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A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are α and β, respectively. Prove that the height of the tower is

\left ( \frac{h\tan \alpha }{\tan \beta -\tan \alpha } \right )

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According to question

Here h is the height of flagstaff AD.
Let \l is the height of the tower
\alpha and \beta be the angle of elevation of the bottom and the top of the flagstaff.
In \bigtriangleupBDC
\tan \alpha = \frac{\l }{BC}  \left [ \because \tan \theta = \frac{Perpendicular}{Base} \right ]
BC= \frac{\l }{\tan \alpha }\cdots \left ( 1 \right )
In \bigtriangleupABC
\tan \beta = \frac{AB}{BC}
\tan \beta = \frac{h+\l }{BC}
BC= \frac{h+\l }{\tan \beta }\cdots \left ( 2 \right )
equate equation (1) and (2) we get
\frac{\l }{\tan \alpha }=\frac{h+\l }{\tan \beta }
\l \tan \beta = h\tan \alpha +\l \tan \alpha [by cross multiplication]
\l \tan \beta -\l \tan \alpha = h\tan \alpha
\l \left ( \l \tan \beta - \tan \alpha \right ) = h\tan \alpha
\l = \frac{h\tan \alpha }{ \tan \beta - \tan \alpha}
Hence Proved

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infoexpert27

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