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 = 40° (Verticallly opposite angles)
∠MJQ
= 40° - 11°
= 29°
∠JLQ
= 180° - 107°
= 73° (Angles on a straight line)
∠LQP
= 73° + 29°
= 102° (Exterior angle of a triangle)
(b)
∠JKQ
= 102° - 40°
= 62° (Exterior angle of a triangle)
Answer(s): (a) 102°; (b) 62°