In a narrow alley of width $w$ a ladder of length $a$ is placed with its foot at point $P$ between the walls. 
Resting against one wall at $Q$, the distance $k$ above the ground makes a $45^\circ$ angle with the ground. 
Resting against the other wall at $R$, a distance $h$ above the ground, the ladder makes a $75^\circ$ angle with the ground. The width $w$ is equal to 
$\textbf{(A)}\ a\qquad \textbf{(B)}\ RQ\qquad \textbf{(C)}\ k\qquad \textbf{(D)}\ \frac{h+k}{2}\qquad \textbf{(E)}\ h$
We know that $m\angle QPL=45^{\circ}$ and $m\angle RPT=75^{\circ}.$ Therefore,  $m\angle QPR=60^{\circ}.$ \[\qquad\]
Because the two ladders are the same length, we know that
\[RP=PQ=a.\]
Since $\triangle QPR$ is isosceles with vertex angle $60^{\circ},$ we can conclude that it must be equilateral.\[\qquad\]
Now,  since $\triangle PTR$ is a right triangle and $m\angle TPR=75^{\circ},$ we can conclude that $m\angle PRT=15^{\circ}.$ Because $\triangle QPR$ is equilateral, we know that $m\angle QRP = 60^{\circ}.$
It then follows that \[m\angle QRS= m\angle QRP + m\angle PRT = 60^{\circ} + 15^{\circ} = 75^{\circ}.\] 
Because of ASA,  $\triangle QRS\cong\triangle RPT.$ From there, it follows that $QS=TR=h.$ Since $QS$ is the width of the alley, the answer is $\boxed{\text{E) }h}.$
~ Saumya Singhal