The figure is not drawn to scale. FHKL is a parallelogram, EFL and FGL are isosceles triangles. LK//NU and ∠KTU= 101°. EH is straight line.
- Find ∠ELF.
- Find ∠HGL.
(a)
∠KTU = 101°
∠LRS = 101° (Corresponding angles)
∠PRL
= 180° - 101°
= 79° (Angles on a straight line)
∠ELF
= 180° - 79° - 79°
= 22° (Isosceles triangle)
(b)
∠PRL = ∠EFL = 79° (Corresponding angles)
∠LFG
= 180° - 79°
= 101° (Angles on a straight line)
∠FGS
= (180° - 101°) ÷ 2
= 39.5° (Isosceles triangle)
∠HGL
= 180° - 39.5°
= 140.5° (Angles on a straight line)
Answer(s): (a) 22°; (b) 140.5°