A man is moving away from a 40 m high tower at a speed of 2m/s . find the rate at which angle of elevation of the top of the tower  is changing when he is at a distance of 30m from foot of tower. Assume that eye level of the man is 1.6 m from the ground.

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Asked by haroonrashidgkp | 8th Sep, 2018, 04:16: PM

Expert Answer:

begin mathsize 16px style PD equals straight x space straight m
AQ space equals space 40 space straight m comma space AB equals 38.4 space straight m comma space PQ space equals space 30 space straight m
fraction numerator negative dx over denominator dt end fraction equals 2 space straight m divided by straight s
Let space straight theta space be space the space angle space of space elevation
tanθ equals AB over PD equals fraction numerator 38.4 over denominator straight x end fraction
straight theta equals tan to the power of negative 1 end exponent open parentheses fraction numerator 38.4 over denominator straight x end fraction close parentheses
dθ over dt equals negative fraction numerator 76.8 over denominator straight x squared plus 38.4 squared end fraction
straight x space equals space 30
dθ over dt equals negative 0.032
Angle space of space elevation space is space decreasing space at space the space rate space 0.032 space degree divided by sec
end style

Answered by Sneha shidid | 10th Sep, 2018, 09:52: AM