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 = 43° (Verticallly opposite angles)
∠KGN
= 43° - 14°
= 29°
∠GJN
= 180° - 109°
= 71° (Angles on a straight line)
∠JNM
= 71° + 29°
= 100° (Exterior angle of a triangle)
(b)
∠GHN
= 100° - 43°
= 57° (Exterior angle of a triangle)
Answer(s): (a) 100°; (b) 57°