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The angle of elevation of the top of a tower, $300\text{ m}$ high, from a point on the ground is observed as $30^\circ$. At an instant a hot air balloon passes vertically above the tower and at that instant its angle of elevation from same point on the ground is $60^\circ$. Find height of the balloon from the ground and distance of tower from point of observation. (Use $\sqrt{3} = 1.73$)
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Solution:
Let $AB$ be the tower, $D$ be the position of balloon and $x$ be the distance of tower from the point of observation.
In $\Delta ABC, \tan 30^\circ = \frac{300}{x} \Rightarrow x = 300\sqrt{3}$
$\Rightarrow x = 519$
In $\Delta DBC, \tan 60^\circ = \frac{DB}{x} \Rightarrow DB = x\sqrt{3}$
$\Rightarrow DB = 897.87$
The distance of the tower from the point of observation is $519\text{ m}$
And height of the balloon from the ground is $897.87\text{ m}$
Let $AB$ be the tower, $D$ be the position of balloon and $x$ be the distance of tower from the point of observation.
In $\Delta ABC, \tan 30^\circ = \frac{300}{x} \Rightarrow x = 300\sqrt{3}$
$\Rightarrow x = 519$
In $\Delta DBC, \tan 60^\circ = \frac{DB}{x} \Rightarrow DB = x\sqrt{3}$
$\Rightarrow DB = 897.87$
The distance of the tower from the point of observation is $519\text{ m}$
And height of the balloon from the ground is $897.87\text{ m}$