This is a classical problem, that will surely bring the best of your skills in trigonometry and geometry. Try to find different solutions working with different techniques (computation of areas of different pieces, pure trigonometric approach, similarity of triangles, etc):

We need to find the width of an alley. We have placed two ladders cross-wise, as the figure suggests. The length of the ladders are respectively two and three meters. They cross each other at a point that is exactly one meter above the ground.

The fastest approach I have found is by trying to establish one single relationship between two entities, each of them belonging to different triangles. In this case, the one that worked best for me was the relationship between the heights of both triangles. Let us simplify the figure above and label all entities appropriately:

By similar triangles, it must be \dfrac{a}{1}=\dfrac{x}{h'} and \dfrac{x-a}{1}=\dfrac{x}{h}

In that case, as x = x-a+a, we can write:

x = (x-a) + a = \dfrac{x}{h} + \dfrac{x}{h'} = x \bigg( \dfrac{1}{h} + \dfrac{1}{h'} \bigg),

which reduces to the relation I was seeking:

1= \dfrac{1}{h} + \dfrac{1}{h'}.

A simple application of Pythagoras Theorem gives h^2 = 4-x^2 and (h')^2=9-x^2. At this point, we have reduced the problem to a single equation with one variable, which represents the entity that we are trying to measure:

1 = \dfrac{1}{\sqrt{4-x^2}} + \dfrac{1}{\sqrt{9-x^2}}.

All is left is to find the algebraic solution of the equation, that I leave to the reader. Any mathematics package should not have any trouble finding (an approximation to) the numerical value of the solution. For instance, sage finds it quickly:



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