In the figure, EFGH is a parallelogram and JG = JH.
- Find ∠k.
- Find ∠l.
(a)
∠EJF = ∠EFJ (Isosceles triangle)
∠k
= 180° - 45° - 45° - 68°
= 22° (Interior angles)
(b)
∠EJF
= 180° - 45° - 45°
= 90° (Angles sum of triangle)
∠l
= 360° - 90°
= 270° (Angles at a point)
Answer(s): (a) 22°; (b) 270°