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 = 47° (Verticallly opposite angles)
∠MJQ
= 47° - 15°
= 32°
∠JLQ
= 180° - 106°
= 74° (Angles on a straight line)
∠LQP
= 74° + 32°
= 106° (Exterior angle of a triangle)
(b)
∠JKQ
= 106° - 47°
= 59° (Exterior angle of a triangle)
Answer(s): (a) 106°; (b) 59°
