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If x, y, z are perpendiculars drawn from the vertices of a triangle having sides a, b and c, then bx/c + cy/a + az/b is equal to

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If x, y, z are perpendiculars drawn from the vertices of a triangle having sides a, b and c, then bx/c + cy/a + az/b  is equal to

(A)  (a2 + b2 + c2)/2R

(B)  (a2 + b2 + c2)/R

(C)  (a2 + b2 + c2)/4R

(A)  2(a2 + b2 + c2)/R

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Correct option  (A)  a2 + b2 + c2/2R

 

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