A vertical tower stands at an arbitrary point inside a rectangular field ABCD . The angles of elevation of the top of the tower from the corners A , B , and C are 30^ , 45^ , and 60^ respectively. The angle of elevation of the top of the tower from the fourth corner D is:

Step-by-step solution

Let the height of the vertical tower be h and its base be at point P in the plane of the rectangle ABCD . The horizontal distances from P to the corners A , B , and C can be found using the angles of elevation: PA = h 30^ = h 3 PB = h 45^ = h PC = h 60^ = h 3 For any point P in the plane of a rectangle ABCD , the distances to the vertices satisfy the relation: PA^2 + PC^2 = PB^2 + PD^2 Substituting the values of PA , PB , and PC : (h 3 )^2 + ( h 3 )^2 = h^2 + PD^2 3h^2 + h^2 3 = h^2 + PD^2 PD^2 = 3h^2 + h^2 3 - h^2