Let's solve the problem step by step.
### Given Values
- The airplane travels at a speed of [tex]\( 400 \sqrt{3} \)[/tex] km/hr.
- The time it takes to look to the right after seeing the lake is [tex]\( 15 \)[/tex] minutes.
- The angle they have to tilt their head is [tex]\( 30^\circ \)[/tex].
### Convert Time to Hours
First, convert the time from minutes to hours because the speed is given in km/hr.
[tex]\[ \text{Time} = \frac{15 \text{ minutes}}{60} = 0.25 \text{ hours} \][/tex]
### Calculate the Distance Traveled
To find the distance the airplane has traveled in those 15 minutes, we use the formula:
[tex]\[ \text{Distance} = \text{Speed} \times \text{Time} \][/tex]
Substituting the given values:
[tex]\[ \text{Distance} = 400 \sqrt{3} \, \text{km/hr} \times 0.25 \, \text{hr} \][/tex]
[tex]\[ \text{Distance} = 100 \sqrt{3} \, \text{km} \][/tex]
[tex]\[ \text{Distance} \approx 173.205 \, \text{km} \][/tex]
### Calculate the Height of the Airplane
The person tilts their head by [tex]\( 30^\circ \)[/tex] to see the lake, meaning we can use trigonometry to find the height of the airplane.
We will use the tangent function since we know the adjacent side (distance traveled) and we want to find the opposite side (height):
[tex]\[ \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \][/tex]
Rearranging to solve for the opposite side (height):
[tex]\[ \text{Height} = \text{Distance} \times \tan(30^\circ) \][/tex]
Substituting the values:
[tex]\[ \text{Height} = 173.205 \, \text{km} \times \tan(30^\circ) \][/tex]
Since [tex]\( \tan(30^\circ) = \frac{1}{\sqrt{3}} \)[/tex]:
[tex]\[ \text{Height} = 173.205 \, \text{km} \times \frac{1}{\sqrt{3}} \][/tex]
[tex]\[ \text{Height} = 173.205 \, \text{km} \times \frac{\sqrt{3}}{3} \][/tex]
[tex]\[ \text{Height} = 100 \, \text{km} \][/tex]
### Summary of Results
- Distance traveled by airplane (horizontal distance): approximately [tex]\( 173.205 \)[/tex] km
- Height of the airplane above the ground: approximately [tex]\( 100 \)[/tex] km
So, the airplane is flying at a height of approximately 100 km above the ground, and the lake is approximately 173.205 km away horizontally from the point where the person initially saw it.
