The top of a tower, the angle of an object are on the same side of the tower end to be $x$ and $b (x > b)$ . If the distance objects is 'h' meters, show that the height of 'h' is given by:

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Solution

Let $A B$ be the tower of height $h$ meters and C and D are two objects which are $y$ meter apart

Then,

$A B = h m e t e r$

$B C = x m e t e r$

$C D = y m e t e r$

Then,

$I n Δ A B C,$

$tan α = \frac{A B}{B C}$

$tan α = \frac{h}{x}$

$x = \frac{h}{tan α} . . . . . . . (1)$

Now, $I n Δ A B D$ $tan β = \frac{h}{x + y}$

$h = (x + y) tan β$

$h = x tan β + y tan β . . . . . . . (2)$

By equation (1) and we get,

$h = y tan β + \frac{h tan β}{tan α}$

$h - \frac{h tan β}{tan α} = y tan β$

$h (1 - \frac{tan β}{tan α}) = y tan β$

$h (\frac{tan α - tan β}{tan α}) = y tan β$

$h = \frac{y tan α tan β}{tan α - tan β}$

$h = \frac{C D tan α tan β}{tan α - tan β}$
Hence, this is the answer.

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