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. $P\cos \theta =Q\cos \alpha$
• B. $P\tan \theta =Q\tan \alpha$
• C. $Q\cos \theta =P\cos \alpha$
• D. $Q\tan \theta =P\tan \alpha$

Answer 1

The correct option is B $P\tan \theta =Q\tan \alpha$

If $OACB$ be a parallelogram. Then,
$\angle AOB=\alpha -\theta$
$\angle ABO=\alpha +\theta$$In$\triangle OAB$\left(bysinerule\right)$\dfrac {OA}{\sin (\alpha + \theta)} = \dfrac {AB}{\sin (\alpha - \theta)}\Rightarrow \dfrac {P + Q}{\sin (\alpha + \theta)} = \dfrac {P - Q}{\sin (\alpha - \theta)}\Rightarrow \dfrac {P + Q}{P - Q} = \dfrac {\sin (\alpha + \theta)}{\sin (\alpha - \theta)}\Rightarrow \dfrac {P}{Q} = \dfrac {\sin (\alpha + \theta) + \sin (\alpha - \theta)}{\sin (\alpha + \theta) - \sin (\alpha - \theta)}\Rightarrow \dfrac {P}{Q} = \dfrac {2\sin \alpha \cos \theta}{2\cos \alpha \sin \theta}\Rightarrow P\tan \theta = Q\tan \alpha$.