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 = 47° (Verticallly opposite angles)
∠KGN
= 47° - 13°
= 34°
∠GJN
= 180° - 108°
= 72° (Angles on a straight line)
∠JNM
= 72° + 34°
= 106° (Exterior angle of a triangle)
(b)
∠GHN
= 106° - 47°
= 59° (Exterior angle of a triangle)
Answer(s): (a) 106°; (b) 59°