Question

If $AD$ is perpendicular to $BC$ of $\Delta ABC$ such that $BD:CD$: $AD=2:3:6$, then $A=$

• A. ${45}^0$
• B. ${60}^0$
• C. ${30}^0$
• D. ${90}^0$

Answer 1

Answer: (A)

${45}^0$

$BD:CD:AD=2:3:6$
$BD=2K$
$CD=3K$
$AD=6K$

$\tan B=\frac{6}{2},\tan C=\frac{AD}{CD}=\frac{6}{3}$

$\tan B=3,\tan C=2$
$A+B+C=\pi$
$A=\pi -(B+C)$
$\tan A=-\tan (B+C)$
$=-\frac{3+2}{1-(3\cdot 2)}$
$\tan A=1$
$A={45}^0$