Question

In a triangle $ABC$,$\angle A,\angle B,\angle C$ are acute, the distance of the orthocentre from the sides are in proportional to

• A. $\sin A:\sin B:\sin C$
• B. $\tan A:\tan B:\tan C$
• C. $\sec A:\sec B:\sec C$
• D. $\cos A:\cos B:\cos C$

Answer 1

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

In $△ABE$ , $\cos A=\frac{AE}{AB}\Rightarrow AE=c\cos A$

In $△AHE$ , $\tan HAE=\frac{HE}{AE}\Rightarrow HE=AE\ast \tan HAE$

Therefore
$HE=c\cos A\tan (\frac{\pi }{2}-C)=c\cos A\cot C$

Using $\frac{c}{\sin C}=2R$

We get
$HE=\frac{2R\sin C\cos A\cos C}{\sin C}=2R\cos A\cos C=\frac{2R\cos A\cos B\cos C}{\cos B}$

Similarly,
$HD=\frac{2R\cos A\cos B\cos C}{\cos A},HF=\frac{2R\cos A\cos B\cos C}{\cos C}$

Hence,the distance from $a,b,c$ is in proportion

$\frac{1}{\cos A},\frac{1}{\cos B},\frac{1}{\cos C}$