Question

Find the angle of depression , of a point on the ground at a distance of 10√3 from the base of tower ,from the top of the tower whose height is 10m .

Answer 1

Given ,

• The distance of the point from the tower (PQ) = 10√3 m
• Height of the tower (PR) = 10 m

As we know that ,

$\sf \star \: \: Angle \: of \: elevation = Angle \: of \: depression \\ \\ \sf \star \: \: </p><p>Tan( \theta) = \frac{Perpendicular}{Base}$

Substitute the known values , we obtain

$\sf \Rightarrow Tan( \theta) = \frac{10}{10 \sqrt{3} } \\ \\ \sf \Rightarrow Tan( \theta) = \frac{1}{ \sqrt{3} } \\ \\ \sf \Rightarrow \theta = {Tan}^{ - 1}( \frac{1}{ \sqrt{3} }) \\ \\ \sf \Rightarrow \theta =30°$

$\therefore \bold{ \underline{\sf Angle \: of \: depression = 30°}}$