The angle of elevation, observed from ground to the top of a building $60 \; m$ high, is $30^\circ$. On walking '$x$' meters towards the building, the angle of elevation to the top changes to $60^\circ$. Find '$x$' to the nearest meter.
$AB =$ Building of height $60 \; m$
$D = $ Initial position of observation
$C = $ Final position of observation
$CD = x \;$ meter
In $\triangle ABC$,
$\tan 60^\circ = \dfrac{AB}{BC}$
i.e. $\;$ $BC = \dfrac{60}{\sqrt{3}}$ $\;\;\; \cdots \; (1)$
In $\triangle ABD$,
$\tan30^\circ = \dfrac{AB}{BD} = \dfrac{AB}{BC + CD} = \dfrac{AB}{BC + x}$
i.e. $\;$ $BC + x = \dfrac{60}{1 / \sqrt{3}} = 60 \sqrt{3}$
i.e. $\; $ $x = 60 \sqrt{3} - BC$ $\;\;\; \cdots \; (2)$
In view of equation $(1)$ equation $(2)$ becomes,
$x = 60 \sqrt{3} - \dfrac{60}{\sqrt{3}} = \dfrac{180 - 60}{\sqrt{3}} = \dfrac{120}{\sqrt{3}}= 40 \sqrt{3} = 69.28 \;$ meter
i.e. $\;$ $x = 69 \; m$ (to the nearest meter)