Derivation for the 1d wave equation

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So I'm working on PDEs, and currently trying to understand the derivation of the 1d wave equation. I can follow most of this derivation just fine, but when I try it myself I run into a snag I'm not sure how to conceptually address. In the text (PDEs by Strauss), the author uses this diagram. This diagram is meant to demonstrate the directions of the tension force along a string, such that newton's laws can be used to derive the PDE.

The diagram shows how to determine the magnitude of the tension in both the x and u (horizontal and vertical respectively) directions. This should just be some basic trig, and so it represents this with a triangle that has a vertical side of $u_x$ and a base of unit length (and thus the hypotenuse is given as $\sqrt{1 + u_x^2}$)

But when I try to approach this problem, without following along, I end up drawing a similar triangle, but this time I make my base a $dx$, and the vertical a $du$. Following this through I'd find:

$\sqrt{dx^2 + du^2} = \sqrt{dx^2 + \frac{dx^2 du^2}{dx^2}} = \sqrt{dx^2 (1 + u_x^2)} = \sqrt{1 + u_x^2}dx$

(Thats just the integral form of the arc length equation, which makes sense).

What I'm doing feels conceptually correct. I'm interested in the tension at two points arbitrarily close together (separated horizontally by $dx$ and vertically by $du$). But that additional $dx$ in my result leaves me a bit stuck on how to proceed. And I don't understand how to justify just setting it equal to 1 (which appears to be what the author does).

Am I overlooking something simple here, or do I have some fundamental misunderstanding of what the author is actually doing? I'm just confused as to the justification of this single operation. Any help on this would be greatly appreciated!

Notation note:

$u_x = \frac{\partial u}{\partial x}$

EDIT: So I decided to pick the problem back up today and just decided to work through my confusion a bit. I noticed that I end up arriving at the same solution in the end. The reason being that the $\cos{(\theta)}$ and $\sin{(\theta)}$ for Strauss work out to directly be:

$\frac{1}{\sqrt{1 + u_x^2}} \ \ and \ \ \frac{u_x}{\sqrt{1+u_x^2}}$

respectively. Mine on the other hand, do indeed simplify to that as:

$\frac{dx}{\sqrt{1 + u_x^2}dx} = \frac{1}{\sqrt{1 + u_x^2}} \ \ and \ \ \frac{du}{\sqrt{1 + u_x^2}dx} = \frac{u_x}{\sqrt{1+u_x^2}}$

So there really is no problem. I'm still a bit conceptually confused as to how the author knew to make such a substitution, but I suppose thats fine for now. I'm updating the post just in case someone else comes across this same issue (unlikely, but possibly still useful to someone!)