Consider the following for the two (02) items that follow: The top...

Let $h$ be the height of the tower $MN$.

In $\triangle MNP$, \tan 30^{}^{\circ} = \dfrac{h}{PN} \Rightarrow PN = h\sqrt{3}

In $\triangle MNQ$, \tan 45^{}^{\circ} = \dfrac{h}{QN} \Rightarrow QN = h

In $\triangle MNR$, \tan 60^{}^{\circ} = \dfrac{h}{RN} \Rightarrow RN = \dfrac{h}{\sqrt{3}}

Since the angles of elevation increase as we move closer to the tower, the points $P, Q, R$ lie on the same side of $N$ in the order $P, Q, R, N$.

Given $PQ = a$, we have $PN - QN = a$.

$h\sqrt{3} - h = a \Rightarrow h(\sqrt{3} - 1) = a$

$h = \dfrac{a}{\sqrt{3} - 1}$

Rationalizing the denominator, we get:

$h = \dfrac{a(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \dfrac{a(\sqrt{3} + 1)}{2}$

We need to find $PN$, which is $h\sqrt{3}$:

$PN = \dfrac{a(\sqrt{3} + 1)}{2} \times \sqrt{3} = \left(\dfrac{3 + \sqrt{3}}{2}\right)a