The horizontal distance between two towers is $150 \; m$. The angle of depression of the top of one tower as observed from the top of the second tower, which is $120 \; m$ in height, is $30^\circ$. Find the height of the first tower. Take $\sqrt{3} = 1.732$.
$O_1 T_1, \; O_2 T_2 = $ two towers
Height of tower $O_1 T_1 = 120 \; m$
Height of tower $O_2 T_2 = h$ meter
Distance between the two towers $= O_1 O_2 = 150 \; m$
From the figure, $\;$ $XT_2 = O_1 O_2 = 150 \; m$
In $\triangle T_1 X T_2$, $\;$ $\tan 30^\circ = \dfrac{X T_1}{X T_2}$
$\implies$ $X T_1 = X T_2 \times \tan 30^\circ = 150 \times \dfrac{1}{\sqrt{3}} = 50 \sqrt{3} \; m$
But $\;$ $X T_1 = O_1 T_1 - O_2 T_2$
i.e. $\;$ $50 \sqrt{3} = 120 - h$
$\implies$ $h = 120 - 50 \sqrt{3} = 120 - 86.6 = 33.4 \;m$
$\therefore \;$ Height of tower $O_2 T_2 = 33.4 \; m$