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 = 47° (Verticallly opposite angles)
∠LHP
= 47° - 15°
= 32°
∠HKP
= 180° - 103°
= 77° (Angles on a straight line)
∠KPN
= 77° + 32°
= 109° (Exterior angle of a triangle)
(b)
∠HJP
= 109° - 47°
= 62° (Exterior angle of a triangle)
Answer(s): (a) 109°; (b) 62°