Description for Correct answer:


In In \( \Large \triangle ABD, \tan \alpha = \frac{h}{2h} \)

=> \( \Large \tan \alpha = \frac{1}{2} \)

In \( \Large \triangle ABC, \tan 2\alpha = \frac{h+p}{2h} \)

=> \( \Large \frac{2 \tan \alpha }{1 - \tan^{2}a} = \frac{h+p}{2h} \)

=> \( \Large \frac{2 \left(\frac{1}{2}\right) }{1- \left(\frac{1}{2}\right)^{2} } = \frac{h+p}{2h} \)

=> \( \Large \frac{4}{3} = \frac{h+p}{2h} => 8h = 3h + 3p \)

=> \( \Large 5h = 3p =>p = \frac{5h}{3} m \)