In the adjoining figure, the shadow of a vertical tower on the level ground increases by 10 m, when the altitude of the sun changes from 45° to 30°. Find the height of the tower and give your answer, correct to $\dfrac{1}{10}$ of a metre.

Let the initial length of shadow be x meters and height of tower be h meters.

Considering right angled △DBC, we get

$\Rightarrow \text{tan 45°} = \dfrac{BC}{DB} \\[1em] \Rightarrow 1 = \dfrac{h}{x} \\[1em] \Rightarrow x = h \text{ ……(Eq 1)}$

Considering right angled △ABC we get,

$\Rightarrow \text{tan 30°} = \dfrac{BC}{AB} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{x + 10} \\[1em] \Rightarrow x + 10 = \sqrt{3}h \\[1em] \Rightarrow h + 10 = \sqrt{3}h \text{ …..(Using Eq 1)} \\[1em] \Rightarrow \sqrt{3}h - h = 10 \\[1em] \Rightarrow 0.732h = 10 \\[1em] \Rightarrow h = \dfrac{10}{0.732} \\[1em] \Rightarrow h = 13.7$

Hence, the height of tower is 13.7 meters.