From the top of a cliff, the angle of depression of a stationary boat on the sea is 60^ . An observer walks 200 m down a straight path inclined at 30^ to the horizontal, moving away from the sea in the vertical plane containing the boat and the cliff. From this new position, the angle of depression of the boat is 45^ . The initial horizontal distance of the boat from the top of the cliff is :

Step-by-step solution

Let the top of the cliff be A(0, H) and the boat be B(x, 0) . The angle of depression of the boat from A is 60^ , so 60^ = H x H = x 3 . The observer moves 200 m down a path inclined at 30^ to the horizontal, away from the sea. The horizontal distance moved is 200 30^ = 100 3 m (away from the boat). The vertical distance moved is 200 30^ = 100 m (downwards). The new position C has coordinates (-100 3 , H - 100) . From C , the angle of depression of the boat is 45^ . The horizontal distance from C to the boat is x -