ABC is a triangular park with a right angle at B . A vertical tower is situated at the corner B . The length of the hypotenuse AC is 150 m. If the angles of elevation of the top of the tower from A and C are ⁻¹ ( 1 17 ) and cosec ⁻¹( 10 ) respectively, then the height of the tower (in m) is

Step-by-step solution

Let the height of the tower at B be h . The angle of elevation of the top of the tower from A is _A = ⁻¹ ( 1 17 ) . This gives _A = 1 17 , which implies _A = 4 . The horizontal distance AB = h _A = 4h . The angle of elevation from C is _C = cosec ⁻¹( 10 ) . This gives cosec _C = 10 , which implies _C = 3 . The horizontal distance BC = h _C = 3h . In the horizontal right-angled triangle ABC , applying the Pythagoras theorem: AB^2 + BC^2 = AC^2 (4h)^2 + (3h)^2 = 150^2 16h^2 + 9h^2 = 22500 25h^2 = 22500 h^2 = 900 h =