Vertex A of the acute angled triangle A B C is equidistant from its circumcentre O and orthocentre H , then the possible value of ∠ A is

Step-by-step solution

Let, the position of circumcentre ( O ) be origin, position vector of orthocentre ( H ) is x → and position vectors of A , B , C are a → , b → , c → respectively Centroid of Δ A B C G = 1 3 a → + b → + c → x → + 0 → 3 = a → + b → + c → 3 ⇒ x → = a → + b → + c → O A → = H A → (given) a → = b → + c → a → = b → = c → = R a → 2 = b → 2 + c → 2 + 2 b → c → cos ⁡ θ ⇒ R 2 = R 2 + R 2 + 2 R 2 cos ⁡ θ ⇒ cos ⁡ θ = - 1 2 ⇒ θ = 12 0 o From diagram, ∠ A = 60 o