Question
The ratio of the distances of the orthocentre of an acute-angled $$\vartriangle ABC$$ from the sides $$BC, AC$$ and $$AB$$ is
A.
$$\cos A:\cos B:\cos C$$
B.
$$\sin A:\sin B:\sin C$$
C.
$$\sec A:\sec B:\sec C$$
D.
None of these
Answer :
$$\sec A:\sec B:\sec C$$
Solution :
$$\eqalign{ & HD = BD \cdot \tan \angle EBC = c\cos B \cdot \tan \left( {{{90}^ \circ } - C} \right) \cr & HD = \frac{{c\cos B \cdot \cos C}}{{\sin C}} = 2R\cos B\cos C \cr & HD = \frac{{2R\cos A\cos B\cos C}}{{\cos A}}. \cr} $$
Similarly for others. So, the ratio of the distances of the orthocentre from the sides $$ = \frac{1}{{\cos A}}:\frac{1}{{\cos B}}:\frac{1}{{\cos C}}.$$
$$\eqalign{ & HD = BD \cdot \tan \angle EBC = c\cos B \cdot \tan \left( {{{90}^ \circ } - C} \right) \cr & HD = \frac{{c\cos B \cdot \cos C}}{{\sin C}} = 2R\cos B\cos C \cr & HD = \frac{{2R\cos A\cos B\cos C}}{{\cos A}}. \cr} $$
Similarly for others. So, the ratio of the distances of the orthocentre from the sides $$ = \frac{1}{{\cos A}}:\frac{1}{{\cos B}}:\frac{1}{{\cos C}}.$$