Calculus - related rates?

Question

A 13 foot ladder is leaning against a house when its base starts to slide away. By the time the base is 12 feet from the house, the base is moving at the rate of 5 ft/sec.
At what rate is the angle between the ladder and the ground changing at that moment?

explanation please..

Answer 1

Let
x = distance of base of ladder from wall
θ = angle between ladder and ground

Given
dx/dt = 5 ft/sec

Find dθ/dt when x = 12 feet.
__________

cosθ = x/13 = 12/13
when x = 12

sinθ = √(1 - x²)/13 = √(1 - 144/169) = √(25/169) = 5/13
when x = 12
_____________

cosθ = x/13

Differentiate implicitly.

(-sinθ)(dθ/dx) = 1/13
dθ/dx = -1 / [13sinθ] = -1 / [13(5/13)] = -1/5
when x = 12

dθ/dt = (dθ/dx)(dx/dt) = (-1/5)(5) = -1 radians/sec

Answer 2

The distance of the base from the house is just 13*cosine(angle).

You are told that d(13cosA)/dt = 5.

But d(13cosA)/dt = 13 * (-sinA) * dA/dt

It turns out that sinA = 5/13. (That was easy because 5-12-13 is a well-known Pythagorean Triple).

So 5 = -5 * dA/dt, and the answer is -1 radians per second (i.e., it's decreasing at 1 radian/second, where 2 pi radians is of the course the same as 360 degrees).