question_answer

The tops of two towers of height x and y, standing on level ground, subtend angles of \[30{}^\circ \] and \[60{}^\circ \] respectively at the centre of the line joining their feet, then find \[x:y\].

Answer:

Let AB and CD be two towers of height x and y respectively.

M is the mid-point of BC i.e., \[BM=MC\]

In \[\Delta \text{ }ABM\], we have

\[\frac{AB}{BM}=\tan \,30{}^\circ \]

\[BM=\frac{x}{\tan \,30{}^\circ }\]                                            ?(i)

In \[\Delta \text{ }CDM\], we have

\[\frac{DC}{MC}=\tan \,\,60{}^\circ \]

\[\frac{y}{MC}=\tan \,\,60{}^\circ \]

\[MC=\frac{y}{\tan \,\,60{}^\circ }\]                                          ?(ii)

From eq. (i) and (ii), we get

\[\frac{x}{\tan 30{}^\circ }=\frac{y}{\tan 60{}^\circ }\]

\[\frac{x}{y}=\frac{\tan 30{}^\circ }{\tan 60{}^\circ }\]

\[\frac{x}{y}=\frac{1/\sqrt{3}}{\sqrt{3}}=\frac{1}{3}\]

\[\therefore x:y=1:3\]