A B C is a triangular park with A B = A C = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at, A and B are c o t - 1 3 2 and c o s e c - 1 2 2 respectively, then the height of the tower (in metres) is

Step-by-step solution

Let A D = y   &   D E = h (Tower) ∴   y 2 = 100 2 - x 2 Now cot θ = 3 2 ⇒ tan θ = 1 3 2 = h y ⇒ y = 3 2 h ⇒ y 2 = 100 2 - x 2 = 18 h 2     … 1 ⇒ tan ∅ = 1 7 = h x ⇒ x = h 7         … 2 Form 1 & 2 100 2 - 7 h 2 = 18 h 2 ⇒   25 h 2 = 100 2 ⇒ h = 20