The figure is not drawn to scale. ∠KJL = 41°, KJ = KL and JP//KN. Given that ∠e is
13 of ∠c and ∠e is 4 times of ∠d, find
- ∠e
- ∠b
(a)
e : c = 1 : 3
c = 3 e
d : e = 1 : 4
c = 3 x 4= 12
c : d : e = 12 : 1 : 4
∠KLJ = 41° (Isosceles triangle)
∠JLM
= 180° - 41°
= 139° (Angles on a straight line)
1 u + 4 u + 12 u + 139° = 360° (Sum of angles in a quadrilateral)
17 u = 360° - 139° = 221°
1 u = 221 ÷ 17 = 13°
∠e
= 4 x 1 u
= 4 x 13°
= 52°
(b)
∠KLJ = ∠KJL = 41° (Isosceles triangle)
∠LQN = ∠LJP = 52° (Corresponding angles, JP//KN)
∠b
= 52° - 41°
= 11° (Exterior angle of a triangle)
Answer(s): (a) 52°; (b) 11°
