PSLE EFGH is a rhombus. HKF and HJGL are straight lines.
- Find ∠EFH.
- Find ∠HGL.
- Find ∠HKJ.
(a)
∠EFH
= (180° - 110°) ÷ 2
= 35° (Isosceles triangle)
(b)
∠FGH
= ∠FEH
= 110° (Rhombus)
∠HGL
= 180° - 110°
= 70° (Angles on a straight line)
(c)
∠JFK
= ∠EFK
= 35° (Rhombus)
∠FJK
= 180° - 66°
= 114° (Angles on a straight line)
∠HKJ
= 35° + 114°
= 149° (Exterior angle of a triangle)
Answer(s): (a) 35° (b) 70° (c) 149°