The figure is not drawn to scale. It shows a parallelogram GHJK and an isosceles triangle DGK next to it. ∠FKD = 14°. FE and DC are straight lines. Find
- ∠n
- ∠l + ∠m
- ∠m
(a)
∠n
= 180° - ∠FKD
= 180° - 14°
= 166° (Angles on a straight line)
(b)
∠l = ∠GKJ (Parallelogram)
∠l + ∠m
= 180° - 14° - 61°
= 105° (Angles on a straight line)
(c)
∠DGK = ∠l (Corresponding angles)
∠DGK = ∠KDG (Isosceles triangle)
∠l
= 180° - (∠l + ∠m)
= 180° - 105°
= 75°
∠m
= 180° - ∠l - ∠l
= 180° - 75° - 75°
= 30° (Angles sum of triangle)
Answer(s): (a) 166°; (b) 105°; (c) 30°