The figure is not drawn to scale. It shows a parallelogram FGHJ and an isosceles triangle CFJ next to it. ∠EJC = 14°. ED and CB are straight lines. Find
- ∠m
- ∠k + ∠l
- ∠l
(a)
∠m
= 180° - ∠EJC
= 180° - 14°
= 166° (Angles on a straight line)
(b)
∠k = ∠FJH (Parallelogram)
∠k + ∠l
= 180° - 14° - 58°
= 108° (Angles on a straight line)
(c)
∠CFJ = ∠k (Corresponding angles)
∠CFJ = ∠JCF (Isosceles triangle)
∠k
= 180° - (∠k + ∠l)
= 180° - 108°
= 72°
∠l
= 180° - ∠k - ∠k
= 180° - 72° - 72°
= 36° (Angles sum of triangle)
Answer(s): (a) 166°; (b) 108°; (c) 36°