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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] $sqrt1+x^2$~~~~~~~ (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
usepackageamsmath, amssymb, tikz
newcommandpythagwidth3cm
newcommandpythagheight2cm
begindocument
beginfigure[h]
centering
begintikzpicture[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] $sqrt1+x^2$~~~~~~~ (B) --
node[right] ? (C) --
node[below] ?
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
endtikzpicture
captionCaption
labelfig:my_label
endfigure
enddocument
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] $sqrt1+x^2$~~~~~~~ (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
usepackageamsmath, amssymb, tikz
newcommandpythagwidth3cm
newcommandpythagheight2cm
begindocument
beginfigure[h]
centering
begintikzpicture[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] $sqrt1+x^2$~~~~~~~ (B) --
node[right] ? (C) --
node[below] ?
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
endtikzpicture
captionCaption
labelfig:my_label
endfigure
enddocument
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] $sqrt1+x^2$~~~~~~~ (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
usepackageamsmath, amssymb, tikz
newcommandpythagwidth3cm
newcommandpythagheight2cm
begindocument
beginfigure[h]
centering
begintikzpicture[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] $sqrt1+x^2$~~~~~~~ (B) --
node[right] ? (C) --
node[below] ?
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
endtikzpicture
captionCaption
labelfig:my_label
endfigure
enddocument
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] $sqrt1+x^2$~~~~~~~ (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
usepackageamsmath, amssymb, tikz
newcommandpythagwidth3cm
newcommandpythagheight2cm
begindocument
beginfigure[h]
centering
begintikzpicture[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] $sqrt1+x^2$~~~~~~~ (B) --
node[right] ? (C) --
node[below] ?
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
endtikzpicture
captionCaption
labelfig:my_label
endfigure
enddocument
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] $sqrt1+x^2$
, where inner sep=0.5pt
controls the distance.
documentclass[hidelinks,14pt, letterpaper]extarticle
usepackageamsmath, amssymb, tikz
newcommandpythagwidth3cm
newcommandpythagheight2cm
begindocument
beginfigure[h]
centering
begintikzpicture[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] $sqrt1+x^2$ (B) --
node[right] ? (C) --
node[below] ?
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
endtikzpicture
captionCaption
labelfig:my_label
endfigure
enddocument
ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture[scale=1.25]
draw (-1.5,-1) coordinate [label=left:$A$] (A) --
node[midway,above,sloped] $sqrt1+x^2$
(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);
endtikzpicture
enddocument
yes that is it thanks! It seems like simply havingnode [midway,above left=0pt] $sqrt1+x^2$ (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:
documentclassarticle
usepackagepst-eucl%,
usepackageauto-pst-pdf
begindocument
beginpostscript
pssetunit=2, linejoin=1, PointSymbol=none,
pstTriangle(-1.5,-1)A(1.5,1)B(1.5,-1)C
ncline[linestyle=none]ABnaput*[nrot=:U]$ sqrt1 + x^2$
pssetPointName=none
pstMiddleABACIuput[d](I)?
pstMiddleABBCJuput[r](J)?
endpostscript
enddocument
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Something like this? I use node[midway,above left=0pt,inner sep=0.5pt] $sqrt1+x^2$
, where inner sep=0.5pt
controls the distance.
documentclass[hidelinks,14pt, letterpaper]extarticle
usepackageamsmath, amssymb, tikz
newcommandpythagwidth3cm
newcommandpythagheight2cm
begindocument
beginfigure[h]
centering
begintikzpicture[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] $sqrt1+x^2$ (B) --
node[right] ? (C) --
node[below] ?
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
endtikzpicture
captionCaption
labelfig:my_label
endfigure
enddocument
ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture[scale=1.25]
draw (-1.5,-1) coordinate [label=left:$A$] (A) --
node[midway,above,sloped] $sqrt1+x^2$
(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);
endtikzpicture
enddocument
yes that is it thanks! It seems like simply havingnode [midway,above left=0pt] $sqrt1+x^2$ (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] $sqrt1+x^2$
, where inner sep=0.5pt
controls the distance.
documentclass[hidelinks,14pt, letterpaper]extarticle
usepackageamsmath, amssymb, tikz
newcommandpythagwidth3cm
newcommandpythagheight2cm
begindocument
beginfigure[h]
centering
begintikzpicture[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] $sqrt1+x^2$ (B) --
node[right] ? (C) --
node[below] ?
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
endtikzpicture
captionCaption
labelfig:my_label
endfigure
enddocument
ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture[scale=1.25]
draw (-1.5,-1) coordinate [label=left:$A$] (A) --
node[midway,above,sloped] $sqrt1+x^2$
(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);
endtikzpicture
enddocument
yes that is it thanks! It seems like simply havingnode [midway,above left=0pt] $sqrt1+x^2$ (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] $sqrt1+x^2$
, where inner sep=0.5pt
controls the distance.
documentclass[hidelinks,14pt, letterpaper]extarticle
usepackageamsmath, amssymb, tikz
newcommandpythagwidth3cm
newcommandpythagheight2cm
begindocument
beginfigure[h]
centering
begintikzpicture[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] $sqrt1+x^2$ (B) --
node[right] ? (C) --
node[below] ?
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
endtikzpicture
captionCaption
labelfig:my_label
endfigure
enddocument
ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture[scale=1.25]
draw (-1.5,-1) coordinate [label=left:$A$] (A) --
node[midway,above,sloped] $sqrt1+x^2$
(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);
endtikzpicture
enddocument
Something like this? I use node[midway,above left=0pt,inner sep=0.5pt] $sqrt1+x^2$
, where inner sep=0.5pt
controls the distance.
documentclass[hidelinks,14pt, letterpaper]extarticle
usepackageamsmath, amssymb, tikz
newcommandpythagwidth3cm
newcommandpythagheight2cm
begindocument
beginfigure[h]
centering
begintikzpicture[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] $sqrt1+x^2$ (B) --
node[right] ? (C) --
node[below] ?
(A);
draw
(1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
endtikzpicture
captionCaption
labelfig:my_label
endfigure
enddocument
ADDENDUM: Just for fun: an even simpler and shorter code with TikZ...
documentclass[tikz,border=3.14mm]standalone
begindocument
begintikzpicture[scale=1.25]
draw (-1.5,-1) coordinate [label=left:$A$] (A) --
node[midway,above,sloped] $sqrt1+x^2$
(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);
endtikzpicture
enddocument
edited 45 mins ago
answered 1 hour ago
marmotmarmot
118k6153288
118k6153288
yes that is it thanks! It seems like simply havingnode [midway,above left=0pt] $sqrt1+x^2$ (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] $sqrt1+x^2$ (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] $sqrt1+x^2$ (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] $sqrt1+x^2$ (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:
documentclassarticle
usepackagepst-eucl%,
usepackageauto-pst-pdf
begindocument
beginpostscript
pssetunit=2, linejoin=1, PointSymbol=none,
pstTriangle(-1.5,-1)A(1.5,1)B(1.5,-1)C
ncline[linestyle=none]ABnaput*[nrot=:U]$ sqrt1 + x^2$
pssetPointName=none
pstMiddleABACIuput[d](I)?
pstMiddleABBCJuput[r](J)?
endpostscript
enddocument
add a comment |
Just for fun: with pstricks
, a very short code to have this figure:
documentclassarticle
usepackagepst-eucl%,
usepackageauto-pst-pdf
begindocument
beginpostscript
pssetunit=2, linejoin=1, PointSymbol=none,
pstTriangle(-1.5,-1)A(1.5,1)B(1.5,-1)C
ncline[linestyle=none]ABnaput*[nrot=:U]$ sqrt1 + x^2$
pssetPointName=none
pstMiddleABACIuput[d](I)?
pstMiddleABBCJuput[r](J)?
endpostscript
enddocument
add a comment |
Just for fun: with pstricks
, a very short code to have this figure:
documentclassarticle
usepackagepst-eucl%,
usepackageauto-pst-pdf
begindocument
beginpostscript
pssetunit=2, linejoin=1, PointSymbol=none,
pstTriangle(-1.5,-1)A(1.5,1)B(1.5,-1)C
ncline[linestyle=none]ABnaput*[nrot=:U]$ sqrt1 + x^2$
pssetPointName=none
pstMiddleABACIuput[d](I)?
pstMiddleABBCJuput[r](J)?
endpostscript
enddocument
Just for fun: with pstricks
, a very short code to have this figure:
documentclassarticle
usepackagepst-eucl%,
usepackageauto-pst-pdf
begindocument
beginpostscript
pssetunit=2, linejoin=1, PointSymbol=none,
pstTriangle(-1.5,-1)A(1.5,1)B(1.5,-1)C
ncline[linestyle=none]ABnaput*[nrot=:U]$ sqrt1 + x^2$
pssetPointName=none
pstMiddleABACIuput[d](I)?
pstMiddleABBCJuput[r](J)?
endpostscript
enddocument
answered 1 hour ago
BernardBernard
176k778210
176k778210
add a comment |
add a comment |
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