Given as a ladder 13 m long leans against a wall.

The foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5m/sec

As to find how fast is the angle θ between the ladder and the ground is changing when the foot of the ladder is 12 m away from the wall

Suppose AC be the position of the ladder initially, then AC = 13m.

DE be the position of the ladder after being pulled at the rate of 1.5m/sec, then DE = 13m as shown in the below figure.

Therefore it is given that foot of the ladder is pulled along the ground away from the wall, at the rate of 1.5 m/sec

So, \( \dfrac{dx}{dt}\) = 1.5 m/sec ...(i)

Considering the ΔABC, it is fight angled triangle, therefore on applying pythagoras theorem

And in same for triangle secθ =\( \dfrac{13}{12}\) ...(iii)

Differetiating equation (ii) with respect to time

Substitute the value of x,y,h and \( \dfrac{dx}{dt}\),

Value of h is always constant as the ladder is not increasing in size, thus the equation becomes

And consider the same triangle tanθ =\( \dfrac{y}{x}\)

Differentiate the above equation with respect to time

Substitute the value of secθ, x,y,h and \( \dfrac{dx}{dt}\), the equation becomes

Thus the angle θ between the ladder and the ground is changing at the rate of 0.3 rad/sec (because angle cannot be negative) when the foot of the ladder is 12 m away from the wall.

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