ of the desired point from our "base point" $(2,3)$. The base point $(2,3)$, the midpoint of the line segment between the two given points, and the desired point form a $30°$-$60°$-$90°$ triangle. The short leg has half the length of the line segment that you already calculated. Let's call that $\frac L2$, where $L$ is your calculated length. We then use the $30°$ angle in the right triangle to find the desired distance $d$:
$$\cos 30°=\frac{\frac L2}{d}$$ $$d=\frac L{2\cos 30°}=1.82574185835$$
Now we want to find the angle of inclination from the base point to the desired point. The angle of inclination from $(2,3)$ to $(5,2)$ is $\tan^{-1}\frac{-1}3=-18.4349488229°$. We add $30°$ to that to get our desired angle of inclination $\theta=11.5650511771°$.
We now find the desired point by its position relative to the base point.
$$\begin{align} (x,y) &= (2+d\cos\theta,3+d\sin\theta) \\ &= (3.78867513459,3.36602540378) \end{align}$$
This answer checks in Geogebra.
As I state in a comment to this answer, the method above for finding $d$, the distance of the desired point from the base point, works only for isosceles triangles where you know the endpoints of the base as well as the angles there. If the angles at the two known vertices differ, there is another way.
Let's say the length of the known triangle side is $L$, the known angle at the base point is $\alpha$ and the angle at the other known vertex is $\beta$. We can use the law of sines to get
$$\frac{\sin(180°-\alpha-\beta)}{L}=\frac{\sin(\beta)}{d}$$
so
$$d=\frac{L\sin(\beta)}{\sin(180°-\alpha-\beta)}$$
The rest of my method above works after this. This more general method also works and checks in your particular problem. Let me know if you want me to edit my answer above to only show the more general method: that might be easier for you.
Let us fix some notations:
let $\alpha_1,\alpha_2,\alpha_3$ be "angle1,angle2,angle3" resp.
let length of $p_1p_2 = a_3$ and the other sides' lengthes names by cyclic permutation.
let $u=x_2-x_1, v=y_2-y_1$. Thus $a_3=\sqrt{u^2+v^2}$.
First of all: $\alpha_3=\pi-(\alpha_1+\alpha_2)$.
Then, using the law of sines (https://en.wikipedia.org/wiki/Law_of_sines):
$$\dfrac{a_1}{\sin \alpha_1}=\dfrac{a_2}{\sin \alpha_2}=\dfrac{a_3}{\sin \alpha_3}$$
one obtains in particular $a_2=a_3\dfrac{\sin \alpha_2}{\sin \alpha_3}$ where where $\alpha_2,\alpha_3$ and $a_3$ are known quantities.
Let us now express
the dot product $\vec{p_1p_2}.\vec{p_1p_3}=a_2 a_3 \cos \alpha_1$ and
the norm of the cross product $\|\vec{p_1p_2}\times\vec{p_1p_3}\|=a_2 a_3 \sin \alpha_1$
by using coordinates:
$$\begin{cases} u(x_3-x_1)+v(y_3-y_1)&=&a_2 a_3 \cos \alpha_1\\ u(y_3-y_2)-v(x_3-x_2)&=&a_2 a_3 \sin \alpha_1\end{cases}$$
One obtains a linear system of 2 equations with the two unknowns $x_3$ and $y_3$ ; the solution of this system is without difficulty.
Here is a complete Matlab program with explicit formulas for $x_3$ and $y_3$:
x1=0;y1=0;x2=6;y2=0; % initial data
alp1=2*pi/3;alp2=pi/6; % initial data
u=x2-x1;v=y2-y1;a3=sqrt(u^2+v^2);
alp3=pi-alp1-alp2;
a2=a3*sin(alp2)/sin(alp3);
RHS1=x1*u+y1*v+a2*a3*cos(alp1);
RHS2=y2*u-x2*v-a2*a3*sin(alp1);
x3=(1/a3^2)*(u*RHS1-v*RHS2);
y3=(1/a3^2)*(v*RHS1+u*RHS2);
Best Answer
You know the following two identities: $$AC = \sqrt{(C_x-A_x)^2+(C_y-A_y)^2}\quad\quad(1)$$ $$BC = \sqrt{(C_x-B_x)^2+(C_y-B_y)^2}\quad\quad(2)$$ So you have two equations with two unknowns. You can solve using any method your familiar with. Here's the first few steps using substitution (to get you started).
Solve $(1)$ for $C_x$ to obtain:
$$ C_x=\sqrt{AC^2-(C_y-A_y)^2} +A_x\quad\quad(3)$$ (We ignored the minus square root option, why?)
Now plug $(3)$ into $(2)$ and solve for $C_y$. (This will give you a numeric answer for $C_y$, which you can then plug into $(3)$ to get a numeric answer for $C_x$, make sense?)
Aside: Note that we didn't use $AB$. Since we already know $A_x,A_y,B_x,$ and $B_y$, $AB$ doesn't give us any new information.