Question

please answer this, it's urgent​

Answer 1

Step-by-step explanation:

Let height of the tower be h'

and distance if the point from the bottom building be b

then,

$\tan( \alpha ) = \frac{h}{b} \\ = > b = \frac{h}{ \tan( \alpha ) }$

___eq(1)

$\tan( \beta ) = \frac{h + {h}^{l} }{b} \\ = > b = \frac{h + {h}^{l} }{ \tan( \beta ) }$

____eq(2)

Equating eq(1) and (2)----

$\frac{h}{ \tan( \alpha ) } = \frac{h + {h}^{l} }{ \tan( \beta ) } \\ = > \frac{h \tan( \beta ) }{ \tan( \alpha ) } = h + {h}^{l} \\ = > \frac{h \tan( \beta ) }{ \tan( \alpha ) } - h = {h}^{l} \\ = > \frac{h \tan( \beta ) - h \tan( \alpha ) }{ \tan( \alpha ) } = {h}^{l} \\ = > \frac{h( \tan( \beta ) - \tan( \alpha ) )}{ \tan( \alpha ) } = {h}^{l}$

Hence Proved

Hope it helps :D