The figure is not drawn to scale. It shows a parallelogram GHJK and an isosceles triangle DGK next to it. ∠FKD = 12°. FE and DC are straight lines. Find
- ∠n
- ∠l + ∠m
- ∠m
(a)
∠n
= 180° - ∠FKD
= 180° - 12°
= 168° (Angles on a straight line)
(b)
∠l = ∠GKJ (Parallelogram)
∠l + ∠m
= 180° - 12° - 60°
= 108° (Angles on a straight line)
(c)
∠DGK = ∠l (Corresponding angles)
∠DGK = ∠KDG (Isosceles triangle)
∠l
= 180° - (∠l + ∠m)
= 180° - 108°
= 72°
∠m
= 180° - ∠l - ∠l
= 180° - 72° - 72°
= 36° (Angles sum of triangle)
Answer(s): (a) 168°; (b) 108°; (c) 36°
