The figure is not drawn to scale. It shows a parallelogram EFGH and an isosceles triangle BEH next to it. ∠DHB = 9°. DC and BA are straight lines. Find
- ∠l
- ∠j + ∠k
- ∠k
(a)
∠l
= 180° - ∠DHB
= 180° - 9°
= 171° (Angles on a straight line)
(b)
∠j = ∠EHG (Parallelogram)
∠j + ∠k
= 180° - 9° - 63°
= 108° (Angles on a straight line)
(c)
∠BEH = ∠j (Corresponding angles)
∠BEH = ∠HBE (Isosceles triangle)
∠j
= 180° - (∠j + ∠k)
= 180° - 108°
= 72°
∠k
= 180° - ∠j - ∠j
= 180° - 72° - 72°
= 36° (Angles sum of triangle)
Answer(s): (a) 171°; (b) 108°; (c) 36°
