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CBSE Class 12-science Answered

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. explain in great detail
Asked by haroonrashidgkp | 08 Sep, 2018, 04:16: PM
answered-by-expert 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 | 10 Sep, 2018, 09:52: AM

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CBSE 12-science - Maths
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. explain in great detail
Asked by haroonrashidgkp | 08 Sep, 2018, 04:16: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
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