Drawing a triangle with a certain angle at a vertex

Question

I want to draw triangle ABC so that the measure of the angle at B is 75 degrees. I use \coordinate (A) at (0,0); to put vertex A at the origin, and I use \coordinate (B) at (60:6); to position vertex B. How do I declare a point C on the horizontal line through A so that the angle at B is 75 degrees?

I placed point S on line segment AB. I want to place T on BC so that angle AST is 110 degrees. (This involves the same procedure as positioning point C.)

\documentclass{amsart}
\usepackage{tikz}
\usetikzlibrary{calc,angles,positioning,intersections,quotes,decorations.markings}

\begin{document}


\begin{tikzpicture}

\coordinate (A) at (0,0);
\coordinate (B) at (60:6);
\draw (A) -- (B);

%These commands have vertex C placed on the horizontal line through A so that angle ABC is $30^{\circ}$.
\coordinate (P) at (0,5);
\path[name path=horizontal] (A) -- (P);
%\coordintate (Q) at 

%These commands are to inscribe a triangle in $\triangle{ABC}$.
\node[circle,fill,inner sep=1.5pt,outer sep=0pt,label={left:$S$}] at ($(A)!0.75!(B)$) {};

\end{tikzpicture}
\end{document}

Answer 1

Edit: I have slightly modified the code in order to have more precise angles and calculation of T on the BC side. Additionally, now it uses the original AB side length provided in the OP (which was 60:6).

I used \tkzFindAngle and \tkzGetAngle (loaded by the tkz-euclide package above) to show you the angles (they are automatically calculated).

Output

Code

\documentclass[margin=10pt]{standalone}
\usepackage{amsmath,tkz-euclide}
\usepackage{tikz}
\usetkzobj{all}

\usetikzlibrary{calc,angles,positioning,intersections,quotes,decorations.markings,shapes}

\tikzset{
    myangle/.style={fill=green!20!white, draw=green!50!black,size=.3,opacity=.3},
    intnode/.style={circle,fill=black,inner sep=1pt,outer sep=0pt}
}

\begin{document}
\begin{tikzpicture}

% Drawing the triangle and the coordinates
\draw coordinate[label=left:A] (a) --++(60:6) coordinate[label=above:B] (b);

\path[name path=ac] (a)--++(0:8.5);
\path[name path=bc] (b)--++(-45:8);
\path[name intersections={of = ac and bc, by=c}];
\node[anchor=west] at (c) {C};

\draw[use as bounding box] (a)--(b)--(c)--cycle;

% Drawing the coordinates S and T
\coordinate (s) at ($(a)!0.75!(b)$);

\path[name path=incls] (s) --++ (-10:5);
\path[name path=altbc] (b) -- (c);
\path[name intersections={of = incls and altbc, by=t}]; 

\draw[dashed] (s) -- (t) node[intnode,label={right:{\color{black}\scriptsize $T$}}] (t) {};

% Angles
\tkzFindAngle(a,b,c)
\tkzGetAngle{angleABC};
\FPround\angleABC\angleABC{0}
\tkzFindAngle(c,a,b)
\tkzGetAngle{angleCAB};
\FPround\angleCAB\angleCAB{0}
\tkzFindAngle(b,c,a)
\tkzGetAngle{angleBCA};
\FPround\angleBCA\angleBCA{0}
\tkzFindAngle(a,s,t)
\tkzGetAngle{angleAST};
\FPround\angleAST\angleAST{0}

\tkzMarkAngle[myangle](a,b,c)
\tkzLabelAngle[pos=.4](a,b,c){\tiny $\angleABC^\circ$}

\tkzMarkAngle[myangle](c,a,b)
\tkzLabelAngle[pos=0.5](c,a,b){\tiny $\angleCAB^\circ$}

\tkzMarkAngle[myangle](b,c,a)
\tkzLabelAngle[pos=0.45](b,c,a){\tiny $\angleBCA^\circ$}

\tkzMarkAngle[myangle](a,s,t)
\tkzLabelAngle[pos=0.4](a,s,t){\tiny $\angleAST^\circ$}

\node[intnode,label={left:\scriptsize $S$}] at (s) {};
\end{tikzpicture}
\end{document}

Answer 2

For the record and for whom it may interest, here is a quick way to do it with MetaPost.

I've made use of MetaPost's trademark implicit linear equation solver. For example, the following line

C = whatever[A, A+right] = whatever[B, A rotatedaround (B, 75)];

tells MetaPost almost literally that C must be somewhere on the horizontal straight line starting from A and somewhere on the straight line starting from B and forming an angle of 75 degree with (BA). It is enough for it to determine exactly where C lies.

T is computed the same way:

T = whatever[B, C] = whatever[S, A rotatedaround (S, 110)];

Here is the complete code, included in a LuaLaTeX program for typesetting convenience.

\documentclass[border=2mm]{standalone}
\usepackage{luamplib, gensymb}
  \mplibsetformat{metafun}
  \mplibtextextlabel{enable}
\begin{document}
  \begin{mplibcode}
    pair A, B, C, S, T; numeric u;
    u = cm; A = origin; B = 6u*dir 60; S = .75[A, B];
    C = whatever[A, A+right] = whatever[B, A rotatedaround (B, 75)];
    T = whatever[B, C] = whatever[S, A rotatedaround (S, 110)];
    beginfig(1);
      draw A--B--C--cycle;
      draw S--T dashed evenly;
      label.llft("$A$", A); label.top("$B$", B); label.lrt("$C$", C);
      label.lft("$S$", S); label.urt("$T$", T);
      anglelength := 12bp;
      draw anglebetween(A--C, A--B, "$60\degree$");
      draw anglebetween(B--A, B--C, "$75\degree$");
      draw anglebetween(S--A, S--T, "$110\degree$");
    endfig;
  \end{mplibcode}
\end{document}

For those who wish to use standalone MetaPost, here is an adequate version, producing the same result.

input latexmp
setupLaTeXMP(textextlabel=enable, mode=rerun, packages="gensymb");
pair A, B, C, S, T; numeric u;
u = cm; A = origin; B = 6u*dir 60; S = .75[A, B];
C = whatever[A, A+right] = whatever[B, A rotatedaround (B, 75)];
T = whatever[B, C] = whatever[S, A rotatedaround (S, 110)];
beginfig(1);
  draw A--B--C--cycle;
  draw S--T dashed evenly;
  label.llft("$A$", A); label.top("$B$", B); label.lrt("$C$", C);
  label.lft("$S$", S); label.urt("$T$", T);
  anglelength := 12bp;
  draw anglebetween(A--C, A--B, "$60\degree$");
  draw anglebetween(B--A, B--C, "$75\degree$");
  draw anglebetween(S--A, S--T, "$110\degree$");
  setbounds currentpicture to boundingbox currentpicture enlarged 2mm;
endfig;
end.

From this code, and with a Unix command line (I don't know anything about Windows!), a PDF figure can be obtained with these instructions, assuming that the code has been saved under the name mytriangle.mp:

mpost --mem=metafun mytriangle.mp
mptopdf mytriangle.1

Answer 3

For straight lines, I'd use the intersection cs: or its implicit version intersection of <p1>--<p2> and <p3>--<p4>.

The turn key is useful for more complicated examples/calculations.

Code

\documentclass[tikz]{standalone}
\usetikzlibrary{angles,quotes}
\begin{document}
\begin{tikzpicture}[declare function={alpha=60; beta=75; sigma=110;}]
\path (0,0) coordinate (A) + (right:1) coordinate (aC)
      (A) -- coordinate[pos=.75] (S) ++ (alpha:6) coordinate (B)
                                  ([turn]beta:-1) coordinate (bC)
      (A) -- (S) ([turn]sigma:-1) coordinate (s);

\draw (A) -- (B) -- (intersection of A--aC and B--bC) coordinate (C) -- cycle;
\draw[dashed] (S) -- (intersection of S--s and B--bC) coordinate (T);

\foreach \p/\dir in {A/180, B/90, C/0, S/alpha+90, T/45}
  \node[circle,inner sep=+1pt,fill,label=\dir:$\p$] at (\p) {};

\foreach \angle/\val in {C--A--B/alpha,          A--B--C/beta,
                         B--C--A/180-beta-alpha, A--S--T/sigma}
  \pic["\scriptsize\pgfmathparse{\val}$\pgfmathprintnumber{\pgfmathresult}^\circ$",
      draw, angle radius=7.5mm] {angle/.expanded=\angle};
\end{tikzpicture}
\end{document}

Output

Answer 4

Here is another TikZ suggestion:

\documentclass[margin=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,intersections,angles, quotes}
\begin{document}
\begin{tikzpicture}[
  dot/.style={circle,fill,inner sep=1.5pt,outer sep=0pt},
  constr/.style={overlay},% % help lines not enlarge the bounding box
  %constr/.append style={draw=red,very thick},% to show the help lines
]
%
\newcommand\dist{6}% distance A to B
\newcommand\CAB{60}% angle CAB
\newcommand\ABC{75}% angle ABC
\newcommand\AST{110}% angle AST
\newcommand\factor{3}% factor for the construction of the help lines
%
\coordinate[label=left:$A$] (A) at (0,0);
\coordinate[label=above:$B$] (B) at (\CAB:\dist);
\coordinate (S) at ($(A)!0.75!(B)$);
% C
\path[name path=horizontal,constr](A) -- ++({\factor*\dist},0);
\path[name path=leg,constr] (B) -- ++({\CAB-180+\ABC}:\factor*\dist);
\path[name intersections={of=horizontal and leg,by=C}](C)coordinate[label=right:$C$];
% T
\path[name path=st,constr] (S) -- ++({\CAB-180+\AST}:\factor*\dist);
\coordinate[name intersections={of=st and leg,by=T}];
% angles
\pic foreach \t/\a/\b/\c in {\CAB/C/A/B,\ABC/A/B/C,\AST/A/S/T}
  ["$\t^\circ$",draw=orange!80!black,angle radius=1.1cm,fill=orange!20]
  {angle=\a--\b--\c};
% triangle
\draw(A)--(B)--(C)--cycle;
\draw[blue](S)node[dot,label=left:$S$]{} -- (T)node[dot,label=right:$T$]{};
\end{tikzpicture}
\end{document}

This works because the angle CAB is given.

---

But maybe A is not at the origin or B is not given by polarcoordinates. Then we can use the turn key as explained in the pgfmanual in subsection "Rotational Relative Coordinates" (in the current manual version on page 142):

The effect of this key is to locally shift the coordinate system so that the last point reached is at the origin and the coordinate system is "turned" so that the x-axis points in the direction of a tangent entering the last point.

So I define an additional coordinate H1 by

\path[constr](A)--(B)--([turn]{-180+\ABC}:\dist)coordinate(H1);

That means: go from A to B and then change your direction relatively by -180°+\ABC° (in the example -180+75=-105) and go \dist (in the example 18) units. The resulting angle ABH1 is then \ABC° (in the example 75°).

In the following example A is at (-2,0), B is at (4,5), C is on the horizontal line through A and the angle at B is again 75° and at S 110°.

\documentclass[margin=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,intersections,angles, quotes}
\begin{document}
\begin{tikzpicture}[
  dot/.style={circle,fill,inner sep=1.5pt,outer sep=0pt},
  constr/.style={overlay},% % help lines not enlarge the bounding box
 % constr/.append style={draw=red,very thick},% to show the help lines
]
%
\newcommand\dist{18}% distance A to B
\newcommand\ABC{75}% angle ABC
\newcommand\AST{110}% angle AST
%
\coordinate[label=left:$A$] (A) at (-2,0);
\coordinate[label=above:$B$] (B) at (4,5);
\coordinate (S) at ($(A)!0.75!(B)$);
% C
\path[name path=horizontal,constr](A) -- ++(\dist,0);
\path[constr](A)--(B)--([turn]{-180+\ABC}:\dist)coordinate(H1);
\path[name path=leg,constr] (B)--(H1);
\path[name intersections={of=horizontal and leg,by=C}](C)coordinate[label=right:$C$];
% T
\path[constr](A) -- (S) -- ([turn]{-180+\AST}:\dist)coordinate(H2);
\path[name path=st,constr] (S) -- (H2);
\coordinate[name intersections={of=st and leg,by=T}];
%% angles
\pic foreach \t/\a/\b/\c in {\ABC/A/B/C,\AST/A/S/T}
  ["$\t^\circ$",draw=purple!80!black,angle radius=1.1cm,fill=purple!20]
  {angle=\a--\b--\c};
%% triangle
\draw(A)--(B)--(C)--cycle;
\draw[blue](S)node[dot,label=left:$S$]{} -- (T)node[dot,label=right:$T$]{};
\end{tikzpicture}
\end{document}