Question

A man is moving away from a tower 41.6 m high at a rate of 2 m/sIf the eye level of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance 30 m from the foot of the tower is:

• A. \( -\frac{4}{125} \) rad/sec
• B. \( \frac{4}{625} \) rad/sec
• C. \( -\frac{2}{125} \) rad/sec
• D. \( \frac{1}{625} \) rad/sec

Hint:

In related rates with angles, always express in terms of \( \tan \theta \) or \( \sin \theta \)Then differentiate implicitly and substitute values carefully.

Answer 1

The Correct Option is A