An airplane in australia is flying 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.

We model the problem as a rectangle triangle.
 We have on the one hand:
 hypotenuse = z = 3
 On the other hand:
 vertical side = y = 2
 The speed of the plane is:
 dz / dt = 600
 The flight height is constant, then:
 dy / dt = 0
 The angle with the horizontal will be:
 sin (θ) = 2/3
 θ = arcsin (2/3) = 0.729728 radians
 On the other hand: 
 sin (θ) = y / z
 z sin (θ) = y
 We derive both sides with respect to time:
 dz / dt sin (θ) + z cos (θ) dθ / dt = dy / dt
 Substituting values:
 (600) (2/3) + (3) cos (0.729728) dθ / dt = 0
 Rewriting:
 400 + [3 / cos (0.729728)] dθ / dt = 0
 400+ 4.0249 dθ / dt = 0
 We cleared dθ / dt:
 dθ / dt = -98.38 radians / hour
 = -99.38 / 60 radians / min
 = -1.66 radians / min = 2PI - 1.66 radians / min
 = 4.62 radians / min

 
 The angle of elevation of the Kangaroo's line of sight is increasing at:
 = 4.62 radians / min

Sin(a)+cos(a)+cot(a)-1=2(csc(x)+1)