A vertical tower stands on a horizontal plane. A car is driving along a straight road in the same horizontal plane at a uniform speed. The road does not pass through the base of the tower. At a point A on the road, the angle of elevation of the top of the tower is 30^ . After 10 minutes, the car reaches a point B where the angle of elevation is 45^ . As the car continues its journey, the maximum angle of elevation it

Step-by-step solution

Let the height of the tower be h and its base be O . The maximum angle of elevation occurs when the car is at the point of closest approach to the tower. Let this point be C . The line OC is perpendicular to the road. The horizontal distances from the base O to the points A , B , and C are: OA = h 30^ = h 3 OB = h 45^ = h OC = h 60^ = h 3 In the horizontal plane, OCA is a right-angled triangle at C . The distance AC along the road is: AC = OA^2 - OC^2 = 3h^2 - h^2 3 = 8 3 h Similarly, OCB is right-angled at C . The