a rectangle has w=40, L=63 and a straight line joining the diagonal corners resulting in 2 righted angled triangles. A straight line 16 units long perpendicular to the diagonal line touches the sides of the rectangle.
What would the coordinates be where the 16 unit line touches the sides of the rectangle and crosses the diagonal line?
I tried to solve it, but I get the vertical lenght (or coordinates) as nearly 16 units [which cannot be since the height of the rectangle is 4 units], since the angle between the sides of the diagonal and the side of the rectangle is so small. Are you sure the dimensions of your rectangle are good?
Ok, I took the diagonal from the upper left corver to the lower right corner, and the 16 units line crossing the diagonal near the lower right corner.
You can find the angle that the line makes with the horizontal, using the tan ratio.
You should get: 32.4 degrees
The vertical height to which the line reaches is then 16 cos (32.4) = 13.6 units (3sf)
The horizontal distance from the origin (I take it as the lower left corner) = 63 - 16 sin (32.4) = 54.4 units.
So, coordinates are:
(54.4, 0) and (63, 13.6)
Well here's what I get. Refer to attached figure (to scale).
This part could be achieved by just considering similar triangles.
Assuming that the bottom left corner of the rectangle is at (0,0) and working out some other stuff, I get that the line segment goes from approximately (21.27416, 13.50740) to (29.85029, 0).