Problem source

A circle of radius $a$
is centred at the origin $O$. A rectangle $PQRS$ lies in the 
minor sector $OMN$ of this circle 
where $M$ is $(a,0)$ and $N$ is $(a \cos \beta, a \sin \beta)$, 
and $\beta$ is a constant with $0 < \beta < \frac{\pi}{2}\,$. 
Vertex $P$ lies on the positive $x$-axis at $(x,0)$; 
vertex $Q$ lies on $ON$; vertex $R$ lies on the arc of the circle 
between $M$ and $N$; and vertex $S$ lies on the positive $x$-axis at $(s,0)$.
Show that the area $A$ of the rectangle can be written in the form
\[
A= x(s-x)\tan\beta
\,.
\]
Obtain an expression for $s$ in terms of $a$, $x$ and $\beta$, and use it to
show  that
\[
\frac{\d A}{\d x} =
(s-2x) \tan \beta  - \frac {x^2} s \tan^3\beta
\,.
\]
Deduce that  the greatest possible area of 
rectangle $PQRS$ occurs when
$s= x(1+\sec\beta)$ and show that this greatest area is $\tfrac12 a^2 \tan \frac12 \beta\,$.
Show also that this greatest area occurs when $\angle ROS = \frac12\beta\,$.

Solution source

\begin{center}
    
\begin{tikzpicture}

\tikzset{
    % 1. Define Styles for Consistency
    axis/.style={thick, -{Stealth[scale=1.0]}, draw=black!40},
    main/.style={thick, draw=black},
    geometry/.style={thick, draw=blue!70!black},
    shaded/.style={fill=blue!10, draw=blue!70!black, thick},
    label node/.style={font=\small, inner sep=2pt}
}
    % Variables
    \def\a{3}    % Radius
    \def\b{40}   % Angle Beta
    \def\x{1.2}  % x-coordinate for P

    % 2. Coordinate Definitions (using calc for cleaner math)
    \coordinate (O) at (0,0);
    \coordinate (M) at (\a,0);
    \coordinate (N) at (\b:\a); % Using polar coordinates (angle:radius)
    \coordinate (P) at (\x, 0);
    \coordinate (Q) at (\x, {tan(\b)*\x});
    \coordinate (R) at ({sqrt(\a*\a - (tan(\b)*\x)^2)}, {tan(\b)*\x});
    \coordinate (S) at ({sqrt(\a*\a - (tan(\b)*\x)^2)}, 0);

    % 3. Background Layer (Axes should be behind the main drawing)
    \begin{scope}
        \draw[axis] ({-1.2*\a}, 0) -- ({1.2*\a}, 0) node[right] {$x$};
        \draw[axis] (0, {-1.2*\a}) -- (0, {1.2*\a}) node[above] {$y$};
    \end{scope}

    % 4. Main Geometry
    \draw[main] (O) circle (\a);
    \draw[main] (O) -- (N) node[label node, anchor=south west] {$N$};
    
    % 5. The Inscribed Rectangle (Shaded for emphasis)
    \filldraw[shaded] (P) -- (Q) -- (R) -- (S) -- cycle;

    % 6. Labels and Markings
    \node[label node, below left] at (O) {$O$};
    \node[label node, below right] at (M) {$M$};
    \node[label node, below] at (P) {$P$};
    \node[label node, above left] at (Q) {$Q$};
    \node[label node, above right] at (R) {$R$};
    \node[label node, below] at (S) {$S$};

    % Angle Label (using quotes library)
    \pic [draw, "$\beta$", angle radius=0.6cm, angle eccentricity=1.3] {angle = M--O--N};

    % Dimension indicator for x
    \draw[<->, >=Stealth, shorten <=1pt, shorten >=1pt, draw=gray] 
        ($(O)+(0,-0.4)$) -- ($(P)+(0,-0.4)$) 
        node[midway, fill=white, inner sep=1pt, font=\scriptsize] {$x$};

\end{tikzpicture}
\end{center}

Clearly the distance $PS$ is $s - x$, so it remains to determine the heigh $PQ$. Notice that $\tan \beta = \frac{PQ}{OP}$ so the height is $x \tan \beta$ and the area is $x(s-x)\tan \beta $


Notice that $R$ has a $y$-coordinate of $x \tan \beta$, but is a distance $a$ from the origin, so

$s^2 + x^2 \tan^2 \beta = a^2 \Rightarrow s = \sqrt{a^2-x^2 \tan^2 \beta}$

\begin{align*}
    && \frac{\d A}{\d x} &= (s-x)\tan \beta + x \left (\frac{\d s}{\d x} - 1 \right) \tan \beta \\
    &&&= (s-x) \tan \beta + \left (\tfrac12(a^2-x^2\tan^2 \beta)^{-1/2} \cdot (-2x \tan^2 \beta) - 1\right) x \tan \beta \\
    &&&= (s-x) \tan \beta + \left ( \frac{-x \tan^2 \beta}{s} -1\right)x \tan \beta \\
    &&&= (s-2x) \tan \beta - \frac{x^2}{s}\tan^3\beta \\
    \\
    \frac{\d A}{\d x} = 0: && 0 &= s(s-2x)-x^2 \tan^2 \beta \\
    &&&= s^2-(2x)s-x^2\tan^2 \beta \\
    &&&= (s-x)^2-(1+\tan^2\beta)x^2 \\
    \Rightarrow && s &= x + x \sec \beta \\
    &&&= (1+\sec \beta)x \\
    \\
    && a^2 &= x^2(1+\sec\beta)^2 + x^2 \tan^2 \beta \\
    &&&= x^2(2\sec \beta +2\sec^2 \beta ) \\
    &&&= 2x^2 \sec \beta(1+\sec \beta) \\
    \\
    && A &= x^2\sec \beta \tan \beta \\
    &&&= \frac12 a^2 \frac{\sec \beta \tan \beta}{\sec \beta(1+\sec \beta)} \\
    &&&= \frac12 a^2 \frac{\tan \beta}{1+\sec \beta}  = \frac12 a^2 \tan \frac{\beta}{2}\\
\end{align*}

This occurs when 

\begin{align*}
    && \frac{RS}{SO} &= \frac{x \tan \beta}{s} \\
    &&&= \frac{\tan \beta}{1+\sec \beta} = \tan \frac{\beta}2 \\
    \Rightarrow&& \angle ROS &= \frac{\beta}2
\end{align*}