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^{0} and 30^{0} respectively. Find the height of the balloon above the ground.

Given that the height of lower window is OR = 2m

Height of upper window is RQ = 4 m

Let height of balloon above the ground is BO = H meter & PO = x meter

∠Bw_{2}R = 60° and ∠Bw_{1}Q = 30°

In ∆Bw_{2}R,

In ∆Bw_{1}Q,

⇒ x = √3 (H - 6) …eq. 2

Comparing eq. 1 and eq. 2, we get

⇒ H – 2 = 3 (H – 6)

⇒ H – 2 = 3H – 18

⇒ 3H – H = 18 – 2

⇒ 2H = 16

⇒ H = 8

Hence, height of the balloon above the ground is 8 meter.

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