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° - 15°
= 22°
∠GJN
= 180° - 103°
= 77° (Angles on a straight line)
∠JNM
= 77° + 22°
= 99° (Exterior angle of a triangle)
(b)
∠GHN
= 99° - 37°
= 62° (Exterior angle of a triangle)
Answer(s): (a) 99°; (b) 62°