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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.
Solution
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.
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