A vertical tower stands at the center of a circular park. Three points A , B , and C are located on the boundary of the park such that AB = AC = 40 m and BC = 48 m. If the angle of elevation of the top of the tower from point A is ⁻¹(2) , then the height of the tower (in m) is

Step-by-step solution

The center of the circular park is the circumcenter of the horizontal triangle ABC . The distance from any point on the boundary (like A ) to the center is the circumradius R . In ABC , AB = AC = 40 and BC = 48 . Draw the altitude AD to BC . Since ABC is isosceles, D is the midpoint of BC , so BD = 24 . By Pythagoras theorem in ABD : AD = AB^2 - BD^2 = 40^2 - 24^2 = 32 Let O be the circumcenter, which lies on AD . Let OA = R , then OD = 32 - R . In right-angled OBD : OB^2 = OD^2 + BD^2 R^2 = (32 - R)^2 + 24^2 R^2 =