In the figure, BEHL is a rectangle, HJLM is a rhombus and EFGH is a parallelogram. ∠HJL = 72° and ∠GHJ = 96°.
- Find ∠DHE
- Find ∠HGF
(a)
∠LMH = ∠LJH = 72°
∠MHL
= (180° - 72°) ÷ 2
= 108° ÷ 2
= 54° (Isosceles triangle)
∠DHE
= 90° - 54°
= 36°
(b)
∠EHG
= 360° - 96° - 90° - 54°
= 120° (Angles at a point)
∠FGH
= 180° - 120°
= 60° (Interior Angles, EH//FF)
Answer(s): (a) 36°; (b) 60°