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° - 12°
= 25°
∠HKP
= 180° - 98°
= 82° (Angles on a straight line)
∠KPN
= 82° + 25°
= 107° (Exterior angle of a triangle)
(b)
∠HJP
= 107° - 37°
= 70° (Exterior angle of a triangle)
Answer(s): (a) 107°; (b) 70°