A vertical tower stands on a horizontal plane. Two points A and B are located on the plane. The angles of elevation of the top of the tower from A and B are 30^ and 45^ , respectively. If the angle subtended by the line segment AB at the foot of the tower is 150^ and the distance AB is 7 7 meters, then the height of the tower (in meters) is

Step-by-step solution

Let the height of the tower be h and its foot be O . The distance of point A from the foot of the tower is OA = h 30^ = h 3 . The distance of point B from the foot of the tower is OB = h 45^ = h . In AOB , the angle subtended by AB at O is AOB = 150^ . Applying the Cosine Rule in AOB : AB^2 = OA^2 + OB^2 - 2(OA)(OB) 150^ Substitute the known values: (7 7 )^2 = (h 3 )^2 + h^2 - 2(h 3 )(h) (- 3 2 ) 343 = 3h^2 + h^2 + 3h^2 343 = 7h^2 h^2 = 49 h = 7 Thus, the height of the tower is 7 meters. Answer: 7