Question

1. A 12-foot ladder is leaning against a wall that makes a 90-degree angle with the ground. A troublemaker is slowly pulling the bottom of the ladder along the ground, away from the wall at a rate of 2 feet per minute when the top of the ladder is 6 feet above the ground.

b. At that moment, what is the rate of change of the angle between the ladder and the ground?

Answer 1

You need to use the derivative of the Pythagorean theorem.
a^2+b^2=c^2

2a da + 2b db = 0
C is a constant, so it goes to 0
db or change along the ground is 2 ft/min

2(6) (da) + 2(b) (2) =0

I then need to find b
6^2+b^2=12^2
b=10.39 ft

so da = -4(10.39)/12
da=-3.46 ft/min
I think thats what you were looking for with question a

Answer 2

for b
cos(angle)=b/h
-sin(angle) dangle=db/h
-sin(angle)=1/2
-(1/2) dangle = 2/12
dangle=-1/3