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Question
Two vertical poles of heights 6 m and 3 m are erected above a horizontal ground AC. Find the value of y
Solution
In the ∆PAC and ∆BQC
∠PAC = ∠QBC = 90°
∠C is common
∆PAC ~ QBC
`"AP"/"BQ" = "AC"/"BC"`
`6/y = "AC"/"BC"`
∴ `"BC"/"AC" = y/6` ...(1)
In the ∆ACR and ∆QBC
∠ACR = ∠QBC = 90°
∠A is common
∆ACR ~ ABQ
`"RC"/"QB" = "AC"/"AB"`
`3/y = "AC"/"AB"`
`"AB"/"AC" = y/3` ...(2)
By adding (1) and (2)
`"BC"/"AC" + "AB"/"AC" = y/6 + y/3`
1 = `(3y + 6y)/18`
9y = 18 ⇒ y = `18/9` = 2
The Value of y = 2 m
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