You don't actually need trigonometry — just basic arithmetic. The zoomed-in focal length gives a field of view as if you'd cropped the image by the ratio of the old focal length over the new: that is, if you have an image taken at 50mm, you can see the field of view of a 75mm lens simply by cropping by ⁵⁰⁄₇₅ths — which is ⅔.
This simple relationship is why the "crop factor" (sometimes, unfortunately, called "focal length multiplier") works. If your sensor is ⅔ the width of a full-frame sensor, that's cropping by a factor of 1.5 (the inverse of ⅔). So, you get the field of view of a lens with 1.5× the focal length on full frame — a 50mm lens on APS-C gives you the same field of view as a 75mm lens on full frame.
To put some numbers to it: if your 50mm-focal-length starting point is a 6-megapixel 3000×2000 image, cropping it to 2000×1333 will give you the field of view of a 75mm lens: in pixels, 3000 × 50 ÷ 75 horizontally, and 2000 × 50 ÷ 75 vertically. (A tangent, if you pardon the trig pun: You'll notice that this is quite a big hit in resolution — you loose a number of pixels equal to the crop factor — the ratio between focal lengths — squared. This is why optical zoom is usually preferable to "digital zoom", which is just cropping. And, generally, smaller sensors cram more pixels into the smaller sensor in order to compensate for the crop, which, depending on the level of technology used works to some degree. But that's a whole different discussion.)
You can use simple (non-trig) geometry to demonstrate this.
You'll need a ruler with millimeter markings, and a blank sheet of paper. I could make some graphics showing all this, but I really strongly believe that it's an exercise which works better if you actually go through it in on actual physical paper. So, if you will humor me and work along....
Along the very bottom edge of the paper, centered in the middle, draw a horizontal line 24mm long. This represents an APS-C sensor.
Measure 50mm up from the very center of that line, and put a dot. This represents the gathering of light within an idealized 50mm lens. (Imagine it as a pinhole camera, if you like.)
Now, draw a line from the left edge of the sensor through the "lens" dot, and continue up through to the top of the paper. Do the same from the right edge, giving you an X shape with the lens point at the center of the X. The top cone of the X represents the horizontal field of view of a 50mm lens on your APS-C sensor.
You can measure the angle with a protractor, if you happen to have one — it should be about 27°. And you can measure the horizontal field of view in millimeters a given distance away from your "camera", by measuring across the cone at the top of the X. (At 10cm away from the idealized lens dot, it should be about 4.8cm.)
Now, measure up 75mm from the middle of your "sensor" line and put another dot, representing an idealized 75mm lens.
Draw an X from the sensor edges through this dot as well. If you measure this angle, it should be about 18.2 degrees, and again, 10cm up from the lens dot, if you measure across, it should be about 3.2cm.
And hey presto: 4.8mm × ⁵⁰⁄₇₅ = 3.2mm. (Of course, your lines are not at the exact same distance from the sensor itself, since you're measuring from the dot representing the lens in order to get the math to come out so nicely. But here we're working with unusually close focusing distances — when you're talking about a subject at normal distances the difference is negligible.)
So anyway, you can then extend your sensor to be 36mm across instead of 24mm — changing it from APS-C to full-frame. Now, draw lines from that new larger sensor through the existing 75mm "lens" point.
Even without measuring, you should be able to see that the angle of view with the larger sensor through the 75mm lens is the same as that with the smaller sensor through the 50mm lens. So there's the "crop factor" equivalence right in front of you. Cool, huh?
Note that this only covers angle of view. Perspective won't change, because you're standing in the same place, but depth of field (and the distribution of the depth of field) will. And of course actual real-world different lenses will have different properties (like distortion) which aren't modeled by this.
But in terms of field of view, that's all there is to it. Nothing beyond middle-school math required.