Question

Answer ittt !!!!................

with full explanation!!!!!​

Answer 1

ANSWER:-

Let the height of the tower BC be x and CD be the flagstaff.

Let CD = h

Let distance AB = y ,

$\angle BAC = \alpha \: \: \: and \\ \\</strong></p><p></p><p><strong>[tex] \angle BAC = \alpha \: \: \: and \\ \\\angle BAD = \beta$

In ∆ABC right angled at B, we have,

$tan \: \alpha = BC/AB \\ \\ \implies \: tan \: \alpha = \frac{x}{y} \\ \\ \implies \: y = \frac{x}{ \tan( \alpha ) } ...........(1)$

In ∆ABD right angled at B, we have,

$\tan \beta = BD/AB \\ \\ \implies \: \tan( \beta ) = \frac{h + x}{y} \\ \\ \implies \: y = \frac{h + x}{ \tan( \beta ) } .........(2)$

From (1) and (2) , we get

$\implies \: \frac{x}{ \tan( \alpha ) } = \frac{h + x}{ \tan( \beta ) } \\ \\ \implies \: \frac{x}{ \tan( \alpha ) } = \frac{h}{ \tan( \beta ) } + \frac{x}{ \tan( \beta ) } \\ \\ \implies \: \frac{h}{ \tan( \beta ) } = x( \frac{1}{ \tan( \alpha ) } - \frac{1}{ \tan( \beta ) } ) \\ \\ \implies \:x = \frac{h \tan( \alpha ) . \tan( \beta ) }{ \tan( \beta ) ( \tan( \beta ) - \tan( \alpha ) )} \\ \\ \implies \:x = \frac{h \: \tan( \alpha ) }{ \tan( \beta ) - \tan( \alpha ) } \\ \\ \huge \bf \: height \: \: of \: \: the \: \: \: tower \: \: : x = \frac{h \: \tan( \alpha ) }{ \tan( \beta ) - \tan( \alpha ) }$