PSLEIn the figure, FGHJ is a straight line, GJKL is a parallelogram and KH = KL. ∠LFG is a right angle, ∠FLG = 27° and ∠JKH = 17°.
- Find ∠m.
- Find ∠n.
(a)
∠LFG = 90° (Right angle)
∠m
= 90° + 27°
= 117° (Exterior angle of a triangle)
(b)
∠JKL = ∠JGL = 117° (Parallelogram)
∠HKL
= 117° - 17°
= 100°
∠n
= (180° - 100°) ÷ 2
= 40° (Isosceles triangle)
Answer(s): (a) 117°; (b) 40°