SSC CGL 2025 PYQ — Question 72

The correct answer is option 4 i.e.  $\frac{15(3 + \sqrt{3})}{2}$ m.

Let the height of the pole = h meters

Let the initial distance from the man to the pole = d meters

From the first observation: $\tan 45^\circ = \frac{h}{d}$

⇒ 1 = h/d

⇒ h = d

After walking 15 m towards the pole: New distance = d - 15 meters

Second observation: $\tan 60^\circ = \frac{h}{d - 15}$

⇒ $\sqrt{3} = \frac{h}{d - 15}$

⇒ $\sqrt{3} = \frac{d}{d - 15}$

⇒ $\sqrt{3}(d - 15) = d$

⇒ $\sqrt{3}d - 15\sqrt{3} = d$

⇒ √3d - d = 15√3

⇒ d(√3 - 1) = 15√3

⇒ $d = \frac{15\sqrt{3}}{\sqrt{3} - 1}= \frac{15\sqrt{3}(\sqrt{3} + 1)}{3 - 1}= \frac{15(3 + \sqrt{3})}{2}$ m