SOLUTION:
Let AB be the tower of height x metre, surmounted by a vertical flagstaff AD. Let C be a point on the plane such that ∠ACB = α, ∠ACB = β and AD = h.
In ∆ABC,
![\[\begin{array}{*{35}{l}} AB/BC\text{ }=\text{ }tan\text{ }\alpha \\ BC\text{ }=\text{ }x/\text{ }tan\text{ }\alpha \text{ }\ldots \ldots .\text{ }\left( i \right) \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b8408e9476856669d297373e89cc2c58_l3.png "Rendered by QuickLaTeX.com")
In ∆DBC,
![\[\begin{array}{*{35}{l}} BD/BC\text{ }=\text{ }tan\text{ }\beta \\ BD\text{ }=\text{ }\left( x/tan\text{ }\alpha \right)\text{ }x\text{ }tan\text{ }\beta \text{ }\ldots \ldots \text{ }\left[ From\text{ }\left( i \right) \right] \\ \left( h\text{ }+\text{ }x \right)\text{ }tan\text{ }\alpha \text{ }=\text{ }x\text{ }tan\text{ }\beta \\ x\text{ }tan\text{ }\beta \text{ }-\text{ }x\text{ }tan\text{ }\alpha \text{ }=\text{ }h\text{ }tan\text{ }\alpha \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-80f81c75cdb21fa5c2138f38610c4ca2_l3.png "Rendered by QuickLaTeX.com")
Therefore, height of the tower is h tan α/ (tan β – tan α)
SOLUTION:
Let AB be the tower of height x metre, surmounted by a vertical flagstaff AD. Let C be a point on the plane such that ∠ACB = α, ∠ACB = β and AD = h.
In ∆ABC,
In ∆DBC,
Therefore, height of the tower is h tan α/ (tan β – tan α)