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Question
There are two poles, one each on either bank of a river just opposite to each other. One pole is 60 m high. From the top of this pole, the angle of depression of the top and foot of the other pole are 30° and 60° respectively. Find the width of the river and height of the other pole.
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Solution
Let the width of the river be w.
In ΔABC,
tan 60° = `"AB"/"BC"`
⇒ `sqrt3` = `60/w`
⇒ w = `60/(sqrt3) = (60sqrt3)/3= 20sqrt3`
In △AED,
tan30° = `"AE"/"ED"`
⇒ `1/(sqrt3) = "AE"/w`
⇒ `1/(sqrt3) = "AE"/(20sqrt3)`
⇒ AE = 20
Height of pole CD = AB − AE
= 60 − 20 = 40 m.
Thus, width of river is `20sqrt3` = 20 x 1.732 = 34.64 m
Height of pole = 40 m.
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