A small smooth ring $R$ of weight $8.5{\text{ }}N$ 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 $15.5{\text{ }}N$ acts on $R$ so that the ring is in equilibrium with angle $ARB = {90^ \circ }$. The part $AR$ of the string makes an angle $\theta $ with the horizontal and the part $BR$ makes an angle $\theta $ with the vertical (see diagram).
The tension in the string is $T{\text{ }}N$. Show that $T\sin \theta = 12$ and $T\cos \theta = 3.5$ and hence find $\theta $.

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