PSLE DEFG and EGHJ are rhombuses. FKG is a straight line.
- Find ∠GEK.
- Find ∠FGH.
(a)
∠EGF
= (180° - 96°) ÷2
= 84° ÷ 2
= 42° (Isosceles triangle)
∠GEK
= 180° - 131° - 42°
= 7° (Angles sum of triangle)
(b)
∠EJH
= 360° - 303°
= 57° (Angles at a point)
∠EGH = ∠EJH = 57° (Rhombus)
∠FGH
= 57° - 42°
= 15°
Answer(s): (a) 7°; (b) 15°