Drawing rectangle triangles with labels only in its sizes

Question

I'm doing an exercise of integral calculus. The integral is $\int \dfrac{dx}{4-x^2}$. This integral can be solved doing $x = 2\sin\,\alpha$, and its solution is $\dfrac{1}{2}\ln\,|\sec\,\alpha + \tan\,\alpha|+C$. I need to draw a rectangle triangle to avoid doing all the tedious maths, trying to find $\sec\,\alpha$ and $\tan\,\alpha$ in terms of $x$. In the image is what I'm trying to draw.

Answer 1

With tkz-euclide:

\documentclass[border=5mm]{standalone}
\usepackage{tkz-euclide}
\usetkzobj{all}
\begin{document}
\begin{tikzpicture}[scale=.8]
\tkzInit[xmax=5,ymax=3] %\tkzClip[space=.5]
\tkzDefPoint(0,0){A} \tkzDefPoint(4,0){B}
\tkzDrawTriangle[pythagore](A,B)
\tkzGetPoint{C}
\tkzMarkRightAngle(A,B,C)
\tkzDrawPoints(A,B,C)
\tkzLabelPoints[below](A,B)
\tkzLabelPoints[above](C)
\tkzLabelSegment[below,font=\footnotesize](A,B){$\sqrt{4 - x^2}$}
\tkzLabelSegment[above,font=\footnotesize](A,C){$2$}
\tkzLabelSegment[right,font=\footnotesize](B,C){$x$}
\tkzMarkAngle[fill= blue!40,size=1.4cm,opacity=.5](B,A,C)
\tkzLabelAngle[pos=0.8](B,A,C){$\alpha$}
\end{tikzpicture}
\end{document}

Answer 2

An attempt with MetaPost, in LuaLaTeX.

Given a real x (supposed to be greater than 0 and lower than 2), it uses the Pythagorean subtraction (built in MetaPost) to compute the length of the other side adjacent to the angle.

\documentclass[border=2bp]{standalone}
\usepackage{luamplib}
    \mplibtextextlabel{enable}
    \mplibsetformat{metafun}
\begin{document}
    \begin{mplibcode}
        numeric u; u = 2.5cm;
        beginfig(1);
            x = 1; y = 2 +-+ x; % Pythagorean subtraction
            pair A, B, C; A = origin; B = (u*y, 0); C = (0, u*x);
            label.bot("$\sqrt{4-x^2}$", .5[A,B]);
            label.lft("$x$", .5[A,C]);
            label.urt("$2$", .5[B,C]);
            draw ((1, 0) -- (1, 1) -- (0, 1)) scaled 2mm; % mark right angle
            path angle_mark; 
            angle_mark = anglebetween(B--A, B--C, "\alpha"); % angle \alpha
            fill B -- angle_mark -- cycle withcolor .8white;
            draw A -- B -- C -- cycle;
            draw angle_mark; 
        endfig;
    \end{mplibcode}
\end{document}

Output with x = 1:

With x = 0.5:

With x = 1.5:

Answer 3

Here is a TikZ version:

\documentclass[border=5mm,tikz]{standalone}
\usepackage{mwe}
\usepackage{tikz}
\begin{document}

  \begin{tikzpicture}[scale=2]
    \draw(0,0)--node[below]{$\sqrt{4-x^2}$} (1.7,0)
              --node[above]{$2$}(0,1)
              --node[left]{$x$}(0,0);
    \draw[very thin,<->](1.4,0) arc [start angle=180,end angle=150, radius=0.3];
    \node at (1.3,0.1){$\alpha$};
  \end{tikzpicture}

\end{document}

Answer 4

Andrew's solution, but a bit more symbolic

\documentclass[border=5mm,tikz]{standalone}
\usetikzlibrary{angles,arrows.meta}
\begin{document}
\begin{tikzpicture}[scale=2]
  \coordinate (A) at (0,0);
  \coordinate (B) at (0,1);
  \coordinate (C) at (1.7,0);
  \draw (A) -- node[auto] {$x$}
        (B) -- node[auto] {$2$}
        (C) -- node[auto] {$\sqrt{4 - x^2}$}
          cycle;
  \draw pic[draw,<->,angle radius=1cm,pic text=$\alpha$,angle eccentricity=1.2] {angle = B--C--A};
\end{tikzpicture}
\end{document}

Answer 5

If you want to consider using pstricks, this is a short code for your figure:

\documentclass[pdf]{standalone}
\usepackage[utf8]{inputenc}

\usepackage{pst-node}

\begin{document}
\footnotesize
\psset{shortput=nab, arrows=c-c, linearc = 0.005}%
\begin{psmatrix}[colsep=3.8cm, rowsep=1.8cm]
    [name = A]\\
    [name = B] & [name = C]
    \pspolygon(A)(B)(C)
    \psset{linewidth=0pt}
    \ncline{A}{C}^{$2$}
    \ncline{A}{B}\nbput{$x$}
    \ncline{B}{C}_{$\sqrt{4-x^2}$}
\end{psmatrix}

\end{document} 

Answer 6

A PSTricks solution:

\documentclass{article}

\usepackage{pst-eucl}
\usepackage{xfp}

%%% parameters
% 0 < \side < 2
\def\side{1}
% \scale > 0
\def\scale{3}

\begin{document}

\begin{pspicture}[unit = \scale]%
(-0.35,-0.5)(\fpeval{\scale*sqrt(4-\side^2)},\fpeval{\scale*\side})
  \pnodes{P}(0,0)(0,0)(0,\side)(\fpeval{sqrt(4-\side^2)},0)
  \pspolygon(P1)(P2)(P3)
  \pstMarkAngle[MarkAngleRadius = 0.2, LabelSep = 0.3]{P2}{P3}{P1}{$\alpha$}
  \pstRightAngle[RightAngleSize = 0.1]{P3}{P1}{P2}
  \psset{linestyle = none, offset = 7pt}
  \pcline(P2)(P3)
  \ncput{$2$}
  \pcline(P1)(P2)
  \ncput{$x$}
  \pcline[offset = 9pt](P3)(P1)
  \ncput{$\sqrt{4 - x^{2}}$}
\end{pspicture}

\end{document}

All you have to do is change the values of the parameters and the drawing will be adjusted accordingly.