The figure is not drawn to scale. It shows a parallelogram FGHJ and an isosceles triangle CFJ next to it. ∠EJC = 11°. ED and CB are straight lines. Find
- ∠m
- ∠k + ∠l
- ∠l
(a)
∠m
= 180° - ∠EJC
= 180° - 11°
= 169° (Angles on a straight line)
(b)
∠k = ∠FJH (Parallelogram)
∠k + ∠l
= 180° - 11° - 51°
= 118° (Angles on a straight line)
(c)
∠CFJ = ∠k (Corresponding angles)
∠CFJ = ∠JCF (Isosceles triangle)
∠k
= 180° - (∠k + ∠l)
= 180° - 118°
= 62°
∠l
= 180° - ∠k - ∠k
= 180° - 62° - 62°
= 56° (Angles sum of triangle)
Answer(s): (a) 169°; (b) 118°; (c) 56°
