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

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a

60°

b

30°

c

45°

d

221°2

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detailed solution

Correct option is A

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∘

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If an triangle $ABC, A \equiv (1, 10),$ circumcentre $\equiv (- 1 / 3, 2 / 3),$ and orthocentre $\equiv (- 11 / 3, 4 / 3),$ then the coordinates of the midpoint of the side opposite to A are

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