The figure is not drawn to scale. FHKL is a parallelogram, EFL and FGL are isosceles triangles. LK//NU and ∠KTU= 106°. EH is straight line.
- Find ∠ELF.
- Find ∠HGL.
(a)
∠KTU = 106°
∠LRS = 106° (Corresponding angles)
∠PRL
= 180° - 106°
= 74° (Angles on a straight line)
∠ELF
= 180° - 74° - 74°
= 32° (Isosceles triangle)
(b)
∠PRL = ∠EFL = 74° (Corresponding angles)
∠LFG
= 180° - 74°
= 106° (Angles on a straight line)
∠FGS
= (180° - 106°) ÷ 2
= 37° (Isosceles triangle)
∠HGL
= 180° - 37°
= 143° (Angles on a straight line)
Answer(s): (a) 32°; (b) 143°