Question

Two vertical poles of heights 6 m and 3 m are erected above a horizontal ground AC. Find the value of y

Answer 1

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