3. The plane continued to climb in altitude for 19 seconds at an angle of \( 2^{\circ} \), reaching a maximum height of 3060 feet. How much further was the plane able to travel during this time frame? (Hint: 1 mile \( =5,280 \) feet)

To find out how far the plane traveled in the horizontal direction during its climb, we can use some basic trigonometry. The vertical height can be related to the angle of climb. Given that the maximum height reached was 3060 feet and the climb angle was \( 2^{\circ} \), we can use the sine function: \[ \text{Vertical height} = \text{Distance traveled} \times \sin(\text{angle}) \] So, rearranging for the distance traveled (D): \[ D = \frac{\text{Vertical height}}{\sin(2^{\circ})} \approx \frac{3060}{0.0349} \approx 87,600 \text{ feet} \] Next, to find the distance in miles: \[ \text{Distance in miles} = \frac{87,600 \text{ feet}}{5280 \text{ feet/mile}} \approx 16.6 \text{ miles} \] Thus, the plane was able to travel approximately 16.6 miles during its ascent.