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 = 45° (Verticallly opposite angles)
∠KGN
= 45° - 12°
= 33°
∠GJN
= 180° - 97°
= 83° (Angles on a straight line)
∠JNM
= 83° + 33°
= 116° (Exterior angle of a triangle)
(b)
∠GHN
= 116° - 45°
= 71° (Exterior angle of a triangle)
Answer(s): (a) 116°; (b) 71°