我有两个角度的测量值和三个其他角度的标签x、y和。我使用包来获取它们。如何让它们与顶点的距离相同？zangles
\documentclass{amsart}
\usepackage{tikz}
\usetikzlibrary{calc,angles,positioning,intersections}
\begin{document}
\begin{tikzpicture}
\coordinate (A) at (0,0);
\coordinate (B) at (-4,0);
%These commands position vertex C so that the line through A and C is inclined at an angle of 95 degrees
%and the line through B and C is inclined at an angle of 58 degrees. Since there are 180 degrees in a
%triangle, the measure of angle ACB is 37 degrees.
\path[name path=A-to-C] (A) -- (95:7.5);
\path[name path=B-to-C] (B) -- ++(58:7.5);
\path[name intersections={of= A-to-C and B-to-C, by=C}];
%These commands draws the triangle.
\draw (A) -- (B) -- (C) -- cycle;
%This command draws a line segment horizontally rightwards from A.
\coordinate (D) at (3,0);
\draw (A) -- (D);
%This command draws a line segment upwards from C.
\draw (C) -- ++(95:1.5) coordinate (E);
%This command draws a line segment upwards from C.
\draw (B) -- ++(-122:1.5) coordinate (F);
%These commands draw the angles for the display.
\path pic[angle radius=5mm,"$\scriptstyle{x}$",angle eccentricity=1.25] {angle = D--A--C};
\path pic[angle radius=5mm,"$\scriptstyle{37}$",angle eccentricity=1.25] {angle = B--C--A};
\path pic[angle radius=5mm,"$\scriptstyle{58}$",angle eccentricity=1.25] {angle = A--B--C};
\path pic[angle radius=5mm,"$\scriptstyle{y}$",angle eccentricity=1.25] {angle = E--C--B};
\path pic[angle radius=5mm,"$\scriptstyle{z}$",angle eccentricity=1.25] {angle = F--B--A};
\end{tikzpicture}
\end{document}