What is another way to construct a quadrilateral with the measure of its four angles?

Question

I tried to create a quadrilateral with the measure of its four angles, as shown in the picture:

My code

\documentclass[border=1.5mm,12pt]{standalone}
\usepackage{tkz-euclide}
\begin{document}
\begin{tikzpicture}
\pgfmathsetmacro{\a}{3} 
\tkzDefPoints{0/0/A,\a/0/B}
\tkzDefPointBy[rotation= center A angle 60](B)  
\tkzGetPoint{X}
\tkzDefPointBy[rotation= center B angle -80](A)     
\tkzGetPoint{Y}
\tkzInterLL(A,X)(B,Y)
\tkzGetPoint{Z}
\tkzDefMidPoint(Z,B)
 \tkzGetPoint{C}
\tkzDefPointBy[rotation= center C angle 60](Z)  
\tkzGetPoint{T}
\tkzInterLL(C,T)(A,Z)
\tkzGetPoint{D}
\tkzLabelAngle[pos=.5](Z,B,A){$80^{\circ}$} 
\tkzLabelAngle[pos=.6](B,A,Z){$60^{\circ}$} 
%\tkzLabelAngle[pos=.6](Z,C,D){$30^{\circ}$}    
\tkzLabelAngle[pos=.5](D,C,B){$120^{\circ}$}
\tkzLabelAngle[pos=.5](A,D,C){$100^{\circ}$}
\tkzLabelPoints[](A,B)
%\tkzLabelPoints[above](Z)
\tkzLabelPoints[right](C)
\tkzLabelPoints[left](D)
\tkzDrawPolygon(A,B,C,D)
    \end{tikzpicture}
\end{document}

Is there another way to construct such a quadrilateral?

There are many ways to do that but you'll need at least two lengths to start with. You can use intersections and/or calc libraries to achieve something like that. — SebGlav, Jul 11 '23 at 06:27
Is there something in particular that you are not happy with? Or do you just want to see other ways? — mickep, Jul 11 '23 at 06:50

score 6 · Answer 1 · answered Jul 11 '23 at 09:04

If I'm seeing this right, the point C is half way between B and Z where Z is just the intersection of AD of BC.

The intersection of two straight liens can be easily foudn with the intersection of syntax. The halfway point is found via calc's syntax ($(B)!.5!(Z)$). (You could also do \path[overlay] (B) -- coordinate (C) (Z);.)

Rotating one coordinate around another can be done via rotate around:

([rotate around=<angle>:(A)]B)

or again with the help of calc

($(A)!1!wa:(B)$)

Here we can even change the distance between A and the new point. But for the intersection of syntax the distance is irrelevant (shouldn't be zero, of course).

Code

\documentclass[tikz]{standalone}
\usetikzlibrary{angles, calc, quotes}
\tikzset{math nodes/.style={execute at begin node=$, execute at end node=$}}
\begin{document}
\begin{tikzpicture}[
  declare function={a=3; wa=60; wb=80; wc=120; wd=100;},
  every label/.append style=math nodes,
  angle eccentricity=1, angle radius=4.5mm,
  pics/angle/.append style={/tikz/nodes={font=\footnotesize}},
]
\coordinate["A" below] (A) at (0,0)
 coordinate["B" below] (B) at (right:a)
 coordinate[overlay]  (B') at ([rotate around=wa:(A)]B)
%coordinate[overlay]  (B') at ($(A)!1!wa:(B)$)
 coordinate[overlay]   (Z) at (intersection of A--B' and B--{[rotate around=-wb:(B)]A})
%coordinate[overlay]   (Z) at (intersection of A--B' and B--{$(B)!1!-wb:(A)$})
 coordinate["C" right] (C) at ($(B)!.5!(Z)$)
 coordinate["D" left ] (D) at (intersection of A--B' and C--{[rotate around=-wc:(C)]B})
%coordinate["D" left ] (D) at (intersection of A--B' and C--{$(C)!1!-wc:(B)$})
;
%\draw[help lines] (D) -- (Z) node[above]{$Z$} -- (C)
%  pic["$\pgfmathprint{int(180-wa-wb)}^\circ$" gray, angle radius=7mm]
%    {angle = D--Z--C};
\draw (A) -- (B) -- (C) -- (D) -- cycle
  foreach \v/\a in {wa/B--A--D, wb/C--B--A,
                   wc/D--C--B, wd/A--D--C}{
    pic["$\pgfmathprint{\v}^\circ$"] {angle/.expand once = \a}
  };
\end{tikzpicture}
\end{document}

Output

The placement of the angle nodes isn't very nice when the angles aren't drawn. These can of course be placed differently or with an adjustment to the angle pic or … — Qrrbrbirlbel, Jul 11 '23 at 09:07
@JohnPaulPeter Yes, that is indeed possible. Let me note that instead of PGFMath function you can also simply use macros, as you did in your Q with \pgfmathsetmacro. — Qrrbrbirlbel, Jul 12 '23 at 00:13
There is a bug withdeclare function={a=3; wa=90; wb=90; wc=90; wd=90;} — John Paul Peter, Jul 12 '23 at 01:08
@JohnPaulPeter Not really a bug. Z doesn't exist then (or only in infinity). Where should C lie then? — Qrrbrbirlbel, Jul 12 '23 at 01:56

Alain Matthes · Accepted Answer · 2023-07-12T12:59:47.647

4

There is a much simpler solution with :

\tkzDefTriangle[two angles ...
\documentclass[border=1.5mm,12pt]{standalone}
\usepackage{tkz-euclide}
\begin{document}
\begin{tikzpicture}
\pgfmathsetmacro{\a}{6} 
\tkzDefPoints{0/0/A,\a/0/B}
\tkzDefTriangletwo angles = 60 and 80
 \tkzGetPoint{Z}
\tkzDefMidPoint(Z,B)
 \tkzGetPoint{C}
\tkzDefTriangletwo angles = 60 and 40
\tkzGetPoint{D}
\tkzLabelAnglepos=.5{$80^{\circ}$} 
\tkzLabelAnglepos=.6{$60^{\circ}$} 
\tkzLabelAnglepos=.5{$120^{\circ}$}
\tkzLabelAnglepos=.5{$100^{\circ}$}
\tkzLabelPointsA,B
\tkzLabelPointsright
\tkzLabelPointsleft
\tkzDrawPolygon(A,B,C,D)
\end{tikzpicture}
\end{document}

Update with the new constraints

\documentclass[border=1.5mm,12pt]{standalone}
\usepackage{tkz-euclide}
\begin{document}
\begin{tikzpicture}
\pgfmathsetmacro{\a}{6} 
\pgfmathsetmacro\angleA{60} 
\pgfmathsetmacro\angleB{80}
\pgfmathsetmacro\angleC{120}    
\tkzDefPoints{0/0/A,\a/0/B}
\tkzDefTriangle[two angles = {\angleA} and {\angleB}](A,B)
 \tkzGetPoint{Z}
\tkzDefMidPoint(Z,B)
 \tkzGetPoint{C}
\tkzDefTriangle[two angles = {180-\angleC} and {(180-(\angleA+\angleB))}](C,Z)
\tkzGetPoint{D}
\tkzLabelAngle[pos=.5](Z,B,A){$80^{\circ}$}
\tkzLabelAngle[pos=.6](B,A,Z){$60^{\circ}$}
\tkzLabelAngle[pos=.5](D,C,B){$120^{\circ}$}
\tkzLabelAngle[pos=.5](A,D,C){$100^{\circ}$}
\tkzLabelPoints[](A,B)
\tkzLabelPoints[right](C)
\tkzLabelPoints[left](D)
\tkzDrawPolygon(A,B,C,D)
\end{tikzpicture}
\end{document}

edited Jul 12 '23 at 12:59

answered Jul 11 '23 at 13:14

Alain Matthes

95,075

If A = 100 degrees, and B = 80 deg. How can I use your code? – John Paul Peter Jul 11 '23 at 14:46
@JohnPaulPeter What is the exact data in this case? Do we know AD? BC? and the others angles ? – Alain Matthes Jul 11 '23 at 15:35
AD = 5' andBC= 4`. – John Paul Peter Jul 11 '23 at 15:39
Please make a code automaticaly.
1. \tkzDefTriangle[two angles = 60 and 80](A,B) replace 60 by angleA and 80 by angleB, e.g;
2. \tkzDefTriangletwo angles = 60 and 40 replace 60 by 180 - angle C and 40 by 180 - angleA - angleB, e.g.
– John Paul Peter Jul 11 '23 at 15:42
@JohnPaulPeter Yes it's possible. There's not enough time, but tomorrow I can do it. – Alain Matthes Jul 11 '23 at 18:47
Doing all this automatically will be easier with the new version of tkz-euclide (tkz-elements) and only lua – Alain Matthes Jul 12 '23 at 13:01

score 2 · Answer 3 · answered Jul 11 '23 at 13:55

Here is another way with Asymptote implementation (no worry of redundancy of boundingbox). Also see this for calculating and labelling an angle.

Its translation to TikZ is straightforward.

size(6cm);
// change t to control the length AT = t; 
// change s to move C and D
real t=1.2,s=2.5;
pair A=(0,0), T=A+t*dir(60);
pair Bt=rotate(100,T)*A;
pair B=extension(T,Bt,A,A+dir(0));
pair D=A+s*(T-A),Bs=B+dir(100);
pair Ds=rotate(100,D)*A;
pair C=extension(D,Ds,B,Bs);
draw(B--T,red); label("$T$",T,W,red);
draw(A--B--C--D--cycle);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
real angleA=degrees(D);
real angleB=180-degrees(C-B);
real angleD=-angleA+degrees(T-B);
real angleC=360-angleA-angleB-angleD;
write("The angle A is: ",angleA);
write("The angle B is: ",angleB);
write("The angle C is: ",angleC);
write("The angle D is: ",angleD);
shipout(bbox(5mm,invisible));

score 2 · Answer 4 · answered Jul 11 '23 at 17:24

For fun, here is a MetaPost/MetaFun suggestion. Compile with context.

\startMPpage[offset=1DK]
u = 2cm ;
z0 = origin ;
z1 = (5u,0) ;
z2 = z1 + 3udir(100) ;
z3 = z2 + whateverdir(160) = whatever*dir(60) ;
draw z0 -- z1 -- z2 -- z3 -- cycle ;
label.llft( "$A$", z0 ) ;
label.lrt ( "$B$", z1 ) ;
label.urt ( "$C$", z2 ) ;
label.top ( "$D$", z3 ) ;
draw anglebetween( z0--z3, z0--z1, "$\unit{60  degree}$" ) ;
draw anglebetween( z1--z0, z1--z2, "$\unit{80  degree}$" ) ;
draw anglebetween( z2--z1, z2--z3, "$\unit{120 degree}$" ) ;
draw anglebetween( z3--z2, z3--z0, "$\unit{100 degree}$" ) ;
\stopMPpage

gigiair · Answer 5 · 2023-07-11T09:29:53.410

You can use external software capable of exporting "LaTeX" code. For example geogebra and export tikz. It's a very simple way.

\documentclass[10pt]{standalone}
\usepackage{pgf,tikz}
\usetikzlibrary{arrows}
\pagestyle{empty}
\newcommand{\degre}{\ensuremath{^\circ}}
\begin{document}
\definecolor{uququq}{rgb}{0.25,0.25,0.25}
\definecolor{xdxdff}{rgb}{0.49,0.49,1}
\definecolor{qqwuqq}{rgb}{0,0.39,0}
\definecolor{qqqqff}{rgb}{0,0,1}
\begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm]
\clip(-3.3,-2) rectangle (24.02,9.6);
\draw [shift={(1.18,1.92)},color=qqwuqq,fill=qqwuqq,fill opacity=0.1] (0,0) -- (0.64:0.6) arc (0.64:60.64:0.6) -- cycle;
\draw [shift={(3.98,6.9)},color=qqwuqq,fill=qqwuqq,fill opacity=0.1] (0,0) -- (-119.36:0.6) arc (-119.36:-19.36:0.6) -- cycle;
\draw [shift={(8.38,2)},color=qqwuqq,fill=qqwuqq,fill opacity=0.1] (0,0) -- (100.64:0.6) arc (100.64:180.64:0.6) -- cycle;
\draw [shift={(7.71,5.59)},color=qqwuqq,fill=qqwuqq,fill opacity=0.1] (0,0) -- (160.64:0.6) arc (160.64:280.64:0.6) -- cycle;
\draw (1.18,1.92)-- (8.38,2);
\draw (1.18,1.92)-- (3.98,6.9);
\draw (3.98,6.9)-- (7.71,5.59);
\draw (8.38,2)-- (7.71,5.59);
\begin{scriptsize}
\fill [color=qqqqff] (1.18,1.92) circle (1.5pt);
\draw[color=qqqqff] (0.9,2.02) node {$A$};
\fill [color=qqqqff] (8.38,2) circle (1.5pt);
\draw[color=qqqqff] (8.66,2.1) node {$B$};
\draw[color=qqwuqq] (1.62,2.14) node {$60\textrm{\degre}$};
\fill [color=xdxdff] (3.98,6.9) circle (1.5pt);
\draw[color=xdxdff] (3.82,7.12) node {$C$};
\draw[color=qqwuqq] (4.12,6.62) node {$100\textrm{\degre}$};
\fill [color=uququq] (7.71,5.59) circle (1.5pt);
\draw[color=uququq] (7.86,5.86) node {$D$};
\draw[color=qqwuqq] (8.08,2.28) node {$80\textrm{\degre}$};
\draw[color=qqwuqq] (7.44,5.44) node {$120\textrm{\degre}$};
\end{scriptsize}
\end{tikzpicture}
\end{document}

What is another way to construct a quadrilateral with the measure of its four angles?

5 Answers5

Code

Output