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 = 37° (Verticallly opposite angles)
∠LHP
= 37° - 13°
= 24°
∠HKP
= 180° - 108°
= 72° (Angles on a straight line)
∠KPN
= 72° + 24°
= 96° (Exterior angle of a triangle)
(b)
∠HJP
= 96° - 37°
= 59° (Exterior angle of a triangle)
Answer(s): (a) 96°; (b) 59°
