Question

In $\Delta ABC$, if $AD$ is the altitude and $O$ is the orthocentre of $\Delta ABC$ then $AO:OD=$

• A. $\tan A:\tan B+\tan C$
• B. $\tan B+\tan C:\tan A$
• C. $\cos A:\cos B+\cos C$
• D. $\cot A:\cot B+\cot C$

Answer 1

The correct option is B $\tan B+\tan C:\tan A$


We know,
$AO=2R\cos A,OD=2R\cos B\cos C$
$\therefore AO:OD=\cos A:\cos B\cos C$
$=\frac{1}{\cos B\cos C}:\frac{1}{\cos A}$
$=\frac{\sin A}{\cos B\cos C}:\frac{\sin A}{\cos A}=\frac{\sin (B+C)}{\cos B\cos C}:\frac{\sin A}{\cos A}=\tan B+\tan C:\tan A$