Q188 / 269 · NDA 2025 (Phase 1) · Mathematics
Consider the following for the two (02) items that follow:
The top ($M$) of a tower is observed from three points $P$, $Q$ and $R$ lying in a horizontal straight line which passes directly along the foot ($N$) of the tower. The angles of elevations of $M$ from $P$, $Q$ and $R$ are $30°$, $45°$ and $60°$ respectively. Let $PQ = a$ and $QR = b$.
What is $PN$ equal to?
The top ($M$) of a tower is observed from three points $P$, $Q$ and $R$ lying in a horizontal straight line which passes directly along the foot ($N$) of the tower. The angles of elevations of $M$ from $P$, $Q$ and $R$ are $30°$, $45°$ and $60°$ respectively. Let $PQ = a$ and $QR = b$.
What is $PN$ equal to?
- A$\left(\dfrac{3-\sqrt{3}}{2}\right)a$
- Correct answerB$\left(\dfrac{3+\sqrt{3}}{2}\right)a$
- C$\left(\dfrac{3-\sqrt{3}}{4}\right)a$
- D$\left(\dfrac{3+\sqrt{3}}{4}\right)a$
Solution
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 $h\sqrt{3}$:
$PN = \dfrac{a(\sqrt{3} + 1)}{2} \times \sqrt{3} = \left(\dfrac{3 + \sqrt{3}}{2}\right)a$
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 $h\sqrt{3}$:
$PN = \dfrac{a(\sqrt{3} + 1)}{2} \times \sqrt{3} = \left(\dfrac{3 + \sqrt{3}}{2}\right)a$
Hence, the correct answer is option: 'B'.