2

MWE:

\documentclass{article}
\usepackage{tikz}
\usepackage{tkz-euclide}
\usetkzobj{all}
\begin{document}
\begin{tikzpicture}
\coordinate (O) at (0,0);
\draw (O) circle (3);
\coordinate[label = above left:$A$] (A) at (130:3);
\coordinate[label = above right:$D$] (D) at (58:3);
\coordinate[label = above :$E$] (E) at (98:1.8);
\coordinate[label = right:$C$] (C) at (0:3);
\coordinate[label = left:$B$] (B) at (180:3);
\draw (B) -- (D) -- (C);
\draw (B) -- (A) -- (C);
\draw (B) -- (C);
\tkzMarkAngle[size=0.60cm,%
opacity=.9](B,E,C)
\tkzLabelAngle[pos = 0.9](B,E,C){$120^{\circ}$}
\tkzMarkAngle[size=0.9cm,%
opacity=.9](D,C,E)
\tkzLabelAngle[pos = 1.3](D,C,E){$20^{\circ}$}
\end{tikzpicture}
\end{document}

Question: I want to adjust the arc length of angles 120 and 20. I mean arc of 120 is slightly big while the arc of angle 20 is slightly short.

enter image description here

3

The reason why \tkzMarkAngle(B,E,C) and \tkzMarkAngle(D,C,E) weren't giving you the correct output was because E wasn't actually being defined as the true intersection point of AC and BD.

Using this tkz-euclide answer from TikZ: Intersection of two lines you can calculate and label the true intersection point E using

\tkzInterLL(A,C)(B,D) \tkzGetPoint{E}
\tkzLabelPoints[above](E)

enter image description here

\documentclass{article}
\usepackage{tikz}
\usepackage{tkz-euclide}
\usetkzobj{all}
\begin{document}
    \begin{tikzpicture}
    \coordinate (O) at (0,0);
    \draw (O) circle (3);
    \coordinate[label = above left:$A$] (A) at (130:3);
    \coordinate[label = above right:$D$] (D) at (58:3);
    %\coordinate[label = above :$E$] (E) at (98:1.8);
    \coordinate[label = right:$C$] (C) at (0:3);
    \coordinate[label = left:$B$] (B) at (180:3);
    \draw (B) -- (D) -- (C);
    \draw (B) -- (A) -- (C);
    \draw (B) -- (C);

    \tkzInterLL(A,C)(B,D) \tkzGetPoint{E}
    \tkzLabelPoints[above](E)

    \tkzMarkAngle[size=0.60cm,%
    opacity=.9](B,E,C)
    \tkzLabelAngle[pos = 0.9](B,E,C){$120^{\circ}$}
    \tkzMarkAngle[size=0.9cm,%
    opacity=.9](D,C,E)
    \tkzLabelAngle[pos = 1.3](D,C,E){$20^{\circ}$}
    \end{tikzpicture}
\end{document}

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