PSLE EFGH is a rhombus. HKF and HJGL are straight lines.
- Find ∠EFH.
- Find ∠HGL.
- Find ∠HKJ.
(a)
∠EFH
= (180° - 106°) ÷ 2
= 37° (Isosceles triangle)
(b)
∠FGH
= ∠FEH
= 106° (Rhombus)
∠HGL
= 180° - 106°
= 74° (Angles on a straight line)
(c)
∠JFK
= ∠EFK
= 37° (Rhombus)
∠FJK
= 180° - 61°
= 119° (Angles on a straight line)
∠HKJ
= 37° + 119°
= 156° (Exterior angle of a triangle)
Answer(s): (a) 37° (b) 74° (c) 156°