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 = 41° (Verticallly opposite angles)
∠MJQ
= 41° - 13°
= 28°
∠JLQ
= 180° - 109°
= 71° (Angles on a straight line)
∠LQP
= 71° + 28°
= 99° (Exterior angle of a triangle)
(b)
∠JKQ
= 99° - 41°
= 58° (Exterior angle of a triangle)
Answer(s): (a) 99°; (b) 58°