I know a weak bound that if in a plane there are finite points then atmost $2/3$ of the triangles formed by them will be acute angled.

according to an old IMO problem 6 [the year I dont remember] , at most 70% of the triangles are acute. I dont know f the bound is attainable.

If 3 lines define a triangle, then whether it is acute or not depends only of their slopes. So we can replace a line by a parallel one (avoiding three concurrent lines) without changing the acuteness of the triangles. This means we can assume wlog that all lines are tangents of a common circle. It should be obvious now (note that I haven't said it is obvious ) that the best solution is obtained if the lines are the sides of a regular (2n+1)-gon, (equivalently. to give the whole picture, the longest diagonals of a regular (2n+1)-gon). In such a constellation, there are $\frac13(2n+1)\binom{n+1}{2}=\frac14 \binom{2n+2}{3} $ acute triangles, which makes their proportion $\frac{n+1}{2(2n-1)}$. This is roughly $\frac14$ as $n$ grows, far from the promised 70%. Equally the problem can be stated as follows in terms of the slopes: Given a set $S$ of (2n+1) distinct nonzero reals such that $ab\ne-1$ for all $a,b\in S$, what is the biggest possible number of sets $\{a,b,c\}\subset S$ such that $ab>0>ac>-1>bc$?.