In the figure, BEHL is a rectangle, HJLM is a rhombus and EFGH is a parallelogram. ∠HJL = 74° and ∠GHJ = 91°.
- Find ∠DHE
- Find ∠HGF
(a)
∠LMH = ∠LJH = 74°
∠MHL
= (180° - 74°) ÷ 2
= 106° ÷ 2
= 53° (Isosceles triangle)
∠DHE
= 90° - 53°
= 37°
(b)
∠EHG
= 360° - 91° - 90° - 53°
= 126° (Angles at a point)
∠FGH
= 180° - 126°
= 54° (Interior Angles, EH//FF)
Answer(s): (a) 37°; (b) 54°