question_answer

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is \[45{}^\circ \]. If the tower is 30 m high, find the height of the building.

Answer:

Let AB be the tower and CD be a building of height 30 m and x m respectively.

Let the distance between the two be y m,

Then, in \[\Delta \text{ }ABC\]

\[\frac{30}{y}=\tan \,45{}^\circ \]

\[\frac{30}{y}=1\Rightarrow y=30\]

And, in \[\Delta \,BDC\]

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

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

\[x=30\times \frac{1}{\sqrt{3}}=10\sqrt{3}\]

Hence, the height of the building is \[10\sqrt{3}\,m\].