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:
Answer & explanation
Correct answer: option 4
In Δ ABC,
⇒ tan θ = \(\frac{Perp}{base}\)
⇒ tan 45° = \(\frac{AC}{BC}\)
⇒\(\frac{1}{1}\) = \(\frac{AC}{500}\)
⇒ AC = 500m