2014-II-21

Ans: A
Since $AB=CD=EF$, then the distances between the centre $O$ and the three chords $AB$, $CD$ and $EF$ are the same. Hence, we can draw a circle inscribed in $\mathrm{\Delta }PQR$ as shown in the following figure.

Note that $PQ$, $QR$ and $PR$ are tangents to the smaller circle. Then by the tangent properties, we have

$\begin{array}{rcl}\mathrm{\angle }QOP& =& \mathrm{\angle }OQR\\ \mathrm{\angle }ORQ& =& \mathrm{\angle }ORP\end{array}$

Consider $\mathrm{\Delta }PQR$, we have

$\begin{array}{rcl}2\alpha +2\beta +{38}^{\circ }& =& {180}^{\circ }\\ \alpha +\beta & =& {71}^{\circ }\end{array}$

Consider $\mathrm{\Delta }ORQ$, we have

$\begin{array}{rcl}\mathrm{\angle }QOR& =& {180}^{\circ }–\beta –\alpha \\ \mathrm{\angle }QOR& =& {180}^{\circ }–(\beta +\alpha )\\ \mathrm{\angle }QOR& =& {180}^{\circ }–{71}^{\circ }\\ \mathrm{\angle }QOR& =& {109}^{\circ }\end{array}$