Portada » ¿Cómo etiquetar ángulos en un triángulo rectángulo?
Solución:
Ya casi has llegado. Solo necesitas agregar
usepackage{tkz-euclide}
usetkzobj{all}
al preámbulo.
documentclass[border=10pt,tikz]{standalone}
usepackage{tkz-euclide}
usetkzobj{all}
begin{document}
begin{tikzpicture}[scale=1.25]
coordinate [label=left:$C$] (C) at (-1.5cm,-1.cm);
coordinate [label=right:$A$] (A) at (1.5cm,-1.0cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
draw (C) -- node[above] {$a$} (B) -- node[right] {$c$} (A) -- node[below] {$b$} (C);
draw (1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
tkzMarkAngle[size=1cm,color=cyan,mark=|](A,C,B)
tkzMarkAngle[size=1cm,color=cyan,mark=|](C,B,A)
end{tikzpicture}
end{document}
EDITAR
En un comentario mencionaste querer evitar tkz-euclide
. De hecho, solo usé esto porque el código que publicaste originalmente lo requería. Si desea marcar ángulos usando Ti simplek Z (en la medida en que exista tal cosa), puede usar el angles
y quotes
bibliotecas, que forman parte de PGF / Tik Z en sí.
Por ejemplo:
documentclass[border=10pt,tikz]{standalone}
usetikzlibrary{angles,quotes}
begin{document}
begin{tikzpicture}[
my angle/.style={
every pic quotes/.append style={text=cyan},
draw=cyan,
angle radius=1cm,
}]
coordinate [label=left:$C$] (C) at (-1.5,-1);
coordinate [label=right:$A$] (A) at (1.5,-1);
coordinate [label=above:$B$] (B) at (1.5,1);
draw (C) -- node[above] {$a$} (B) -- node[right] {$c$} (A) -- node[below] {$b$} (C);
draw (A) +(-.25,0) |- +(0,.25);
pic [my angle, "$alpha$"] {angle=A--C--B};
pic [my angle, "$beta$"] {angle=C--B--A};
end{tikzpicture}
end{document}
documentclass{article}
usepackage{pst-eucl}
newcommand*Label[3]{%
pcline[linestyle = none, offset = 6pt](#1)(#2)
ncput{#3}}
begin{document}
begin{pspicture}(-0.43,-0.35)(5.35,3.4) % boundry found manually
pstTriangle[PointSymbol = none](0,0){A}(5,3){B}(5,0){C}
pstRightAngle{B}{C}{A}
pstMarkAngle[LabelSep = 0.6]{C}{A}{B}{$alpha$}
pstMarkAngle[LabelSep = 0.67]{A}{B}{C}{$beta$}
Label{A}{B}{$c$}
Label{B}{C}{$a$}
Label{C}{A}{$b$}
end{pspicture}
end{document}
¡Ahora es correcto, sin errores!
documentclass{article}
usepackage{tikz}
usetikzlibrary{calc}
usepackage{tkz-euclide}
usetkzobj{all}
usepackage{etoolbox}
makeatletter
begin{document}
begin{tikzpicture}[scale=1.25]
coordinate [label=left:$C$] (A) at (-1.5cm,-1.cm);
coordinate [label=right:$A$] (C) at (1.5cm,-1.0cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
draw
(A) --
node[above] {$a$} (B) --
node[right] {$c$} (C) --
node[below] {$b$} (A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
tkzMarkAngle[size=1cm,color=cyan,mark=|](C,A,B);
end{tikzpicture}
end{document}
Para la segunda pregunta, puede utilizar una cuadrícula.
documentclass{article}
usepackage[utf8]{inputenc}
usepackage{tikz}
usetikzlibrary{arrows}
begin{document}
begin{tikzpicture}[line cap=round,line join=round,>=triangle
45,x=1.0cm,y=1.0cm]
draw[step=1cm,gray,very thin] (-4,-4) grid (4,4);
draw[color=black] (-4.,0.) -- (0.,0.);
draw[color=black] (0.,0.) -- (0.,3.);
clip(-4.,-0.4) rectangle (0.4,3.4);
draw [shift={(-4.,0.)},color=gray,fill=darkgray,fill opacity=0.1] (0,0) -- (0.:0.6) arc (0.:36.9:0.6) -- cycle;
draw[color=gray,fill=darkgray,fill opacity=0.1] (0.,0.42) -- (-0.42,0.42) -- (-0.42,0.) -- (0.,0.) -- cycle;
draw (0.,3.)-- (-4.,0.);
draw (-4.,0.)-- (0.,0.);
draw (0.,0.)-- (0.,3.);
draw[color=gray] (-3.2,0.4) node {large $theta$};
draw[color=gray] (0.34,0.33);
end{tikzpicture}
end{document}
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