Part a
[tex]\angle POR=\frac{\pi}{2}[/tex] radians, so [tex]\theta=\frac{3}{7}(\pi/2)=\boxed{\frac{3\pi}{14}}[/tex]
Part b
The area of sector QOR is [tex]\frac{1}{2}(15^2)\left(\frac{3\pi}{14} \right)=\frac{675\pi}{28}[/tex]
The area of triangle POQ is [tex]\frac{1}{2}(15)(15) \sin \frac{2\pi}{7}=\frac{225}{2} \sin \frac{2\pi}{7}[/tex]
So, the total area, in square centimeters, is [tex]\frac{675\pi}{28}+\frac{225}{2} \sin \frac{2\pi}{7}[/tex]
