Estimating the angle in which two lines are intersecting

[stathisgar]

Is there any mathematical type through which I can estimate the angle that line I crosses line II?
It can be considered that line II is parallel to the ground and that the beginning of line I intersects line II at a certain spot of line II (attached picture)

 

[lev888]

What do you mean by "mathematical type"?

 

[Dr.Peterson]

What information will be given, from which to determine the angle?

If you only have what you've shown us, then I'd say it's around 60°, because I can imagine something close to an equilateral triangle. I think you want something more than that.

 

[stathisgar]

What do you mean by "mathematical type"?

Mathematical type, I mean a math expression.

What information will be given, from which to determine the angle?

If you only have what you've shown us, then I'd say it's around 60°, because I can imagine something close to an equilateral triangle. I think you want something more than that.

I have the distance AB, BC, and DB. And I can also draw a line from D to C and have the related measurement. What I want is to estimate the angle of DBC

 

[Deleted member 4993]

Mathematical type, I mean a math expression.



I have the distance AB, BC, and DB. And I can also draw a line from D to C and have the related measurement. What I want is to estimate the angle of DBC

Do you know the co-ordinates of the points D, B & C?

Do you know vector dot product?

 

[HallsofIvy]

Cosine rule: Given the lengths of the three sides of a triangle, a, b, and c, then we can calculate the angles, A, B, and C, (c is the side opposite angle C, b is the side opposite angle B and c is the side opposite angle C) from the "cosine rule".

$\displaystyle c^2= a^2+ b^2- 2ab cos(C)$ so $\displaystyle cos(C)= \frac{c^2- a^2- b^2}{2ab}$
$\displaystyle b^2= a^2+ c^2- 2ac cos(B)$ so $\displaystyle cos(B)= \frac{b^2- a^2- c^2}{2ac}$
$\displaystyle a^2= b^2+ c2- 2bc cos(A)$ so $\displaystyle cos(A)= \frac{a^2- b^2- c^2}{2bc}$

In your problem, "a" can be taken as the distance from B to C, "b" can be taken as the distance from B to D, and "c" can be taken as the distance from C to D. Then your desired angle is "C" so use $\displaystyle cos(C)= \frac{c^2- a^2- b^2}{2ab}$.