The figure is not drawn to scale. It shows a parallelogram FGHJ and an isosceles triangle CFJ next to it. ∠EJC = 12°. ED and CB are straight lines. Find
- ∠m
- ∠k + ∠l
- ∠l
(a)
∠m
= 180° - ∠EJC
= 180° - 12°
= 168° (Angles on a straight line)
(b)
∠k = ∠FJH (Parallelogram)
∠k + ∠l
= 180° - 12° - 57°
= 111° (Angles on a straight line)
(c)
∠CFJ = ∠k (Corresponding angles)
∠CFJ = ∠JCF (Isosceles triangle)
∠k
= 180° - (∠k + ∠l)
= 180° - 111°
= 69°
∠l
= 180° - ∠k - ∠k
= 180° - 69° - 69°
= 42° (Angles sum of triangle)
Answer(s): (a) 168°; (b) 111°; (c) 42°