How many convex angles, not including straight angles, are formed by n straight lines that intersect in one point?
Example: A convex angle is an angle between zero and 180. If 2 lines intersect at a point they create four convex angles and if n = 3, you can count 12 different convex angles and at n = 4, there is 24 convex angle and so on.
When n straight lines intersect at a single point, they create 2n vertical angles and combinations of these angles result in new angles. However, not all of these angles are convex. Some angles might be larger than 180, forming concave angles or straight angles. To find the number of convex angles, we need to subtract the number of non-convex angles (concave angles and straight angles) from the total number of angles.
I am pretty much stuck and the only formula that's is working is:
the number of convex angles formed by n straight lines that intersect at a single point is $2(n^2 -n)$, not including straight angles.
but i have no idea how to derive it.
When $n$ lines intersects at a point, you can basically choose any two of the lines and those two lines gives you four angles as you already pointed out.
That means we basically have ${n \choose 2} \cdot 4$, which gives your proposed formula.
If you are not familiar with combinatorics yet, ${n \choose 2}$ is the number of ways to choose two things from $n$ things, and equals $n(n-1)\over 2$. You can think of it this way: first you choose the first element, which has $n$ ways, then you choose the second element, which has $n-1$ ways, so in total there are $n(n-1)$ ways. However we double counted the result because the order we choose the two elements doesn't matter, so we divide by $2$ at last.