(d) \(\sqrt3\big(10+\frac{20}{\sqrt3}\big)\)m
Let PQ be the total length of the tree. Let A be the point at which the tree breaks and B be the point where the top of the tree touches the ground. ∴ AP = AB
Given, ∠ABQ = 60°, BQ = 10 m.
In rt. Δ AQB, \(\frac{AQ}{BQ}\) = tan 60°
AQ = BQ tan 60° = 10√3 m
Also, \(\frac{BQ}{AB}\) = cos 60°
⇒ AB = BQ sec 60° = (10 × 2) m = 20 m
∴ PQ = AQ + AP = AQ + AB = (10√3 + 20) m (∴AP = AB)
= \(\sqrt3\big(10+\frac{20}{\sqrt3}\big)\)m.