# Lorentz transformations in relativity [closed]

I'm trying to solve a particular problem given in Introduction to Special Relativity by R. Resnick, which goes like this:

Two observers in the S frame, $$A$$ and $$B$$, are separated by a distance of 60 m. Let $$S'$$ move at a speed $$\dfrac{3}{5}c$$ relative to $$S$$, the origins of the two systems, $$O$$ and $$O'$$, being coincident at $$t = t'= 3 *10^{-7}$$ sec ($$90/c$$). The S frame has two observers, one at $$A'$$ and the other at B' such that, according to clocks in the $$S$$ frame, $$A'$$ is opposite $$A$$ at the same time that $$B'$$ is opposite B. $$O$$ and $$O'$$ lie at the centre of $$A$$ and $$B$$ and $$A'$$ and $$B'$$ respectively. a) What is the reading on the clock of $$B'$$ when $$B'$$ is opposite $$B$$?

My question: I do not really get how is it possible for an observer in the S frame to observe A' and A being coincident at the same time as B and B' are. Isn't the length, and hence the distance between A' and B' contracting in S' w.r.t S?

Before this, I tried the following:

At $$t=t'=\dfrac{90}{c}$$, $$x=x'=0$$ where $$x$$ and $$x'$$ are the locations of the origins O and O' respectively. We want to find $$t'$$ at which $$b=b'$$. Using Lorentz transformations, we know that $$t' = \dfrac{t-\dfrac{xv}{c^2}}{\sqrt{1-\left (\dfrac{v}{c} \right )^2}}$$. Put $$t=90/c, v=\dfrac{3}{5}c$$ and $$x = +30$$ (since that is the location of B' in S frame at the given $$t$$), and we obtain, $$t' = \dfrac{90}{c}$$, which does not surprise me, because conditions, in away, imply that A & A', B and B', and O & O' are all coincident at the same instant of time (in S, I think, but then what is this t'?). The answer, however, is $$\dfrac{45}{c}$$, which is curiously what comes out if we put $$x=+60$$ instead of $$30$$. I do not understand where is it that I'm going wrong. I believe that there is a large conceptual hole in my understanding of time dilation, length contraction, Lorentz transforms, and relativity of simultaneity.