A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 20 ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 15 ft from the point on the path closest to the searchlight

0.128rad/sec

Step-by-step explanation:

Let x represent the between the man and the point on the path

θ = the angle

dx/dt = 4 ft/s

dθ/dt = rate is the searchlight rotating when the man is 15 ft from the point on the path closest to the searchlight

tan θ = x/20 ft

Cross Multiply

20tan θ = x

dx/dt = 20sec² θ dθ/dt

dθ/dt = 1/20 × cos² θ dx/dt

dθ/dt= 1/20 × cos² θ × 4

dθ/dt = 1/5 × cos² θ

Note : cos θ = 4/5

dθ/dt = 1/5 × (4/5)²

dθ/dt = 16/125

dθ/dt = 0.128rad/sec

its 69.420

step-by-step explanation:

2 over 5 and 4 over 15

don’t know last one sorry

step-by-step explanation: