The shadow of a vertical tower on a level ground increases by 10 m when the altitude of the sun changes from 45° to 30°. Find the height of the tower, correct to two decimal places.

Let the height of the tower BD be h metres and the length of its shadow be d metres when the sun's altitude is 45°. When the sun's altitude is 30°, then the length of shadow of tower is 10 m longer,
i.e., BD = h meters, AB = d meters and CA = 10 metres.

From right angled △ABD, we get

$\Rightarrow \text{tan 45°} = \dfrac{BD}{AB} \\[1em] \Rightarrow 1 = \dfrac{h}{d} \\[1em] \Rightarrow h = d$

From right angled △BCD, we get

$\Rightarrow \text{tan 30°} = \dfrac{BD}{BC} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{AC + AB} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{10 + d} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{10 + h} [\because h = d] \\[1em] \Rightarrow 10 + h = \sqrt{3}h \\[1em] \Rightarrow \sqrt{3}h - h = 10 \\[1em] \Rightarrow h(\sqrt{3} - 1) = 10 \\[1em] \Rightarrow 0.732\text{ h} = 10 \\[1em] \Rightarrow h = \dfrac{10}{0.732} \\[1em] \Rightarrow h = 13.66$

Hence, the height of the tower is 13.66 meters.