The figure is not drawn to scale. It shows a parallelogram EFGH and an isosceles triangle BEH next to it. ∠DHB = 8°. DC and BA are straight lines. Find
- ∠l
- ∠j + ∠k
- ∠k
(a)
∠l
= 180° - ∠DHB
= 180° - 8°
= 172° (Angles on a straight line)
(b)
∠j = ∠EHG (Parallelogram)
∠j + ∠k
= 180° - 8° - 57°
= 115° (Angles on a straight line)
(c)
∠BEH = ∠j (Corresponding angles)
∠BEH = ∠HBE (Isosceles triangle)
∠j
= 180° - (∠j + ∠k)
= 180° - 115°
= 65°
∠k
= 180° - ∠j - ∠j
= 180° - 65° - 65°
= 50° (Angles sum of triangle)
Answer(s): (a) 172°; (b) 115°; (c) 50°