PSLE EFGH is a rhombus. HKF and HJGL are straight lines.
- Find ∠EFH.
- Find ∠HGL.
- Find ∠HKJ.
(a)
∠EFH
= (180° - 104°) ÷ 2
= 38° (Isosceles triangle)
(b)
∠FGH
= ∠FEH
= 104° (Rhombus)
∠HGL
= 180° - 104°
= 76° (Angles on a straight line)
(c)
∠JFK
= ∠EFK
= 38° (Rhombus)
∠FJK
= 180° - 68°
= 112° (Angles on a straight line)
∠HKJ
= 38° + 112°
= 150° (Exterior angle of a triangle)
Answer(s): (a) 38° (b) 76° (c) 150°