The angle of elevation $\theta$ of the top of a light-house as seen by a person on the ground is such that $\tan \theta =\frac{5}{12}$. When the person moves a distance $240m$ towards the light-house, the angle of elevation become $\varphi$ such that $\tan \varphi =\frac{3}{4}$, Find the height of the light house.

Let $AB$ be the position of the light house whose height is $h$ metres.
Required: To find the height of the light house.
$AB=$Height $h=?$
Let $P$ be the position of the person with the angle of elevation of $P$ be $\theta$, $\angle APB=\theta$ such that $\tan \theta =\frac{5}{12}$, Let $Q$ be the position of the person when he covered a distance $240$ metres towards th elight house then $PQ=240m,QB=xmetres,\angle AQB=\varphi$.
In $△ABP$,
$\tan \theta =\frac{h}{240+x}$
$\frac{5}{12}=\frac{h}{240+x}$,
$12h=1200+5x......(1)$
In $△AQB$, $\tan \varphi =\frac{3}{4}$
$\frac{h}{x}=\frac{3}{4}$
$4h=\frac{3}{4}x.........(2)$
From $(1)$ $12\times \frac{3}{4}=1200+5x$
$9x=1200+5x,4x=1200,x=300$
$h=\frac{3}{4}\times 300=225m$