The figure is not drawn to scale. FHKL is a parallelogram, EFL and FGL are isosceles triangles. LK//NU and ∠KTU= 109°. EH is straight line.
- Find ∠ELF.
- Find ∠HGL.
(a)
∠KTU = 109°
∠LRS = 109° (Corresponding angles)
∠PRL
= 180° - 109°
= 71° (Angles on a straight line)
∠ELF
= 180° - 71° - 71°
= 38° (Isosceles triangle)
(b)
∠PRL = ∠EFL = 71° (Corresponding angles)
∠LFG
= 180° - 71°
= 109° (Angles on a straight line)
∠FGS
= (180° - 109°) ÷ 2
= 35.5° (Isosceles triangle)
∠HGL
= 180° - 35.5°
= 144.5° (Angles on a straight line)
Answer(s): (a) 38°; (b) 144.5°