The shadow of a vertical tower AB on level ground is increased by 10 m, when the altitude of the sun changes from 45° to 30°, as shown in the figure. Find the height of the tower and give your answer correct to $\Big(\dfrac{1}{10}\Big)$ of a metre.

Given,

The height of the tower AB = h metres and the length of its shadow = x 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., BA = h meters, AD = x meters and CD = 10 metres.

From right angled △ABD, we get

$\Rightarrow \tan 45° = \dfrac{BA}{AD} \\[1em] \Rightarrow 1 = \dfrac{h}{x} \\[1em] \Rightarrow h = x$

From right angled △BCA, we get

$\Rightarrow \tan 30° = \dfrac{BA}{CA} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{CD + AD} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{10 + x} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{10 + h} [\because h = x] \\[1em] \Rightarrow 10 + h = \sqrt3h \\[1em] \Rightarrow 10 = \sqrt3h - h \\[1em] \Rightarrow 10 = h(1.732 - 1) \\[1em] \Rightarrow 10 = 0.732h \\[1em] \Rightarrow h = \dfrac{10}{0.732} \\[1em] \Rightarrow h = 13.66 \approx 13.7 \text{ m}.$

Hence, the height of the tower is 13.7 m.