Practicing Success
Practicing Success
CUET Preparation Today
If H is orthocenter of the triangle ABC, then AH is equal to: |
$a \cot A$ $a \cot B$ $b \cot A$ $c \cot A$ |
$a \cot A$ |
from $ΔAHB$ $\frac{AH}{\sin(90-A)}=\frac{AB}{\sin(A+B)}=\frac{BH}{sin(90-B)}$ $⇒\frac{AH}{\sin(90-A)}=\frac{AB}{\sin(A+B)}=\frac{C}{sin(180-(A+B))}$ $⇒\frac{AH}{\cos A}=\frac{c}{\sin C}=\frac{a}{\sin A}$ $⇒AH=a\cot A$ |
