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 = 38° (Verticallly opposite angles)
∠LHP
= 38° - 15°
= 23°
∠HKP
= 180° - 110°
= 70° (Angles on a straight line)
∠KPN
= 70° + 23°
= 93° (Exterior angle of a triangle)
(b)
∠HJP
= 93° - 38°
= 55° (Exterior angle of a triangle)
Answer(s): (a) 93°; (b) 55°