P, Q and R are three coplanar forces acting a | vedclass

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Question

(C) According to Lami's Theorem,for three forces in equilibrium:
$\frac{P}{\sin 	heta_R} = \frac{Q}{\sin 	heta_P} = \frac{R}{\sin 	heta_Q}$
F

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(C) According to Lami's Theorem,for three forces in equilibrium:
$\frac{P}{\sin 	heta_R} = \frac{Q}{\sin 	heta_P} = \frac{R}{\sin 	heta_Q}$
From the figure,the angle opposite to $P$ is $	heta_1$,the angle opposite to $Q$ is $	heta_2$,and the angle opposite to $R$ is $150^\circ$.
Thus,$\frac{P}{\sin 	heta_1} = \frac{R}{\sin 150^\circ}$.
Given $P = 1.9318 \, kg \, wt$ and $\sin 	heta_1 = 0.9659$.
Substituting the values:
$\frac{1.9318}{0.9659} = \frac{R}{\sin 150^\circ}$
Since $\sin 150^\circ = \sin(180^\circ - 30^\circ) = \sin 30^\circ = 0.5$.
$2 = \frac{R}{0.5}$
$R = 2 	imes 0.5 = 1 \, kg \, wt$.

$P, Q$ and $R$ are three coplanar forces acting at a point and are in equilibrium. Given $P = 1.9318 \, kg \, wt$,$\sin {\theta _1} = 0.9659$,the value of $R$ is (in $kg \, wt$):

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