An airplane flys at a constant altitude of 2 miles and a constant speed of 600 miles per hour on a straight course that will take it directly over a kangaroo on the ground. how fast is the angle of elevation of the kangaroo's line of sight increasing when the distance from the kangaroo to the plane is 3 miles? give your answer in radians per minute.

Let us say that y = vertical height and z = hypotenuse 
dz/dt = 600 
dy/dt = 0 since y (altitude) is constant 
sin(θ) = 2/3 
Calculating for the angle: θ = arcsin(2/3) = 0.729728 radians 
sin(θ) = y/z 
z sin(θ) = y 
We then differentiate the two sides with respect to time t:
dz/dt sin(θ)+ z cos(θ) dθ/dt = dy/dt 
(600) (2/3) + (3) cos (0.729728) dθ/dt = 0 
400 + [3 / cos (0.729728) ] dθ/dt = 0 
400+ 4.0249 dθ/dt = 0 
dθ/dt = -98.38 radians/hour 
= -99.38/60 radians / min 
= -1.66 radians / min = 2PI - 1.66 radians / min 
= 4.62 radians / min    (ANSWER)

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step-by-step explanation:

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step-by-step explanation: