calculus
posted by ron .
A 13 foot ladder leaning against a wall makes an angle of degree radians with the ground. The base of the ladder is pulled away from the wall at a rate of 2ft/sec. How fast is degree changing when the base of the ladder is 5 feet way.

if the base is x from the wall,
cosθ = x/13
sinθ dθ/dt = 1/13 dx/dt
when x=5, sinθ = 12/13
12/13 dθ/dt = 1/13 (2)
dθ/dt = 1/6
note the "" sign: the angle decreases as the ladder slips down. 
5, 12, 13 right triangle
call angle ladder to ground T
cos T = x/13 where x is base of ladder to base of wall
sin T dT/dt = (1/13) dx/dt
at x = 5, t = 0, dx/dt = 5
cos T = 5/13 so T = 67.4 degrees or 1.18 radians
sin T = sin 67.4 = 12/13 = .923
.923 dT/dt = (1/13)(2)
so
dT/dt = .167 radians/second
times 180/pi = 9.55 degrees/second