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. \[\therefore \tan \theta =\frac{AB}{BC}\] \[[\because \,\,tan\theta =\frac{Perpendicular}{Base}]\] \[\Rightarrow \tan \theta =\frac{20}{20\sqrt{3}}=\frac{1}{\sqrt{3}}\] \[\Rightarrow \tan \theta =tan30{}^\circ \,\,[\because \,\,tan\,30{}^\circ =\frac{1}{\sqrt{3}}]\] \[\Rightarrow \theta =30{}^\circ \]
You need to login to perform this action.
You will be redirected in
3 sec