3 \includegraphics[max width=\textwidth, alt={}, center]{d5acfe31-8614-4508-ac5b-865e15a1f539-2_661_565_1069_790} A small smooth ring \(R\) of weight 8.5 N is threaded on a light inextensible string. The ends of the string are attached to fixed points \(A\) and \(B\), with \(A\) vertically above \(B\). A horizontal force of magnitude 15.5 N acts on \(R\) so that the ring is in equilibrium with angle \(A R B = 90 ^ { \circ }\). The part \(A R\) of the string makes an angle \(\theta\) with the horizontal and the part \(B R\) makes an angle \(\theta\) with the vertical (see diagram). The tension in the string is \(T \mathrm {~N}\). Show that \(T \sin \theta = 12\) and \(T \cos \theta = 3.5\) and hence find \(\theta\).

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\includegraphics[max width=\textwidth, alt={}, center]{d5acfe31-8614-4508-ac5b-865e15a1f539-2_661_565_1069_790}

A small smooth ring $R$ of weight 8.5 N is threaded on a light inextensible string. The ends of the string are attached to fixed points $A$ and $B$, with $A$ vertically above $B$. A horizontal force of magnitude 15.5 N acts on $R$ so that the ring is in equilibrium with angle $A R B = 90 ^ { \circ }$. The part $A R$ of the string makes an angle $\theta$ with the horizontal and the part $B R$ makes an angle $\theta$ with the vertical (see diagram). The tension in the string is $T \mathrm {~N}$. Show that $T \sin \theta = 12$ and $T \cos \theta = 3.5$ and hence find $\theta$.

\hfill \mbox{\textit{CAIE M1 2011 Q3 [6]}}