PSLE EFGH is a rhombus. HKF and HJGL are straight lines.
- Find ∠EFH.
- Find ∠HGL.
- Find ∠HKJ.
(a)
∠EFH
= (180° - 108°) ÷ 2
= 36° (Isosceles triangle)
(b)
∠FGH
= ∠FEH
= 108° (Rhombus)
∠HGL
= 180° - 108°
= 72° (Angles on a straight line)
(c)
∠JFK
= ∠EFK
= 36° (Rhombus)
∠FJK
= 180° - 63°
= 117° (Angles on a straight line)
∠HKJ
= 36° + 117°
= 153° (Exterior angle of a triangle)
Answer(s): (a) 36° (b) 72° (c) 153°