A vertical pole of height 15 m stands at the centre of a circular park. A straight path forms a chord AB of the circular boundary of the park. The angle of elevation of the top of the pole from point A is 30^ . While walking along the path from A to B , the maximum angle of elevation of the top of the pole is observed to be 45^ . The square of the length of the path AB (in m ^2 ) is

Step-by-step solution

Let the height of the pole be h = 15 m and the centre of the park be O . Point A lies on the boundary of the circular park. The horizontal distance OA is the radius R of the park. Since the angle of elevation from A is 30^ : R = h 30^ = 15 3 m The angle of elevation of the top of the pole is maximum when the observer is at the minimum horizontal distance from the pole. This occurs at the foot of the perpendicular from O to the chord AB . Let this minimum distance be d . Since the maximum angle of elevation is 45^ :