Find the distance of the plane from the observer 10 seconds after the plane has passed directly overhead.?

Question

The plane is going 180 mph passes over 400 ft directly above an observer. The plane is traveling along a straight path with an angle of elevation of 14 degrees.

Answer 1

10 seconds later the plane has changed its horizontal position by

x = ( 180 miles / hour ) ( cos 14 ) ( 10 sec ) ( 1 hour / 3600 sec ) ( 5280 feet / mile )

and its vertical position by

y = ( 180 miles / hour ) ( sin 14 ) ( 10 sec ) ( 1 hour / 3600 ) ( 5280 feet / mile )

The original vertical position was 400 ft, so the position is now

horizontal: x
vertical: 400 + y

The distance is therefore sqrt ( x^2 + ( 400 + y )^2 )

Answer 2

sin14=400/x
=>x=400/sin14 (x is the number in ft. u need to find, i dont have calculator sorry)