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How to align text above triangle figure
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Rotate a node but not its content: the case of the ellipse decorationHow to define the default vertical distance between nodes?TikZ scaling graphic and adjust node position and keep font sizealign text in tikz figureNumerical conditional within tikz keys?Relating tree nodes in forest to content in a tableTikZ: Node position in draw environmentLatex and Game Theory: Combining an Extensive and Normal Form for a Three Players GameDrawing graph with Tikz: Link it with main text without overlapping with textTikZ: define arrow starting position based on style and format node label
I managed to align my hypotenuse text with the hypotenuse side of my triangle, but I feel like it was done inefficiently using a lot of ~~~~~~
in this line node[above] {$sqrt{1+x^2}$~~~~~~~} (B) --
.
Is there a better way to get the same alignment that I have now without the excessive use of ~
?
documentclass[hidelinks,14pt, letterpaper]{extarticle}
usepackage{amsmath, amssymb, tikz}
newcommand{pythagwidth}{3cm}
newcommand{pythagheight}{2cm}
begin{document}
begin{figure}[h]
centering
begin{tikzpicture}[scale=1.25]
coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
draw
(A) --
node[above] {$sqrt{1+x^2}$~~~~~~~} (B) --
node[right] {?} (C) --
node[below] {?}
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
end{tikzpicture}
caption{Caption}
label{fig:my_label}
end{figure}
end{document}
tikz-pgf
add a comment |
I managed to align my hypotenuse text with the hypotenuse side of my triangle, but I feel like it was done inefficiently using a lot of ~~~~~~
in this line node[above] {$sqrt{1+x^2}$~~~~~~~} (B) --
.
Is there a better way to get the same alignment that I have now without the excessive use of ~
?
documentclass[hidelinks,14pt, letterpaper]{extarticle}
usepackage{amsmath, amssymb, tikz}
newcommand{pythagwidth}{3cm}
newcommand{pythagheight}{2cm}
begin{document}
begin{figure}[h]
centering
begin{tikzpicture}[scale=1.25]
coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
draw
(A) --
node[above] {$sqrt{1+x^2}$~~~~~~~} (B) --
node[right] {?} (C) --
node[below] {?}
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
end{tikzpicture}
caption{Caption}
label{fig:my_label}
end{figure}
end{document}
tikz-pgf
add a comment |
I managed to align my hypotenuse text with the hypotenuse side of my triangle, but I feel like it was done inefficiently using a lot of ~~~~~~
in this line node[above] {$sqrt{1+x^2}$~~~~~~~} (B) --
.
Is there a better way to get the same alignment that I have now without the excessive use of ~
?
documentclass[hidelinks,14pt, letterpaper]{extarticle}
usepackage{amsmath, amssymb, tikz}
newcommand{pythagwidth}{3cm}
newcommand{pythagheight}{2cm}
begin{document}
begin{figure}[h]
centering
begin{tikzpicture}[scale=1.25]
coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
draw
(A) --
node[above] {$sqrt{1+x^2}$~~~~~~~} (B) --
node[right] {?} (C) --
node[below] {?}
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
end{tikzpicture}
caption{Caption}
label{fig:my_label}
end{figure}
end{document}
tikz-pgf
I managed to align my hypotenuse text with the hypotenuse side of my triangle, but I feel like it was done inefficiently using a lot of ~~~~~~
in this line node[above] {$sqrt{1+x^2}$~~~~~~~} (B) --
.
Is there a better way to get the same alignment that I have now without the excessive use of ~
?
documentclass[hidelinks,14pt, letterpaper]{extarticle}
usepackage{amsmath, amssymb, tikz}
newcommand{pythagwidth}{3cm}
newcommand{pythagheight}{2cm}
begin{document}
begin{figure}[h]
centering
begin{tikzpicture}[scale=1.25]
coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
draw
(A) --
node[above] {$sqrt{1+x^2}$~~~~~~~} (B) --
node[right] {?} (C) --
node[below] {?}
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
end{tikzpicture}
caption{Caption}
label{fig:my_label}
end{figure}
end{document}
tikz-pgf
tikz-pgf
asked 1 hour ago
Evan KimEvan Kim
1453
1453
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Something like this? I use node[midway,above left=0pt,inner sep=0.5pt] {$sqrt{1+x^2}$}
, where inner sep=0.5pt
controls the distance.
documentclass[hidelinks,14pt, letterpaper]{extarticle}
usepackage{amsmath, amssymb, tikz}
newcommand{pythagwidth}{3cm}
newcommand{pythagheight}{2cm}
begin{document}
begin{figure}[h]
centering
begin{tikzpicture}[scale=1.25]
coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
draw
(A) --
node[midway,above left=0pt,inner sep=0.5pt] {$sqrt{1+x^2}$} (B) --
node[right] {?} (C) --
node[below] {?}
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
end{tikzpicture}
caption{Caption}
label{fig:my_label}
end{figure}
end{document}
ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}[scale=1.25]
draw (-1.5,-1) coordinate [label=left:$A$] (A) --
node[midway,above,sloped] {$sqrt{1+x^2}$}
(1.5,1) coordinate [label=above:$B$] (B) --
node[right] {?}
(1.5,-1)coordinate [label=below right:$C$] (C) --
node[below] {?} cycle;
draw ([xshift=-0.25cm]C) |- ([yshift=0.25cm]C);
end{tikzpicture}
end{document}
yes that is it thanks! It seems like simply havingnode [midway,above left=0pt] {$sqrt{1+x^2}$} (B) --
does the trick too without usinginner_sept
– Evan Kim
1 hour ago
add a comment |
Just for fun: with pstricks
, a very short code to have this figure:
documentclass{article}
usepackage{pst-eucl}%,
usepackage{auto-pst-pdf}
begin{document}
begin{postscript}
psset{unit=2, linejoin=1, PointSymbol=none,}
pstTriangle(-1.5,-1){A}(1.5,1){B}(1.5,-1){C}
ncline[linestyle=none]{A}{B}naput*[nrot=:U]{$ sqrt{1 + x^2}$}
psset{PointName=none}
pstMiddleAB{A}{C}{I}uput[d](I){?}
pstMiddleAB{B}{C}{J}uput[r](J){?}
end{postscript}
end{document}
add a comment |
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2 Answers
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oldest
votes
Something like this? I use node[midway,above left=0pt,inner sep=0.5pt] {$sqrt{1+x^2}$}
, where inner sep=0.5pt
controls the distance.
documentclass[hidelinks,14pt, letterpaper]{extarticle}
usepackage{amsmath, amssymb, tikz}
newcommand{pythagwidth}{3cm}
newcommand{pythagheight}{2cm}
begin{document}
begin{figure}[h]
centering
begin{tikzpicture}[scale=1.25]
coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
draw
(A) --
node[midway,above left=0pt,inner sep=0.5pt] {$sqrt{1+x^2}$} (B) --
node[right] {?} (C) --
node[below] {?}
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
end{tikzpicture}
caption{Caption}
label{fig:my_label}
end{figure}
end{document}
ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}[scale=1.25]
draw (-1.5,-1) coordinate [label=left:$A$] (A) --
node[midway,above,sloped] {$sqrt{1+x^2}$}
(1.5,1) coordinate [label=above:$B$] (B) --
node[right] {?}
(1.5,-1)coordinate [label=below right:$C$] (C) --
node[below] {?} cycle;
draw ([xshift=-0.25cm]C) |- ([yshift=0.25cm]C);
end{tikzpicture}
end{document}
yes that is it thanks! It seems like simply havingnode [midway,above left=0pt] {$sqrt{1+x^2}$} (B) --
does the trick too without usinginner_sept
– Evan Kim
1 hour ago
add a comment |
Something like this? I use node[midway,above left=0pt,inner sep=0.5pt] {$sqrt{1+x^2}$}
, where inner sep=0.5pt
controls the distance.
documentclass[hidelinks,14pt, letterpaper]{extarticle}
usepackage{amsmath, amssymb, tikz}
newcommand{pythagwidth}{3cm}
newcommand{pythagheight}{2cm}
begin{document}
begin{figure}[h]
centering
begin{tikzpicture}[scale=1.25]
coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
draw
(A) --
node[midway,above left=0pt,inner sep=0.5pt] {$sqrt{1+x^2}$} (B) --
node[right] {?} (C) --
node[below] {?}
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
end{tikzpicture}
caption{Caption}
label{fig:my_label}
end{figure}
end{document}
ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}[scale=1.25]
draw (-1.5,-1) coordinate [label=left:$A$] (A) --
node[midway,above,sloped] {$sqrt{1+x^2}$}
(1.5,1) coordinate [label=above:$B$] (B) --
node[right] {?}
(1.5,-1)coordinate [label=below right:$C$] (C) --
node[below] {?} cycle;
draw ([xshift=-0.25cm]C) |- ([yshift=0.25cm]C);
end{tikzpicture}
end{document}
yes that is it thanks! It seems like simply havingnode [midway,above left=0pt] {$sqrt{1+x^2}$} (B) --
does the trick too without usinginner_sept
– Evan Kim
1 hour ago
add a comment |
Something like this? I use node[midway,above left=0pt,inner sep=0.5pt] {$sqrt{1+x^2}$}
, where inner sep=0.5pt
controls the distance.
documentclass[hidelinks,14pt, letterpaper]{extarticle}
usepackage{amsmath, amssymb, tikz}
newcommand{pythagwidth}{3cm}
newcommand{pythagheight}{2cm}
begin{document}
begin{figure}[h]
centering
begin{tikzpicture}[scale=1.25]
coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
draw
(A) --
node[midway,above left=0pt,inner sep=0.5pt] {$sqrt{1+x^2}$} (B) --
node[right] {?} (C) --
node[below] {?}
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
end{tikzpicture}
caption{Caption}
label{fig:my_label}
end{figure}
end{document}
ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}[scale=1.25]
draw (-1.5,-1) coordinate [label=left:$A$] (A) --
node[midway,above,sloped] {$sqrt{1+x^2}$}
(1.5,1) coordinate [label=above:$B$] (B) --
node[right] {?}
(1.5,-1)coordinate [label=below right:$C$] (C) --
node[below] {?} cycle;
draw ([xshift=-0.25cm]C) |- ([yshift=0.25cm]C);
end{tikzpicture}
end{document}
Something like this? I use node[midway,above left=0pt,inner sep=0.5pt] {$sqrt{1+x^2}$}
, where inner sep=0.5pt
controls the distance.
documentclass[hidelinks,14pt, letterpaper]{extarticle}
usepackage{amsmath, amssymb, tikz}
newcommand{pythagwidth}{3cm}
newcommand{pythagheight}{2cm}
begin{document}
begin{figure}[h]
centering
begin{tikzpicture}[scale=1.25]
coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
coordinate [label=above:$B$] (B) at (1.5cm,1.0cm);
coordinate [label=below right:$C$] (C) at (1.5cm,-1.0cm);
draw
(A) --
node[midway,above left=0pt,inner sep=0.5pt] {$sqrt{1+x^2}$} (B) --
node[right] {?} (C) --
node[below] {?}
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
end{tikzpicture}
caption{Caption}
label{fig:my_label}
end{figure}
end{document}
ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...
documentclass[tikz,border=3.14mm]{standalone}
begin{document}
begin{tikzpicture}[scale=1.25]
draw (-1.5,-1) coordinate [label=left:$A$] (A) --
node[midway,above,sloped] {$sqrt{1+x^2}$}
(1.5,1) coordinate [label=above:$B$] (B) --
node[right] {?}
(1.5,-1)coordinate [label=below right:$C$] (C) --
node[below] {?} cycle;
draw ([xshift=-0.25cm]C) |- ([yshift=0.25cm]C);
end{tikzpicture}
end{document}
edited 27 mins ago
answered 1 hour ago
marmotmarmot
118k6153288
118k6153288
yes that is it thanks! It seems like simply havingnode [midway,above left=0pt] {$sqrt{1+x^2}$} (B) --
does the trick too without usinginner_sept
– Evan Kim
1 hour ago
add a comment |
yes that is it thanks! It seems like simply havingnode [midway,above left=0pt] {$sqrt{1+x^2}$} (B) --
does the trick too without usinginner_sept
– Evan Kim
1 hour ago
yes that is it thanks! It seems like simply having
node [midway,above left=0pt] {$sqrt{1+x^2}$} (B) --
does the trick too without using inner_sept
– Evan Kim
1 hour ago
yes that is it thanks! It seems like simply having
node [midway,above left=0pt] {$sqrt{1+x^2}$} (B) --
does the trick too without using inner_sept
– Evan Kim
1 hour ago
add a comment |
Just for fun: with pstricks
, a very short code to have this figure:
documentclass{article}
usepackage{pst-eucl}%,
usepackage{auto-pst-pdf}
begin{document}
begin{postscript}
psset{unit=2, linejoin=1, PointSymbol=none,}
pstTriangle(-1.5,-1){A}(1.5,1){B}(1.5,-1){C}
ncline[linestyle=none]{A}{B}naput*[nrot=:U]{$ sqrt{1 + x^2}$}
psset{PointName=none}
pstMiddleAB{A}{C}{I}uput[d](I){?}
pstMiddleAB{B}{C}{J}uput[r](J){?}
end{postscript}
end{document}
add a comment |
Just for fun: with pstricks
, a very short code to have this figure:
documentclass{article}
usepackage{pst-eucl}%,
usepackage{auto-pst-pdf}
begin{document}
begin{postscript}
psset{unit=2, linejoin=1, PointSymbol=none,}
pstTriangle(-1.5,-1){A}(1.5,1){B}(1.5,-1){C}
ncline[linestyle=none]{A}{B}naput*[nrot=:U]{$ sqrt{1 + x^2}$}
psset{PointName=none}
pstMiddleAB{A}{C}{I}uput[d](I){?}
pstMiddleAB{B}{C}{J}uput[r](J){?}
end{postscript}
end{document}
add a comment |
Just for fun: with pstricks
, a very short code to have this figure:
documentclass{article}
usepackage{pst-eucl}%,
usepackage{auto-pst-pdf}
begin{document}
begin{postscript}
psset{unit=2, linejoin=1, PointSymbol=none,}
pstTriangle(-1.5,-1){A}(1.5,1){B}(1.5,-1){C}
ncline[linestyle=none]{A}{B}naput*[nrot=:U]{$ sqrt{1 + x^2}$}
psset{PointName=none}
pstMiddleAB{A}{C}{I}uput[d](I){?}
pstMiddleAB{B}{C}{J}uput[r](J){?}
end{postscript}
end{document}
Just for fun: with pstricks
, a very short code to have this figure:
documentclass{article}
usepackage{pst-eucl}%,
usepackage{auto-pst-pdf}
begin{document}
begin{postscript}
psset{unit=2, linejoin=1, PointSymbol=none,}
pstTriangle(-1.5,-1){A}(1.5,1){B}(1.5,-1){C}
ncline[linestyle=none]{A}{B}naput*[nrot=:U]{$ sqrt{1 + x^2}$}
psset{PointName=none}
pstMiddleAB{A}{C}{I}uput[d](I){?}
pstMiddleAB{B}{C}{J}uput[r](J){?}
end{postscript}
end{document}
answered 47 mins ago
BernardBernard
176k778210
176k778210
add a comment |
add a comment |
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