A plane is flying at 500 km/h. The angle of depression to a lighthouse on an island in the distance is 6 degrees. After 12 min, as the plane continues to approach teh lighthouse, the angle of...
A plane is flying at 500 km/h. The angle of depression to a lighthouse on an island in the distance is 6 degrees. After 12 min, as the plane continues to approach teh lighthouse, the angle of depression is 15 degrees. How long will it take for the plane to be directly above the light house?
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 tan15^@ - d tan 6^@=100 tan 15^@`
`d=(100tan15^@)/(tan15^@ - tan6^@)`
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.