(a) A regular polygon of n sides is such that each interior angle is 120° greater than the exterior angle. Find :
(i) the value of n ; (ii) the sum of all the interior angles.
(b) A boy walks 6km from a point P to a point Q on a bearing of 065°. He then walks to a point R, a distance of 13km, on a bearing of 146°.
(i) Sketch the diagram of his movement. (ii) Calculate, correct to the nearest kilometre, the distance PR.
Explanation
(a) (i) Let the exterior angle be x
Let the interior angle be x + 120°.
Exterior angle + Interior angle = 180°
(therefore x + x + 120 = 180 implies 2x + 120 = 180)
(2x = 60 implies x = 30°)
Number of sides = (frac{360}{text{exterior angle}})
= (frac{360}{30} = 12 sides)
(ii) Sum of interior angle = ((2n – 4) times 90°)
= ((2(12) – 4) times 90°)
= (20 times 90)
= (1800°)
(b) Note : < PQR = 65° + 34° = 99°)
(|PR|^{2} = |PQ|^{2} + |QR|^{2} – 2(|QR|)(|PQ|) cos 99)
(|PR|^{2} = 6^{2} + 13^{2} – 2(6)(13) cos 99)
(|PR|^{2} = 205 – 156 cos 99)
(|PR|^{2} = 205 + 24.40 = 229.40)
(|PR| = sqrt{229.40})
= (15.146 km)
(approxeq 15 km)