Question

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

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

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\tan \theta =3\tan B$
B $\tan \theta =3\tan C$
C $\tan A=\frac{6\tan \theta }{{\tan }^2\theta -9}$
D $9{\cot }^2\frac{A}{2}={\tan }^2\theta$