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)

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

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