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The lower window of a house is at height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At a 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.

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Answer

Let A be the position of balloon, F be the lower window and G be the upper window.

EF = 2 m and FG = 4 m

The lower window of a house is at height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At a 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. Model Paper 5, Concise Mathematics Solutions ICSE Class 10.

From figure,

BG = CF = x (let)

BC = FG = 4 m

In △ABG,

tan 30°=ABBG13=ABBGBG=3ABx=3AB …………(1)\Rightarrow \text{tan 30°} = \dfrac{AB}{BG} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BG} \\[1em] \Rightarrow BG = \sqrt{3}AB \\[1em] \Rightarrow x = \sqrt{3}AB \text{ …………(1)}

In △ACF,

tan 60°=ACCF3=ACCFAC=3CFAB+BC=3xx=AB+BC3x=AB+43 …………(2)\Rightarrow \text{tan 60°} = \dfrac{AC}{CF} \\[1em] \Rightarrow \sqrt{3} = \dfrac{AC}{CF} \\[1em] \Rightarrow AC = \sqrt{3}CF \\[1em] \Rightarrow AB + BC = \sqrt{3}x \\[1em] \Rightarrow x = \dfrac{AB + BC}{\sqrt{3}} \\[1em] \Rightarrow x = \dfrac{AB + 4}{\sqrt{3}}\text{ …………(2)}

From (1) and (2), we get :

3AB=AB+433AB=AB+42AB=4AB=42=2 m.\Rightarrow \sqrt{3}AB = \dfrac{AB + 4}{\sqrt{3}} \\[1em] \Rightarrow 3AB = AB + 4 \\[1em] \Rightarrow 2AB = 4 \\[1em] \Rightarrow AB = \dfrac{4}{2} = 2 \text{ m}.

From figure,

AD = AB + BC + CD = 2 m + 4 m + 2 m = 8 m.

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

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