Q189 / 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 $MN$ equal to?

  • Correct answer
    A
    $\left(\dfrac{3+\sqrt{3}}{2}\right)b$
  • B
    $\left(\dfrac{3-\sqrt{3}}{2}\right)b$
  • C
    $\left(\dfrac{3-\sqrt{3}}{4}\right)b$
  • D
    $\left(\dfrac{3+\sqrt{3}}{4}\right)b$

Solution

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

The distances of points $P$, $Q$, and $R$ from the foot of the tower $N$ are given by:

$PN = h \cot 30^{\circ} = h\sqrt{3}$

$QN = h \cot 45^{\circ} = h$

$RN = h \cot 60^{\circ} = \dfrac{h}{\sqrt{3}}$

Given that $P$, $Q$, $R$ lie on a straight line passing through $N$, the distance $QR$ is:

$QR = QN - RN$

$b = h - \dfrac{h}{\sqrt{3}}$

$b = h\left(\dfrac{\sqrt{3}-1}{\sqrt{3}}\right)$

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

Rationalizing the denominator:

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

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

$h = \left(\dfrac{3+\sqrt{3}}{2}\right)b$

Answer: $\left(\dfrac{3+\sqrt{3}}{2}\right)b$

Hence, the correct answer is option: 'A'.

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