The figure is not drawn to scale. FHKL is a parallelogram, EFL and FGL are isosceles triangles. LK//NU and ∠KTU= 119°. EH is straight line.
- Find ∠ELF.
- Find ∠HGL.
(a)
∠KTU = 119°
∠LRS = 119° (Corresponding angles)
∠PRL
= 180° - 119°
= 61° (Angles on a straight line)
∠ELF
= 180° - 61° - 61°
= 58° (Isosceles triangle)
(b)
∠PRL = ∠EFL = 61° (Corresponding angles)
∠LFG
= 180° - 61°
= 119° (Angles on a straight line)
∠FGS
= (180° - 119°) ÷ 2
= 30.5° (Isosceles triangle)
∠HGL
= 180° - 30.5°
= 149.5° (Angles on a straight line)
Answer(s): (a) 58°; (b) 149.5°