The figure is not drawn to scale. It shows a parallelogram JKLM and an isosceles triangle FJM next to it. ∠HMF = 14°. HG and FE are straight lines. Find
- ∠q
- ∠n + ∠p
- ∠p
(a)
∠q
= 180° - ∠HMF
= 180° - 14°
= 166° (Angles on a straight line)
(b)
∠n = ∠JML (Parallelogram)
∠n + ∠p
= 180° - 14° - 56°
= 110° (Angles on a straight line)
(c)
∠FJM = ∠n (Corresponding angles)
∠FJM = ∠MFJ (Isosceles triangle)
∠n
= 180° - (∠n + ∠p)
= 180° - 110°
= 70°
∠p
= 180° - ∠n - ∠n
= 180° - 70° - 70°
= 40° (Angles sum of triangle)
Answer(s): (a) 166°; (b) 110°; (c) 40°