JKLM is a rhombus, FGM and KGL are straight lines. If KG = GM, ∠GML = 16° and ∠HKL = 29°, find
- ∠GLM
- ∠HFG
(a)
∠GLM
= 81° - 16°
= 65° (Exterior angle of a triangle)
(b)
∠KFG
= 81° - 29°
= 52° (Exterior angle of a triangle)
∠HFG
= 180° - 52°
= 128° (Angles on a straight line)
Answer(s): (a) 65°; (b) 128°