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 = 39° (Verticallly opposite angles)
∠LHP
= 39° - 10°
= 29°
∠HKP
= 180° - 97°
= 83° (Angles on a straight line)
∠KPN
= 83° + 29°
= 112° (Exterior angle of a triangle)
(b)
∠HJP
= 112° - 39°
= 73° (Exterior angle of a triangle)
Answer(s): (a) 112°; (b) 73°