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 = 38° (Verticallly opposite angles)
∠KGN
= 38° - 10°
= 28°
∠GJN
= 180° - 104°
= 76° (Angles on a straight line)
∠JNM
= 76° + 28°
= 104° (Exterior angle of a triangle)
(b)
∠GHN
= 104° - 38°
= 66° (Exterior angle of a triangle)
Answer(s): (a) 104°; (b) 66°