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Question

A man of height 6 ft is moving away from a building of height 106 ft at the rate of 25 ft/sec. The absolute value of the rate at which the angle of elevation of the top of the building is changing, when he is at a distance of 50 ft from the foot of the building is rad/sec

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Solution


AB=Man, CD=Building
Let BC=x ft dxdt=25 ft/sec
tanθ=100xsec2θ dθdtdtdx=100x2
When x=50 ft
tanθ=10050=2secθ=1+4=5
Now,
5 dθdt25=100502dθdt=2 rad/s

Hence the absolute value of the rate will be,
dθdt=2 rad/s

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