Triangles and intercepts

[.... deleted; based on over-simplified assumptions that were never specified ....]

Assuming that the right angle vertex is at the origin and the base of the triangle extends into the positive x axis:

Formula for the new vector is (please don't use x for an angle, it's for an independent variable. Use Greek letters for angles.):
y = Tan(θ)

Formula for the hypotenuse is:
y = -4/3x + 4

If we set the two expressions equal we can determine the intersection:
tan(θ) = -4/3x + 4

Solve for x
x = -(3/4)*(tan(θ) - 4)

Now you know x, so you can plug x back into either of the first two equations to solve for y.

Please disregard this solution and see Croweater's post for a correct solution.

If you want an implemented Excel solution, please attach an Excel file to show your sample data.

Assuming the orientation of the triangle is as in post #3, the line will intersect at a point (x,y) such that.

Tan(θ) = y/x (This is where I disagree with post #3)

ie. x = y/Tan(θ)
y = xTan(θ)

Also (as stated in post#3 above) we know that x and y satisfy the equation of the hypotenuse. ie.

y = -4x/3 + 4 (Another point I disagree with post#3)

So we have 2 equations and 2 unknowns. Solve them simultaneously.

xTan(θ) = -4x/3+4
x(Tan(θ)+4/3) = 4

x=4/(Tan(θ)+4/3)

y = 4Tan(θ)/(Tan(θ)+4/3)

Originally Posted by Croweater

y = -4x/3 + 4 (Another point I disagree with post#3)

My notation was poor. I should have written
-(4/3)x + 4
which is effectively the same as yours.

I'll look at the other bit when I get some time, but I think you are correct.