The figure is not drawn to scale. FHKL is a parallelogram, EFL and FGL are isosceles triangles. LK//NU and ∠KTU= 117°. EH is straight line.
- Find ∠ELF.
- Find ∠HGL.
(a)
∠KTU = 117°
∠LRS = 117° (Corresponding angles)
∠PRL
= 180° - 117°
= 63° (Angles on a straight line)
∠ELF
= 180° - 63° - 63°
= 54° (Isosceles triangle)
(b)
∠PRL = ∠EFL = 63° (Corresponding angles)
∠LFG
= 180° - 63°
= 117° (Angles on a straight line)
∠FGS
= (180° - 117°) ÷ 2
= 31.5° (Isosceles triangle)
∠HGL
= 180° - 31.5°
= 148.5° (Angles on a straight line)
Answer(s): (a) 54°; (b) 148.5°