Question

The angle between two active forces $P+Q$ and $P-Q$ is $2\alpha$ . If their resultant make angle $\theta$ with bisector of angle. Then

• A. $Pcos\theta =Qcos\alpha$
• B. $Ptan\theta =Qtan\alpha$
• C. $Qcos\theta =Pcos\alpha$
• D. $Qtan\theta =Ptan\alpha$

Answer 1

Answer: (B)

If $OACB$ be a parallelogram. Then,

$A\cup B=\alpha =$
$\angle ABO=\alpha +\theta$
In $\Delta OAB$ (by sine rule)
$\frac{OA}{\sin (\alpha +\theta )}=\frac{AB}{\sin (\alpha -\theta )}$
$\Rightarrow \frac{P+Q}{\sin (\alpha +\theta )}=\frac{P-Q}{\sin (\alpha -\theta )}$
$\Rightarrow \frac{P+Q}{P-Q}=\frac{\sin (\alpha +\theta )}{\sin (\alpha -\theta )}$
$\Rightarrow \frac{P}{Q}=\frac{\sin (\alpha +\theta )+\sin (\alpha -\theta )}{\sin (\alpha +\theta )-\sin (\alpha -\theta )}$
$\Rightarrow \frac{P}{Q}=\frac{2\sin \alpha \cos \theta }{2\cos \alpha \sin \theta }$
$\Rightarrow P\tan \theta =Q\tan \alpha$