For a calculated result with an 18-270 lens on a 15.5 X 23.2 sensor, I get 5°55’ and 75°33’ which matches what your program computes. You seem to have typed the telephoto result wrong in your posting as 5°51’, though.

There’s a small error in the size you’ve stated for a 135 film image on your calculator page and in the code. It is not 24mm X 35mm. It’s 24mm X 36mm.

I can’t find an area for the sensor size noted in the Casio manual. I did find “1/2.5-inch square pixel” sensor which means that the pixels are square (which is normal for most cameras) not that the 1/2.5-inch value refers to the sensor’s area.
Unfortunately, the 1/2.5-inch notation isn’t even remotely associated with reality. It’s an arcane and absurd way of describing a digital image sensor’s size. It goes back to vidicon tubes referring to the size of the glass tube. In my view-- and I’ve got lots of company-- using that to describe a digital camera sensor doesn’t make sense but manufacturers sometimes use it. But I digress.
The size of a 1/2.5-inch digital camera sensor is actually 5.76mm X 4.29mm. So, the diagonal is 7.18mm and its area is 24.71mm^2. This area is almost 25mm^2 which explains why using 1/25 for the area in square inches gives you a result that makes some sense since an inch is 25.4mm-- but it’s just a coincidence that it almost works in this case.

Focal length does not depend on the sensor size. It’s the angle of view that depends on the sensor size. Focal length is independent of sensor. Focal length is where an object at infinity forms an image. The size of what’s “seeing” that image doesn’t change where the image is formed but only how much (angle of view) of it is seen.