In the figure, HBDG and HBCF are parallelograms. Given that ∠DBE = 15°, ∠HGF = 73° and HFG is an isosceles triangle where HF = HG. Find
- ∠BED
- ∠HBE
(a)
∠BDG
= 180° - 73°
= 107° (Interior angles)
∠BED
= 180° - 107° - 15°
= 58° (Angles sum of triangle)
(b)
∠HBE
= ∠BED
= 58° (Alternate angles)
Answer(s): (a) 58°; (b) 58°