In any triangle $A B C$ , the distance of the orthocentre from the vertices $A, B, C$ are in the ratio

sin A : sin B : sin C

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cos A : cos B : cos C

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tan A : tan B : tan C

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none of these.

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Solution

The correct option is D $cos A : cos B : cos C$
$∵ O A = 2 R cos A$
$O B = 2 R cos B$
and $O C = 2 R cos C$
$∴ O A : O B : O C = 2 R cos A : 2 R cos B : 2 R cos C$
$= cos A : cos B : cos C$

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In any triangle ABC, the distance of the orthocentre from the vertices A,B,C are in the ratio

The correct option is D cosA:cosB:cosC∵OA=2RcosAOB=2RcosBand OC=2RcosC∴OA:OB:OC=2RcosA:2RcosB:2RcosC =cosA:cosB:cosC

In any triangle $A B C$ , the distance of the orthocentre from the vertices $A, B, C$ are in the ratio

The correct option is D $cos A : cos B : cos C$
$∵ O A = 2 R cos A$
$O B = 2 R cos B$
and $O C = 2 R cos C$
$∴ O A : O B : O C = 2 R cos A : 2 R cos B : 2 R cos C$
$= cos A : cos B : cos C$