(c) 100 (√3 - 1)m
Let BC be the incomplete pillar and BD be the complete pillar.
In ΔABC,
tan 45° = \(\frac{BC}{AB}\)
⇒ \(\frac{BC}{100}\) = 1 ⇒ BC = 100 ...(i)
In ΔABD,
tan 60° = \(\frac{BD}{AB}\) ⇒ √3 = \(\frac{BC+x}{100}\) ⇒ \(x\) = 100√3 - BC
= 100√3 - 100 = 100 (√3 - 1)m. (∵ BC = 100 m)