27. Aviation A plane is flying at an elevation of 10,000 meters and a speed of 300 kilometers per hour. The pilot sees a lake at an angle of depression of \( 60^{\circ} \). In how much time, to the nearest minute, will the plane be directly over the lake?

To find out how long it will take for the plane to be directly over the lake, we first need to determine the horizontal distance from the plane to the lake at an elevation of 10,000 meters and an angle of depression of \( 60^{\circ} \). Using the tangent of the angle of depression, we can set up the equation: \[ \tan(60^{\circ}) = \frac{\text{opposite}}{\text{adjacent}} = \frac{10,000}{d} \] Where \( d \) is the horizontal distance. Since \( \tan(60^{\circ}) = \sqrt{3} \), the equation becomes: \[ \sqrt{3} = \frac{10,000}{d} \] Rearranging gives: \[ d = \frac{10,000}{\sqrt{3}} \approx 5773.50 \text{ meters} \] Now, to find the time it takes for the plane to cover this horizontal distance at a speed of 300 km/h, we first convert the speed to meters per second: \[ 300 \text{ km/h} = \frac{300,000 \text{ meters}}{3600 \text{ seconds}} \approx 83.33 \text{ m/s} \] Now, we can find the time \( t \): \[ t = \frac{d}{\text{speed}} = \frac{5773.50}{83.33} \approx 69.28 \text{ seconds} \] Finally, converting seconds to minutes gives: \[ t \approx 1.15 \text{ minutes} \] Rounding to the nearest minute, the time for the plane to be directly over the lake is approximately **1 minute**.