Everything In The US Has To Be Supersized

$$ Let $x$ be the distance Susie stands from the the screen. Take a look the picture below.

View Angle

$\begin{aligned} \tan\theta &=\tan(\alpha-\beta)\\ &=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}\\ &=\frac{\frac{25}{x}-\frac{5}{x}}{1+\frac{25}{x}\cdot\frac{5}{x}}\\ &=\frac{\frac{20}{x}}{1+\frac{125}{x}}\\ &=\frac{20x}{x^2+125}\\ \end{aligned}$ $$ Since we want to maximize $\theta$ , we set $\dfrac{d\theta}{dx}=0$ . $$ $\begin{aligned} \sec^2\theta\frac{d\theta}{dx} &=\frac{20(x^2+125)-2x(20x)}{(x^2+125)^2}\\ 0&=\frac{20(125-x^2)}{(x^2+125)^2}\\ 125-x^2&=0\\ x&=\pm5\sqrt{5}. \end{aligned}$ $$ Note that, you can easily check $\theta$ will be maximum for $x=5\sqrt{5}$ and will be minimum for $x=-5\sqrt{5}$ . In order to get the largest viewing angle, Susie should stand $5\sqrt{5}\text{ m}$ in front of the screen. Hence, we plug in $x=5\sqrt{5}$ into $\tan\theta$ to obtain a maximum value of $\theta$ . $$ $\begin{aligned} \tan\theta&=\frac{20(5\sqrt{5})}{(5\sqrt{5})^2+125}\\ &=\frac{2}{\sqrt{5}}\\ \theta&=\tan^{-1}\left(\frac{2}{\sqrt{5}}\right)\\ &\approx\boxed{\color{#3D99F6}{41.8^\circ}} \end{aligned}$

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$\Large\color{#3D99F6}{\text{\# }\mathbb{Q.E.D.}\text{ \#}}$