A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.8 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall?

Answer: -0.133 radians/s is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall.Step-by-step explanation:Let x= distance between the wall and ladderФ = angle between the ladder and groundThe rate of change is: dx/dt = 0.8 ft/sWe need to find dФ/dt when x = 8 ft/sFrom the triangle in the figure we see that:cos Ф = x/10=> x = 10 cosФNow, the rate of change will bed(x)/dt = d/dt (10cosФ)dx/dt = -10 sin Ф dФ/dtdx/dt = 0.8 (given)to find dФ/dtwe need to find sin Ф when x = 8By Pythagoras theorem(hypotenuse)^2 = (Perpendicular)^2+(base)^2(10)^2 = (8)^2+ (Perpendicular)^2100 = 64 +  (Perpendicular)^2=>  (Perpendicular)^2 = 100-64 (Perpendicular)^2 = 36 Perpendicular = 6sin Ф = Perpendicular/hypotenusesin Ф = 6/10 = 3/5Putting values in:dx/dt = -10 sin Ф dФ/dt0.8 = -10(3/5) dФ/dt0.8 = -6 dФ/dt=> dФ/dt = 0.8 / -6dФ/dt = - 0.133 radians/sSo, -0.133 radians/s is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall.