I am having a hard time here. I am trying to make this triangle look like the one I draw on the Paper. But but I am not sure how i can implemantate the missing lines and angles?


\coordinate [label=left:$A$] (A) at (-1.5cm,-1.cm);
\coordinate [label=right:$B$] (C) at (1.5cm,-1.0cm);
\coordinate [label=above:$C$] (B) at (1.5cm,1.0cm);
\draw (A) -- node[sloped,above] {b} (B) -- node[right] {a = Rundetårn} (C) -- node[below] {$c=\num{20}$} (A);

\draw (1.25cm,-1.0cm) rectangle (1.5cm,-0.75cm);
\tkzMarkAngle[fill= gray,size=0.8cm,opacity=.2](C,A,B)
\tkzLabelAngle[pos = 0.6](C,A,B){$61^0$}


enter image description here

  • the package tkz-euclide is the perfect package to do this. – Arne Timperman Nov 17 '15 at 12:14
  • What are the numbers without degrees? – Tom Bombadil Nov 17 '15 at 13:26
  • The numbers without degrees os meters – Daniel Guldberg Aaes Nov 17 '15 at 13:44
  • 1
    how come 90 = 126? – CroCo Nov 17 '15 at 13:47
  • So something on the right is 56 Meters? But what exactly is? The dashed line or the solid line to the left and right? Or the solid one across from it? – Tom Bombadil Nov 17 '15 at 13:52

Here's haw it can be done with the help of the calc library and the pic mechanism (and therefore the angles and quotes libraries):


\documentclass[tikz, border=2mm]{standalone}
\usetikzlibrary{calc, angles, quotes}


    % Define A, B and C
    \draw (0,0) coordinate (A) -- ++ (8,0) coordinate(B) -- ++ (0,6) coordinate (C) -- cycle;

    % go to the point 40% on the way from C to A, name it D; from there go to the intersection from a horizontal line through A and a vertical line through D, call it E
    \draw[dashed] ($(C)!0.4!(A)$) coordinate (D) -- (A -| D) coordinate (E);

    % connect E and C
    \draw (E) -- (C);

    % draw the angle at A; point are specified in negative direction (clockwise)
    \path pic [draw, "$46^{\circ}$", angle radius=1.2cm] {angle=B--A--C};

    % draw the angle at E
    \path pic [draw, "$54^{\circ}$", angle radius=1.2cm] {angle=B--E--C};




enter image description here

  • Wow thanks! do you know how to get the 90 angle and 126 angle on too ? – Daniel Guldberg Aaes Nov 17 '15 at 14:04
  • The same as the other paths; for 90° the points are D--E--A, for 126° its C--E--A. Probably you should alter the angle radius for one of them, otherwise they're partially drawn on top of each other. – Tom Bombadil Nov 17 '15 at 14:15

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