In the figure, BEHL is a rectangle, HJLM is a rhombus and EFGH is a parallelogram. ∠HJL = 78° and ∠GHJ = 91°.
- Find ∠DHE
- Find ∠HGF
(a)
∠LMH = ∠LJH = 78°
∠MHL
= (180° - 78°) ÷ 2
= 102° ÷ 2
= 51° (Isosceles triangle)
∠DHE
= 90° - 51°
= 39°
(b)
∠EHG
= 360° - 91° - 90° - 51°
= 128° (Angles at a point)
∠FGH
= 180° - 128°
= 52° (Interior Angles, EH//FF)
Answer(s): (a) 39°; (b) 52°