In the figure, CDE and EFG are isosceles triangles where CE = DE and EF = EG. Given that FH, GK and CM are straight lines, find
- ∠JNM
- ∠GHN
(a)
∠CDE = ∠DCE = ∠EFG = ∠EGF (Isosceles triangle)
∠EGF = ∠HGN = 36° (Verticallly opposite angles)
∠KGN
= 36° - 10°
= 26°
∠GJN
= 180° - 110°
= 70° (Angles on a straight line)
∠JNM
= 70° + 26°
= 96° (Exterior angle of a triangle)
(b)
∠GHN
= 96° - 36°
= 60° (Exterior angle of a triangle)
Answer(s): (a) 96°; (b) 60°