PSLEIn the figure, FGHJ is a straight line, GJKL is a parallelogram and KH = KL. ∠LFG is a right angle, ∠FLG = 38° and ∠JKH = 20°.
- Find ∠m.
- Find ∠n.
(a)
∠LFG = 90° (Right angle)
∠m
= 90° + 38°
= 128° (Exterior angle of a triangle)
(b)
∠JKL = ∠JGL = 128° (Parallelogram)
∠HKL
= 128° - 20°
= 108°
∠n
= (180° - 108°) ÷ 2
= 36° (Isosceles triangle)
Answer(s): (a) 128°; (b) 36°