The angle of depression of a point situa

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Target Exam

CUET

Subject

General Test

Chapter

Numerical Ability

Topic

Time, Speed and Distance

Question:

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:

Options:

\(\frac{500}{3}\sqrt {3}\) m

\( 500(\sqrt{3} + 1) \) m

\( 500\sqrt{3} \) m

500 m

Correct Answer:

500 m

Explanation:

In Δ ABC,

⇒ tan θ = \(\frac{Perp}{base}\)

⇒ tan 45° = \(\frac{AC}{BC}\)

⇒\(\frac{1}{1}\) = \(\frac{AC}{500}\)

⇒ AC = 500m