The figure is not drawn to scale. ∠KJL = 41°, KJ = KL and JP//KN. Given that ∠c is
12 of ∠a and ∠c is 4 times of ∠b, find
- ∠c
- ∠z
(a)
c : a = 1 : 2
a = 2 c
b : c = 1 : 4
a = 2 x 4= 8
a : b : c = 8 : 1 : 4
∠KLJ = 41° (Isosceles triangle)
∠JLM
= 180° - 41°
= 139° (Angles on a straight line)
1 u + 4 u + 8 u + 139° = 360° (Sum of angles in a quadrilateral)
13 u = 360° - 139° = 221°
1 u = 221 ÷ 13 = 17°
∠c
= 4 x 1 u
= 4 x 17°
= 68°
(b)
∠KLJ = ∠KJL = 41° (Isosceles triangle)
∠LQN = ∠LJP = 68° (Corresponding angles, JP//KN)
∠z
= 68° - 41°
= 27° (Exterior angle of a triangle)
Answer(s): (a) 68°; (b) 27°