The angle of depression of a point situated at a distance 500 mtr from the top of a tower is 45°. The height of tower is: |
\(\frac{500}{3}\sqrt {3}\) m \( 500(\sqrt{3} + 1) \) m \( 500\sqrt{3} \) m 500 m |
500 m |
In Δ ABC, ⇒ tan θ = \(\frac{Perp}{base}\) ⇒ tan 45° = \(\frac{AC}{BC}\) ⇒\(\frac{1}{1}\) = \(\frac{AC}{500}\) ⇒ AC = 500m |