Two Ladders

Let the width of the alleyway be $a$ , the angles between the ladders and the floor be $\alpha$ and $\beta$ , the bottom of the left wall be the origin $O$ of an $xy$ -plane and the crossing point of the ladders be $P(x,3)$ as shown. Then we have:

$\begin{cases} \tan \alpha = \dfrac 3x & \implies \tan \left(\cos^{-1} \dfrac a9\right) = \dfrac 3x & \implies \dfrac {\sqrt{81-a^2}}a = \dfrac 3x & \implies x = \dfrac {3a}{\sqrt{81-a^2}} \\ \tan \beta = \dfrac 3{a-x} & \implies \tan \left(\cos^{-1} \dfrac a6\right) = \dfrac 3{a-x} & \implies \dfrac {\sqrt{36-a^2}}a = \dfrac 3{a-x} & \implies x = a - \dfrac {3a}{\sqrt{36-a^2}} \end{cases}$

Therefore,

$\begin{aligned} \frac {3a}{\sqrt{81-a^2}} & = a - \frac {3a}{\sqrt{36-a^2}} & \small \color{#3D99F6} \text{Since }a \ne 0 \\ \frac {3}{\sqrt{81-a^2}} & = 1 - \frac {3}{\sqrt{36-a^2}} \end{aligned}$

Solving the equation numerically, we have $a \approx \boxed{3.694}$ .

After writing up the equations using Pythagoras's theorem and intersections of two lines.

We just need to solve for these two equations to get the width x

y = (sqrt(81-x^2)*sqrt(36-x^2))/(sqrt(81-x^2) + sqrt(36-x^2)) .............. (1)

y = 3 ..............(2)

We get the solution for x and y as (3.694,3)