Calculate Third Point of Triangle I'm trying to calculate the third point of a triangle:
I know two points, $(2,3)$ and $(5,2)$ and the angles at this sides, both of them angles of $30^\circ$:
I made a drawing showing my example case:

I was able to get the distance between $(2,3)$ and $(5,2)$ = $3,1622776601683795$
But I don`t know how to get the third point! Do you maybe have an idea?  Thanks
 A: 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.
A: Just a hint : when calculating analytically you should exact value of the distance, Which is $\sqrt{10}$, and not use the approximation.
To solve your problem:
A  vector space is equipped with an inner product in order to being able to calculate angles between elements of that space. For $\mathbb{R}^2$ (and higher dimensions) one can use the relation
$$\frac{\langle x,y\rangle}{\|x\|\,\|y\|} = cos(\alpha)$$
The vector connecting your two given points is $$x=\begin{bmatrix}5-2\\2-3\end{bmatrix}=\begin{bmatrix}3\\-1\end{bmatrix}$$
Im gonna give you the construction for tasks like yours:


*

*Use the angle relation above to find some vector y that satisfies the angle condition ($\alpha=30$) with x, where x is the vector connecting your given points. 

*Construct a line starting at the point (5,2) in direction of y

*Proceed same as in 1. Step, but with -x to get a vector y'

*Same as step 2, now starting at the point (2,3) in direction of y'

*Connect the intersection of the 2 lines (see your sketch). This unique point is the third point if the triangle you were looking for 

