The figure is not drawn to scale. FHKL is a parallelogram, EFL and FGL are isosceles triangles. LK//NU and ∠KTU= 107°. EH is straight line.
- Find ∠ELF.
- Find ∠HGL.
(a)
∠KTU = 107°
∠LRS = 107° (Corresponding angles)
∠PRL
= 180° - 107°
= 73° (Angles on a straight line)
∠ELF
= 180° - 73° - 73°
= 34° (Isosceles triangle)
(b)
∠PRL = ∠EFL = 73° (Corresponding angles)
∠LFG
= 180° - 73°
= 107° (Angles on a straight line)
∠FGS
= (180° - 107°) ÷ 2
= 36.5° (Isosceles triangle)
∠HGL
= 180° - 36.5°
= 143.5° (Angles on a straight line)
Answer(s): (a) 34°; (b) 143.5°