Sara,
Due to lack of complete information, I need to make several guesses in order to answer this question.
Let's suppose the right triangle contains points A, B, C with BC being the horizontal base and C is to the right of B and AC is the hypotenuse of the right triangle. ∠BCA=θ.
You said θ is changing at π/180 radians/sec. This tells me that C is moving towards B. Since the distance between B and C is changing, let's call this variable x
With the above information, we can write the following equation: dθ/dt = π/180. We also know that the height of the triangle is 4500'. Using that information we can write the following equation.
tan(θ) = 4500/BC = 4500/x. Or x = 4500cot(θ)
dx/dt = -4500csc2(θ).dθ/dt. The negative value for dx/dt indicates that x is decreasing. From this equation it is clear that dx/dt changes with θ and dθ/dt. But since dθ/dt is constant, dx/dt only depends on θ. The speed is variable. The last missing piece of information that is essential for calculating the ground speed is the value of θ. If we assume that the value of θ is π/2 (when C coincides with B and moving to the left) the ground speed is -4500csc2(θ).dθ/dt=-25π feet/sec
Imtiazur S.
01/02/15