Circle

In the given figure, O is the center of the circle. If $\angle POR = 150^\circ$ and $\angle ORQ = 60^\circ$, find the measure of:

1. $\angle PQR$
2. $\angle OPR$
3. $\angle QRP$
4. $\angle RPQ$

1. Given: $\angle POR = 150^\circ$
   
   $\therefore \;$ Reflex $\angle POR = 360^\circ - 150^\circ = 210^\circ$
   
   Now, $\;$ $\angle PQR = \dfrac{1}{2} \times \text{ Reflex } \angle POR$
   
   [$\because \;$ angle at the center is twice the angle at the remaining circumference]
   
   $\therefore \;$ $\angle PQR = \dfrac{210^\circ}{2} = 105^\circ$
   
   
2. 
   Join $PR$.
   
   In $\triangle POR$, $\;$ $OP = OR = $ radii of the same circle
   
   $\therefore \;$ $\angle OPR = \angle ORP$ $\;\;\;$ [$\because \;$ angles opposite to equal sides are equal]
   
   Now, $\;$ $\angle POR + \angle OPR + \angle ORP = 180^\circ$ $\;\;\;$ [$\because \;$ sum of angles of a $\triangle$ equal $180^\circ$]
   
   i.e. $\;$ $150^\circ + 2 \angle OPR = 180^\circ$
   
   i.e. $\;$ $2 \angle OPR = 30^\circ$ $\implies$ $\angle OPR = 15^\circ$
   
   
3. $\because \;$ $\angle OPR = \angle ORP$ $\implies$ $\angle ORP = 15^\circ$
   
   From the figure, $\;$ $\angle ORQ = \angle ORP + \angle QRP$
   
   Given: $\;$ $\angle ORQ = 60^\circ$
   
   $\therefore \;$ We have, $\;$ $60^\circ = 15^\circ + \angle QRP$
   
   $\implies$ $\angle QRP = 60^\circ - 15^\circ = 45^\circ$
   
   
4. In $\triangle PQR$, $\;$ $\angle RPQ + \angle PQR + \angle QRP = 180^\circ$ $\;\;$ [$\because \;$ sum of angles of a $\triangle$ equal $180^\circ$]
   
   i.e. $\;$ $\angle RPQ + 105^\circ + 45^\circ = 180^\circ$
   
   $\implies$ $\angle RPQ = 30^\circ$