In the figure, EFG and GHJ are isosceles triangles where EG = FG and GH = GJ. Given that HK, JM and EP are straight lines, find
- ∠LQP
- ∠JKQ
(a)
∠EFG = ∠FEG = ∠GHJ = ∠GJH (Isosceles triangle)
∠GJH = ∠KJQ = 36° (Verticallly opposite angles)
∠MJQ
= 36° - 12°
= 24°
∠JLQ
= 180° - 100°
= 80° (Angles on a straight line)
∠LQP
= 80° + 24°
= 104° (Exterior angle of a triangle)
(b)
∠JKQ
= 104° - 36°
= 68° (Exterior angle of a triangle)
Answer(s): (a) 104°; (b) 68°