The figure is not drawn to scale. FHKL is a parallelogram, EFL and FGL are isosceles triangles. LK//NU and ∠KTU= 103°. EH is straight line.
- Find ∠ELF.
- Find ∠HGL.
(a)
∠KTU = 103°
∠LRS = 103° (Corresponding angles)
∠PRL
= 180° - 103°
= 77° (Angles on a straight line)
∠ELF
= 180° - 77° - 77°
= 26° (Isosceles triangle)
(b)
∠PRL = ∠EFL = 77° (Corresponding angles)
∠LFG
= 180° - 77°
= 103° (Angles on a straight line)
∠FGS
= (180° - 103°) ÷ 2
= 38.5° (Isosceles triangle)
∠HGL
= 180° - 38.5°
= 141.5° (Angles on a straight line)
Answer(s): (a) 26°; (b) 141.5°