Let AB be the first pole such that AB = 60 m. Let the weight of the second pole be n metre. It is given that the angles of depression of the top A and the bottom B of the pole AB are 30° and 60° respectively.
∴ ∠ACE = 30° and ∠ADB = 60°
Let BD = EC = x
Now, in right triangle ACE, we have
In right triangle ADB, we have
⇒ x = 20 × 1.732 = 34.64 m
Hence, width of the river = 34.64 m.
Comparing (i) and (ii), we gel
Hence,height of other pole = 40 m.