A 25 -ft ladder is leaning against a house when its base starts to slide away. By the time the base is $7 \mathrm{ft}$ from the house, the base is moving away at the rate of $24 \mathrm{ft} / \mathrm{sec}$. a. What is the rate of change of the height of the top of the ladder? b. At what rate is the area of the triangle formed by the ladder, wall, and ground changing then? c. At what rate is the angle between the ladder and the ground changing then?

Step 1 ：Given that the base of the ladder is moving away at a rate of 24 ft/sec, we need to find the rate of change of the height of the ladder, the area of the triangle formed by the ladder, wall, and ground, and the angle between the ladder and the ground.

Step 2 ：We can use the Pythagorean theorem to relate the base and height of the ladder to the length of the ladder. The equation is \(x^2 + y^2 = 625\).

Step 3 ：Differentiating this equation with respect to time, we get \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\).

Step 4 ：Substituting the given values, we can solve for \(\frac{dy}{dt}\).

Step 5 ：For the area of the triangle, we can use the formula \(A = 0.5xy\).

Step 6 ：Differentiating this equation with respect to time, we get \(\frac{dA}{dt} = 0.5x\frac{dy}{dt} + 0.5y\frac{dx}{dt}\).

Step 7 ：Substituting the given values, we can solve for \(\frac{dA}{dt}\).

Step 8 ：For the angle between the ladder and the ground, we can use the trigonometric relationship \(\tan(\theta) = \frac{y}{x}\).

Step 9 ：Differentiating this equation with respect to time, we get \(\frac{d\theta}{dt} = \frac{1}{\cos^2(\theta)}\left(\frac{y}{x^2}\frac{dx}{dt} - \frac{1}{x}\frac{dy}{dt}\right)\).

Step 10 ：Substituting the given values, we can solve for \(\frac{d\theta}{dt}\).

Step 11 ：Final Answer: The rate of change of the height of the ladder is \(\boxed{24}\) ft/sec, the rate of change of the area of the triangle formed by the ladder, wall, and ground is \(\boxed{24}\) sq ft/sec, and the rate of change of the angle between the ladder and the ground is \(\boxed{24}\) rad/sec.