I try to find a way to calculate value of one of the isosceles triangle angles when I have given values of its height h = 200 and base x = 200. Values of those can vary depend on the condition, so I need an universal solution. I'll be very grateful for your help.
Best Answer
If you have a computer, you can say that
Alternatively, you can use the simpler form
a = arctan(x/(2h))
If you type this into google, for particular values of $x$ and $h$, you'll get the answer you want. For instance, searching for
returned the answer
The query
produced the same thing.
A bit more detail.
Computing sine, cosine, and tangent isn't as easy as square root, but it's not insanely hard. It turns out that if you know $\tan(a)$ and $\tan(b)$ you can use "addition formulas" to compute $\tan(a \pm b)$ and $\tan(a/2)$. Starting from one known value, like $\tan(45^{\deg}) = 1$, you can compute many other values, enough to build a pretty complete table, and then fill in by interpolating. This takes time and energy, but that's life. Once you have a table of tangents, you can swap the columns to give you "inverse tangent" -- a function that says "what angle has this tangent?" That's called "arctan".
In fact, you can build an arctan table using addition rules as well, which is a bit more direct. Proving the addition laws? That's what trigonometry is all about.
It turns out that $\arctan(x)$ can be expressed as a polynomial in $x$...but one with infinitely many terms. Fortunately, for small $x$, most of the terms are very small, so you can get away with evaluating just the first few (i.e., the larger) terms without making much error. This is in fact what the Java Math library's authors do, albeit in a somewhat more sophisticated way. The proof that arctan can be written as a polynomial comes up in calculus, under the general heading of "Approximation by Polynomials", or more specifically, "Taylor Series". Textbooks have multiple chapters about these, so I can't explain it all here, of course. Wish I could, but...