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From the top of a tower h m high, the angles of depression of two objects, which are in line with the foot of the tower are α and β (β > α). Find the distance between the two objects

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Let x and y are two objects and \beta and \alpha use the angles of depression of two objects.
In \bigtriangleupAOX
\tan \beta = \frac{h}{Ox}        \because \tan \theta = \frac{Perpendicular}{Base}
Ox= \frac{h}{\tan \beta }
Ox= h\cot \beta \cdots \left ( 1 \right )
In \bigtriangleupAOY
 \tan \alpha = \frac{h}{Oy}= \frac{h}{Ox+xy}\; \; \left ( \because Oy= Ox+xy \right )
Ox+xy= \frac{h}{\tan \alpha }
Ox+xy= h\cot \alpha
xy= h\cot \alpha- Ox
xy= h\cot \alpha- h\cot \beta  (from equation (1))
xy= h\left ( \cot \alpha- \cot \beta \right )
Hence the distance between two objects is

h\left ( \cot \alpha- \cot \beta \right ) .

 

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