2003 - WAEC Mathematics Past Questions and Answers - page 3

< QPS = < PRS = 65° (angles in the same segment)

< PSR + 40° + 65° = 180°

< PSR + 105° = 180°

< PSR = 75°

< PSR = < PSQ + < QSR

75° = < PSQ + 20° \(\implies\) < PSQ = 75° - 20° = 55°

\((111_{2})^2 - (101_{2})^2\)

Difference of two squares

\((111 - 101)(111 + 101)\)

= \((10)(1100)\)

= \(11000_{2}\)

Using the formula, \((n - 2) \times 180°\) to get the sum of the interior angles. Then we can have

\((n - 2) \times 180° = 108n\) ... (1)

\((n - 2) \times 180° = 116n\) ... (2)

\((n - 2) \times 180° = 120n\) ... (3)

Solving the above given equations, where n is not a positive integer then that angle is not the interior for a regular polygon.

(1): \(180n - 360 = 108n \implies 72n = 360\)

 \(n = 5\) (regular pentagon)

(2): \(180n - 360 = 116n \implies 64n = 360\)

 \(n = 5.625\)

(3): \(180n - 360 = 120n \implies 60n = 360\)

 \(n = 6\) (regular hexagon)

Hence, 116° is not an angle of a regular polygon.