$\cos \theta=\frac{Q-P}{Q}$ $\cos \theta=1-\frac{P}{Q}$ $1-2 \sin ^{2} \frac{\theta}{2}=1-\frac{P}{Q}$ $\therefore 2 \sin ^{2} \frac{\theta}{2}=\fr

$2{\sin ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)^{\frac{1}{2}}$

$\cos \theta=\frac{Q-P}{Q}$ $\cos \theta=1-\frac{P}{Q}$ $1-2 \sin ^{2} \frac{\theta}{2}=1-\frac{P}{Q}$ $\therefore 2 \sin ^{2} \frac{\theta}{2}=\frac{P}{Q}$ $\sin ^{2} \frac{\theta}{2}=\frac{P}{2 Q}$ $\Rightarrow \sin \frac{\theta}{2}=\left(\frac{P}{2 Q}\right)^{1 / 2}$ $\Rightarrow \theta=2 \sin ^{-1}\left(\frac{P}{2 Q}\right)^{1 / 2}$

English

3-1.VectorsDifficult

Two forces of magnitude $P$ & $Q$ acting at a point have resultant $R$. The resolved part of $R$ in the direction of $P$ is of magnitude $Q$. Angle between the forces is :

A
$2{\sin ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)$
B
$2{\sin ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)^{\frac{1}{2}}$
C
$2{\cos ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)$
D
$2{\cos ^{ - 1}}\left( {\frac{P}{{2Q}}} \right)^{\frac{1}{2}}$

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