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 = 39° (Verticallly opposite angles)
∠MJQ
= 39° - 12°
= 27°
∠JLQ
= 180° - 97°
= 83° (Angles on a straight line)
∠LQP
= 83° + 27°
= 110° (Exterior angle of a triangle)
(b)
∠JKQ
= 110° - 39°
= 71° (Exterior angle of a triangle)
Answer(s): (a) 110°; (b) 71°