The angle of depression of a boat from the top of a cliff 10m high is 30. How far is the boat from the foot of the cliff?

The given scenario can be visualized as follows:

          A (top of cliff)
          /|
         / |
        /  |
       /   | 10m
      /    |
     /θ    |   
    /      |
   /       |
  B--------C (boat on water surface)

Here, the angle of depression of point B from point A is given as 30 degrees. We are required to find the distance between point B and point C, denoted by BC.

We know that the tangent of an angle is the ratio of the opposite side to the adjacent side. In this case, the opposite side is AB and the adjacent side is BC.

Thus, we have:

tan 30° = AB / BC

AB is the height of the cliff, which is given as 10m.

Hence, we have:

1/√3 = 10 / BC

Solving for BC, we get:

BC = 10√3 meters

Therefore, the boat is 10√3 meters away from the foot of the cliff.

Hence, the answer is 10√3m.