Question

In a triangle ABC, points D and E are taken on sides BC such that $DB=DE=EC$. If $\angle ADE=\angle AED=\theta$, then

• A. $\tan \theta =3\tan B$
• B. $\tan \theta =3\tan C$
• C. $\tan A=\frac{6\tan \theta }{{\tan }^2\theta }$
• D. $9{\cot }^2\frac{A}{2}={\tan }^2\theta$

Answer 1

The correct option is A $\tan \theta =3\tan B$

$BD=DE-EC$

$\angle ADE=\angle AED=\theta$

$AD=DE$

$DF=\frac{1}{2}DE$

Let $BD=x$$AF=y$

$\tan \theta =\frac{2y}{x}$

$\tan B=\frac{AF}{BF}=\frac{2y}{3x}$

$\tan B=\frac{\tan \theta }{3}$

$\Rightarrow \tan \theta =3\tan B$