Question

A ladder rests against a tree on one side of a street. The foot of the ladder makes an angle of 50° with the ground. When the ladder is turned over to rest against another tree on the other side of the street it makes an angle of 40° with the ground. If the length of the ladder is 60m, find the width of the street. 

Answer 1

Let AB and CD be two trees and P be a point on the street AC between the two trees. 

PD and PB denotes the ladder at the two instants. 

In ΔPCD,

cos 50° = `"PC"/"PD"` 

0.6428 = `"PC"/60`

⇒ PC = 0.6428 × 60 = 38.568

In ΔABP,

cos 40° = `"AP"/"BP"`

⇒ `0.7660 = ("AP")/60`

⇒ AP = 0.7660 × 60 = 45.96

∴ AC = AP +PC = 38.568 m + 45.96m = 84.528 m ≈ 84.53 m.

Thus , the width of the street is 84.53 m.