Question

An airplane is flying at an elevation of 15,000 feet. The airport is 6 miles away from a point directly below the airplane on the ground. How far is the airplane from the airport?

Answer 1

The distance between the airplane and the airport is:

`c=6.64\: miles`

Explanation:

We can use the Pythagoras theorem to find the distance from the airplane to the airport.

The equation is:

`c^(2)=a^(2)+b^(2)`

Where:

• a represents the elevations of the airplane (15000 feet)
• b is the distance from the ground directly below the airplane to the airport (6 miles)
• c is the distance between the airplane and the airport.

We need first to convert 15000 feet in miles.

`15000\: feet * (0.00019\: miles)/(1\: feet)=2.85\: miles`

Therefore, c will be:

`c^(2)=a^(2)+b^(2)`

`c=\sqrt{a^(2)+b^(2)}`

`c=\sqrt{2.85^(2)+6^(2)}`

`c=6.64\: miles`

I hope it helps you!