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 = 46° (Verticallly opposite angles)
∠MJQ
= 46° - 13°
= 33°
∠JLQ
= 180° - 107°
= 73° (Angles on a straight line)
∠LQP
= 73° + 33°
= 106° (Exterior angle of a triangle)
(b)
∠JKQ
= 106° - 46°
= 60° (Exterior angle of a triangle)
Answer(s): (a) 106°; (b) 60°