In the figure, DEF and FGH are isosceles triangles where DF = EF and FG = FH. Given that GJ, HL and DN are straight lines, find
- ∠KPN
- ∠HJP
(a)
∠DEF = ∠EDF = ∠FGH = ∠FHG (Isosceles triangle)
∠FHG = ∠JHP = 42° (Verticallly opposite angles)
∠LHP
= 42° - 14°
= 28°
∠HKP
= 180° - 103°
= 77° (Angles on a straight line)
∠KPN
= 77° + 28°
= 105° (Exterior angle of a triangle)
(b)
∠HJP
= 105° - 42°
= 63° (Exterior angle of a triangle)
Answer(s): (a) 105°; (b) 63°