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\markboth{Mochammad  Idris \textit{et al}} %Jika lebih dari dua penulis, tuliskan sebagai Nama Penulis Pertama dkk.
{An Area of Right Triangle for Trigonometry}

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\title{AN AREA OF RIGHT TRIANGLE FOR TRIGONOMETRY}
 
\author{MOCHAMMAD IDRIS$^{a}$\footnote{penulis korespondensi}, MOHAMMAD MAHFUZH SHIDDIQ$^{b}$, ERIDANI$^{c}$}

\address{$^{a}$ Universitas Lambung Mangkurat, Faculty of Mathematics and Natural Sciences, Department of
Mathematics, Banjarbaru 70714, INDONESIA,\\
$^{b}$ Universitas Lambung Mangkurat, Faculty of Mathematics and Natural Sciences, Department of
Mathematics, Banjarbaru 70714, INDONESIA,\\
$^{c}$ Universitas Airlangga, Faculty of Sciences and Technologies, Department of Mathematics, Campus
C Mulyorejo Surabaya 60115, INDONESIA.\\
email : \email{moch.idris@ulm.ac.id, mmahfuzhs@ulm.ac.id, eridani.dinadewi@gmail.com}}

\maketitle
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\begin{abstract}
\begin{center}
Diterima ..... \quad Direvisi ..... \quad Dipublikasikan ..... %tanggal-tanggal tersebut \textbf{dikosongkan} saja
\end{center}

\bigskip

\textbf{Abstrak}. %Dalam bahasa Indonesia
Kita akan menyelidiki proses pembentukan formula fungsi tangent dalam kajian trigonometri. Untuk hal tersebut, kita memanfaatkan 
hasil perhitungan area suatu segitiga siku-siku dan persegi panjang. Akibatnya kita juga mendapatkan formula fungsi sinus dan fungsi cosinus. Di samping itu, teorema Pythagoras juga dibuktikan kembali di sini.

\bigskip

\textbf{Abstract}. % Dalam bahasa Inggris
\textit{In this paper, we shall investigate formula of tangent function through determination of area of right triangle and rectangle. The consequence of this formula is used to obtain the other two trigonometric function, i.e. sine and cosine functions. Moreover, we also reprove Pythagorean theorem.}

\end{abstract}

\keywords{Pythagorean Theorem, Right Triangle, Trigonometry}

\section{Introduction}
One  of popular branch of mathematics is trigonometry. From many reference books, it is mentioned that the origin of the word  "trigonometry"  comes from two Greek  words. In \cite{mishra}, the first,   trigon  means "triangle" and the second,   metria/ metron  means "measure". 
Specifically, in  literature,  
trigonometry is the study of measuring triangles.  

Since in the middle school, we have been introduced to understand about sine, cosine and tangent functions.
The history of trigonometry spans a time period of more than two millennia and
crosses several cultural spheres, including the Hellenistic world, India, and Islam   (see \cite{brum}). The study of trigonometry still exists. In the recent era, research on trigonometry can be seen in  \cite{dattoli1, dattoli2}.

Many ways to know about trigonometry. Usually teachers directly define sine, cosine and tangent functions while they firstly introduce it to their students. On the other hand, some researcher used algebra to understand about trigonometry. Essentially in study about trigonometry, we can associate an angle (formed by two sides which are taken from  three sides of right triangle). Most trigonometry defined by taking a ratio of two side of right triangle. Considering a similar of right triangles, we knew that these ratio don't change whatever their side changed as long as the triangle are similar. The question: how to prove it? In this article, we will find the answer. After that we will define a function that represent this ratio that depend on the angle measure. Then we come to three well known trigonometric function that is sine, cosine and tangent functions. Furthermore, the results will be used to prove Pythagorean theorem.

\section{Concept Area of  Rectangle and Right Triangle}


It is well known that  a rectangle is a quadrilateral in $\mathbb{R}^2$,  with four sides and four  vertices. Check that  All  sides  are right sides (equal to $90^o $) and the  opposite sides   are equal and resemblant to each other.  
How about 
the area  of the rectangle? The area of a rectangle depends on its length ($L$) and width ($W$). The formula  is 
 $$\textrm{Area}_\textrm{rectangle } = L\times W.$$ 


A rectangle can be divided into two equal parts by cutting through one of its diagonals. The result is two congruent right triangles. The length and width of the rectangle become the legs of the right triangle.
How about 
the area   of the right triangle? The formula is 
 $$\textrm{Area}_\textrm{right triangle} =\frac{ L\times W}{2}.$$ 

 

\section{Main Results}

The discussion of angles begins by investigating the ratio of the two sides of a right triangle. We will use the formula for calculating the area of ​​a rectangle and the area of ​​a right triangle.  This investigation was carried out to convince us that the ratio of two sides of the right triangle does not depend  the size of the sides of the right  triangle.


\subsection{Three Trigonometry Functions}

\begin{figure}[h]
	 \centering 
	\includegraphics[width=0.6\linewidth]{gambar00.jpg} \caption{A right triangle $AED$}
	\label{gambar0}
\end{figure}


In this subsection, we want to obtain definitions of  three trigonometry functions. The definitions  are developed through area of right triangle that more simple than before. With this definition, trigonometry can introduce earlier, even, at elementary school. We use \(\overline{AB}\) to denote a segment from \(A\) to \(B\) and \(AB\) denote length of \(\overline{AB}\).

 
 \begin{proposition}\label{teo.01} 
Let $ \triangle AED $ be a right triangle with $ \angle AED = 90^{0} $ (see Figure \ref{gambar0}). If $B, C $ and $ O $ located on $\overline{AE}, \overline{ED} $ and $ \overline{AD} $, respectively, such that $ \overline{OC} $ is parallel to  $ \overline{AE} $ and $ \overline{OB} $ is perpendicular to $ \overline{AE} $, then we have\[\frac{OB}{AB}=\frac{CD}{OC}=\frac{ED}{AE}.\]
 \end{proposition}

\begin{proof}
Let $ AB=a, BE= b, EC=c $ and $ CD=d $. The area of right triangle $ \triangle AED $  is
\begin{eqnarray*}
\textrm{Area}_{AED}&=&\frac{(a+b)(c+d)}{2}
\end{eqnarray*} 
or
\begin{eqnarray*}
\textrm{Area}_{AED}&=&\frac{ac}{2}+\frac{bd}{2}+bc.
\end{eqnarray*}
So we have
\begin{eqnarray*}
 \frac{(a+b)(c+d)}{2}&=&  \frac{ac }{2}+ \frac{bd}{2} + bc\\
ad &=&bc \\
 \frac{c}{a}&=&\frac{d}{b}.
\end{eqnarray*}
From the equation above, we also obtain  
\begin{eqnarray*}
ad &=& bc \\
ad + bd&=& bc + bd.
\end{eqnarray*}
Consequently,  $\frac{c+d}{a+b}=\frac{d}{b}$. Hence  $\frac{c}{a}=\frac{d}{b}=\frac{c+d}{a+b}$ or \[\frac{OB}{AB}=\frac{CD}{OC}=\frac{ED}{AE}.\]The proof is complete.
\end{proof}

It follow from Proposition \ref{teo.01} that whenever we construct a segment that parallel with a leg of right triangle we can get those ratio. In other words, we have a constant $K \in  \mathbb{R}^+$ such that
\begin{equation}\label{rasio01}
K=\frac{c}{a}=\frac{d}{b}=\frac{c+d}{a+b} 
\end{equation}
holds.

Numerator of those ratios in \eqref{rasio01} is length of opposite side of angle of three similar right triangle and denominator is length of adjacent side of the angle. In fact, these ratios do not depend on the triangle formed but depend on the measure of angle $ \angle EAD $.




Now, we can define $ K $ as \textit{tangent function} of angle $ \theta $. Formally,$$\tan \theta:=\frac{y}{x},$$ 
where $x$ is adjacent side of $ \theta $ and $y$ is opposite side of $ \theta $.
So far, in trigonometry books, the authors immediately define the tangent function, but \textbf{without investigating the truth of the ratio}.

%Selama ini, dalam buku-buku trigonometry (lihat....),  langsung saja penulis mendifinisikan fungsi tangen, namun tanpa menyelidiki kebenaran rasio tersebut.


Moreover, we can find other ratio that involve hypotenuse from the right triangle. In the following proposition we determine it through area of right triangle again. 


 \begin{figure}[h]
	\centering
	\includegraphics[width=0.6\linewidth]{gambar11.jpg}
	\caption{A right triangle $AED$ with   a description of the angles and side lengths}\label{gambar1}
 \end{figure}

\begin{proposition}\label{teo.02} 
Let $ \triangle AED $ be a right triangle with $ \angle AED = 90^{0} $. If $B, C $ and $ O $ located on $\overline{AE}, \overline{ED} $ and $ \overline{AD} $, respectively, such that $ \overline{OC} $ is parallel to  $ \overline{AE} $ and $ \overline{OB} $ is perpendicular to $ \overline{AE} $, then we have\[\frac{AB}{AO}=\frac{OC}{OD}=\frac{AE}{AD}\]and\[\frac{BO}{AO}=\frac{CD}{OD}=\frac{ED}{AD}.\]
\end{proposition}

\begin{proof}
Let $ AO=e, OD=f, OC_{1}=a_{1}, C_{1}O_{1}=c_{1} $ and $ OO_{1}=e_{1} $ (see Figure \ref{gambar1}).

If $ O=O_1$, then $a_1=c_1=e_1=0$. 
The area of right triangle $\triangle AED$ is $$\frac{(e+f)v}{2}=\frac{(a+b)(c+d)}{2}.$$
We also determine area of right triangle $\triangle AOE$ 
 \begin{align*}  
\textrm{Area}_{AOE} =& \frac{ve}{2}= \frac{(a+b)c}{2}.   
\end{align*}
Next,  we have  $\frac{c}{e}= \frac{v}{a+b}.$   
Meanwhile $\frac{v}{a+b}=\frac{c+d}{e+f}$, so $\frac{c}{e}=\frac{c+d}{e+f}$. Moreover we obtain
\begin{align}\label{per.t021}   
\frac{c}{e}=&\frac{d}{f}=\frac{c+d}{e+f}    
\end{align}
or\[\frac{BO}{AO}=\frac{CD}{OD}=\frac{ED}{AD}.\]   

On the other side, the area of right triangle $\triangle DOE$ is
\begin{align*}  
\textrm{Area}_{DOE}=& \frac{vf}{2}= \frac{(c+d)b}{2}.   
\end{align*}
So,   $\frac{b}{f}= \frac{v}{c+d}.$ We also have   $\frac{v}{c+d}=\frac{a+b}{e+f}$. Consequently
\begin{align}\label{per.t022}   
\frac{a}{e}= & \frac{b}{f}=\frac{a+b}{e+f} 
\end{align}
or\[\frac{AB}{AO}=\frac{OC}{OD}=\frac{AE}{AD}.\]   


If  $O \neq O_1$, then $a_1, c_1, e_1 >0$. By  (\ref{per.t021}), and (\ref{per.t022}), we have
$$\frac{c+c_1}{e+e_1}= \frac{d-c_1}{f-e_1}=\frac{c+d}{e+f},$$ 
and  
$$\frac{a+a_1}{e+e_1}= \frac{b-a_1}{f-e_1}=\frac{a+b}{e+f}.$$
Since $\frac{c+c_1}{e+e_1}= \frac{d-c_1}{f-e_1}$ and $\frac{a+a_1}{e+e_1}= \frac{b-a_1}{f-e_1}$, then  
\begin{align*}  
f(d-c_1)+e(d-c_1)=& f(c+d)-e_1(c+d)\\   
 d( e+e_1) =& f( c+c_1)+(ec_1-ce_1)
 \end{align*}
and
\begin{align*}  
f(b-a_1)+e(b-a_1)=& f(a+b)-e_1(a+b)\\   
 b( e+e_1) =& f(a+a_1)+(ea_1-ae_1).\\
 \end{align*}   
Recall Proposition  \ref{teo.01},   
we have
$$\frac{c+c_1}{a+a_1}=\frac{d}{b}=\frac{f( c+c_1)+(ec_1-ce_1)}{f(a+a_1)+(ea_1-ae_1)}.$$
Here,   $ec_1-ce_1$ and $ea_1-ae_1$   must be equal $0$.  Thus 
$$\frac{c_1}{e_1}=\pmb{\frac{c}{e}}=\frac{c+c_1}{e+e_1} = \frac{d-c_1}{f-e_1}=\pmb{\frac{c+d}{e+f}= \frac{d }{f} }$$ and 
$$\frac{a_1}{e_1}=\pmb{\frac{a}{e}}=\frac{a+a_1}{e+e_1}=\frac{b-a_1}{f-e_1}=\pmb{\frac{a+b}{e+f}=\frac{b }{f} }.$$
 Our conclusions are $\frac{BO}{AO}=\frac{CD}{OD}=\frac{ED}{AD}$ and $\frac{AB}{AO}=\frac{OC}{OD}=\frac{AE}{AD}$. Finish.
\end{proof}

Based on Propostion \ref{teo.02}, there are two constants $L, M \in  \mathbb{R}^+$ such that
\begin{equation}\label{rasiocos}
L=\frac{a}{e}=\frac{b}{f}=\frac{a+b}{e+f}
\end{equation}
and
\begin{equation}\label{rasiosin}
M=\frac{c}{e}=\frac{d}{f}=\frac{c+d}{e+f}
\end{equation}
hold.  
Then, we can define cosine and sine function of angle $ \theta $ based on \eqref{rasiocos} and \eqref{rasiosin} as 
$$\cos \theta:=\frac{x}{z} $$
and $$\sin \theta:=\frac{y}{z}, $$ where $x$ is adjacent of right triangle,  $y$ is opposite of right  triangle, and $y$ is hypotenuse of right  triangle. Consequently, we obtain  other formula of tangent function as follow
 \[ \tan \theta =\frac{\sin \theta}{\cos \theta} .\]


\subsection{Pythagorean Theorem}

Lets  a right triangle. We have known a famous statement: the sum of the area of two square on the legs equals the area of the square on the hypotenuse. It is called Pythagorean Theorem. Until now, there have been many ways to prove it (there are hundreds). Here, we try to prove it using the results above. 


\begin{figure}[h]
	\centering
	\includegraphics[width=0.6\linewidth]{gambar22.jpg}
	\caption{A right triangles}\label{gambar2}
\end{figure}
 
 \begin{theorem}\label{teo.03} 
Lets  a right triangle with  legs $p$ and $q$ and hypotenuse $r$.  Then we have
$$p^2+q^2=r^2.$$
\end{theorem}


\begin{proof}
See Figure \ref{gambar2} and we know that the area of a right triangle (with legs $p$ and $q$)  is 
\begin{align}\label{per.t031}  
\textrm{Area}_{\textrm{ right   triangle}}=&\frac{pq}{2}=  \frac{rv_1}{2}.
 \end{align}   
Proposition \ref{teo.01} says that $$\frac{ v_1}{v_2}=\frac{ r-v_2}{v_1}=\frac{ p}{q},$$ 
so we get $v_1 =\frac{p}{q}v_2$ and  $v_1^2=v_2 r -v_2^2$. Consequently, we obtain $$\frac{ p^2}{q^2}v_2 =r-v_2$$ 
or 
$$r=v_2(1+\frac{ p^2}{q^2}).$$ 
Meanwhile  by (\ref{per.t031}),   
$$q^2=rv_2$$ holds. 
We conclude that  $$r=\frac{q^2}{r}(1+\frac{ p^2}{q^2})$$ 
or $p^2+q^2=r^2$. Finish.
\end{proof}

\section{Aplications}

Sparks said, in \cite{sparks},  that trigonometry rests on five pillars constructed by Pythagorean principle. We, now, apply Theorem \ref{teo.03} to construct two well known pillars of trigonometry, that is identity, sine and cosine law.



\begin{corollary}\label{teo.04} 
	Let   a right triangle (see Figure \ref{gambar1}). Then we have trigonometry identity
	$$\cos^2 \theta+\sin^2 \theta=1.$$
\end{corollary}


\begin{proof}
	From Figure \ref{gambar2} and Theorem  \ref{teo.03}, we have
	\begin{align*}  
	\frac{p^2}{r^2}+\frac{q^2}{r^2}=& \frac{r^2}{r^2}=1.   
	\end{align*}   
	Since $\cos  \theta=\frac{p }{r }$ and $\sin  \theta=\frac{q }{r }$ , then
	$\cos^2 \theta+\sin^2 \theta=1.$
\end{proof}



Next, Let \(\triangle ABC\) is an acute triangle (Figure \ref{gambar3}). By definition of sine function,
\begin{align*}   
CC_1=&AC \sin \beta= BC \sin \alpha\\
BB_1=&BC \sin \gamma= AB  \sin \beta.
\end{align*}   
We conclude that $\frac{AC}{\sin\alpha}=\frac{BC}{\sin\beta}=\frac{AB}{\sin\gamma}$. It is called \textit{sine law}.
We also show \textit{cosine law} using Theorem  \ref{teo.03} (Pythagorean formula) and Figure \ref{gambar3}
\begin{align*}   
(CC_1)^2 + (AC_1)^2=&(AC)^2  \\
(CC_1)^2 + (AB)^2+(C_1B)^2 -2(AB) (C_1B)=&(AC)^2
\end{align*} 
and
\begin{align*}   
(CC_1)^2 + (C_1B)^2=&(BC)^2. 
\end{align*} 
Thus, $(AB)^2  -2(AB)(C_1B)= (AC)^2-(BC)^2$ holds.
Recall definition of cosine to get  $$(C_1B)=(BC) \cos \beta.$$  We obtain  cosine law $$(AC)^2=(BC)^2+(AB)^2  -2(AB)(BC) \cos \beta .$$

\begin{figure}[h]
	\centering
	\includegraphics[width=0.6\linewidth]{gambar33.jpg}
	\caption{An acute triangle}\label{gambar3}
\end{figure}


\section{Concluding Remarks}

We have constructed only a tangent function instead three basic trigonometric functions. Consequently, construction of five pillars of trigonometry as told in   \cite{sparks} can reduce to just one tangent function through area of right triangle. Hence, other pillars of trigonometry can be seen as corollary of propostions and theorem.
Furthermore, until the recent era, angle calculations can also be performed on two subspaces (see \cite{gun1,nur1}). We have konwn that  cosine function has a very important role.  Additionally, in a normed space, several types of angles can be calculated. The result can be checked ini \cite{bale, gun2,nur2}.


 
 

 
\section*{Acknowledgment}
 The research is supported by PNBP FMIPA ULM. 
The author would like to thank the anonymous reviewer(s) for useful comments and
suggestions to improve this article.



    
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\end{document}
