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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.

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Solution
                  

Let y be the height of the balloon from the second window
In \bigtriangleupAOB
\tan 30= \frac{y}{d} \; \; \left ( \because \tan \theta = \frac{Perpendicular}{Base} \right )
d= y\sqrt{3}\; \cdots \left ( 1 \right )           \left ( \because \tan 30= 1\sqrt{3} \right )
In \bigtriangleupOCD
\tan 60= \frac{4+y}{d}
\sqrt{3}d= 4+y
d= \frac{4+y}{\sqrt{3}}\cdots \left ( 2 \right )
Equating equation (1) & (2) we get
y\sqrt{3}= \frac{4+y}{\sqrt{3}}
y\sqrt{3}\times \sqrt{3}= 4+y
3y-y= 4
2y= 4
y= \frac{4}{2}
y = 2
Height of balloon = 2 + 4 + y
= 2 + 4 + 2
= 8m

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