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 = 46° (Verticallly opposite angles)
∠KGN
= 46° - 14°
= 32°
∠GJN
= 180° - 106°
= 74° (Angles on a straight line)
∠JNM
= 74° + 32°
= 106° (Exterior angle of a triangle)
(b)
∠GHN
= 106° - 46°
= 60° (Exterior angle of a triangle)
Answer(s): (a) 106°; (b) 60°