A plane flies at a speed 580 km/hr at a constant height of 16 km. How rapidly is the angle of elevation to the plane changing when the plane is directly above a point 125 km away from the observer?

-0.584 rad/hr, rounded to the nearest 1000th  

Step-by-step explanation:

Let D = D(t) = the horizontal distance from the observer to the plane at time t in units of km.  

Let E = E(t) = the angle of elevation at time t in units of Radians.

Then, by geometry,

tan(E) = 16 / D, and furthermore, we have that

tan(E) = 16 * D^(-1)

d/dt (tan(E)) = d/dt ( 16 * D^(-1))

sec^2 (E) * dE/dt = 16(-1)(D^(-2)) * dD/dt

( sqrt(D^2 + 16^2) / D )^2 * dE/dt = -16 (D^(-2)) * dD/dt

(D^2 + 16^2)/D^2  *  dE/dt = -16 (D^(-2)) * dD/dt

dE/dt = -16 (D^(-2)) * dD/dt * D^2 / (D^2 + 16^2)

dE/dt = -16 (dD/dt) / (D^2 + 16^2)

dE/dt = -16 (580 km/hr) / (125^2 + 16^2), at time t such that D(t) = 125 km,

dE/dt = (-9280 / 15881) rad/hr

dE/dt = -0.584 rad/hr, rounded to the nearest 1000th

x = 17

step-by-step explanation:

in the given figure, two triangles are inscribed in a circle where ∠bae ≅ ∠dae which makes the side be and de equal so we can write it as:

3x - 24 = x + 10

solving this equation for x to get:

3x - x = 10 + 24

2x = 34

x = 34/2

x = 17

therefore, the value of x = 17.