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A tree is broken by wind, its upper part touches the ground at a point 10 metres from the foot of the tree and makes an angle of 60° with the ground. The entire length of the tree is

(a) 15 m

(b) \(\big(10+\frac{\sqrt3}{2}\big)\)m

(c) \(\sqrt2\big(10+\frac{\sqrt5}{3}\big)\)m

(d) \(\sqrt3\big(10+\frac{20}{\sqrt3}\big)\)m

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(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.

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