Triangles of different size; determining the height difference

Question

I want to find how far back I should move to take a 35mm equivalent photo on a 1.6x crop sensor camera. Effective focal lengths are multiplied by 1.6x when using this crop sensor, so a photo shot at 20mm is instead 32mm, 50mm is 80mm, etc. To try and figure this out, I've made a diagram of two triangles:

The two triangles have the same base width but have varying heights. The topmost angle of the green triangle is $25°$ and the black triangle is $40°$. I've chosen these arbitrarily, the only important bit is the black triangle's angle is $1.6$ times the green one. $40 = 25 * 1.6$.

To help illustrate, I've drawn a red box between the two triangle bases, representing the distance needed to achieve the same focal length.

What is the height of the red box, or rather the difference between heights of the two triangles?

Answer 1

Draw the vertical from the top point, which will bisect the angle, and the bases of the two triangles.

Then the height $h$ of the green triangle is $h=\frac{50/2}{\tan(25^{\circ}/2)} \simeq 112.8$.

The height $h'$ of the black triangle is $h'=\frac{31.25/2}{\tan(40^{\circ}/2)} \simeq 42.93$.

The height of the red box is the difference between the two $h-h'\simeq 69.9$.

The end result will not vary linearly with the ratio of the angles $\frac{40}{25}=1.6$

Answer 2

The image size is the object size multiplied by the distance to the image divided by the distance to the object. If you want your image size reduced by a factor $1.6$, keeping the focal length the same, for a far object you need to be at a distance $1/1.6$ as far. This will be true as long as you use the same focal length lens and the object is far enough away that you can consider the distance from the lens to the sensor to be the lens focal length.