Question

only legends can solve thiss​

Answer 1

Step-by-step explanation:

hope this helps you mate ✨

Answer 2

$\large\underline{\bold{Solution-}}$

Given that,

• Height of the first house, WB = 'h' meter.

and

• ∠QWM = α and ∠OWM = β

Now,

• ∠OWM = ∠WOB = β (Alternate angles)

• Let the height of the other house, OQ = 'H' meter.

and

• Let OB = WM = 'x' meter.

Now,

In right triangle WOB,

$\sf \: tan \beta = \dfrac{WB}{OB}$

$\sf \: tan \beta = \dfrac{h}{x}$

$\therefore \: \boxed{ \bf{ \: h \: = \: x \: tan \beta }} - - - - (1)$

Now,

In right triangle QWM,

$\sf \: tan \alpha \: = \dfrac{QM}{WM}$

$\sf \: tan \alpha = \dfrac{H - h}{x}$

$\sf \: x \: tan \alpha = H - xtan \beta$

$\sf \: xtan \alpha + xtan \beta = H$

$\sf \: x(tan \beta + tan \alpha ) = H$

$\therefore \bf \: \: x \: = \: \dfrac{H}{tan \beta + tan \alpha } - - - (2)$

On substituting equation (2) in equation (1), we get

$\sf \: h = \dfrac{H}{tan \beta + tan \alpha } \times tan \beta$

$\sf \: H = h \: \bigg( \dfrac{tan \beta + tan \alpha }{tan \beta } \bigg)$

$\sf \: H = h \: \bigg(1 + \dfrac{tan \alpha }{tan \beta } \bigg)$

$\therefore \bf\: H \: = \: h \: (1 + tan \alpha \: cot \beta )$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Note :-

Angles of Elevation and Depression

The angle of elevation of an object as seen by an observer is the angle between the horizontal and the line from the object to the observer's eye (the line of sight).

If the object is below the level of the observer, then the angle between the horizontal and the observer's line of sight is called the angle of depression.