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 = 42° (Verticallly opposite angles)
∠KGN
= 42° - 11°
= 31°
∠GJN
= 180° - 99°
= 81° (Angles on a straight line)
∠JNM
= 81° + 31°
= 112° (Exterior angle of a triangle)
(b)
∠GHN
= 112° - 42°
= 70° (Exterior angle of a triangle)
Answer(s): (a) 112°; (b) 70°
