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 = 37° (Verticallly opposite angles)
∠KGN
= 37° - 12°
= 25°
∠GJN
= 180° - 99°
= 81° (Angles on a straight line)
∠JNM
= 81° + 25°
= 106° (Exterior angle of a triangle)
(b)
∠GHN
= 106° - 37°
= 69° (Exterior angle of a triangle)
Answer(s): (a) 106°; (b) 69°