In the figure, BEHL is a rectangle, HJLM is a rhombus and EFGH is a parallelogram. ∠HJL = 76° and ∠GHJ = 96°.
- Find ∠DHE
- Find ∠HGF
(a)
∠LMH = ∠LJH = 76°
∠MHL
= (180° - 76°) ÷ 2
= 104° ÷ 2
= 52° (Isosceles triangle)
∠DHE
= 90° - 52°
= 38°
(b)
∠EHG
= 360° - 96° - 90° - 52°
= 122° (Angles at a point)
∠FGH
= 180° - 122°
= 58° (Interior Angles, EH//FF)
Answer(s): (a) 38°; (b) 58°