If (0,0) is orthocentre of triangle formed by A(cos⁡α,sin⁡α) , B(cos⁡β,sin⁡β) and C(cos⁡γ,sin⁡γ) then ∠BAC is

Points A(cos⁡α,sin⁡α),B(cos⁡β,sin⁡β),C(cos⁡γ,sin⁡γ) are equidistant from origin O(0, 0). OA=sin2α+cos2α=1 similarly OB=1, OC=1Thus, circumcenter is origin. Also orthocenter is origin. circumcenter and orthocenter are coincident,  So triangle is equilateral∴ ∠BAC=60∘