We assume that the plane maintains the same altitude and velocity.

Let A be the initial position of the plane; let B be the point on the ground directly beneath the plane.

Let C be the position of the plane 12min from the initial position and let D be the point directly beneath C on the ground.

Let E be the lighthouse. Let AB=CD=h. Let BE=d.

Consider the plane at A. We can form right triangle ABE. We are given that the angle of depression from the plane to the lighthouse is 6 degrees. Then the angle of elevation from the lighthouse to the plane is also 6 degrees so `m/_AEB=6^@` . ` `Then `tan6^@=h/d`

Consider the plane at C. The plane has traveled at 500km/hr for 12min or 1/5 of an hour, so it has traveled 100km. Then DE=d-100. The angle of depression is given as 15 degrees, so the angle of elevation from the lighthouse is also 15 degrees. So `tan15^@=h/(d-100)`

Solving each equation for h we get `h=d tan6^@,h=(d-100)tan15^@` Setting the expressions for h equal to each other we get:

`d tan6^@=(d-100)tan15^@`

`d tan15^@ - d tan 6^@=100 tan 15^@`

`d=(100tan15^@)/(tan15^@ - tan6^@)`

`d~~164.54`

So the inital position of the plane is approximately 164.54km from the lighthouse. At 500 km/hr it would take the plane about .32908hr or about 19.75 min to be directly over the lighthouse from the initial position.

**After 12 minutes the plane's position is approximately 64.54km from the lighthouse. It would take approximately 7.75 min to be directly over the lighthouse from this position.**