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 = 39° (Verticallly opposite angles)
∠KGN
= 39° - 15°
= 24°
∠GJN
= 180° - 102°
= 78° (Angles on a straight line)
∠JNM
= 78° + 24°
= 102° (Exterior angle of a triangle)
(b)
∠GHN
= 102° - 39°
= 63° (Exterior angle of a triangle)
Answer(s): (a) 102°; (b) 63°