The angle of elevation of the top of an unfinished tower from a point at a distance of 80 m from its base is 30°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60° ?

Let AB be the unfinished tower and C be the point 80 m from base of tower.

From figure,

In △ABC,

$\Rightarrow \text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BC} \\[1em] \Rightarrow AB = \dfrac{BC}{\sqrt{3}} \\[1em] \Rightarrow AB = \dfrac{80}{\sqrt{3}} \\[1em] \Rightarrow AB = \dfrac{80}{1.732} \\[1em] \Rightarrow AB = 46.19\text{ meters}.$

Let tower be raised to point D in order to make angle of elevation 60°.

From figure,

In △DBC,

$\Rightarrow \text{tan 60°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \sqrt{3} = \dfrac{BD}{BC} \\[1em] \Rightarrow BD = BC\sqrt{3} \\[1em] \Rightarrow BD = 80\sqrt{3} \\[1em] \Rightarrow BD = 138.56\text{ metres}.$

AD = BD - AB = 138.56 - 46.19 = 92.37 meters.

Hence, the tower must be raised by 92.37 meters.