Question

The distance of the circumcentre of the acute angled $\angle ABC$ from the sides $BC,CA$ and $AB$ are in the ratio ?

• A. $a\sin A:b\sin B:c\sin C$
• B. $\cos A$$:\cos B$$:\cos C$
• C. $a\cot A:b\cot B:c\cot C$
• D. $noneofthese$

Answer 1

The correct option is B $\cos A$$:\cos B$$:\cos C$

$LetAD,BC,CFbethealtitudes$
$\angle HCD={90}^0-B$
$\angle DHC=B$
$\sec B=\frac{HC}{HD}.........\left(i\right)$
$in\Delta CDH$
$\angle HCF=90-A$
$\angle EHC=A$
$\sec A=\frac{HC}{HE}.......\left(ii\right)$
$from\left(i\right)and\left(ii\right)$
$HE=\sec HD=HE\sec A$
$\frac{HD}{HE}=\frac{\sec A}{\sec B}$
$\frac{HE}{HF}=\frac{\sec B}{\sec C}$
$\therefore HD:HE:HF=\sec A:\sec B:\sec C$
$=\cos A:\cos B:\cos C$