The figure is not drawn to scale. FHKL is a parallelogram, EFL and FGL are isosceles triangles. LK//NU and ∠KTU= 105°. EH is straight line.
- Find ∠ELF.
- Find ∠HGL.
(a)
∠KTU = 105°
∠LRS = 105° (Corresponding angles)
∠PRL
= 180° - 105°
= 75° (Angles on a straight line)
∠ELF
= 180° - 75° - 75°
= 30° (Isosceles triangle)
(b)
∠PRL = ∠EFL = 75° (Corresponding angles)
∠LFG
= 180° - 75°
= 105° (Angles on a straight line)
∠FGS
= (180° - 105°) ÷ 2
= 37.5° (Isosceles triangle)
∠HGL
= 180° - 37.5°
= 142.5° (Angles on a straight line)
Answer(s): (a) 30°; (b) 142.5°