Question

From the top of a building AB, 60 m high, the angles of depression of the top and bottom of a vertical lamp post CD are observed to be 30° and 60°, respectively. Find
(i) the horizontal distance between AB and CD,
(ii) the height of the lamp post,
(iii) the difference between the heights of the building and the lamp post. [CBSE 2009]

Answer 1

We have,

$$\begin{gathered} \mathrm{AB}=60\mathrm{m},\angle \mathrm{ACE}=30°\mathrm{and}\angle \mathrm{ADB}=60° \\ \mathrm{Let}\mathrm{BD}=\mathrm{CE}=x\mathrm{and}\mathrm{CD}=\mathrm{BE}=y \\ \Rightarrow \mathrm{AE}=\mathrm{AB}-\mathrm{BE}=60-y \\ \mathrm{In}∆\mathrm{ACE}, \\ \tan 30°=\frac{\mathrm{AE}}{\mathrm{CE}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{60-y}{x} \\ \Rightarrow x=60\sqrt{3}-y\sqrt{3}.....\left(\mathrm{i}\right) \\ \mathrm{Also},\mathrm{in}∆\mathrm{ABD}, \\ \tan 60°=\frac{\mathrm{AB}}{\mathrm{BD}} \\ \Rightarrow \sqrt{3}=\frac{60}{x} \\ \Rightarrow x=\frac{60}{\sqrt{3}} \\ \Rightarrow x=\frac{60}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ \Rightarrow x=\frac{60\sqrt{3}}{3} \\ \Rightarrow x=20\sqrt{3} \\ \mathrm{Substituting}x=20\sqrt{3}\mathrm{in}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get} \\ 20\sqrt{3}=60\sqrt{3}-y\sqrt{3} \\ \Rightarrow y\sqrt{3}=60\sqrt{3}-20\sqrt{3} \\ \Rightarrow y\sqrt{3}=40\sqrt{3} \\ \Rightarrow y=\frac{40\sqrt{3}}{\sqrt{3}} \\ \Rightarrow y=40\mathrm{m} \\ \left(\mathrm{i}\right)\mathrm{the}\mathrm{horizontal}\mathrm{distance}\mathrm{between}\mathrm{AB}\mathrm{and}\mathrm{CD}=\mathrm{BD}=x \\ =20\sqrt{3} \\ =20\times 1.732 \\ =34.64\mathrm{m} \\ \left(\mathrm{ii}\right)\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{lamp}\mathrm{post}=\mathrm{CD}=y=40\mathrm{m} \\ \left(\mathrm{iii}\right)\mathrm{the}\mathrm{difference}\mathrm{between}\mathrm{the}\mathrm{heights}\mathrm{of}\mathrm{the}\mathrm{building}\mathrm{and}\mathrm{the}\mathrm{lamp}\mathrm{post}=\mathrm{AB}-\mathrm{CD}=60-40=20\mathrm{m} \end{gathered}$$