A man standing on top of a vertical cliff looking at a canoe out at sea. The angle of depression of the canoe is 25° . If the cliff is 100 meter high. How far from the base of the cliff is the canoe?
Let x be the distance from the base of the cliff to the canoe.
The angle of depression of the canoe is the angle between the line of sight from the man to the canoe and the horizontal line.
Since the man is standing on top of a vertical cliff, the line of sight from the man to the canoe is also vertical.
Therefore, the angle between the line of sight and the horizontal line is 90° - 25° = 65°.
In the right triangle formed by the man, the cliff, and the line of sight to the canoe, the opposite side is the height of the cliff and the adjacent side is the distance from the base of the cliff to the canoe.
Using the tangent function, we can write the following equation:
tan(65°) = opposite/adjacent
tan(65°) = 100/x
x = 100/tan(65°)
x ≈ 46.07 meters
Hence, the canoe is approximately 46.07 meters away from the base of the cliff. Answer: \boxed{46.07}.