A vertical tower stands at a point O on horizontal ground. A point A is located due East of O , and a point B is located due North of O . The distance between A and B is 20 meters. If the angle of elevation of the top of the tower from A is 45^ and from B is 30^ , then the height of the tower (in meters) is

Step-by-step solution

Let the height of the tower be h meters. Since the angle of elevation of the top of the tower from A is 45^ , the distance OA = h 45^ = h . Since the angle of elevation from B is 30^ , the distance OB = h 30^ = h 3 . Point A is due East of O and point B is due North of O , so AOB is a right-angled triangle at O . Using the Pythagorean theorem in AOB : AB^2 = OA^2 + OB^2 Substitute the given values: (20)^2 = h^2 + (h 3 )^2 400 = h^2 + 3h^2 400 = 4h^2 h^2 = 100 Therefore, h = 10 meters. Answer: 10