GHJK is a rhombus, DEK and HEJ are straight lines. If HE = EK, ∠EKJ = 17° and ∠FHJ = 36°, find
- ∠EJK
- ∠FDE
(a)
∠EJK
= 84° - 17°
= 67° (Exterior angle of a triangle)
(b)
∠HDE
= 84° - 36°
= 48° (Exterior angle of a triangle)
∠FDE
= 180° - 48°
= 132° (Angles on a straight line)
Answer(s): (a) 67°; (b) 132°