The figure is not drawn to scale. FHKL is a parallelogram, EFL and FGL are isosceles triangles. LK//NU and ∠KTU= 110°. EH is straight line.
- Find ∠ELF.
- Find ∠HGL.
(a)
∠KTU = 110°
∠LRS = 110° (Corresponding angles)
∠PRL
= 180° - 110°
= 70° (Angles on a straight line)
∠ELF
= 180° - 70° - 70°
= 40° (Isosceles triangle)
(b)
∠PRL = ∠EFL = 70° (Corresponding angles)
∠LFG
= 180° - 70°
= 110° (Angles on a straight line)
∠FGS
= (180° - 110°) ÷ 2
= 35° (Isosceles triangle)
∠HGL
= 180° - 35°
= 145° (Angles on a straight line)
Answer(s): (a) 40°; (b) 145°
