Answer:
Step-by-step explanation:
Let d be the perpendicular distance of the target from the person. His angle of view is a random variable [tex]\Theta[/tex] which uniformly distributed (given). [tex]\Theta \sim U\left [-\pi /2,\pi /2 \right ][/tex] ., since his field view is from [tex]-\pi /2[/tex] to [tex]\pi /2[/tex] . That is
[tex]f_{\Theta }\left ( \theta \right )=\frac{1}{\pi };-\pi /2<\theta <\pi /2 [/tex]
Now the distance X where the arrow strikes from the target is
[tex]\tan \theta =\frac{x}{d}[/tex] Or [tex]x=d\tan \theta . [/tex]
Now we haver to find the distribution of [tex]X=d\tan\left | \Theta \right |[/tex] which is a function of known [tex]RV \Theta[/tex] .
We know the distribution of the transformation [tex]X=g\left (\Theta \right )[/tex] is
[tex]f_X\left (x \right )=\sum f_{\Theta }\left ( g^{-1}\left (y\right ) \right )\left | \frac{\mathrm{d} g^{-1}\left (y\right ) }{\mathrm{d} x} \right |[/tex].
The sum is because if we have more than one inverse functions of [tex]\Theta[/tex] .
Now, the inverse function is [tex]\left |\Theta \right |=\tan^{-1}\left ( \frac{X}{d} \right )[/tex] . So the PDF of [tex]X=d\tan\left | \Theta \right |[/tex] is
[tex]f_X\left (x \right )=\sum f_{\Theta }\left ( \tan^{-1}\left ( \frac{x}{d} \right )\right )\left | \frac{\mathrm{d} \tan^{-1}\left ( \frac{x}{d} \right ) }{\mathrm{d} x} \right |\\ f_X\left (x \right )=2\frac{1}{\pi }\frac{d}{d^2+x^2};0<x<\infty \\ {\color{Blue} f_X\left (x \right )=\frac{2}{\pi }\frac{d}{d^2+x^2};0<x<\infty } [/tex]
I do not understand "perches". If you want you can set d=1 and your distribution becomes,
[tex]{\color{Blue} f_X\left (x \right )=\frac{2}{\pi }\frac{1}{1+x^2};0<x<\infty } [/tex]
Then X measured in so called perches.
