A surface ship is moving (horizontally) in a straight line at \( 19 \mathrm{~km} / \mathrm{hr} \). At the same time, an enemy submarine maintains a position directly below the ship while diving at an angle that is \( 25^{\circ} \) below the horizontal. How fast is the submarine's altitude decreasing? The altitude of the submarine is decreasing at a rate of about (Do not round until the final answer. Then round to two decimal places as needed.)

To find the rate at which the submarine's altitude is decreasing, we start by using trigonometry. The horizontal speed of the ship is \( 19 \, \mathrm{km/hr} \). Since the submarine is diving at an angle of \( 25^{\circ} \) below the horizontal, we can use the sine function to relate the horizontal speed to the vertical speed of the submarine. The vertical component of the submarine's speed can be calculated using the sine of the angle: \[ \text{Vertical Speed} = \text{Ship Speed} \times \sin(\theta) \] Where: - \(\text{Ship Speed} = 19 \, \mathrm{km/hr}\) - \(\theta = 25^{\circ}\) Calculating this gives: \[ \text{Vertical Speed} = 19 \times \sin(25^{\circ}) \] Using a calculator: \[ \sin(25^{\circ}) \approx 0.4226 \] \[ \text{Vertical Speed} \approx 19 \times 0.4226 \approx 8.02 \, \mathrm{km/hr} \] Since the altitude is decreasing, we consider this rate as negative: Thus, the altitude of the submarine is decreasing at a rate of approximately \(-8.02 \, \mathrm{km/hr}\). So, the altitude of the submarine is decreasing at a rate of about **8.02 km/hr** when rounded to two decimal places.