Question

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

Answer 1

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

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

The correct answer is A, B and C

ABC is a triangle, point D and E are taken on sides BC.

such that$BD=DE=EC$ and $\angle ADE=\angle ADE=\theta$

then,

using $m-n$

$3cot\theta =2cotB-cotC--------------\left(1\right)$

$3cot(\pi -\theta )=cotB-2cotC----------\left(2\right)$

from (1) and (2)

$cotB=cotC$---------(3)

using equation (1) & (3)

$\Rightarrow 3tanB=tan\theta$

similarlly equation (2)&(3)

$\Rightarrow 3tanC=tan\theta$

Draw a perpendicular line from A and midpoint of BC.

if is median of triangle ABC is isosceles

$B=90-\frac{A}{2}=ta{n}^2\theta$

now,

$tanA=\frac{2tan\frac{A}{2}}{1-ta{n}^2\frac{A}{2}}$

$=\frac{6tan\theta }{ta{n}^2\theta -9}$