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$P, Q$ and $R$ are three coplanar forces acting at a point and are in equilibrium. Given $P=1.931 .8$ $\mathrm{kg} \omega t, \sin \theta_1=0.9659$, the value of $R$ is (in $\mathrm{kg} w t$ )
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$\frac{P}{\sin \theta_1}=\frac{Q}{\sin \theta_2}=\frac{R}{\sin 150^{\circ}}$
$\Rightarrow \frac{1.93}{\sin \theta_1}=\frac{R}{\sin 150^{\circ}}$
$\Rightarrow R=\frac{1.93 \times \sin 150^{\circ}}{\sin \theta_1}=\frac{1.93 \times 0.5}{0.9659}=1$
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