The figure is not drawn to scale. FHKL is a parallelogram, EFL and FGL are isosceles triangles. LK//NU and ∠KTU= 113°. EH is straight line.
- Find ∠ELF.
- Find ∠HGL.
(a)
∠KTU = 113°
∠LRS = 113° (Corresponding angles)
∠PRL
= 180° - 113°
= 67° (Angles on a straight line)
∠ELF
= 180° - 67° - 67°
= 46° (Isosceles triangle)
(b)
∠PRL = ∠EFL = 67° (Corresponding angles)
∠LFG
= 180° - 67°
= 113° (Angles on a straight line)
∠FGS
= (180° - 113°) ÷ 2
= 33.5° (Isosceles triangle)
∠HGL
= 180° - 33.5°
= 146.5° (Angles on a straight line)
Answer(s): (a) 46°; (b) 146.5°