(a) The ratio of the interior angle to the exterior angle of a regular polygon is 5 : 2, Find the number of sides of the polygon.
(b) 
The diagram shows a circle PQRS with centre O, < UQR = 68°, < TPS = 74° and < QSR = 40°. Calculate the value of < PRS.
Explanation
(a) Let the diagram below represent a section of the polygon
Also, let :
\(i\) represent the size of an interior angle; \(e\) represent the size of an exterior angle.
Then \(\frac{i}{e} = \frac{5}{2}\)
\(\implies i = \frac{5}{2}e\)
But \(i + e = 180°\) (sum of angles on a straight line)
Substitute \(\frac{5}{2}e\) for \(i\) in the equation
\(\frac{5}{2}e + e = 180°\)
\(\frac{7}{2}e = 180°\)
\(e = \frac{180° \times 2}{7} = \frac{360°}{7}\)
Number of sides of the polygon = \(\frac{360°}{\text{size of one exterior angle}}\)
= \(360° \div \frac{360°}{7}\)
\(360° \times \frac{7}{360°} = 7 sides\).
(b) 
In the diagram above, < PSR = 68° (interior angle of a cyclic quad = opp exterior angle)
< PSQ = 68° - 40° = 28°
< PRQ = 28° (angles in same segment)
< SRQ = 74° (interior angle of a cyclic quad = opp exterior angle)
hence, < PRS = 74° - 28° = 46°