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In the figure, PQ = RS, \(\angle QOR = 48\),  Arc PS = 12cm and  Arc QS = 7cm.
(a) Find \(\angle POR\).
(b) Find the length of Arc RQ.?

\(\begin{array}{|rcll|} \hline PQ = RS \\ \hline x+48^\circ &=& 48^\circ +\varphi \\ x &=& \varphi \qquad \text{so}\quad \text{Arc}\ PR = 7cm \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 7 = 2\pi r \dfrac{x}{360^\circ} \qquad \text{or} \quad \mathbf{x = \dfrac{7*360^\circ}{2\pi r}} \\\\ 12 = 2\pi r \dfrac{\theta}{360^\circ} \qquad \text{or} \quad \mathbf{\theta = \dfrac{12*360^\circ}{2\pi r}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{360^\circ} &=& \mathbf{x + 48^\circ+\varphi+\theta} \quad & | \quad \varphi= x \\\\ 360^\circ &=& x + 48^\circ+x+\theta \\ 312^\circ &=& 2x+\theta \quad | \quad \mathbf{x = \dfrac{7*360^\circ}{2\pi r}},\ \mathbf{\theta = \dfrac{12*360^\circ}{2\pi r}} \\\\ 312^\circ &=& 2*\dfrac{7*360^\circ}{2\pi r}+\dfrac{12*360^\circ}{2\pi r} \\\\ 312^\circ &=& \dfrac{14*360^\circ}{2\pi r}+\dfrac{12*360^\circ}{2\pi r} \\\\ 312^\circ &=& \dfrac{14*360^\circ+12*360^\circ}{2\pi r} \\\\ 312^\circ &=& \dfrac{26*360^\circ}{2\pi r} \\\\ 2\pi r &=& \dfrac{26*360^\circ}{312^\circ} \\\\ \mathbf{ 2\pi r } &=& \mathbf{30\ cm} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{7*360^\circ}{2\pi r}} \quad | \quad \mathbf{ 2\pi r=30 } \\\\ x &=& \dfrac{7*360^\circ}{30} \\\\ \mathbf{x} &=& \mathbf{84\ cm} \qquad \angle POR = 84\ cm \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{Arc}\ RQ &=& 2\pi r \dfrac{48^\circ}{360^\circ} \quad | \quad \mathbf{ 2\pi r=30 } \\\\ \text{Arc}\ RQ &=& \dfrac{30*48^\circ}{360^\circ} \\\\ \mathbf{\text{Arc}\ RQ} &=& \mathbf{ 4\ cm } \\ \hline \end{array}\)