question_answer

In figure 1, a tower AB is 20 m high and BC, its shadow on the ground, is \[20\sqrt{3}\,m\] long. Find the sun's altitude.

Answer:

Given, AB is the tower and BC is its shadow.

\[\therefore \tan \theta =\frac{AB}{BC}\]

\[[\because \,\,tan\theta =\frac{Perpendicular}{Base}]\]

\[\Rightarrow \tan \theta =\frac{20}{20\sqrt{3}}=\frac{1}{\sqrt{3}}\]

\[\Rightarrow \tan \theta =tan30{}^\circ \,\,[\because \,\,tan\,30{}^\circ =\frac{1}{\sqrt{3}}]\]

\[\Rightarrow \theta =30{}^\circ \]