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3-1.Vectors
hard
Two forces $P + Q$ and $P -Q$ make angle $2 \alpha$ with each other and their resultant make $\theta$ angle with bisector of angle between them. Then :
A
$P\, tan\, \theta = Q\, tan \, \alpha$
B
$P\, sin\, \theta = Q\, sin\, \alpha$
C
$P\, cos\, \alpha = Q\, sin\, \theta$
D
$P\, sin\, \alpha = Q\, sin\, \theta$
Solution
As shown in figure resolve vector of magnitude $(P+Q)$ and $(P-Q)$ along line of resultant and perpendicular to it then sine components cancel each other.
$(P+Q) \sin (\alpha-\theta)=(P-Q) \sin (\alpha+\theta)$
$\mathrm{P}[\sin (\alpha+\theta)-\sin (\alpha-\theta)]=Q[\sin (\alpha-\theta)+\sin (\alpha+\theta)]$
$\mathrm{P} .2 \cos \alpha \sin \theta=\mathrm{Q} \cdot 2 \sin \alpha \cos \theta$
$\mathrm{P} \tan \theta=\theta \tan \alpha$
Standard 11
Physics
