Each side of a square subtends an angle of 60 o at the top of a tower 5 meters high standing in the center of the square. If a meters is the length of each side of the square, then a is equal to

Step-by-step solution

Let A B C D be a square of each side of length a meters. It is given that ∠ B P C = 60 o . Let M be the midpoint of B C . Then, ∠ B P M = ∠ C P M = 30 o In ∆ B M P , right angled at M , We have tan ⁡ ∠ B P M = B M P M ⇒ tan ⁡ 30 ° = a / 2 P M = 1 3 ⇒ P M = 3 a 2 In ∆ O P M , P M 2 = O M 2 + O P 2 3 a 2 4 = a 2 4 + 5 2 ∴ a 2 = 50 ⇒ a = 5 2   m