Consider the following code:
\DocumentMetadata{} \documentclass[ a4paper, 12pt, danish ]{article} \usepackage{fontspec} \setmainfont[ NFSSFamily = tgp, Extension = .otf, UprightFont = *-Regular, BoldFont = *-Bold, ItalicFont = *-Italic, BoldItalicFont = *-BoldItalic, Ligatures = { TeX, CommonOff } ]{texgyrepagella} \usepackage[ math-style = TeX ]{unicode-math} \setmathfont{texgyrepagella-math.otf} \setmathfont[ version = bold, FakeBold = 4 ]{texgyrepagella-math.otf} \DeclareSymbolFont{textdigits}{TU}{tgp}{m}{n} \SetSymbolFont{textdigits}{bold}{TU}{tgp}{b}{n} \Umathcode`0="7 \symtextdigits `0 \Umathcode`1="7 \symtextdigits `1 \Umathcode`2="7 \symtextdigits `2 \Umathcode`3="7 \symtextdigits `3 \Umathcode`4="7 \symtextdigits `4 \Umathcode`5="7 \symtextdigits `5 \Umathcode`6="7 \symtextdigits `6 \Umathcode`7="7 \symtextdigits `7 \Umathcode`8="7 \symtextdigits `8 \Umathcode`9="7 \symtextdigits `9 \usepackage{polyglossia} \setdefaultlanguage{danish} \usepackage[ hmargin = 2.4cm, vmargin = 3cm ]{geometry} \usepackage{ multirow, array } \usepackage{siunitx} \usepackage{pst-eucl} \sisetup{ locale = DE } \psset{ dimen = m, PointSymbol = none, PointName = none, MarkAngleRadius = 0.25, RightAngleSize = 0.2, LabelSep = -0.3 } \def\mlr{1} \newcommand*\saenk[1]{\raisebox{\fpeval{-5/9*\mlr}ex}{#1}} \newcommand*\nyLinje[2][c]{\begin{tabular}[#1]{@{}l@{}}#2\end{tabular}} \begin{document} \begin{center} \bfseries {\fontsize{30}{36}\selectfont Retvinklede trekanter}\\[1ex] {\fontsize{20}{24}\selectfont (Pythagoras og trigonometri)} \end{center} \vspace{\fill} \begin{alignat*}{2} &\text{Pythagoras:}&\qquad a^{2} + b^{2} &= c^{2}\\ &\text{Trigonometri:}& \sin(v) &= \frac{\text{modstående karate}}{\text{hypotenusen}}\\ &\text{Trigonometri:}& \cos(v) &= \frac{\text{hosliggende karate}}{\text{hypotenusen}}\\ &\text{Trigonometri:}& \tan(v) &= \frac{\text{modstående katete}}{\text{hosliggende katete}} \end{alignat*} \vspace{\fill} \begin{center} \begin{tabular}{ |p{4.95cm}| >{\centering\arraybackslash}p{4.95cm}| >{\centering\arraybackslash}p{4.95cm}| } \hline \multicolumn{1}{|c|}{\multirow{2}{*}{\raisebox{\fpeval{-1.2*\mlr}ex}{Du kender \dots}}} & \multicolumn{2}{ c|}{\saenk{Du kan beregne \dots}} \\[\mlr ex] \cline{2-3} & \saenk{sider} & \saenk{vinkler} \\[\mlr ex] \hline Siderne~$a$ og~$b$\hspace*{2em}% \raisebox{\fpeval{-4.5*\mlr}ex}{% \begin{pspicture}(0,-0.38)(1.33,1.5) \small \pstGeonode(0,0){A}(1,1.5){B}(1,0){C} \pspolygon(A)(B)(C) \pstRightAngle{B}{C}{A} \uput{0.1}[315](C){$C$} {\psset{ linestyle = none } \pcline[offset = 6pt](B)(C) \ncput{$a$} \pcline[offset = 7pt](C)(A) \ncput{$b$}} \end{pspicture}} & $c = \sqrt{a^{2} + b^{2}}$ & \parbox{4cm}{\vspace*{-1ex}\begin{align*} A &= \tan^{-1}{\mkern -7mu}\left(\frac{a}{b}\right){\mkern -7mu}\\ B &= \tan^{-1}{\mkern -7mu}\left(\frac{b}{a}\right){\mkern -7mu} % \\ &= \ang{90} - A \end{align*}\vspace*{-1ex}} \\[\mlr ex] \hline Siderne~$a$ og~$c$\hspace*{2.06em}% \raisebox{\fpeval{-4.5*\mlr}ex}{% \begin{pspicture}(0,-0.35)(1.33,1.5) \small \pstGeonode(0,0){A}(1,1.5){B}(1,0){C} \pspolygon(A)(B)(C) \pstRightAngle{B}{C}{A} \uput{0.1}[315](C){$C$} {\psset{ linestyle = none } \pcline[offset = 6pt](B)(C) \ncput{$a$} \pcline[offset = 6pt](A)(B) \ncput{$c$}} \end{pspicture}} & $b = \sqrt{c^{2} - a^{2}}$ & \parbox{4cm}{\begin{align*} A &= \sin^{-1}{\mkern -7mu}\left(\frac{a}{c}\right){\mkern -7mu}\\ B &= \cos^{-1}{\mkern -7mu}\left(\frac{a}{c}\right){\mkern -7mu} % \\ &= \ang{90} - A \end{align*}} \\[\mlr ex] \hline Siderne~$b$ og~$c$\hspace*{2.05em}% \raisebox{\fpeval{-4.5*\mlr}ex}{% \begin{pspicture}(0,-0.38)(1.33,1.5) \small \pstGeonode(0,0){A}(1,1.5){B}(1,0){C} \pspolygon(A)(B)(C) \pstRightAngle{B}{C}{A} \uput{0.1}[315](C){$C$} {\psset{ linestyle = none } \pcline[offset = 7pt](C)(A) \ncput{$b$} \pcline[offset = 6pt](A)(B) \ncput{$c$}} \end{pspicture}} & $a = \sqrt{c^{2} - b^{2}}$ & \parbox{4cm}{\vspace*{-1ex}\begin{align*} A &= \cos^{-1}{\mkern -7mu}\left(\frac{b}{c}\right){\mkern -7mu}\\ B &= \sin^{-1}{\mkern -7mu}\left(\frac{b}{c}\right){\mkern -7mu} % \\ &= \ang{90} - A \end{align*}\vspace*{-1ex}} \\[\mlr ex] \hline \raisebox{\fpeval{-0.3*\mlr}ex}{\nyLinje[l]{Siden~$a$ og\\ vinklen~$A$}}% \hspace*{2.71em}% \raisebox{\fpeval{-4.5*\mlr}ex}{% \begin{pspicture}(-0.38,-0.35)(1.33,1.5) \small \pstGeonode(0,0){A}(1,1.5){B}(1,0){C} \pspolygon(A)(B)(C) \pstMarkAngle{C}{A}{B}{$A$} \pstRightAngle{B}{C}{A} \uput{0.1}[315](C){$C$} {\psset{ linestyle = none } \pcline[offset = 6pt](B)(C) \ncput{$a$}} \end{pspicture}} & \parbox{4cm}{\begin{align*} b &= \frac{a}{\tan(A)}\\ c &= \frac{a}{\sin(A)} \end{align*}} & \parbox{4cm}{\begin{equation*} B = \ang{90} - A \end{equation*}} \\[\fpeval{\mlr+3.5} ex] \hline \raisebox{\fpeval{-0.6*\mlr}ex}{\nyLinje[c]{Siden~$a$ og\\ vinklen~$B$}}% \hspace*{3.61em}% \raisebox{\fpeval{-5.3*\mlr}ex}{% \begin{pspicture}(0,-0.33)(1.33,1.95) \small \pstGeonode(0,0){A}(1,1.5){B}(1,0){C} \pspolygon(A)(B)(C) \pstMarkAngle{A}{B}{C}{$B$} \pstRightAngle{B}{C}{A} \uput{0.1}[330](C){$C$} {\psset{ linestyle = none } \pcline[offset = 6pt](B)(C) \ncput{$a$}} \end{pspicture}} & \parbox{4cm}{\begin{align*} b &= a \cdot \tan(B)\\ c &= \frac{a}{\cos(B)} \end{align*}} & \raisebox{\fpeval{-0.3*\mlr}ex}{\parbox{4cm}{\begin{equation*} A = \ang{90} - B \end{equation*}}} \\[\fpeval{\mlr+3.5} ex] \hline \end{tabular} \end{center} \vspace*{-1.5ex} \hspace*{\fill} \emph{fortsættes på næste side \dots} \newpage \noindent \emph{\dots{} fortsat fra sidste side} \vspace*{-3ex} \begin{center} \begin{tabular}{ |p{4.96cm}| >{\centering\arraybackslash}p{4.96cm}| >{\centering\arraybackslash}p{4.96cm}| } \hline \multicolumn{1}{|c|}{\multirow{2}{*}{\raisebox{\fpeval{-1.2*\mlr}ex}{Du kender \dots}}} & \multicolumn{2}{ c|}{\saenk{Du kan beregne \dots}} \\[\mlr ex] \cline{2-3} & \saenk{sider} & \saenk{vinkler} \\[\mlr ex] \hline \raisebox{\fpeval{-0.2*\mlr}ex}{\nyLinje[c]{Siden~$b$ og\\ vinklen~$A$}}% \hspace*{3.59em}% \raisebox{\fpeval{-4*\mlr}ex}{% \begin{pspicture}(0,-0.33)(1.33,1.5) \small \pstGeonode(0,0){A}(1,1.5){B}(1,0){C} \pspolygon(A)(B)(C) \pstMarkAngle{C}{A}{B}{$A$} \pstRightAngle{B}{C}{A} \uput{0.1}[330](C){$C$} {\psset{ linestyle = none } \pcline[offset = 6pt](C)(A) \ncput{$b$}} \end{pspicture}} & \parbox{4cm}{\begin{align*} a &= b \cdot \tan(A)\\ c &= \frac{b}{\cos(A)} \end{align*}} & \raisebox{\fpeval{-0.3*\mlr}ex}{\parbox{4cm}{\begin{equation*} B = \ang{90} - A \end{equation*}}} \\[\fpeval{\mlr+3.5} ex] \hline \raisebox{\fpeval{-0.2*\mlr}ex}{\nyLinje[c]{Siden~$b$ og\\ vinklen~$B$}}% \hspace*{3.59em}% \raisebox{\fpeval{-5*\mlr}ex}{% \begin{pspicture}(0,-0.33)(1.33,1.5) \small \pstGeonode(0,0){A}(1,1.5){B}(1,0){C} \pspolygon(A)(B)(C) \pstMarkAngle{A}{B}{C}{$B$} \pstRightAngle{B}{C}{A} \uput{0.1}[330](C){$C$} {\psset{ linestyle = none } \pcline[offset = 6pt](C)(A) \ncput{$b$}} \end{pspicture}} & \parbox{4cm}{\begin{align*} a &= \frac{b}{\tan(A)}\\ c &= \frac{b}{\sin(B)} \end{align*}} & \raisebox{\fpeval{-0.3*\mlr}ex}{\parbox{4cm}{\begin{equation*} A = \ang{90} - B \end{equation*}}} \\[\fpeval{\mlr+3.5} ex] \hline \raisebox{\fpeval{-0.3*\mlr}ex}{\nyLinje[l]{Siden~$c$ og\\ vinklen~$A$}}% \hspace*{2.75em}% \raisebox{\fpeval{-4.5*\mlr}ex}{% \begin{pspicture}(-0.38,-0.35)(1.33,1.5) \small \pstGeonode(0,0){A}(1,1.5){B}(1,0){C} \pspolygon(A)(B)(C) \pstRightAngle{B}{C}{A} \uput{0.1}[315](C){$C$} \pstMarkAngle{C}{A}{B}{$A$} {\psset{ linestyle = none } \pcline[offset = 6pt](A)(B) \ncput{$c$}} \end{pspicture}} & \parbox{4cm}{\begin{align*} a &= c \cdot \sin(A)\\ b &= c \cdot \cos(A) \end{align*}} & \raisebox{\fpeval{-0.3*\mlr}ex}{\parbox{4cm}{\begin{equation*} B = \ang{90} - A \end{equation*}}} \\[\fpeval{\mlr+3.5} ex] \hline \raisebox{\fpeval{0.5*\mlr}ex}{\nyLinje[l]{Siden~$c$ og\\ vinklen~$B$}}% \hspace*{3.4em}% \raisebox{\fpeval{-5*\mlr}ex}{ \begin{pspicture}(0,-0.4)(1.33,2.15) \small \pstGeonode(0,0){A}(1,1.5){B}(1,0){C} \pspolygon(A)(B)(C) \pstMarkAngle{A}{B}{C}{$B$} \pstRightAngle{B}{C}{A} \uput{0.1}[330](C){$C$} {\psset{ linestyle = none } \pcline[offset = 6pt](A)(B) \ncput{$c$}} \end{pspicture}} & \raisebox{\mlr ex}{\parbox{4cm}{\begin{align*} a &= c \cdot \cos(B)\\ b &= c \cdot \sin(B) \end{align*}}} & \raisebox{\mlr ex}{\parbox{4cm}{\begin{equation*} A = \ang{90} - B \end{equation*}}} \\[\fpeval{\mlr+3.5} ex] \hline \end{tabular} \end{center} \end{document} The code more or less produces the desired output, but it is a mess. Problems, I would like help dealing with are the following:
How do I automatically left-align the text in the first column and right-align he drawings? (All the manually added space is really ugly.)
How do I make all the table cells (except the header) have the same height?
How do I color the table header on both pages, say, grey?
Also, I would like a general clean-up of my code, if possible.
