From a point on the ground, the angle of elevation of the bottom and top of a transmission tower fixed at the top of…

1 for correct figure
Height of building AB = $30$ m
BP = transmission tower = h(say)
$\angle ACB = 30^\circ$, $\angle ACP = 60^\circ$
In $\triangle ABC$, $\tan 30^\circ = \frac{AB}{AC}$
$\Rightarrow \frac{1}{\sqrt{3}} = \frac{30}{AC} \Rightarrow AC = 30\sqrt{3}$
In $\triangle APC$, $\tan 60^\circ = \frac{AP}{AC}$
$\sqrt{3}= \frac{30+ h}{30\sqrt{3}}$
$\Rightarrow 30\sqrt{3} \times \sqrt{3}= 30 + h$
$\Rightarrow h = 30 (3 - 1)$
$\Rightarrow h = 60$
$\therefore$ Height of transmission tower = $60$ m