Trigonometric Functions in Right-Angle Triangle


\begin{tikzpicture}
  % grid
  \draw[help lines] (-2,-2) grid (2,2);
  
  % origin
  %\draw[red, line width=.1mm] (-0.1,-0.1) -- (0.1,0.1)
  %  (0.1,-0.1) -- (-0.1,0.1);
  %\coordinate[label={[red]above:$O$}] (O) at (0,0);
  
  % coordinates
  \coordinate[label={[black]left:$A$}] (A) at (-1,1);
  \coordinate[label={[black]below:$B$}] (B) at (-1,-1);
  \coordinate[label={[black]right:$C$}] (C) at (1,-1);
  
  % triangle 
  \draw[black, line width=.1mm] (A) -- (B) -- (C) -- cycle;
  
  % alpha 
  \markangle{A}{B}{C}{3mm}{3mm}{$\alpha$}{cyan}{north}
  
  % braces
  %\drawbrace{B}{C}{2mm}{blue}{$a$}{0}{-4mm}{mirror}
  %\drawbrace{A}{B}{2mm}{green}{$c$}{-4mm}{0}{mirror}
  %\drawbrace{A}{C}{2mm}{red}{$b$}{3mm}{3mm}{}
  
\end{tikzpicture}

\begin{eqnarray*}
\sin{{\color{cyan}\alpha}} &=& \frac{\color{blue}{a}}{\color{red}{b}} = \frac{\overline{BC}}{\overline{AC}} & \\
\cos{{\color{cyan}\alpha}} &=& \frac{\color{green}{c}}{\color{red}{b}} = \frac{\overline{AB}}{\overline{AC}} & \\
\tan{{\color{cyan}\alpha}} &=& \frac{\color{blue}{a}}{\color{green}{c}} = \frac{\overline{BC}}{\overline{AB}}
\end{eqnarray*}

\begin{eqnarray*}
\csc{{\color{cyan}\alpha}} &=& \frac{\color{red}{b}}{\color{blue}{a}} = \frac{\overline{AC}}{\overline{BC}} & \\
\sec{{\color{cyan}\alpha}} &=& \frac{\color{red}{b}}{\color{green}{c}} = \frac{\overline{AC}}{\overline{AB}} & \\
\cot{{\color{cyan}\alpha}} &=& \frac{\color{green}{c}}{\color{blue}{a}} = \frac{\overline{AB}}{\overline{BC}} 
\end{eqnarray*}

\begin{eqnarray*}
\tan{{\color{cyan}\alpha}} &=& \frac{\sin{{\color{cyan}\alpha}}}{\cos{{\color{cyan}\alpha}}}
\end{eqnarray*}

Pythagoras (Law of Cosines)



\begin{tikzpicture}
  % grid
  \draw[help lines] (-2,-2) grid (2,2);
  
  % origin
  %\draw[red, line width=.1mm] (-0.1,-0.1) -- (0.1,0.1)
  %  (0.1,-0.1) -- (-0.1,0.1);
  %\coordinate[label={[red]above:$O$}] (O) at (0,0);
  
  % coordinates
  \coordinate[label={[black]left:$A$}] (A) at (-1,1);
  \coordinate[label={[black]below:$B$}] (B) at (-1,-1);
  \coordinate[label={[black]right:$C$}] (C) at (1,0);
  
  % triangle 
  \draw[black, line width=.1mm] (A) -- (B) -- (C) -- cycle;
  
  % alpha 
  \markangle{A}{B}{C}{3mm}{3mm}{$\alpha$}{cyan}{north}
  
  % braces
  \drawbrace{B}{C}{2mm}{blue}{$a$}{0}{-4mm}{mirror}
  \drawbrace{A}{B}{2mm}{green}{$c$}{-4mm}{0}{mirror}
  \drawbrace{A}{C}{2mm}{red}{$b$}{3mm}{3mm}{}
  
\end{tikzpicture}

\begin{eqnarray*}
{\color{blue}a}^{2} &=& {\color{green}c}^{2} + {\color{red}b}^{2} - 2{\color{green}c}{\color{red}b}\cos{{\color{cyan}\alpha}}
\end{eqnarray*}

in a right-angle triangle, $\alpha=90^{\circ}$ such that $\cos{{\color{cyan}\alpha}}=0$:

\begin{eqnarray*}
{\color{blue}a}^{2} &=& {\color{green}c}^{2} + {\color{red}b}^{2}
\end{eqnarray*}

Medians in a Triangle


\begin{tikzpicture}
  % grid
  \draw[help lines] (-5,-5) grid (5,5);
  
  % coordinates
  \coordinate[label={[black]left:$A$}] (A) at (-3,3);
  \coordinate[label={[black]below:$B$}] (B) at (-3,-3.5);
  \coordinate[label={[black]right:$C$}] (C) at (3,0);
  \coordinate[label={[black]above:$M_{a}$}] (Ma) at ($ (B) !.5! (C) $);
  \coordinate[label={[black]below left:$M_{b}$}] (Mb) at ($ (A) !.5! (C) $);
  \coordinate[label={[black]below right:$M_{c}$}] (Mc) at ($ (A) !.5! (B) $);
  
  % triangle 
  \draw[black, line width=.1mm] (A) -- (B) -- (C) -- cycle;
  
  % median for a
  \path[draw, black, line width=.1mm, dotted, name path=median from A] (A) -- (Ma);
  % median for b
  \path[draw, black, line width=.1mm, dotted, name path=median from B] (B) -- (Mb);
  % median for c
  \path[draw, black, line width=.1mm, dotted, name path=median from C] (C) -- (Mc);
  
  % point
  \drawpoint{Ma}{.5mm}{black}
  \drawpoint{Mb}{.5mm}{black}
  \drawpoint{Mc}{.5mm}{black}

  % braces for segment medians a
  \drawbrace{B}{Ma}{3mm}{blue}{$a'$}{0}{-5mm}{mirror}
  \drawbrace{Ma}{C}{3mm}{blue}{$a''$}{0}{-5mm}{mirror}
  \drawbrace{B}{C}{14mm}{blue}{$a$}{8mm}{-14mm}{mirror}
  % braces for segment medians b
  \drawbrace{A}{Mb}{3mm}{red}{$b'$}{3mm}{5mm}{}
  \drawbrace{Mb}{C}{3mm}{red}{$b''$}{3mm}{5mm}{}
  \drawbrace{A}{C}{16mm}{red}{$b$}{8mm}{17mm}{}
  % braces for segment medians c
  \drawbrace{A}{Mc}{3mm}{green}{$c'$}{-5mm}{0}{mirror}
  \drawbrace{Mc}{B}{3mm}{green}{$c''$}{-5mm}{0}{mirror}
  \drawbrace{A}{B}{18mm}{green}{$c$}{-20mm}{0}{mirror}
  
  % median braces
  \drawbrace{A}{Ma}{1mm}{black}{$m_{a}$}{-2mm}{-2mm}{mirror}
  \drawbrace{B}{Mb}{1mm}{black}{$m_{b}$}{3.5mm}{-1mm}{mirror}
  \drawbrace{C}{Mc}{1mm}{black}{$m_{c}$}{0}{-3mm}{}
  
  % gravity centre
  \path[draw, red, name intersections={of=median from A and median from B, by=G}];
  \drawpoint{G}{.5mm}{red}
  \coordinate[label={[black]above:$G$}] (G) at (G);
  
\end{tikzpicture}


\begin{eqnarray*}
m_{a} &=& \sqrt{\frac{2{\color{red}b}^{2}+2{\color{green}c}^{2}-{\color{blue}a}^{2}}{4}} 
\end{eqnarray*}

\begin{eqnarray*}
a' &\equiv& a'' & \\
b' &\equiv& b'' & \\
c' &\equiv& c''
\end{eqnarray*}

\begin{eqnarray*}
\frac{3}{4}({\color{blue}a}^{2} + {\color{red}b}^{2} + {\color{green}c}^{2}) &=& m_{{\color{blue}a}}^{2} + m_{{\color{red}b}}^{2} + m_{{\color{green}c}}^{2}
\end{eqnarray*}

\begin{itemize}
  \item $G$ represents the centre of mass also called the centroid. The entire weight of the triangle can be balanced in $G$ by pinning a pivot in $G$.
  \item The medians $m_{a}$, $m_{b}$ and $m_{c}$ divide the triangle $\widehat{ABC}$ into $6$ smaller triangles of equal surface.
\end{itemize}

Area of a Triangle



\begin{tikzpicture}
  % grid
  \draw[help lines] (-2,-2) grid (2,2);
  
  % origin
  %\draw[red, line width=.1mm] (-0.1,-0.1) -- (0.1,0.1)
  %  (0.1,-0.1) -- (-0.1,0.1);
  %\coordinate[label={[red]above:$O$}] (O) at (0,0);
  
  %\draw[black,line width=5mm] (-3, {3 * sqrt(3) - sqrt(3)}) -- ((-3,0);
  %\node [square,rotate={30},minimum size=10mm] at (-3, {3 * sqrt(3) - sqrt(3)}) [draw] (d2) [orange,fill,text=white] {$d_{2}$};
  %\draw[orange,line width=.1mm] (-3, {3 * sqrt(3) - sqrt(3)}) -- (-2,0); 
  %\draw[orange,line width=.1mm] (-3, {3 * sqrt(3) - sqrt(3)}) -- (0,2);
  
  % coordinates
  \coordinate[label={[black]left:$A$}] (A) at (-1,1.5);
  \coordinate[label={[black]left:$B$}] (B) at (-1.5,-1);
  \coordinate[label={[black]right:$C$}] (C) at (2,-1);
  \coordinate[label={[black]below:$P$}] (P) at (-1,-1);
  
  % mark P
  \drawpoint{P}{.5mm}{black}
  
  % triangle 
  \draw[black, line width=.1mm] (A) -- (B) -- (C) -- cycle;
  
  % perpendicular
  \draw[black, line width=.1mm] (A) -- (P);
  
  % alpha 
  \node [square,minimum size=1mm,dotted] at (-1.14,-0.85) [draw] (d2) [black] {};
  
  % braces
  \drawbrace{B}{C}{2mm}{blue}{$a$}{0}{-4mm}{mirror}
  \drawbrace{A}{P}{2mm}{red}{$h$}{4mm}{0}{}
  \drawbrace{A}{B}{2mm}{green}{$c$}{-4mm}{0}{mirror}
  \drawbrace{A}{C}{2mm}{red}{$b$}{3mm}{3mm}{}
  
\end{tikzpicture}

\begin{eqnarray*}
A_{\widehat{ABC}} &=& \frac{{\color{red}h}{\color{blue}a}}{2}
\end{eqnarray*}

Using Heron's formula:

\begin{eqnarray*}
A_{\widehat{ABC}} &=& \sqrt{p(p-{\color{blue}a})(p-{\color{red}b})(p-{\color{green}c})} \text{ and } p=\frac{{\color{blue}{a}}+{\color{red}{b}}+{\color{green}{c}}}{2}
\end{eqnarray*}

A complete derivation of the Heron formula can be found in the mathematics section on triangles.

Theorem of Ceva


\begin{tikzpicture}
  % grid
  \draw[help lines] (-2,-2.5) grid (2,2);
  
  % coordinates
  \coordinate[label={[black]left:$A$}] (A) at (-2,2);
  \coordinate[label={[black]below:$B$}] (B) at (-2,-2.5);
  \coordinate[label={[black]right:$C$}] (C) at (2,0);
  \coordinate[label={[black]above:$D$}] (D) at ($ (B) !.52! (C) $);
  \coordinate[label={[black]below left:$E$}] (E) at ($ (A) !.4! (C) $);
  \coordinate[label={[black]below right:$F$}] (F) at ($ (A) !.4! (B) $);
  
  % triangle 
  \draw[black, line width=.1mm] (A) -- (B) -- (C) -- cycle;
  
  % line from a
  \path[draw, black, line width=.1mm, dotted, name path=line from A] (A) -- (D);
  % line from b
  \path[draw, black, line width=.1mm, dotted, name path=line from B] (B) -- (E);
  % line from c
  \path[draw, black, line width=.1mm, dotted, name path=line from C] (C) -- (F);
  
  % point
  \drawpoint{D}{.5mm}{black}
  \drawpoint{E}{.5mm}{black}
  \drawpoint{F}{.5mm}{black}

  % braces for segment medians a
  \drawbrace{A}{F}{3mm}{green}{$c$}{-5mm}{0}{mirror}
  \drawbrace{F}{B}{3mm}{green}{$c'$}{-5mm}{0}{mirror}
  % braces for segment medians b
  \drawbrace{B}{D}{3mm}{blue}{$a$}{3mm}{-5mm}{mirror}
  \drawbrace{D}{C}{3mm}{blue}{$a'$}{3mm}{-5mm}{mirror}
  % braces for segment medians c
  \drawbrace{C}{E}{3mm}{red}{$b$}{0}{5mm}{mirror}
  \drawbrace{E}{A}{3mm}{red}{$b'$}{0}{5mm}{mirror}
  
  % gravity centre
  \path[draw, red, name intersections={of=line from A and line from C, by=O}];
  \drawpoint{O}{.5mm}{red}
  \coordinate[label={[black]above:$O$}] (O) at (O);
  
\end{tikzpicture}


\begin{eqnarray*}
\frac{{\color{green}c}}{{\color{green}c'}} * \frac{{\color{blue}a}}{{\color{blue}a'}} * \frac{{\color{red}b}}{{\color{red}b'}} &=& 1
\end{eqnarray*}

The point $O$ represents the intersection of three arbitrarily drawn lines from the vertices ($A$, $B$ or $C$) of the triangle \widehat{ABC}.

Theorem of Menelaus



\begin{tikzpicture}
  % grid
  \draw[help lines] (-3,-3) grid (5,3);
  
  % coordinates
  \coordinate[label={[black]left:$A$}] (A) at (-2,2.5);
  \coordinate[label={[black]left:$B$}] (B) at (-2.5,-2);
  \coordinate[label={[black]right:$C$}] (C) at (3,-2);
  
  \coordinate[label={[black]south north:$P$}] (P) at ($ (A) !.3! (B) $);
  \drawpoint{P}{.5mm}{red}
  \coordinate[label={[black]south north:$Q$}] (Q) at ($ (A) !.7! (C) $);
  \drawpoint{Q}{.5mm}{red}
  \path[draw, black, line width=.1mm, dotted, name path=line from B] (B) -- (C) -- (3,-1);
  
  \draw[red, solid, line width=.1mm, name path=secant] (P) -- (Q) -- (4,-2);
  
  \draw[draw, red, name intersections={of=line from B and secant, by=R}];
  \coordinate[label={[black]south north:$R$}] (R) at (4,-2);
  \drawpoint{R}{.5mm}{red}
  
  % triangle 
  \draw[black, line width=.1mm] (A) -- (B) -- (C) -- cycle;
  
  % perpendicular
  %\draw[black, line width=.1mm] (A) -- (P);
  
  % braces
  \drawbrace{B}{R}{6mm}{blue}{$a$}{0}{-8mm}{mirror}
  \drawbrace{B}{C}{2mm}{blue}{$a'$}{0}{-4mm}{mirror}
  \drawbrace{C}{R}{2mm}{green}{$r$}{0}{-4mm}{mirror}
  \drawbrace{Q}{C}{2mm}{orange}{$s$}{-4mm}{-1mm}{mirror}
  \drawbrace{A}{Q}{2mm}{violet}{$t$}{4mm}{1mm}{}
  \drawbrace{A}{P}{2mm}{teal}{$u$}{-4mm}{0}{mirror}
  \drawbrace{P}{B}{2mm}{olive}{$v$}{-8mm}{0}{mirror}
  
\end{tikzpicture}

\begin{eqnarray*}
\frac{{\color{blue}a'}}{{\color{blue}a}}*\frac{{\color{orange}s}}{{\color{violet}}t}}*\frac{{\color{teal}u}}{{\color{olive}v}} &=& 1
\end{eqnarray*}

Theorem of Parallels



\begin{tikzpicture}
  % grid
  \draw[help lines] (-3,-3) grid (5,3);
  
  % coordinates
  \coordinate[label={[black]left:$A$}] (A) at (-2,2.5);
  \coordinate[label={[black]left:$B$}] (B) at (-2.5,-2);
  \coordinate[label={[black]right:$C$}] (C) at (3,-2);
  
  \coordinate[label={[black]left:$P$}] (P) at ($ (A) !.3! (B) $);
  \drawpoint{P}{.5mm}{red}
  \coordinate[label={[black]south north:$Q$}] (Q) at ($ (A) !.3! (C) $);
  \drawpoint{Q}{.5mm}{red}
  
  \draw[red, solid, line width=.1mm, name path=secant] (P) -- (Q);
  
  % triangle 
  \draw[black, line width=.1mm] (A) -- (B) -- (C) -- cycle;
  
  % perpendicular
  %\draw[black, line width=.1mm] (A) -- (P);
  
\end{tikzpicture}

If $\overline{PQ}$ is parallel to $\overline{BC}$ then:

\begin{eqnarray*}
\frac{\overline{AP}}{\overline{AB}}=\frac{\overline{AQ}}{\overline{AC}}&=&\frac{\overline{PQ}}{\overline{BC}}
\end{eqnarray*}


fuss/mathematics/geometry/shapes/triangles.txt ยท Last modified: 2017/02/22 18:30 (external edit)

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