(A) True.
$Case-I$: Let the height of a tower be $h$ and the distance of the point of observation from its foot be $x$.
In $\triangle ABC$,$\tan \theta_1 = \frac{h}{x}$,which implies $\theta_1 = \tan^{-1}(\frac{h}{x})$ ... $(i)$
$Case-II$: Now,the height of the tower is increased by $10 \%$,so the new height $h' = h + 0.1h = 1.1h = \frac{11h}{10}$.
The distance of the point of observation from its foot is also increased by $10 \%$,so the new distance $x' = x + 0.1x = 1.1x = \frac{11x}{10}$.
In the new triangle,$\tan \theta_2 = \frac{h'}{x'} = \frac{1.1h}{1.1x} = \frac{h}{x}$.
Thus,$\theta_2 = \tan^{-1}(\frac{h}{x})$ ... $(ii)$
From equations $(i)$ and $(ii)$,we get $\theta_1 = \theta_2$.
Therefore,the angle of elevation remains unchanged.