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Question
The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30° respectively. Find the height of the balloon above the ground.
Solution
Let the height of the balloon from above the ground is H.
A and OP = w2R = w1Q = x
Given that, height of lower window from above the ground = w2P = 2 m = OR
Height of upper window from above the lower window = w1w2 = 4 m = QR
∴ BQ = OB – (QR + RO)
= H – (4 + 2)
= H – 6
And ∠Bw1Q = 30°
⇒ ∠Bw2R = 60°
Now, in ΔBw2R,
tan 60° = `"BR"/("w"_2"R") = ("BQ" + "QR")/x`
⇒ `sqrt(3) = (("H" - 6) + 4)/x`
⇒ `x = ("H" - 2)/sqrt(3)` ...(i)
And in ΔBw1Q,
tan 30° = `"BQ"/("w"_1"Q")`
tan 30° = `("H" - 6)/x = 1/sqrt(3)`
⇒ `x = sqrt(3)("H" - 6)` ...(ii)
From equations (i) and (ii),
`sqrt(3)("H" - 6) = (("H" - 2))/sqrt(3)`
⇒ 3(H – 6) = H – 2
⇒ 3H – 18 = H – 2
⇒ 2H = 16
⇒ H = 8
So, the required height is 8 m.
Hence, the required height of the balloon from above the ground is 8 m.
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