Length of hypotenuse v/s change in height of the opposite I have always struggled to understand mathematical concepts, and have a very different way of thinking about problems.  I suspect this is a very simple problem, but its confusing me a great deal.
I made a right triangle where the "opposite" side was $10$ and the "adjacent side" was $60$.  I then kept increasing the height of the opposite side by $10$ and measuring the length of the hypotenuse.  This rate of change in the hypotenuse length is not constant and reminds me of a sine wave because of its non-linear acceleration.
However, the rate of change appears to become linear as the graph progresses.
I just want to have an intuition for why the rate of change the length of the hypotenuse is not linear, and if there is one or several different ways I could visualize this to have a better intuition for whats going on.
Graph & Triangle: 
 A: Imagine a right-angle triangle with opposite side of $x$, that you'll keep changing. An adjacent side $a$ a constant that you'll leave alone and fix. Then the length of the hypotenuse is given by $$\sqrt{x^2 + a}$$ 
Now, as $x$ grows very large, the constant $a$ becomes kind of irrelevant. Think of adding $10$ to $20,000,000,000,000$ it'll hardly make a difference to the calculuations, so we can say that $$\sqrt{x^2 + a} \approx \sqrt{x^2} = x$$
for large $x$. For the smaller values of $x$, that extra $+a$ will have quite an impact on the size of the $x^2 + a$ term. 
A: Let me try a slightly different intuition.
Suppose you are out in the open somewhere and you see a hungry lion
$60$ meters west of you and $10$ meters south.
The lion runs faster than you, so your only hope for survival is to
put as much distance between you and the lion as quickly as possible
before it starts to chase you,
and hope that a helicopter will arrive 
and rescue you before the lion catches you.
(If you prefer, substitute "velociraptor" or "zombie" everywhere you
see "lion" in this answer.)
At the start of this scenario, you and the lion are at vertices of a right
triangle whose $60$-meter leg is adjacent to the lion and whose
$10$-meter leg is adjacent to you:

Now it should be fairly evident from the picture that while running due
north will put some extra distance between you and the lion,
it does not increase the distance nearly as quickly as running
in the direction labeled "best escape route" (in the same direction as
the hypotenuse of the triangle, slightly north of east).
Your calculations should bear out the fact that the distance to the lion
(the length of the hypotenuse) does not grow nearly as fast as the
distance you have run.
But suppose you were already $600$ meters north of the lion, not $10$,
at the moment you see each other.
This is the situation in the diagram below:

The "best escape route," which will maximize the distance you put
between you and the lion, is still in the direction of the hypotenuse,
but now that direction is nearly (though not quite) due north.
So if you happen to run due north, you will not get quite as far from
the lion as you might, but you will get very nearly as far as possible
as quickly as possible.
The "run due north" strategy gets better and better the farther north
you start, because the difference between due north and "directly away
from the lion" gets less and less. But you can never get away faster
than running directly away, so that is the limit on how fast the
"run due north" strategy can put distance between you and the lion.
In other words, the rate of change of the hypotenuse starts small,
increases as you increase the length of the leg,
but eventually starts to approach a maximum value.
Therefore it cannot increase linearly.
A: Have a look at the Pythagorian theorem. It solves the problem of your hypotenuse.
Furthermore the equation for the length of the hypotenuse as a function of one of the other sides (whichever you choose) is of second degree. Hence it can not be linear.The resemblence with trigonometric formulas depends (probably) on the fact that several of those formulas are rewritings of the Pythagorian theorem,with sin x and cos x for the two shorter sides.
A: You have drawn a hyperbola. At large values ( x >> 60 ) the slope tends to a constant, to that of its asymptote, so it is becoming straight quite fast.
A: We can always try a little tad o' the method o' fluxions (what Newton called calculus):
Letting the adjacent side have length $a$, the opposite side length $x$, and the hypoteneuse length $h$, we can, in concert with Gilbert and Sullivan's Modern Major General, deploy one of his "many cheerful facts about the square of the hypoteneuse", viz., the Pythagorean theorem :
$h^2 = a^2 + x^2; \tag{1}$
then, differentiating with respect to $x$,
$2h\dfrac{dh}{dx} = 2x, \tag{2}$
or
$\dfrac{dh}{dx} = \dfrac{x}{h}; \tag{3}$
since, from (1),
$h = \sqrt{a^2 + x^2}, \tag{4}$
(3) becomes
$\dfrac{dh}{dx} = \dfrac{x}{\sqrt{a^2 + x^2}}; \tag{5}$
and since ourselves, like our friend the Major, are "acquainted with methods mathematical", and "understand equations both the simple and quadratical", we can do a little algebraic fiddling with (5):
$\dfrac{dh}{dx} = \dfrac{x}{\sqrt{a^2 + x^2}} = \dfrac{x}{x \sqrt{(a/x)^2 + 1}}$
$= \dfrac{1}{\sqrt{(a/x)^2 + 1}} \to 1 \tag{6}$
as $x \to \infty$, since $a/x \to 0$.  We see that the rate of change of $h$ with respect to $x$ approaches $1$ with ever-increasing $x$, i.e., $h(x)$ becomes more and more linear the bigger $x$ gets; this in keeping with our OP Cggart's graph, the slope of which is indeed very close to $1$ for even moderately large $x$.  One can also note that
(3) can be cast in the form
$\dfrac{dh}{dx} = \sin \theta, \tag{7}$
where $\theta$ is the angle 'twixt the side of length $a$ and the hypoteneuse; as $x \to \infty$, $\theta \to \pi/2$, so $\sin \theta \to 1$, consistent with what we have seen so far.  And (7) also allows us to see the sine-like behavior near $x = 0$, as Cggart has observed.  
These things being said, I must rush off to a meeting with the Major, who has telegraphed me that "about binomial theorem he is teeming with a lot of news".  
With thanks to Gilbert, Sullivan, Pythagoras, the Major, and last but by no means least, Sir Isaac.
A: This is a very good question and your diagram is helping me to understand a surveying problem im trying to solve, but anyhow i think i might be able to explain this a little bit, i hope haha.
Without getting into the math it might help to look at the problem by understanding the absolute limits of the triangle you are describing ( a right triangle with a fixed horizantal length) by finding the absolute limits of a geometric shape you will see that all real points that are obtainable fit within the absolute limits.  I hope this makes sense.  Absolute limits are 0 and infinty  both are un-attainable limits and everything must fit inside(or around in the case of zero) these limits.  Make sense?
So because of these two conditions you have created limits for the triangle.  When we take the two variables ( vertical length and hypotenuse length) to their limits ( infinity) we see that they both become infinite or in other words when the vertical length goes to infinity the hypotenuse goes to infinity at the same time.  This means that the vertical length is catching up to the hypotenuse length as the vertical gets longer ( hence not a linear ratio) until the vertical length reaches infinity and then it will equal the hypotenuse length of infinity.  Otherwise the vertical length is always less than the hypotenuse because infinty is un attainable.  This is why your graph stables out  vertical length is catching up to hypotenuse length the longer it gets.
Now, look at it another way.
If you take the vertical length to 0  you will see that the hypotenuse reaches a new limit of?.....  the adjacent side length.  And now when the vertical length is 0 it is no longer a triangle it is just a line.   As soon as the vertical length is anything greater than zero you will see that the hypotenuse becomes just greater than the adjacent length but it will never be less than or equal to the adjacent length or else you wont have a right triangle.   
Another limit of this triangle exist within the angle between the hypotenuse and the adjacent side.  This angle can go from anything greater than 0 to anything less than 90 degrees.  It will never be 0 or 90 degrees or else again the triangle breaks down and becomes a line. Even as that vertical line goes onto infinty the angle between the adjacent and hypotenuse will only approach 90 degrees but will never be 90 degrees. 
I hope this sheds some practical light on your mathmatical question. 
Cheers! 
