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

Answer: (B)

If $O A C B$ be a parallelogram. Then,

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