If x, y, z are the distances of the vertices of triangle ABC from its orthocenter, then x+y+z = .
A
r+R
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B
2(r+R)
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C
2r+R
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D
r+2R
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Solution
The correct option is B 2(r+R) We have,
x+y+z = 2R cos A +2R cos B +2R cos C = 2R(cos A + cos B +cos C) =2R{2cosA+B2cosA−B2+1−2sin2C2}=2R{2sinC2cosA−B2−2sin2C2}+2R=4RsinC2{cosA−B2−cosA+B2}+2R=8RsinA2sinB2sinC2+2R=2r+2R=2(r+R)