The figure is not drawn to scale. FHKL is a parallelogram, EFL and FGL are isosceles triangles. LK//NU and ∠KTU= 112°. EH is straight line.
- Find ∠ELF.
- Find ∠HGL.
(a)
∠KTU = 112°
∠LRS = 112° (Corresponding angles)
∠PRL
= 180° - 112°
= 68° (Angles on a straight line)
∠ELF
= 180° - 68° - 68°
= 44° (Isosceles triangle)
(b)
∠PRL = ∠EFL = 68° (Corresponding angles)
∠LFG
= 180° - 68°
= 112° (Angles on a straight line)
∠FGS
= (180° - 112°) ÷ 2
= 34° (Isosceles triangle)
∠HGL
= 180° - 34°
= 146° (Angles on a straight line)
Answer(s): (a) 44°; (b) 146°